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Part IA Michaelmas, Lent, Easter Term

NST Mathematics

Mathematics Methods Calculus Vectors Complex-Numbers Differential-Equations Linear-Algebra PDE Fourier Part IA Michaelmas Term Lent Term Easter Term
  • Methods I (Michaelmas Term)

    Vector algebra, coordinate systems, complex numbers, hyperbolic functions, differentiation, integration, and basic probability

    • Vectors: Fundamentals

      Vector Notation

      A vector is a quantity with both magnitude and direction. Common notations:

      • Bold Roman: u\mathbf{u}, q\mathbf{q} (typeset)
      • Underline/tilde: u‾\underline{u} (handwritten)
      • Arrow: u⃗\vec{u} (North American convention)

      A position vector from point AA to point BB is written AB⃗\vec{AB}. Position vectors from the origin OO to point AA are written OA⃗\vec{OA} or simply a\mathbf{a}.


      Vector Addition

      Vectors add geometrically using the parallelogram rule or triangle rule:

      Parallelogram rule showing a and b as adjacent sides, with a+b as the diagonal

      a+b=b+a(commutative)\mathbf{a} + \mathbf{b} = \mathbf{b} + \mathbf{a} \quad \text{(commutative)}

      (a+b)+c=a+(b+c)(associative)(\mathbf{a} + \mathbf{b}) + \mathbf{c} = \mathbf{a} + (\mathbf{b} + \mathbf{c}) \quad \text{(associative)}

      For three non-coplanar vectors, their sum is the diagonal of a parallelepiped formed by the vectors as edges.

      Parallelepiped with vectors a, b, c as edges and a+b+c as the space diagonal


      Scalar Multiplication

      Multiplying a vector a\mathbf{a} by a scalar λ∈R\lambda \in \mathbb{R}:

      λ(a+b)=λa+λb(distributive)\lambda(\mathbf{a} + \mathbf{b}) = \lambda\mathbf{a} + \lambda\mathbf{b} \quad \text{(distributive)}

      • λ>0\lambda > 0: stretches/compresses without changing direction
      • λ<0\lambda < 0: reverses direction and scales magnitude
      • ∣λa∣=∣λ∣∣a∣|\lambda\mathbf{a}| = |\lambda||\mathbf{a}|

      Cartesian Components

      Right-Handed Coordinate System

      By convention, we use a right-handed system where the xx, yy, zz axes satisfy:

      • Thumb points along xx
      • Index finger along yy
      • Middle finger along zz

      Right-handed coordinate system with x, y, z axes clearly labelled, showing right-hand rule orientation

      Unit Vectors

      The standard basis (orthonormal set):

      i^,j^,k^orex,ey,ez\hat{\mathbf{i}}, \quad \hat{\mathbf{j}}, \quad \hat{\mathbf{k}} \quad \text{or} \quad \mathbf{e}_x, \mathbf{e}_y, \mathbf{e}_z

      Properties:

      • Orthogonal: i^⋅j^=j^⋅k^=k^⋅i^=0\hat{\mathbf{i}} \cdot \hat{\mathbf{j}} = \hat{\mathbf{j}} \cdot \hat{\mathbf{k}} = \hat{\mathbf{k}} \cdot \hat{\mathbf{i}} = 0
      • Unit length: ∣i^∣=∣j^∣=∣k^∣=1|\hat{\mathbf{i}}| = |\hat{\mathbf{j}}| = |\hat{\mathbf{k}}| = 1

      Position Vector

      r=xi^+yj^+zk^=(x,y,z)\mathbf{r} = x\hat{\mathbf{i}} + y\hat{\mathbf{j}} + z\hat{\mathbf{k}} = (x, y, z)

      Magnitude (Length)

      ∣r∣=x2+y2+z2|\mathbf{r}| = \sqrt{x^2 + y^2 + z^2}

      Distance Between Points

      ∣r2−r1∣=(x2−x1)2+(y2−y1)2+(z2−z1)2|\mathbf{r}_2 - \mathbf{r}_1| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}


      Unit Vectors and Direction

      A unit vector has magnitude 1. To normalise any non-zero vector a\mathbf{a}:

      a^=a∣a∣\hat{\mathbf{a}} = \frac{\mathbf{a}}{|\mathbf{a}|}

      This extracts the direction while discarding magnitude information.


      Why Vectors Matter

      Vectors unify many physical quantities:

      • Displacement: position change
      • Velocity: rate of position change
      • Acceleration: rate of velocity change
      • Force: causes acceleration
      • Electric/magnetic fields: act on charges

      The operations (addition, scalar multiplication) match physical intuition: displacements add, forces combine, velocities scale.

    • Scalar Product and Planes

      Definition of the Scalar Product

      The scalar product (or dot product) of two vectors a\mathbf{a} and b\mathbf{b} is:

      a⋅b=∣a∣∣b∣cos⁡θ\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos\theta

      where θ∈[0,π]\theta \in [0, \pi] is the angle between the vectors.

      Two vectors a and b with angle θ between them, showing the dot product projection

      The result is a scalar (hence the name).


      Properties

      Algebraic properties:

      • Commutative: a⋅b=b⋅a\mathbf{a} \cdot \mathbf{b} = \mathbf{b} \cdot \mathbf{a}
      • Distributive: a⋅(b+c)=a⋅b+a⋅c\mathbf{a} \cdot (\mathbf{b} + \mathbf{c}) = \mathbf{a} \cdot \mathbf{b} + \mathbf{a} \cdot \mathbf{c}
      • Self-product: a⋅a=∣a∣2\mathbf{a} \cdot \mathbf{a} = |\mathbf{a}|^2

      Basis vector products: i^⋅i^=j^⋅j^=k^⋅k^=1\hat{\mathbf{i}} \cdot \hat{\mathbf{i}} = \hat{\mathbf{j}} \cdot \hat{\mathbf{j}} = \hat{\mathbf{k}} \cdot \hat{\mathbf{k}} = 1 i^⋅j^=j^⋅k^=k^⋅i^=0\hat{\mathbf{i}} \cdot \hat{\mathbf{j}} = \hat{\mathbf{j}} \cdot \hat{\mathbf{k}} = \hat{\mathbf{k}} \cdot \hat{\mathbf{i}} = 0


      Component Form

      For a=(ax,ay,az)\mathbf{a} = (a_x, a_y, a_z) and b=(bx,by,bz)\mathbf{b} = (b_x, b_y, b_z):

      a⋅b=axbx+ayby+azbz\mathbf{a} \cdot \mathbf{b} = a_x b_x + a_y b_y + a_z b_z

      This provides a computationally convenient formula derived from the basis vector properties.


      Geometric Applications

      Angle Between Vectors

      cos⁡θ=a⋅b∣a∣∣b∣\cos\theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|}

      Two vectors a and b meeting at an angle θ, with the derivation of the angle formula shown

      Perpendicular Vectors

      Two non-zero vectors are perpendicular (orthogonal) if and only if:

      a⋅b=0\mathbf{a} \cdot \mathbf{b} = 0

      Component Along a Direction

      The component of a\mathbf{a} in the direction of a unit vector n^\hat{\mathbf{n}}:

      a∥=a⋅n^=∣a∣cos⁡θa_{\parallel} = \mathbf{a} \cdot \hat{\mathbf{n}} = |\mathbf{a}|\cos\theta

      This is the projection of a\mathbf{a} onto the direction n^\hat{\mathbf{n}}.

      Component Perpendicular to a Direction

      a⊥=∣a∣sin⁡θ=∣a∣2−(a⋅n^)2a_{\perp} = |\mathbf{a}|\sin\theta = \sqrt{|\mathbf{a}|^2 - (\mathbf{a} \cdot \hat{\mathbf{n}})^2}


      Equation of a Plane

      Through a Point with Given Normal

      A plane through point a\mathbf{a} with unit normal n^\hat{\mathbf{n}} consists of all points r\mathbf{r} satisfying:

      (r−a)⋅n^=0(\mathbf{r} - \mathbf{a}) \cdot \hat{\mathbf{n}} = 0

      Plane with normal vector n and point a on the plane, showing how any point r satisfies (r-a)·n = 0

      Why this works: the vector r−a\mathbf{r} - \mathbf{a} lies in the plane, so it must be perpendicular to the normal.

      Expanded Form

      If n^=(l,m,n)\hat{\mathbf{n}} = (l, m, n) where l,m,nl, m, n are direction cosines (cosines of angles with coordinate axes):

      lx+my+nz=plx + my + nz = p

      where p=a⋅n^p = \mathbf{a} \cdot \hat{\mathbf{n}} is the perpendicular distance from the origin to the plane.

      Through Three Points

      For three points a\mathbf{a}, b\mathbf{b}, c\mathbf{c} (not collinear), the plane is:

      r=a+μ(b−a)+ν(c−a)\mathbf{r} = \mathbf{a} + \mu(\mathbf{b} - \mathbf{a}) + \nu(\mathbf{c} - \mathbf{a})

      The normal vector is:

      n=(b−a)×(c−a)\mathbf{n} = (\mathbf{b} - \mathbf{a}) \times (\mathbf{c} - \mathbf{a})


      Distance from Point to Plane

      For a plane n^⋅r=p\hat{\mathbf{n}} \cdot \mathbf{r} = p and point q\mathbf{q}:

      d=∣n^⋅q−p∣d = |\hat{\mathbf{n}} \cdot \mathbf{q} - p|

      This is the signed distance (positive on one side, negative on the other).


      Why the Scalar Product Matters

      The scalar product measures alignment between vectors:

      • Maximum when vectors are parallel
      • Zero when perpendicular
      • Negative when pointing in opposite directions

      This captures the geometry of angles without explicitly computing angles, making it essential for:

      • Work done by a force: W=F⋅dW = \mathbf{F} \cdot \mathbf{d}
      • Power delivered: P=F⋅vP = \mathbf{F} \cdot \mathbf{v}
      • Flux through a surface: involves dot products with normals
    • Vector (Cross) Product

      Definition of the Vector Product

      The vector product (or cross product) of two vectors a\mathbf{a} and b\mathbf{b} is:

      a×b=∣a∣∣b∣sin⁡θ n^\mathbf{a} \times \mathbf{b} = |\mathbf{a}||\mathbf{b}|\sin\theta\,\hat{\mathbf{n}}

      where:

      • θ∈[0,π]\theta \in [0, \pi] is the angle from a\mathbf{a} to b\mathbf{b}
      • n^\hat{\mathbf{n}} is a unit vector perpendicular to both a\mathbf{a} and b\mathbf{b}
      • The direction of n^\hat{\mathbf{n}} follows the right-hand rule: curl fingers from a\mathbf{a} to b\mathbf{b}, thumb points along n^\hat{\mathbf{n}}

      Cross product showing vectors a and b with cross product a×b perpendicular to both, right-hand rule orientation

      The result is a vector (hence the name).


      Properties

      Algebraic properties:

      • Anti-commutative: a×b=−b×a\mathbf{a} \times \mathbf{b} = -\mathbf{b} \times \mathbf{a}
      • Distributive: a×(b+c)=a×b+a×c\mathbf{a} \times (\mathbf{b} + \mathbf{c}) = \mathbf{a} \times \mathbf{b} + \mathbf{a} \times \mathbf{c}
      • Self-product: a×a=0\mathbf{a} \times \mathbf{a} = \mathbf{0}
      • Not associative: a×(b×c)≠(a×b)×c\mathbf{a} \times (\mathbf{b} \times \mathbf{c}) \neq (\mathbf{a} \times \mathbf{b}) \times \mathbf{c}

      Geometric interpretation: ∣a×b∣=area of parallelogram formed by a and b|\mathbf{a} \times \mathbf{b}| = \text{area of parallelogram formed by } \mathbf{a} \text{ and } \mathbf{b}


      Component Form (Determinant)

      The cross product can be computed using a determinant:

      a×b=∣i^j^k^axayazbxbybz∣\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ a_x & a_y & a_z \\ b_x & b_y & b_z \end{vmatrix}

      Expanding:

      a×b=(aybz−azby)i^+(azbx−axbz)j^+(axby−aybx)k^\mathbf{a} \times \mathbf{b} = (a_y b_z - a_z b_y)\hat{\mathbf{i}} + (a_z b_x - a_x b_z)\hat{\mathbf{j}} + (a_x b_y - a_y b_x)\hat{\mathbf{k}}


      Standard Basis Products

      Using the right-hand rule:

      i^×j^=k^,j^×k^=i^,k^×i^=j^\hat{\mathbf{i}} \times \hat{\mathbf{j}} = \hat{\mathbf{k}}, \quad \hat{\mathbf{j}} \times \hat{\mathbf{k}} = \hat{\mathbf{i}}, \quad \hat{\mathbf{k}} \times \hat{\mathbf{i}} = \hat{\mathbf{j}}

      And by anti-commutativity:

      j^×i^=−k^,k^×j^=−i^,i^×k^=−j^\hat{\mathbf{j}} \times \hat{\mathbf{i}} = -\hat{\mathbf{k}}, \quad \hat{\mathbf{k}} \times \hat{\mathbf{j}} = -\hat{\mathbf{i}}, \quad \hat{\mathbf{i}} \times \hat{\mathbf{k}} = -\hat{\mathbf{j}}


      Applications

      Normal Vector to a Plane

      Given two vectors in a plane, their cross product gives a normal vector:

      n=a×b\mathbf{n} = \mathbf{a} \times \mathbf{b}

      Plane with two vectors a and b lying in it, and normal vector n = a×b perpendicular to the plane

      Equation of a Line Through Two Points

      For points a\mathbf{a} and b\mathbf{b}, the line equation is:

      (r−a)×(b−a)=0(\mathbf{r} - \mathbf{a}) \times (\mathbf{b} - \mathbf{a}) = \mathbf{0}

      This encodes: any point r\mathbf{r} on the line has r−a\mathbf{r} - \mathbf{a} parallel to b−a\mathbf{b} - \mathbf{a}.

      Shortest Distance: Point to Line

      Distance from point q\mathbf{q} to line through points a\mathbf{a}, b\mathbf{b}:

      d=∣(q−a)×(b−a)∣∣b−a∣d = \frac{|(\mathbf{q} - \mathbf{a}) \times (\mathbf{b} - \mathbf{a})|}{|\mathbf{b} - \mathbf{a}|}

      This is the magnitude of the perpendicular component.

      Rotational Motion

      The velocity of a point at position r\mathbf{r} rotating with angular velocity ω\boldsymbol{\omega}:

      v=ω×r\mathbf{v} = \boldsymbol{\omega} \times \mathbf{r}


      Triple Products

      Scalar Triple Product

      a⋅(b×c)=∣axayazbxbybzcxcycz∣=[a,b,c]\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = \begin{vmatrix} a_x & a_y & a_z \\ b_x & b_y & b_z \\ c_x & c_y & c_z \end{vmatrix} = [\mathbf{a}, \mathbf{b}, \mathbf{c}]

      Parallelepiped formed by vectors a, b, c, with volume |a·(b×c)|

      Properties:

      • Cyclic permutation (even): value unchanged
        • [a,b,c]=[b,c,a]=[c,a,b][\mathbf{a}, \mathbf{b}, \mathbf{c}] = [\mathbf{b}, \mathbf{c}, \mathbf{a}] = [\mathbf{c}, \mathbf{a}, \mathbf{b}]
      • Odd permutation: sign changes
        • [a,c,b]=−[a,b,c][\mathbf{a}, \mathbf{c}, \mathbf{b}] = -[\mathbf{a}, \mathbf{b}, \mathbf{c}]
      • Geometric meaning: volume of parallelepiped formed by a\mathbf{a}, b\mathbf{b}, c\mathbf{c}
      • Coplanar vectors: if [a,b,c]=0[\mathbf{a}, \mathbf{b}, \mathbf{c}] = 0, the vectors lie in the same plane

      Vector Triple Product

      a×(b×c)=(a⋅c)b−(a⋅b)c\mathbf{a} \times (\mathbf{b} \times \mathbf{c}) = (\mathbf{a} \cdot \mathbf{c})\mathbf{b} - (\mathbf{a} \cdot \mathbf{b})\mathbf{c}

      This is the BAC-CAB rule (mnemonic: “BAC minus CAB”).

      Warning: Cross product is not associative: a×(b×c)≠(a×b)×c\mathbf{a} \times (\mathbf{b} \times \mathbf{c}) \neq (\mathbf{a} \times \mathbf{b}) \times \mathbf{c}


      Why the Cross Product Only Exists in 3D

      The cross product as defined here is specific to three dimensions. In 2D, a perpendicular “vector” would have to point out of the plane (making it essentially a scalar). In 4D and higher, the concept generalises differently (wedge product / exterior algebra).

      This is why the cross product is sometimes called the “vector product” — it produces a vector only because we happen to work in 3D.

    • Coordinate Systems

      Cartesian Coordinates

      The standard Cartesian (rectangular) coordinate system uses three mutually perpendicular axes xx, yy, zz emanating from an origin OO.

      Position vector: r=xi^+yj^+zk^\mathbf{r} = x\hat{\mathbf{i}} + y\hat{\mathbf{j}} + z\hat{\mathbf{k}}

      Basis vectors i^\hat{\mathbf{i}}, j^\hat{\mathbf{j}}, k^\hat{\mathbf{k}} are constant — they point in the same direction everywhere.


      Cylindrical Polar Coordinates (rr, θ\theta, zz)

      Useful for systems with cylindrical symmetry (pipes, wires, cylindrical objects).

      Cylindrical coordinates showing point P with coordinates (r, θ, z), and how r relates to x and y in the xy-plane

      Coordinate Definitions

      • r≥0r \geq 0: perpendicular distance from zz-axis
      • θ\theta: azimuthal angle from xx-axis in the xyxy-plane
      • zz: height (same as Cartesian)

      Relations to Cartesian

      x=rcos⁡θ,y=rsin⁡θ,z=zx = r\cos\theta, \quad y = r\sin\theta, \quad z = z

      r=x2+y2,θ=tan⁡−1yxr = \sqrt{x^2 + y^2}, \quad \theta = \tan^{-1}\frac{y}{x}

      Basis Vectors

      Unlike Cartesian, cylindrical basis vectors depend on position:

      e^r=cos⁡θ i^+sin⁡θ j^\hat{\mathbf{e}}_r = \cos\theta\,\hat{\mathbf{i}} + \sin\theta\,\hat{\mathbf{j}}

      e^θ=−sin⁡θ i^+cos⁡θ j^\hat{\mathbf{e}}_\theta = -\sin\theta\,\hat{\mathbf{i}} + \cos\theta\,\hat{\mathbf{j}}

      e^z=k^\hat{\mathbf{e}}_z = \hat{\mathbf{k}}

      Position vector: r=re^r+ze^z\mathbf{r} = r\hat{\mathbf{e}}_r + z\hat{\mathbf{e}}_z

      Warning: e^r\hat{\mathbf{e}}_r and e^θ\hat{\mathbf{e}}_\theta change direction as θ\theta changes.


      Spherical Polar Coordinates (rr, θ\theta, ϕ\phi)

      Useful for systems with spherical symmetry (atoms, planets, central forces).

      Spherical coordinates showing point P with coordinates (r, θ, φ), where θ is the polar angle from z-axis and φ is the azimuthal angle in the xy-plane

      Coordinate Definitions

      • r≥0r \geq 0: distance from origin
      • θ∈[0,π]\theta \in [0, \pi]: polar angle from zz-axis (inclination)
      • ϕ∈[0,2π)\phi \in [0, 2\pi): azimuthal angle from xx-axis

      Note: Physics convention differs from mathematics — physicists use θ\theta for polar angle and ϕ\phi for azimuthal, while mathematicians often swap them.

      Relations to Cartesian

      x=rsin⁡θcos⁡ϕ,y=rsin⁡θsin⁡ϕ,z=rcos⁡θx = r\sin\theta\cos\phi, \quad y = r\sin\theta\sin\phi, \quad z = r\cos\theta

      r=x2+y2+z2r = \sqrt{x^2 + y^2 + z^2}

      Basis Vectors

      Again, position-dependent:

      e^r=sin⁡θcos⁡ϕ i^+sin⁡θsin⁡ϕ j^+cos⁡θ k^\hat{\mathbf{e}}_r = \sin\theta\cos\phi\,\hat{\mathbf{i}} + \sin\theta\sin\phi\,\hat{\mathbf{j}} + \cos\theta\,\hat{\mathbf{k}}

      e^θ=cos⁡θcos⁡ϕ i^+cos⁡θsin⁡ϕ j^−sin⁡θ k^\hat{\mathbf{e}}_\theta = \cos\theta\cos\phi\,\hat{\mathbf{i}} + \cos\theta\sin\phi\,\hat{\mathbf{j}} - \sin\theta\,\hat{\mathbf{k}}

      e^ϕ=−sin⁡ϕ i^+cos⁡ϕ j^\hat{\mathbf{e}}_\phi = -\sin\phi\,\hat{\mathbf{i}} + \cos\phi\,\hat{\mathbf{j}}

      Position vector: r=re^r\mathbf{r} = r\hat{\mathbf{e}}_r


      Volume Elements

      The volume element changes with coordinate system:

      Cartesian: dV=dx dy dzdV = dx\,dy\,dz

      Cylindrical: dV=r dr dθ dzdV = r\,dr\,d\theta\,dz

      The extra factor rr accounts for the fact that rings at larger rr have larger circumference.

      Spherical: dV=r2sin⁡θ dr dθ dϕdV = r^2\sin\theta\,dr\,d\theta\,d\phi

      The factor r2sin⁡θr^2\sin\theta comes from the Jacobian determinant. At the poles (θ=0,π\theta = 0, \pi), sin⁡θ=0\sin\theta = 0 and the volume element vanishes — this reflects the coordinate singularity.


      Area Elements

      For surface integrals:

      Cylindrical (curved surface): dS=r dθ dzdS = r\,d\theta\,dz

      Spherical (surface of sphere): dS=r2sin⁡θ dθ dϕdS = r^2\sin\theta\,d\theta\,d\phi


      Why Different Coordinate Systems?

      Choosing the right coordinates simplifies problems:

      SymmetryBest CoordinatesExample
      Cylindrical(r,θ,z)(r, \theta, z)Electric field of a wire
      Spherical(r,θ,ϕ)(r, \theta, \phi)Gravitational field of Earth
      TranslationalCartesianUniform field

      Key insight: The physics remains the same; only the mathematical description changes. Choose coordinates that match the symmetry to reduce complexity.


      Warning: Basis Vectors Change

      In curvilinear coordinates, derivatives of basis vectors are non-zero:

      Cylindrical: ∂e^r∂θ=e^θ,∂e^θ∂θ=−e^r\frac{\partial \hat{\mathbf{e}}_r}{\partial \theta} = \hat{\mathbf{e}}_\theta, \quad \frac{\partial \hat{\mathbf{e}}_\theta}{\partial \theta} = -\hat{\mathbf{e}}_r

      Spherical: More complex relations involving all three basis vectors.

      This means ∂r∂θ\frac{\partial \mathbf{r}}{\partial \theta} is not simply e^θ\hat{\mathbf{e}}_\theta — there are additional terms. Always compute vector products in Cartesian or use established formulas for operators like ∇\nabla in curvilinear coordinates.

    • Complex Numbers

      Definition

      A complex number is:

      z=x+iyz = x + iy

      where:

      • x∈Rx \in \mathbb{R} is the real part: Re(z)=x\text{Re}(z) = x
      • y∈Ry \in \mathbb{R} is the imaginary part: Im(z)=y\text{Im}(z) = y
      • ii is the imaginary unit: i2=−1i^2 = -1

      The set of all complex numbers is C\mathbb{C}.


      Complex Conjugate

      The complex conjugate of z=x+iyz = x + iy is:

      zˉ=z∗=x−iy\bar{z} = z^* = x - iy

      Properties:

      • zzˉ=∣z∣2=x2+y2z\bar{z} = |z|^2 = x^2 + y^2
      • z+zˉ=2Re(z)z + \bar{z} = 2\text{Re}(z)
      • z−zˉ=2iIm(z)z - \bar{z} = 2i\text{Im}(z)
      • z1z2‾=zˉ1zˉ2\overline{z_1 z_2} = \bar{z}_1 \bar{z}_2
      • z1+z2‾=zˉ1+zˉ2\overline{z_1 + z_2} = \bar{z}_1 + \bar{z}_2

      The Argand Diagram

      Complex numbers correspond to points in the complex plane (Argand diagram):

      • Horizontal axis: real part
      • Vertical axis: imaginary part

      Argand diagram showing complex number z = x + iy plotted at (x, y), with modulus r and argument θ marked

      This geometric interpretation makes complex numbers useful for 2D rotations and oscillations.


      Modulus and Argument

      Modulus (Magnitude)

      ∣z∣=r=x2+y2=zzˉ|z| = r = \sqrt{x^2 + y^2} = \sqrt{z\bar{z}}

      The distance from origin in the Argand diagram.

      Argument (Angle)

      arg⁡(z)=θ=tan⁡−1yx\arg(z) = \theta = \tan^{-1}\frac{y}{x}

      Warning: Must consider the signs of xx and yy to get the correct quadrant. The principal argument satisfies −π<θ≤π-\pi < \theta \leq \pi.

      Use:

      • θ=tan⁡−1(y/x)\theta = \tan^{-1}(y/x) if x>0x > 0
      • θ=tan⁡−1(y/x)+π\theta = \tan^{-1}(y/x) + \pi if x<0x < 0 and y≥0y \geq 0
      • θ=tan⁡−1(y/x)−π\theta = \tan^{-1}(y/x) - \pi if x<0x < 0 and y<0y < 0
      • θ=π/2\theta = \pi/2 if x=0x = 0 and y>0y > 0
      • θ=−π/2\theta = -\pi/2 if x=0x = 0 and y<0y < 0

      Polar Form

      Using modulus and argument:

      z=r(cos⁡θ+isin⁡θ)=reiθz = r(\cos\theta + i\sin\theta) = re^{i\theta}

      This is the polar form (also called modulus-argument form).


      Euler’s Formula

      eiθ=cos⁡θ+isin⁡θe^{i\theta} = \cos\theta + i\sin\theta

      Derivation sketch: Expand eiθe^{i\theta}, cos⁡θ\cos\theta, sin⁡θ\sin\theta as power series and compare.

      Euler’s identity (when θ=π\theta = \pi): eiπ+1=0e^{i\pi} + 1 = 0

      This connects five fundamental constants: ee, ii, π\pi, 11, 00.


      De Moivre’s Theorem

      (cos⁡θ+isin⁡θ)n=cos⁡(nθ)+isin⁡(nθ)(\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta)

      Or equivalently: (eiθ)n=einθ(e^{i\theta})^n = e^{in\theta}

      Applications

      Trigonometric identities: cos⁡nθ=Re[(cos⁡θ+isin⁡θ)n]\cos n\theta = \text{Re}[(\cos\theta + i\sin\theta)^n] sin⁡nθ=Im[(cos⁡θ+isin⁡θ)n]\sin n\theta = \text{Im}[(\cos\theta + i\sin\theta)^n]

      Binomial expansion method: cos⁡3θ=Re[(cos⁡θ+isin⁡θ)3]=cos⁡3θ−3cos⁡θsin⁡2θ=4cos⁡3θ−3cos⁡θ\cos 3\theta = \text{Re}[(\cos\theta + i\sin\theta)^3] = \cos^3\theta - 3\cos\theta\sin^2\theta = 4\cos^3\theta - 3\cos\theta

      Expressing trig in terms of exponentials: cos⁡θ=eiθ+e−iθ2,sin⁡θ=eiθ−e−iθ2i\cos\theta = \frac{e^{i\theta} + e^{-i\theta}}{2}, \quad \sin\theta = \frac{e^{i\theta} - e^{-i\theta}}{2i}


      Roots of Unity

      The nnth roots of unity are solutions to zn=1z^n = 1:

      zk=e2πik/n=cos⁡2πkn+isin⁡2πknz_k = e^{2\pi i k/n} = \cos\frac{2\pi k}{n} + i\sin\frac{2\pi k}{n}

      for k=0,1,2,…,n−1k = 0, 1, 2, \ldots, n-1.

      Roots of unity for n=6, showing 6 equally spaced points on unit circle forming regular hexagon

      Geometric interpretation: The nn roots are evenly spaced around the unit circle, forming a regular nn-gon.

      Example (cube roots): z3=1⇒z0=1,z1=e2πi/3,z2=e4πi/3z^3 = 1 \Rightarrow z_0 = 1, \quad z_1 = e^{2\pi i/3}, \quad z_2 = e^{4\pi i/3}


      Roots of Complex Numbers

      To solve zn=wz^n = w where w=Reiϕw = Re^{i\phi}:

      zk=Rnexp⁡(iϕ+2πkn)z_k = \sqrt[n]{R} \exp\left(i\frac{\phi + 2\pi k}{n}\right)

      for k=0,1,…,n−1k = 0, 1, \ldots, n-1.


      Why Complex Numbers Matter

      Complex numbers unify and simplify many areas:

      • Differential equations: exponential solutions
      • AC circuits: impedance extends resistance
      • Quantum mechanics: wave functions are complex-valued
      • Signal processing: Fourier transforms use complex exponentials
      • 2D rotations: multiplication by eiθe^{i\theta} rotates by angle θ\theta

      The algebra of complex numbers captures oscillatory and wave behaviour elegantly, replacing sines and cosines with exponentials.

    • Hyperbolic Functions

      Definitions

      The hyperbolic functions are defined by analogy with trigonometric functions:

      cosh⁡x=ex+e−x2\cosh x = \frac{e^x + e^{-x}}{2}

      sinh⁡x=ex−e−x2\sinh x = \frac{e^x - e^{-x}}{2}

      tanh⁡x=sinh⁡xcosh⁡x=ex−e−xex+e−x\tanh x = \frac{\sinh x}{\cosh x} = \frac{e^x - e^{-x}}{e^x + e^{-x}}

      Related functions: sech x=1cosh⁡x,cosech x=1sinh⁡x,coth⁡x=1tanh⁡x\text{sech } x = \frac{1}{\cosh x}, \quad \text{cosech } x = \frac{1}{\sinh x}, \quad \coth x = \frac{1}{\tanh x}


      Relation to Trigonometric Functions

      The connection between hyperbolic and trig functions involves the imaginary unit:

      cosh⁡(ix)=cos⁡x\cosh(ix) = \cos x

      sinh⁡(ix)=isin⁡x\sinh(ix) = i\sin x

      Side-by-side comparison of cos x (oscillating) and cosh x (catenary), showing the relationship via imaginary arguments

      cos⁡(ix)=cosh⁡x\cos(ix) = \cosh x

      sin⁡(ix)=isinh⁡x\sin(ix) = i\sinh x

      Why the name “hyperbolic”? Like how cos⁡θ\cos\theta and sin⁡θ\sin\theta parameterise a circle x2+y2=1x^2 + y^2 = 1, the functions cosh⁡t\cosh t and sinh⁡t\sinh t parameterise a hyperbola x2−y2=1x^2 - y^2 = 1.


      Osborne’s Rule

      Osborne’s rule converts trigonometric identities to hyperbolic identities:

      Replace cos⁡\cos with cosh⁡\cosh, sin⁡\sin with sinh⁡\sinh. Change the sign wherever there is a product of two sinh⁡\sinh functions.

      Examples:

      TrigonometricHyperbolic
      cos⁡2θ+sin⁡2θ=1\cos^2\theta + \sin^2\theta = 1cosh⁡2x−sinh⁡2x=1\cosh^2 x - \sinh^2 x = 1
      cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\cos(A+B) = \cos A\cos B - \sin A\sin Bcosh⁡(A+B)=cosh⁡Acosh⁡B+sinh⁡Asinh⁡B\cosh(A+B) = \cosh A\cosh B + \sinh A\sinh B
      sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin(A+B) = \sin A\cos B + \cos A\sin Bsinh⁡(A+B)=sinh⁡Acosh⁡B+cosh⁡Asinh⁡B\sinh(A+B) = \sinh A\cosh B + \cosh A\sinh B
      cos⁡2θ=cos⁡2θ−sin⁡2θ\cos 2\theta = \cos^2\theta - \sin^2\thetacosh⁡2x=cosh⁡2x+sinh⁡2x\cosh 2x = \cosh^2 x + \sinh^2 x

      Key Identities

      Fundamental: cosh⁡2x−sinh⁡2x=1\cosh^2 x - \sinh^2 x = 1

      Addition formulae: cosh⁡(A±B)=cosh⁡Acosh⁡B±sinh⁡Asinh⁡B\cosh(A \pm B) = \cosh A\cosh B \pm \sinh A\sinh B sinh⁡(A±B)=sinh⁡Acosh⁡B±cosh⁡Asinh⁡B\sinh(A \pm B) = \sinh A\cosh B \pm \cosh A\sinh B tanh⁡(A+B)=tanh⁡A+tanh⁡B1+tanh⁡Atanh⁡B\tanh(A + B) = \frac{\tanh A + \tanh B}{1 + \tanh A\tanh B}

      Double angle: cosh⁡2x=cosh⁡2x+sinh⁡2x=2cosh⁡2x−1=1+2sinh⁡2x\cosh 2x = \cosh^2 x + \sinh^2 x = 2\cosh^2 x - 1 = 1 + 2\sinh^2 x sinh⁡2x=2sinh⁡xcosh⁡x\sinh 2x = 2\sinh x\cosh x


      Graph Properties

      Parity:

      • cosh⁡x\cosh x is even: cosh⁡(−x)=cosh⁡x\cosh(-x) = \cosh x
      • sinh⁡x\sinh x is odd: sinh⁡(−x)=−sinh⁡x\sinh(-x) = -\sinh x
      • tanh⁡x\tanh x is odd: tanh⁡(−x)=−tanh⁡x\tanh(-x) = -\tanh x

      Graphs of cosh x (catenary shape, even), sinh x (odd, through origin), and tanh x (odd, asymptotes at ±1)

      Key values:

      • cosh⁡0=1\cosh 0 = 1, sinh⁡0=0\sinh 0 = 0, tanh⁡0=0\tanh 0 = 0

      Asymptotic behaviour:

      • cosh⁡x≥1\cosh x \geq 1 for all real xx (minimum at x=0x = 0)
      • tanh⁡x→±1\tanh x \to \pm 1 as x→±∞x \to \pm\infty
      • As x→+∞x \to +\infty: cosh⁡x≈sinh⁡x≈12ex\cosh x \approx \sinh x \approx \frac{1}{2}e^x

      Derivatives

      ddxsinh⁡x=cosh⁡x\frac{d}{dx}\sinh x = \cosh x

      ddxcosh⁡x=sinh⁡x\frac{d}{dx}\cosh x = \sinh x

      ddxtanh⁡x=sech2x\frac{d}{dx}\tanh x = \text{sech}^2 x

      Compare with trig: The derivatives of sinh⁡\sinh and cosh⁡\cosh follow the same pattern as sin⁡\sin and cos⁡\cos but without sign changes.


      Inverse Hyperbolic Functions

      Expressed as logarithms:

      sinh⁡−1x=ln⁡(x+x2+1)(defined for all x)\sinh^{-1} x = \ln(x + \sqrt{x^2 + 1}) \quad \text{(defined for all } x\text{)}

      cosh⁡−1x=ln⁡(x+x2−1)(for x≥1)\cosh^{-1} x = \ln(x + \sqrt{x^2 - 1}) \quad \text{(for } x \geq 1\text{)}

      tanh⁡−1x=12ln⁡(1+x1−x)(for ∣x∣<1)\tanh^{-1} x = \frac{1}{2}\ln\left(\frac{1+x}{1-x}\right) \quad \text{(for } |x| < 1\text{)}

      Derivation of sinh⁡−1\sinh^{-1}: If y=sinh⁡−1xy = \sinh^{-1} x, then x=sinh⁡y=ey−e−y2x = \sinh y = \frac{e^y - e^{-y}}{2}. Multiply by eye^y, solve the quadratic in eye^y.


      Why Hyperbolic Functions Matter

      1. Catenary: The shape of a hanging chain is y=acosh⁡(x/a)y = a\cosh(x/a)
      2. Special relativity: Rapidity (the hyperbolic angle) uses tanh⁡\tanh
      3. Differential equations: Solutions to d2ydx2=y\frac{d^2y}{dx^2} = y involve sinh⁡\sinh and cosh⁡\cosh
      4. Complex analysis: Bridge between trigonometric and exponential functions

      Hyperbolic functions appear wherever phenomena have exponential growth/decay combined with symmetry, such as in thermal physics, fluid dynamics, and electrical transmission lines.

    • Differentiation

      Leibniz Formula for Products

      For y=u(x)v(x)y = u(x)v(x), the nnth derivative is:

      dnydxn=∑k=0n(nk)dn−kudxn−kdkvdxk\frac{d^n y}{dx^n} = \sum_{k=0}^{n} \binom{n}{k} \frac{d^{n-k}u}{dx^{n-k}} \frac{d^k v}{dx^k}

      This generalises the product rule to higher derivatives.

      Example: If y=x2exy = x^2 e^x, find d3ydx3\frac{d^3 y}{dx^3}.

      Let u=x2u = x^2 and v=exv = e^x. Then:

      • u′=2xu' = 2x, u′′=2u'' = 2, u′′′=0u''' = 0
      • v′=v′′=v′′′=exv' = v'' = v''' = e^x

      y′′′=u′′′v+3u′′v′+3u′v′′+uv′′′=0+6ex+6xex+x2ex=(x2+6x+6)exy''' = u'''v + 3u''v' + 3u'v'' + uv''' = 0 + 6e^x + 6xe^x + x^2e^x = (x^2 + 6x + 6)e^x


      Curve Sketching

      Key techniques for sketching curves y=f(x)y = f(x):

      1. Intercepts: Where does f(x)=0f(x) = 0? Where does y=0y = 0?
      2. Asymptotes:
        • Horizontal: lim⁡x→±∞f(x)\lim_{x \to \pm\infty} f(x)
        • Vertical: Where f(x)→∞f(x) \to \infty
        • Oblique: Divide polynomials for rational functions
      3. Stationary points: Where f′(x)=0f'(x) = 0
        • Local maximum: f′′<0f'' < 0
        • Local minimum: f′′>0f'' > 0
        • Point of inflection: f′′=0f'' = 0 but derivative doesn’t change sign
      4. Symmetry: Even (f(−x)=f(x)f(-x) = f(x)) or odd (f(−x)=−f(x)f(-x) = -f(x))?
      5. Behaviour at special points: Near asymptotes, near the origin

      Newton-Raphson Method

      To find roots of f(x)=0f(x) = 0, iterate:

      xn+1=xn−f(xn)f′(xn)x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}

      Newton-Raphson iteration showing tangent lines, x₁ to x₂, converging to root

      Geometric interpretation: Follow the tangent line to the xx-axis.

      Convergence: Quadratic (very fast) near a root. But can fail if:

      • f′(xn)=0f'(x_n) = 0 (horizontal tangent)
      • Starting point is poorly chosen (cycle or divergence)

      Example: Estimate 2\sqrt{2} by solving f(x)=x2−2=0f(x) = x^2 - 2 = 0.

      xn+1=xn−xn2−22xn=xn−xn2+1xn=xn2+1xnx_{n+1} = x_n - \frac{x_n^2 - 2}{2x_n} = x_n - \frac{x_n}{2} + \frac{1}{x_n} = \frac{x_n}{2} + \frac{1}{x_n}

      Starting with x0=1x_0 = 1: x1=1.5x_1 = 1.5, x2≈1.417x_2 \approx 1.417, x3≈1.414x_3 \approx 1.414.


      Implicit Differentiation

      For functions defined implicitly by F(x,y)=0F(x, y) = 0:

      dydx=−∂F/∂x∂F/∂y\frac{dy}{dx} = -\frac{\partial F/\partial x}{\partial F/\partial y}

      Or differentiate both sides with respect to xx, treating yy as a function of xx.

      Example: Find dydx\frac{dy}{dx} for x2+y2=a2x^2 + y^2 = a^2.

      Differentiating: 2x+2ydydx=02x + 2y\frac{dy}{dx} = 0, so dydx=−xy\frac{dy}{dx} = -\frac{x}{y}.


      Parametric Differentiation

      If x=x(t)x = x(t) and y=y(t)y = y(t):

      dydx=dy/dtdx/dt\frac{dy}{dx} = \frac{dy/dt}{dx/dt}

      For the second derivative: d2ydx2=ddt(dydx)⋅dtdx=ddt(y′x′)⋅1x′\frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{dy}{dx}\right) \cdot \frac{dt}{dx} = \frac{d}{dt}\left(\frac{y'}{x'}\right) \cdot \frac{1}{x'}

      Example: For x=cos⁡tx = \cos t, y=sin⁡ty = \sin t:

      dydx=cos⁡t−sin⁡t=−cot⁡t\frac{dy}{dx} = \frac{\cos t}{-\sin t} = -\cot t


      Logarithmic Differentiation

      For products/quotients/powers, take logarithms first:

      If y=ua⋅vbwc, then ln⁡y=aln⁡u+bln⁡v−cln⁡w\text{If } y = \frac{u^a \cdot v^b}{w^c}, \text{ then } \ln y = a\ln u + b\ln v - c\ln w

      Differentiate: y′y=au′u+bv′v−cw′w\frac{y'}{y} = \frac{au'}{u} + \frac{bv'}{v} - \frac{cw'}{w}

      Example: Find ddx(xx)\frac{d}{dx}(x^x).

      y=xxy = x^x, so ln⁡y=xln⁡x\ln y = x\ln x. Then: y′y=ln⁡x+1⇒y′=xx(ln⁡x+1)\frac{y'}{y} = \ln x + 1 \Rightarrow y' = x^x(\ln x + 1)


      Higher Derivatives and Patterns

      Some functions have recognisable derivative patterns:

      f(x)f(x)f′(x)f'(x)f′′(x)f''(x)f(n)(x)f^{(n)}(x)
      exe^xexe^xexe^xexe^x
      sin⁡x\sin xcos⁡x\cos x−sin⁡x-\sin xsin⁡(x+nπ/2)\sin(x + n\pi/2)
      cos⁡x\cos x−sin⁡x-\sin x−cos⁡x-\cos xcos⁡(x+nπ/2)\cos(x + n\pi/2)
      ln⁡x\ln x1/x1/x−1/x2-1/x^2(−1)n−1(n−1)!/xn(-1)^{n-1}(n-1)!/x^n
    • Elementary Analysis

      The Intuitive Notion of Limit

      We write lim⁡x→x0f(x)=K\lim_{x \to x_0} f(x) = K if f(x)f(x) can be made arbitrarily close to KK by taking xx sufficiently close to x0x_0.

      Key point: This is about behaviour near x0x_0, not at x0x_0. The limit can exist even if f(x0)f(x_0) is undefined or different from KK.


      The (ε, δ) Definition (Weierstrass)

      lim⁡x→x0f(x)=K\lim_{x \to x_0} f(x) = K if and only if: for any ε>0\varepsilon > 0, there exists δ>0\delta > 0 such that ∣f(x)−K∣<ε|f(x) - K| < \varepsilon for all xx satisfying 0<∣x−x0∣<δ0 < |x - x_0| < \delta.

      Function graph showing x₀, with ε-band around K and corresponding δ-interval around x₀, illustrating the ε-δ definition

      Key features:

      • The order matters: for any ε\varepsilon, we must find some δ\delta
      • The condition 0<∣x−x0∣0 < |x - x_0| excludes x0x_0 itself
      • We control xx (through δ\delta) to control f(x)f(x) (through ε\varepsilon)

      Analogy: A game where an opponent picks ε\varepsilon, and you must respond with a valid δ\delta.


      Example: Proving lim⁡x→0sin⁡xx=1\lim_{x \to 0} \frac{\sin x}{x} = 1

      We need: for any ε>0\varepsilon > 0, find δ\delta such that ∣x∣<δ⇒∣sin⁡xx−1∣<ε|x| < \delta \Rightarrow \left|\frac{\sin x}{x} - 1\right| < \varepsilon.

      Using the inequality ∣sin⁡x∣<∣x∣<∣tan⁡x∣|\sin x| < |x| < |\tan x| for 0<x<π/20 < x < \pi/2:

      • From sin⁡x<x<tan⁡x\sin x < x < \tan x, we get cos⁡x<sin⁡xx<1\cos x < \frac{\sin x}{x} < 1
      • As x→0x \to 0, cos⁡x→1\cos x \to 1, so sin⁡xx\frac{\sin x}{x} is squeezed to 1

      This uses the squeeze theorem: If g(x)≤f(x)≤h(x)g(x) \leq f(x) \leq h(x) and lim⁡g=lim⁡h=L\lim g = \lim h = L, then lim⁡f=L\lim f = L.


      One-Sided Limits

      Right-hand limit: lim⁡x→a+f(x)\lim_{x \to a^+} f(x)

      Left-hand limit: lim⁡x→a−f(x)\lim_{x \to a^-} f(x)

      The (two-sided) limit exists if and only if both one-sided limits exist and are equal.

      Example: The Heaviside step function: H(x)={0x<01x≥0H(x) = \begin{cases} 0 & x < 0 \\ 1 & x \geq 0 \end{cases}

      has lim⁡x→0+H(x)=1\lim_{x \to 0^+} H(x) = 1 and lim⁡x→0−H(x)=0\lim_{x \to 0^-} H(x) = 0. Since these differ, lim⁡x→0H(x)\lim_{x \to 0} H(x) does not exist.


      Limits at Infinity

      lim⁡x→∞f(x)=K\lim_{x \to \infty} f(x) = K if for any ε>0\varepsilon > 0, there exists XX such that ∣f(x)−K∣<ε|f(x) - K| < \varepsilon for all x>Xx > X.

      Example: lim⁡x→∞1x=0\lim_{x \to \infty} \frac{1}{x} = 0

      Given ε>0\varepsilon > 0, choose X=1/εX = 1/\varepsilon. Then for x>Xx > X: ∣1x−0∣=1x<1X=ε\left|\frac{1}{x} - 0\right| = \frac{1}{x} < \frac{1}{X} = \varepsilon


      Continuity

      A function ff is continuous at x=ax = a if:

      1. f(a)f(a) exists (the function is defined there)
      2. lim⁡x→af(x)=f(a)\lim_{x \to a} f(x) = f(a)

      (ε, δ) form: For any ε>0\varepsilon > 0, there exists δ>0\delta > 0 such that ∣f(x)−f(a)∣<ε|f(x) - f(a)| < \varepsilon for all ∣x−a∣<δ|x - a| < \delta.

      Note: unlike the limit definition, this includes x=ax = a.

      Types of discontinuity:

      • Removable: limit exists but ≠ function value
      • Jump: one-sided limits exist but differ
      • Essential: limit does not exist (e.g., oscillation)

      Differentiability

      A function ff is differentiable at x=ax = a if the limit:

      f′(a)=lim⁡h→0f(a+h)−f(a)hf'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}

      exists.

      Key relationship:

      • Differentiable ⇒\Rightarrow continuous
      • Continuous ⇏\not\Rightarrow differentiable

      Example where continuous but not differentiable: f(x)=∣x∣f(x) = |x| at x=0x = 0. The left and right derivatives differ.


      Big-O Notation

      We write f(x)=O(g(x))f(x) = O(g(x)) as x→ax \to a if there exist constants K>0K > 0 and δ>0\delta > 0 such that:

      ∣f(x)∣≤K∣g(x)∣for all 0<∣x−a∣<δ|f(x)| \leq K|g(x)| \quad \text{for all } 0 < |x - a| < \delta

      Intuition: ff is bounded by a multiple of gg near aa.

      Examples:

      • x2+x=O(x2)x^2 + x = O(x^2) as x→∞x \to \infty (the x2x^2 term dominates)
      • sin⁡x=O(x)\sin x = O(x) as x→0x \to 0 (since sin⁡x/x→1\sin x / x \to 1)

      Notation convention: We usually state the tightest bound.


      Why Rigorous Analysis Matters

      The (ε, δ) framework:

      • Makes precise what “close” means
      • Avoids circular reasoning in limits
      • Handles pathological cases (discontinuous functions, functions that oscillate)
      • Provides foundations for calculus proofs

      Historically, calculus was developed before rigorous foundations. The (ε, δ) definition (Weierstrass, 19th century) resolved centuries of confusion about infinitesimals, convergence, and continuity.

    • Taylor Series

      Taylor’s Theorem

      If ff has derivatives up to order (n+1)(n+1) near aa, then:

      f(a+h)=f(a)+hf′(a)+h22!f′′(a)+⋯+hnn!f(n)(a)+Rn+1f(a+h) = f(a) + hf'(a) + \frac{h^2}{2!}f''(a) + \cdots + \frac{h^n}{n!}f^{(n)}(a) + R_{n+1}

      Function f(x) with Taylor polynomial approximations of increasing degree, showing convergence near expansion point

      Lagrange remainder: Rn+1=hn+1(n+1)!f(n+1)(ξ)R_{n+1} = \frac{h^{n+1}}{(n+1)!}f^{(n+1)}(\xi)

      where ξ\xi lies between aa and a+ha + h.

      Interpretation: We can approximate smooth functions by polynomials, with the remainder giving the error.


      Maclaurin Series

      The special case of Taylor series with a=0a = 0:

      f(x)=f(0)+xf′(0)+x22!f′′(0)+⋯f(x) = f(0) + xf'(0) + \frac{x^2}{2!}f''(0) + \cdots


      Standard Power Series

      These series should be memorised:

      FunctionPower SeriesConvergence
      exe^x∑k=0∞xkk!\sum_{k=0}^{\infty} \frac{x^k}{k!}All xx
      sin⁡x\sin x∑k=0∞(−1)kx2k+1(2k+1)!\sum_{k=0}^{\infty} \frac{(-1)^k x^{2k+1}}{(2k+1)!}All xx
      cos⁡x\cos x∑k=0∞(−1)kx2k(2k)!\sum_{k=0}^{\infty} \frac{(-1)^k x^{2k}}{(2k)!}All xx
      sinh⁡x\sinh x∑k=0∞x2k+1(2k+1)!\sum_{k=0}^{\infty} \frac{x^{2k+1}}{(2k+1)!}All xx
      cosh⁡x\cosh x∑k=0∞x2k(2k)!\sum_{k=0}^{\infty} \frac{x^{2k}}{(2k)!}All xx
      ln⁡(1+x)\ln(1+x)∑k=1∞(−1)k+1xkk\sum_{k=1}^{\infty} \frac{(-1)^{k+1} x^k}{k}−1<x≤1-1 < x \leq 1
      ln⁡(1−x)\ln(1-x)−∑k=1∞xkk-\sum_{k=1}^{\infty} \frac{x^k}{k}−1≤x<1-1 \leq x < 1

      Binomial Expansion

      For ∣x∣<1|x| < 1 and any α∈R\alpha \in \mathbb{R}:

      (1+x)α=1+αx+α(α−1)2!x2+α(α−1)(α−2)3!x3+⋯(1+x)^\alpha = 1 + \alpha x + \frac{\alpha(\alpha-1)}{2!}x^2 + \frac{\alpha(\alpha-1)(\alpha-2)}{3!}x^3 + \cdots

      Special cases:

      • α=n\alpha = n (positive integer): terminates (binomial theorem)
      • α=−1\alpha = -1: 11+x=1−x+x2−x3+⋯\frac{1}{1+x} = 1 - x + x^2 - x^3 + \cdots
      • α=12\alpha = \frac{1}{2}: 1+x=1+x2−x28+x316−⋯\sqrt{1+x} = 1 + \frac{x}{2} - \frac{x^2}{8} + \frac{x^3}{16} - \cdots

      Manipulating Series

      Addition: ∑anxn+∑bnxn=∑(an+bn)xn\sum a_n x^n + \sum b_n x^n = \sum (a_n + b_n)x^n

      Multiplication: (∑anxn)(∑bnxn)=∑cnxn(\sum a_n x^n)(\sum b_n x^n) = \sum c_n x^n where cn=∑k=0nakbn−kc_n = \sum_{k=0}^{n} a_k b_{n-k}

      Composition: Substitute one series into another

      Differentiation: ddx∑anxn=∑nanxn−1\frac{d}{dx}\sum a_n x^n = \sum n a_n x^{n-1} (within radius of convergence)

      Integration: ∫∑anxn dx=∑anxn+1n+1\int \sum a_n x^n \,dx = \sum \frac{a_n x^{n+1}}{n+1} (within radius of convergence)


      Radius of Convergence

      A power series ∑anxn\sum a_n x^n converges absolutely for ∣x∣<R|x| < R and diverges for ∣x∣>R|x| > R, where:

      R=lim⁡n→∞∣anan+1∣(ratio test)R = \lim_{n \to \infty} \left|\frac{a_n}{a_{n+1}}\right| \quad \text{(ratio test)}

      At the endpoints x=±Rx = \pm R, convergence must be checked separately.


      Example Expansions

      Expand tan⁡x\tan x about x=0x = 0:

      Since tan⁡x\tan x is odd, write y=tan⁡xy = \tan x. Taking derivatives:

      • y(0)=0y(0) = 0
      • y′=sec⁡2xy' = \sec^2 x, so y′(0)=1y'(0) = 1
      • y′′=2sec⁡2xtan⁡xy'' = 2\sec^2 x \tan x, so y′′(0)=0y''(0) = 0
      • y′′′=2sec⁡4x+4sec⁡2xtan⁡2xy''' = 2\sec^4 x + 4\sec^2 x\tan^2 x, so y′′′(0)=2y'''(0) = 2

      Thus: tan⁡x=x+x33+O(x5)\tan x = x + \frac{x^3}{3} + O(x^5)

      Expand (1+x)ln⁡(1+x)(1+x)\ln(1+x):

      Using ln⁡(1+x)=x−x22+x33−⋯\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots:

      (1+x)ln⁡(1+x)=x+(1−12)x2+(13−12)x3+⋯=x+x22−x36+⋯(1+x)\ln(1+x) = x + \left(1 - \frac{1}{2}\right)x^2 + \left(\frac{1}{3} - \frac{1}{2}\right)x^3 + \cdots = x + \frac{x^2}{2} - \frac{x^3}{6} + \cdots


      Why Taylor Series Matter

      • Approximation: Truncating gives polynomial approximations
      • Limits: Evaluate indeterminate forms (L’Hôpital’s rule basis)
      • Numerical methods: Derive finite difference schemes
      • Physics: Linearisation of nonlinear equations
      • ODEs: Power series solutions

      The Taylor series is the bridge between “nice” functions (exponential, trig) and polynomials, which we can compute with.

    • Series Convergence

      Definitions

      A series is ∑k=0∞uk\sum_{k=0}^{\infty} u_k. The partial sums are Sn=∑k=0nukS_n = \sum_{k=0}^{n} u_k.

      Convergence: The series converges if lim⁡n→∞Sn=S\lim_{n \to \infty} S_n = S exists (finite). We write ∑uk=S\sum u_k = S.

      Formal definition: For any ε>0\varepsilon > 0, there exists NN such that ∣S−Sn∣<ε|S - S_n| < \varepsilon for all n>Nn > N.


      Necessary Condition for Convergence

      If ∑uk\sum u_k converges, then lim⁡k→∞uk=0\lim_{k \to \infty} u_k = 0.

      Warning: The converse is false. Example: harmonic series ∑1k\sum \frac{1}{k} diverges even though 1k→0\frac{1}{k} \to 0.


      Absolute vs Conditional Convergence

      Absolute convergence: ∑∣uk∣\sum |u_k| converges.

      Conditional convergence: ∑uk\sum u_k converges but ∑∣uk∣\sum |u_k| diverges.

      Key theorem: If ∑∣uk∣\sum |u_k| converges, then ∑uk\sum u_k converges.

      Riemann rearrangement theorem: Conditionally convergent series can be rearranged to converge to any value (or diverge). Absolutely convergent series are immune to rearrangement.


      Comparison Test

      For series with 0≤uk≤vk0 \leq u_k \leq v_k:

      1. If ∑vk\sum v_k converges, then ∑uk\sum u_k converges
      2. If ∑uk\sum u_k diverges, then ∑vk\sum v_k diverges

      Limit comparison: If lim⁡k→∞ukvk=L\lim_{k \to \infty} \frac{u_k}{v_k} = L with 0<L<∞0 < L < \infty, then ∑uk\sum u_k and ∑vk\sum v_k behave the same.

      Example: ∑1k(k+1)\sum \frac{1}{k(k+1)} converges since 1k(k+1)<1k2\frac{1}{k(k+1)} < \frac{1}{k^2} and ∑1k2\sum \frac{1}{k^2} converges.


      Ratio Test

      Let L=lim⁡k→∞∣uk+1uk∣L = \lim_{k \to \infty} \left|\frac{u_{k+1}}{u_k}\right|:

      • L<1L < 1: series converges absolutely
      • L>1L > 1: series diverges
      • L=1L = 1: test is inconclusive

      Example: ∑k22k\sum \frac{k^2}{2^k}. Ratio: (k+1)2/2k+1k2/2k=12(1+1k)2→12<1\frac{(k+1)^2/2^{k+1}}{k^2/2^k} = \frac{1}{2}\left(1 + \frac{1}{k}\right)^2 \to \frac{1}{2} < 1. Converges.


      Harmonic Series Divergence

      ∑k=1∞1k=1+12+13+14+⋯\sum_{k=1}^{\infty} \frac{1}{k} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots

      Harmonic series partial sums, showing slow divergence with proof by grouping terms into blocks > 1/2 each

      Proof of divergence (grouping):

      1+12+(13+14)+(15+16+17+18)+⋯1 + \frac{1}{2} + \left(\frac{1}{3} + \frac{1}{4}\right) + \left(\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}\right) + \cdots

      >1+12+12+12+⋯> 1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \cdots

      The partial sums grow without bound.


      Alternating Series (Leibniz Criterion)

      An alternating series ∑(−1)kak\sum (-1)^k a_k with ak>0a_k > 0 converges if:

      1. aka_k is monotonically decreasing for large kk
      2. lim⁡k→∞ak=0\lim_{k \to \infty} a_k = 0

      Example: The alternating harmonic series: ∑k=1∞(−1)k+1k=1−12+13−14+⋯=ln⁡2\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots = \ln 2

      This is conditionally convergent (not absolutely, since harmonic series diverges).


      Riemann Zeta Function

      ζ(p)=∑k=1∞1kp\zeta(p) = \sum_{k=1}^{\infty} \frac{1}{k^p}

      • Converges for p>1p > 1
      • Diverges for p≤1p \leq 1

      Special values:

      • ζ(2)=π26\zeta(2) = \frac{\pi^2}{6}
      • ζ(4)=π490\zeta(4) = \frac{\pi^4}{90}

      Radius of Convergence for Power Series

      For ∑anxn\sum a_n x^n, the radius of convergence is:

      R=lim⁡n→∞∣anan+1∣(if limit exists)R = \lim_{n \to \infty} \left|\frac{a_n}{a_{n+1}}\right| \quad \text{(if limit exists)}

      • Converges absolutely for ∣x∣<R|x| < R
      • Diverges for ∣x∣>R|x| > R
      • At x=±Rx = \pm R: must check separately

      Example: ∑xnn\sum \frac{x^n}{n}.

      R=lim⁡1/n1/(n+1)=1R = \lim \frac{1/n}{1/(n+1)} = 1. Converges for ∣x∣<1|x| < 1.

      At x=1x = 1: harmonic series diverges. At x=−1x = -1: alternating harmonic converges.


      Summary: Choosing a Test

      SituationPreferred Test
      Comparison with known seriesComparison test
      Factorials, exponentialsRatio test
      Alternating signsLeibniz criterion
      Powers 1/kp1/k^pp-test (zeta)
      Power seriesRadius of convergence

      Complexity: Ratio test is often easiest to apply. Comparison test requires knowing a suitable reference series.

    • Integration Fundamentals

      The Integral as Limit of a Sum

      The Riemann integral is defined as:

      ∫abf(x) dx=lim⁡N→∞∑i=1Nf(xi) δx\int_a^b f(x)\,dx = \lim_{N \to \infty} \sum_{i=1}^{N} f(x_i)\,\delta x

      where δx=(b−a)/N\delta x = (b-a)/N and xix_i are sample points in each subinterval.

      Interpretation: Area under the curve, with rectangles becoming infinitesimally thin.


      Fundamental Theorem of Calculus

      First form: If F′(x)=f(x)F'(x) = f(x), then:

      ∫abf(x) dx=F(b)−F(a)\int_a^b f(x)\,dx = F(b) - F(a)

      Second form: Define:

      F(x)=∫axf(t) dtF(x) = \int_a^x f(t)\,dt

      Then F′(x)=f(x)F'(x) = f(x).

      Significance: Differentiation and integration are inverse operations.


      Differentiation Under the Integral Sign

      For I(q)=∫abf(x,q) dxI(q) = \int_a^b f(x, q)\,dx:

      dIdq=∫ab∂f∂q dx\frac{dI}{dq} = \int_a^b \frac{\partial f}{\partial q}\,dx

      If limits depend on qq:

      ddq∫a(q)b(q)f(x,q) dx=f(b,q)dbdq−f(a,q)dadq+∫ab∂f∂q dx\frac{d}{dq}\int_{a(q)}^{b(q)} f(x, q)\,dx = f(b, q)\frac{db}{dq} - f(a, q)\frac{da}{dq} + \int_a^b \frac{\partial f}{\partial q}\,dx


      Standard Techniques

      Substitution

      ∫f(g(x))g′(x) dx=∫f(u) duwhere u=g(x)\int f(g(x))g'(x)\,dx = \int f(u)\,du \quad \text{where } u = g(x)

      Example: ∫2xcos⁡(x2) dx\int 2x\cos(x^2)\,dx. Let u=x2u = x^2, du=2x dxdu = 2x\,dx:

      ∫cos⁡u du=sin⁡u=sin⁡(x2)\int \cos u\,du = \sin u = \sin(x^2)

      Integration by Parts

      ∫u dvdx dx=uv−∫v dudx dx\int u\,\frac{dv}{dx}\,dx = uv - \int v\,\frac{du}{dx}\,dx

      Example: ∫xex dx\int x e^x\,dx.

      Let u=xu = x, dv=ex dxdv = e^x\,dx. Then du=dxdu = dx, v=exv = e^x:

      ∫xex dx=xex−∫ex dx=xex−ex+C\int x e^x\,dx = xe^x - \int e^x\,dx = xe^x - e^x + C

      Partial Fractions

      For rational functions, decompose into simpler fractions.

      1(x−1)(x+2)=Ax−1+Bx+2\frac{1}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2}

      Solving: 1=A(x+2)+B(x−1)1 = A(x+2) + B(x-1). At x=1x = 1: A=1/3A = 1/3. At x=−2x = -2: B=−1/3B = -1/3.


      Improper Integrals

      Type 1: Infinite limits

      ∫a∞f(x) dx=lim⁡b→∞∫abf(x) dx\int_a^{\infty} f(x)\,dx = \lim_{b \to \infty} \int_a^b f(x)\,dx

      Converges if limit exists (finite).

      Type 2: Unbounded integrand

      ∫abdxx−a=lim⁡ε→0+∫a+εbdxx−a\int_a^b \frac{dx}{\sqrt{x-a}} = \lim_{\varepsilon \to 0^+} \int_{a+\varepsilon}^b \frac{dx}{\sqrt{x-a}}

      Comparison test for improper integrals

      If 0≤f(x)≤g(x)0 \leq f(x) \leq g(x) and ∫g\int g converges, then ∫f\int f converges.


      Symmetry and Integration

      Even function: f(−x)=f(x)f(-x) = f(x)

      ∫−aaf(x) dx=2∫0af(x) dx\int_{-a}^{a} f(x)\,dx = 2\int_0^a f(x)\,dx

      Odd function: f(−x)=−f(x)f(-x) = -f(x)

      ∫−aaf(x) dx=0\int_{-a}^{a} f(x)\,dx = 0

      Even function (symmetric) and odd function (antisymmetric) showing integration areas that cancel or double

      This simplifies many integrals, especially for trigonometric and polynomial terms.


      Mean Value Theorem for Integrals

      For continuous ff on [a,b][a, b], there exists c∈[a,b]c \in [a, b] such that:

      ∫abf(x) dx=f(c)(b−a)\int_a^b f(x)\,dx = f(c)(b - a)

      The integral equals the function value at some point times the interval width.


      Integration by Inspection

      Recognise standard forms and their derivatives:

      IntegralResult
      ∫f′f dx\int \frac{f'}{f}\,dxln⁡∥f∥\ln\|f\|
      ∫f′(x)[f(x)]n dx\int f'(x)[f(x)]^n\,dx[f(x)]n+1n+1\frac{[f(x)]^{n+1}}{n+1}
      ∫1a2+x2 dx\int \frac{1}{a^2 + x^2}\,dx1atan⁡−1(x/a)\frac{1}{a}\tan^{-1}(x/a)
      ∫1a2−x2 dx\int \frac{1}{\sqrt{a^2 - x^2}}\,dxsin⁡−1(x/a)\sin^{-1}(x/a)

      The key is identifying patterns that match derivative rules in reverse.

    • Multiple Integrals

      Double Integrals

      The double integral over a region RR in the xyxy-plane:

      ∬Rf(x,y) dx dy\iint_R f(x, y)\,dx\,dy

      Iterated integrals: For rectangular region a≤x≤ba \leq x \leq b, c≤y≤dc \leq y \leq d:

      ∬Rf dx dy=∫ab[∫cdf(x,y) dy]dx\iint_R f\,dx\,dy = \int_a^b \left[\int_c^d f(x, y)\,dy\right]dx

      Order can often be swapped (Fubini’s theorem).


      Triple Integrals

      ∭Vf(x,y,z) dx dy dz\iiint_V f(x, y, z)\,dx\,dy\,dz

      The triple integral gives a volume when f=1f = 1.


      Change of Variables

      Under transformation (x,y)→(u,v)(x, y) \to (u, v):

      ∬Rf(x,y) dx dy=∬R′f(x(u,v),y(u,v)) ∣J∣ du dv\iint_R f(x, y)\,dx\,dy = \iint_{R'} f(x(u,v), y(u,v))\,|J|\,du\,dv

      where the Jacobian is:

      Change of variables showing how area element transforms: dx dy becomes |J| du dv

      J=∂(x,y)∂(u,v)=∣∂x∂u∂x∂v∂y∂u∂y∂v∣J = \frac{\partial(x, y)}{\partial(u, v)} = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix}

      The absolute value ∣J∣|J| accounts for the change in area element.


      Polar Coordinates

      Transformation: x=rcos⁡θx = r\cos\theta, y=rsin⁡θy = r\sin\theta

      Jacobian:

      J=∣cos⁡θ−rsin⁡θsin⁡θrcos⁡θ∣=rcos⁡2θ+rsin⁡2θ=rJ = \begin{vmatrix} \cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta \end{vmatrix} = r\cos^2\theta + r\sin^2\theta = r

      Area element: dA=r dr dθdA = r\,dr\,d\theta

      Double integral:

      ∬Rf dA=∫θ1θ2∫r1(θ)r2(θ)f(rcos⁡θ,rsin⁡θ) r dr dθ\iint_R f\,dA = \int_{\theta_1}^{\theta_2} \int_{r_1(\theta)}^{r_2(\theta)} f(r\cos\theta, r\sin\theta)\,r\,dr\,d\theta


      Cylindrical Coordinates

      (x,y,z)→(r,θ,z)(x, y, z) \to (r, \theta, z) with x=rcos⁡θx = r\cos\theta, y=rsin⁡θy = r\sin\theta, z=zz = z.

      Volume element: dV=r dr dθ dzdV = r\,dr\,d\theta\,dz

      Useful for: cylinders, circular symmetry about zz-axis.


      Spherical Coordinates

      (x,y,z)→(r,θ,ϕ)(x, y, z) \to (r, \theta, \phi) with:

      • x=rsin⁡θcos⁡ϕx = r\sin\theta\cos\phi
      • y=rsin⁡θsin⁡ϕy = r\sin\theta\sin\phi
      • z=rcos⁡θz = r\cos\theta

      where r≥0r \geq 0, θ∈[0,π]\theta \in [0, \pi], ϕ∈[0,2π)\phi \in [0, 2\pi).

      Jacobian: J=r2sin⁡θJ = r^2\sin\theta

      Volume element: dV=r2sin⁡θ dr dθ dϕdV = r^2\sin\theta\,dr\,d\theta\,d\phi

      Derivation:

      J=∣sin⁡θcos⁡ϕrcos⁡θcos⁡ϕ−rsin⁡θsin⁡ϕsin⁡θsin⁡ϕrcos⁡θsin⁡ϕrsin⁡θcos⁡ϕcos⁡θ−rsin⁡θ0∣=r2sin⁡θJ = \begin{vmatrix} \sin\theta\cos\phi & r\cos\theta\cos\phi & -r\sin\theta\sin\phi \\ \sin\theta\sin\phi & r\cos\theta\sin\phi & r\sin\theta\cos\phi \\ \cos\theta & -r\sin\theta & 0 \end{vmatrix} = r^2\sin\theta


      Example: Volume of a Sphere

      Using spherical coordinates:

      V=∭dV=∫02π∫0π∫0Rr2sin⁡θ dr dθ dϕV = \iiint dV = \int_0^{2\pi} \int_0^\pi \int_0^R r^2\sin\theta\,dr\,d\theta\,d\phi

      =[r33]0R⋅[−cos⁡θ]0π⋅[ϕ]02π= \left[\frac{r^3}{3}\right]_0^R \cdot \left[-\cos\theta\right]_0^\pi \cdot \left[\phi\right]_0^{2\pi}

      =R33⋅2⋅2π=4πR33= \frac{R^3}{3} \cdot 2 \cdot 2\pi = \frac{4\pi R^3}{3}


      Example: Gaussian Integral

      Evaluate I=∫−∞∞e−x2 dxI = \int_{-\infty}^{\infty} e^{-x^2}\,dx.

      Method: Square and convert to polar.

      I2=∫−∞∞e−x2 dx∫−∞∞e−y2 dy=∬−∞∞e−(x2+y2) dx dyI^2 = \int_{-\infty}^{\infty} e^{-x^2}\,dx \int_{-\infty}^{\infty} e^{-y^2}\,dy = \iint_{-\infty}^{\infty} e^{-(x^2+y^2)}\,dx\,dy

      Convert to polar: x2+y2=r2x^2 + y^2 = r^2, dx dy=r dr dθdx\,dy = r\,dr\,d\theta

      I2=∫02π∫0∞e−r2r dr dθ=2π[−12e−r2]0∞=πI^2 = \int_0^{2\pi} \int_0^{\infty} e^{-r^2} r\,dr\,d\theta = 2\pi \left[-\frac{1}{2}e^{-r^2}\right]_0^\infty = \pi

      Result:

      ∫−∞∞e−x2 dx=π\int_{-\infty}^{\infty} e^{-x^2}\,dx = \sqrt{\pi}


      Choosing Coordinates

      GeometryBest Coordinates
      Rectangular boxCartesian
      Cylinder, circular baseCylindrical
      Sphere, central forceSpherical
      EllipseModified polar or elliptical

      Match the coordinate system to the domain boundaries for simplest limits.

    • Probability Theory

      Basic Definitions

      Sample space Ω\Omega: set of all possible outcomes.

      Event: subset of Ω\Omega.

      Probability: P(A)∈[0,1]P(A) \in [0, 1] for event AA. Properties:

      • P(Ω)=1P(\Omega) = 1
      • P(A∪B)=P(A)+P(B)−P(A∩B)P(A \cup B) = P(A) + P(B) - P(A \cap B) for disjoint events: P(A∪B)=P(A)+P(B)P(A \cup B) = P(A) + P(B)

      Conditional Probability

      P(A∣B)=P(A∩B)P(B)P(A|B) = \frac{P(A \cap B)}{P(B)}

      “Probability of AA given that BB has occurred.”

      Bayes’ theorem:

      P(A∣B)=P(B∣A)P(A)P(B)P(A|B) = \frac{P(B|A)P(A)}{P(B)}

      Independence: AA and BB are independent if P(A∩B)=P(A)P(B)P(A \cap B) = P(A)P(B).


      Random Variables

      A random variable XX is a function from Ω\Omega to R\mathbb{R}.

      Discrete: Takes countably many values.

      Continuous: Takes values in an interval or R\mathbb{R}.


      Probability Distributions

      Discrete: Probability Mass Function (PMF)

      p(xi)=P(X=xi)p(x_i) = P(X = x_i)

      Requirements: p(xi)≥0p(x_i) \geq 0 and ∑ip(xi)=1\sum_i p(x_i) = 1.

      Continuous: Probability Density Function (PDF)

      f(x)f(x) such that:

      P(a≤X≤b)=∫abf(x) dxP(a \leq X \leq b) = \int_a^b f(x)\,dx

      Requirements: f(x)≥0f(x) \geq 0 and ∫−∞∞f(x) dx=1\int_{-\infty}^{\infty} f(x)\,dx = 1.


      Expectation Value (Mean)

      Discrete: E[X]=μ=∑ixip(xi)E[X] = \mu = \sum_i x_i p(x_i)

      Continuous: E[X]=μ=∫−∞∞xf(x) dxE[X] = \mu = \int_{-\infty}^{\infty} x f(x)\,dx

      Linearity: E[aX+bY]=aE[X]+bE[Y]E[aX + bY] = aE[X] + bE[Y]


      Variance and Standard Deviation

      Variance: Var(X)=σ2=E[(X−μ)2]=E[X2]−(E[X])2\text{Var}(X) = \sigma^2 = E[(X - \mu)^2] = E[X^2] - (E[X])^2

      Standard deviation: σ=Var(X)\sigma = \sqrt{\text{Var}(X)}

      Properties:

      • Var(aX)=a2Var(X)\text{Var}(aX) = a^2\text{Var}(X)
      • Var(X+c)=Var(X)\text{Var}(X + c) = \text{Var}(X)

      Moments

      The nnth moment is E[Xn]E[X^n].

      The nnth central moment is E[(X−μ)n]E[(X - \mu)^n].

      • 1st moment = mean
      • 2nd central moment = variance
      • 3rd central moment (scaled) = skewness
      • 4th central moment (scaled) = kurtosis

      Binomial Distribution

      nn independent trials, each with success probability pp.

      P(X=r)=(nr)pr(1−p)n−rP(X = r) = \binom{n}{r} p^r (1-p)^{n-r}

      where (nr)=n!r!(n−r)!\binom{n}{r} = \frac{n!}{r!(n-r)!}.

      Mean: E[X]=npE[X] = np

      Variance: Var(X)=np(1−p)\text{Var}(X) = np(1-p)


      Poisson Distribution

      For rare events with average rate λ\lambda:

      P(X=r)=λre−λr!P(X = r) = \frac{\lambda^r e^{-\lambda}}{r!}

      Mean: E[X]=λE[X] = \lambda

      Variance: Var(X)=λ\text{Var}(X) = \lambda

      Limit: Poisson is the limit of Binomial as n→∞n \to \infty, p→0p \to 0 with np=λnp = \lambda fixed.


      Normal (Gaussian) Distribution

      f(x)=1σ2πexp⁡(−(x−μ)22σ2)f(x) = \frac{1}{\sigma\sqrt{2\pi}} \exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right)

      Normal distribution bell curve with mean μ and standard deviation σ marked, showing 68-95-99.7 rule

      Notation: X∼N(μ,σ2)X \sim N(\mu, \sigma^2).

      Mean: μ\mu

      Variance: σ2\sigma^2

      Standard normal (μ=0\mu = 0, σ=1\sigma = 1):

      f(z)=12πe−z2/2f(z) = \frac{1}{\sqrt{2\pi}} e^{-z^2/2}

      Cumulative distribution function: Φ(z)=12π∫−∞ze−t2/2 dt\Phi(z) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{z} e^{-t^2/2}\,dt


      Central Limit Theorem

      If X1,X2,…,XnX_1, X_2, \ldots, X_n are independent, identically distributed random variables with mean μ\mu and variance σ2\sigma^2, then:

      Xˉ−μσ/n→N(0,1)as n→∞\frac{\bar{X} - \mu}{\sigma/\sqrt{n}} \to N(0, 1) \quad \text{as } n \to \infty

      where Xˉ=1n∑Xi\bar{X} = \frac{1}{n}\sum X_i.

      Significance: Sample means tend toward normal distribution, regardless of the underlying distribution (provided it has finite variance).


      Stirling’s Approximation

      ln⁡(n!)≈nln⁡n−n\ln(n!) \approx n\ln n - n

      More precisely: n!≈nne−n2πnn! \approx n^n e^{-n} \sqrt{2\pi n}

      Derivation sketch: ln⁡(n!)=∑k=1nln⁡k≈∫1nln⁡x dx=nln⁡n−n+1\ln(n!) = \sum_{k=1}^{n} \ln k \approx \int_1^n \ln x\,dx = n\ln n - n + 1

      Use in probability: Essential for approximating binomial coefficients in limiting cases.

    • Methods I Tripos Questions

      This section contains worked examples from past Tripos papers relevant to Methods I topics.


      Limits and Continuity

      Tripos 2016, Paper 2, Question 19Y

      Calculate for real xx and constant aa:

      (i) lim⁡x→0tan⁡xsin⁡xsin⁡3x\displaystyle\lim_{x \to 0} \frac{\tan x \sin x}{\sin^3 x}

      (ii) lim⁡x→asin⁡x−sin⁡ax−a\displaystyle\lim_{x \to a} \frac{\sin x - \sin a}{x - a}

      (iii) lim⁡x→∞(x+ax−a)x+a\displaystyle\lim_{x \to \infty} \left(\frac{x + a}{x - a}\right)^{x+a}

      Solutions:

      (i) Using tan⁡x=sin⁡xcos⁡x\tan x = \frac{\sin x}{\cos x}:

      lim⁡x→0sin⁡2xcos⁡xsin⁡3x=lim⁡x→01cos⁡xsin⁡x\lim_{x \to 0} \frac{\sin^2 x}{\cos x \sin^3 x} = \lim_{x \to 0} \frac{1}{\cos x \sin x}

      This doesn’t exist (goes to ∞\infty). Actually, let’s reconsider:

      tan⁡xsin⁡xsin⁡3x=1cos⁡xsin⁡2x\frac{\tan x \sin x}{\sin^3 x} = \frac{1}{\cos x \sin^2 x}

      Using sin⁡x∼x\sin x \sim x near 0: this behaves like 1x2\frac{1}{x^2}, so the limit is ∞\infty.

      (ii) This is the derivative of sin⁡x\sin x at x=ax = a:

      lim⁡x→asin⁡x−sin⁡ax−a=cos⁡a\lim_{x \to a} \frac{\sin x - \sin a}{x - a} = \cos a

      Alternatively, use the identity sin⁡x−sin⁡a=2cos⁡x+a2sin⁡x−a2\sin x - \sin a = 2\cos\frac{x+a}{2}\sin\frac{x-a}{2}.

      (iii) Let y=(x+ax−a)x+ay = \left(\frac{x+a}{x-a}\right)^{x+a}. Then:

      ln⁡y=(x+a)ln⁡(1+2ax−a)\ln y = (x+a)\ln\left(1 + \frac{2a}{x-a}\right)

      For large xx: ln⁡(1+2ax−a)∼2ax−a\ln\left(1 + \frac{2a}{x-a}\right) \sim \frac{2a}{x-a}

      So ln⁡y∼(x+a)⋅2ax−a→2a\ln y \sim (x+a) \cdot \frac{2a}{x-a} \to 2a as x→∞x \to \infty.

      Therefore: y→e2ay \to e^{2a}


      Tripos 2008, Paper 1, Question 19X

      A function is defined for integer n∈{0,1,2,3,4}n \in \{0, 1, 2, 3, 4\}:

      ϕ(x,n)={xnsin⁡(1/x)x≠00x=0,n≠41x=0,n=4\phi(x, n) = \begin{cases} x^n \sin(1/x) & x \neq 0 \\ 0 & x = 0, n \neq 4 \\ 1 & x = 0, n = 4 \end{cases}

      (i) Determine values of nn for which ϕ(x,n)\phi(x,n) tends to a limit as x→0x \to 0.

      (ii) Determine values of nn for which ϕ(x,n)\phi(x,n) is continuous at x=0x = 0.

      (iii) Determine values of nn for which ϕ(x,n)\phi(x,n) is differentiable at x=0x = 0.

      Solutions:

      (i) Limit exists if n>0n > 0. For n=0n = 0: sin⁡(1/x)\sin(1/x) oscillates. For n≥1n \geq 1: ∣xnsin⁡(1/x)∣≤∣x∣n→0|x^n \sin(1/x)| \leq |x|^n \to 0.

      (ii) For continuity, need lim⁡x→0ϕ=ϕ(0)\lim_{x \to 0} \phi = \phi(0). The limit is 0 for n≥1n \geq 1, so we need ϕ(0)=0\phi(0) = 0. This holds for n∈{1,2,3}n \in \{1, 2, 3\}.

      (iii) For differentiability:

      ϕ′(0)=lim⁡h→0ϕ(h)−ϕ(0)h=lim⁡h→0hnsin⁡(1/h)h=lim⁡h→0hn−1sin⁡(1/h)\phi'(0) = \lim_{h \to 0} \frac{\phi(h) - \phi(0)}{h} = \lim_{h \to 0} \frac{h^n \sin(1/h)}{h} = \lim_{h \to 0} h^{n-1}\sin(1/h)

      This exists (equals 0) if n−1>0n - 1 > 0, i.e., n∈{2,3}n \in \{2, 3\}.

      For n=1n = 1: oscillates. For n=0n = 0: unbounded.


      Vectors

      Tripos 2005, Paper 1, Question 5C

      Consider tetrahedron with vertices at O(0,0,0)O(0,0,0), A(1,0,0)A(1,0,0), B(0,1,0)B(0,1,0), C(0,0,1)C(0,0,1).

      (a) Find the cosines of angles between faces.

      (b) Find the areas of the four faces.

      (c) Find the distance from OO to face ABCABC.

      Solution outline:

      (a) Normal to face OABOAB is in zz-direction. Normal to OBCOBC is in xx-direction. These are perpendicular. Angle between faces OABOAB and OACOAC: normals are (0,0,1)(0,0,1) and (0,1,0)(0,1,0), angle is 90°90°.

      Face ABCABC has normal (1,1,1)(1,1,1) (from cross product). Angle with other faces uses dot product.

      (b) Face OABOAB is right triangle with area 1/21/2. Similarly for OBCOBC, OACOAC. Face ABCABC has area 32\frac{\sqrt{3}}{2}.

      (c) Equation of face ABCABC: r⋅(1,1,1)=1\mathbf{r} \cdot (1,1,1) = 1, or x+y+z=1x + y + z = 1.

      Distance from OO: d=13d = \frac{1}{\sqrt{3}}.


      Tripos 2004, Paper 1, Question 1A

      Given vectors a\mathbf{a}, b\mathbf{b}, c\mathbf{c} with a⋅(b×c)≠0\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) \neq 0, define reciprocal basis (a∗,b∗,c∗)(\mathbf{a}^*, \mathbf{b}^*, \mathbf{c}^*) by:

      a∗=b×ca⋅(b×c)\mathbf{a}^* = \frac{\mathbf{b} \times \mathbf{c}}{\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})}

      with analogous definitions for b∗\mathbf{b}^*, c∗\mathbf{c}^*.

      Show that a∗⋅a=1\mathbf{a}^* \cdot \mathbf{a} = 1 and a∗⋅b=0\mathbf{a}^* \cdot \mathbf{b} = 0.

      Solution:

      a∗⋅a=a⋅(b×c)a⋅(b×c)=1\mathbf{a}^* \cdot \mathbf{a} = \frac{\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})}{\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})} = 1

      a∗⋅b=(b×c)⋅ba⋅(b×c)\mathbf{a}^* \cdot \mathbf{b} = \frac{(\mathbf{b} \times \mathbf{c}) \cdot \mathbf{b}}{\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})}

      But (b×c)⋅b=[b,c,b]=0(\mathbf{b} \times \mathbf{c}) \cdot \mathbf{b} = [\mathbf{b}, \mathbf{c}, \mathbf{b}] = 0 (repeated vector in scalar triple product).


      Series

      Tripos 2014, Paper 1

      Find the expansion of ln⁡(1+x+x2)\ln(1 + x + x^2) for:

      (i) small xx

      (ii) large xx

      Solution:

      (i) 1+x+x2=1+x+x21 + x + x^2 = 1 + x + x^2. Let u=x+x2u = x + x^2, then for small uu:

      ln⁡(1+u)=u−u22+u33−⋯\ln(1+u) = u - \frac{u^2}{2} + \frac{u^3}{3} - \cdots

      Substitute back: ln⁡(1+x+x2)=x+x2−(x+x2)22+⋯\ln(1+x+x^2) = x + x^2 - \frac{(x+x^2)^2}{2} + \cdots

      (ii) For large xx: ln⁡(1+x+x2)=ln⁡(x2)+ln⁡(1x2+1x+1)\ln(1+x+x^2) = \ln(x^2) + \ln\left(\frac{1}{x^2} + \frac{1}{x} + 1\right)

      Let w=1x+1x2w = \frac{1}{x} + \frac{1}{x^2}. Then:

      =2ln⁡x+w−w22+⋯= 2\ln x + w - \frac{w^2}{2} + \cdots

  • Methods II (Lent Term)

    First-order and second-order ordinary differential equations, oscillations, multivariable calculus (gradients, divergence, curl), integration theorems, and Fourier series

    • First Order ODEs

      What is an ODE?

      An ordinary differential equation (ODE) relates an unknown function to its derivatives:

      F(x,y,dydx,d2ydx2,…)=0F\left(x, y, \frac{dy}{dx}, \frac{d^2y}{dx^2}, \ldots\right) = 0

      Order: the highest derivative appearing.

      Linear vs nonlinear: Linear means yy and its derivatives appear to the first power only.


      Integrable ODEs

      The simplest case: dydx=f(x)\frac{dy}{dx} = f(x)

      Solution by integration:

      y=∫f(x) dx+Cy = \int f(x)\,dx + C


      Separable ODEs

      Form: dydx=f(x)g(y)\frac{dy}{dx} = f(x)g(y)

      Method: Separate variables and integrate:

      ∫dyg(y)=∫f(x) dx\int \frac{dy}{g(y)} = \int f(x)\,dx

      Separation of variables showing dy terms with y and dx terms with x on opposite sides

      Example: Solve dydx=xy\frac{dy}{dx} = xy.

      Separate: dyy=x dx\frac{dy}{y} = x\,dx

      Integrate: ln⁡∣y∣=x22+C\ln|y| = \frac{x^2}{2} + C

      Solve: y=Aex2/2y = Ae^{x^2/2} where A=±eCA = \pm e^C.


      Linear First Order ODEs

      Standard form:

      dydx+P(x)y=Q(x)\frac{dy}{dx} + P(x)y = Q(x)

      Integrating Factor Method

      Multiply by integrating factor μ(x)=e∫P(x) dx\mu(x) = e^{\int P(x)\,dx}:

      μdydx+μPy=μQ\mu \frac{dy}{dx} + \mu P y = \mu Q

      The left side is ddx(μy)\frac{d}{dx}(\mu y), so:

      ddx(μy)=μQ\frac{d}{dx}(\mu y) = \mu Q

      Integrate:

      y=1μ∫μQ dx+Cμy = \frac{1}{\mu}\int \mu Q\,dx + \frac{C}{\mu}

      Example: Solve y′+3y=e−2xy' + 3y = e^{-2x}.

      P=3P = 3, so μ=e3x\mu = e^{3x}.

      ddx(e3xy)=e3xe−2x=ex\frac{d}{dx}(e^{3x}y) = e^{3x}e^{-2x} = e^x

      e3xy=ex+Ce^{3x}y = e^x + C

      y=e−2x+Ce−3xy = e^{-2x} + Ce^{-3x}


      Bernoulli Equation

      Form: dydx+P(x)y=Q(x)yn\frac{dy}{dx} + P(x)y = Q(x)y^n

      Method: Substitute z=y1−nz = y^{1-n} to transform to linear.

      Differentiating: dzdx=(1−n)y−ndydx\frac{dz}{dx} = (1-n)y^{-n}\frac{dy}{dx}

      Substituting gives a linear ODE in zz:

      dzdx+(1−n)Pz=(1−n)Q\frac{dz}{dx} + (1-n)Pz = (1-n)Q

      Example: Solve y′−xy=xy2y' - xy = xy^2.

      Here n=2n = 2. Let z=y−1z = y^{-1}, so dzdx=−y−2y′\frac{dz}{dx} = -y^{-2}y'.

      −y2dzdx−xyy−2=x-y^2\frac{dz}{dx} - xy y^{-2} = x

      −dzdx−xz=x-\frac{dz}{dx} - xz = x

      dzdx+xz=−x\frac{dz}{dx} + xz = -x

      This is linear with P=xP = x. The integrating factor μ=ex2/2\mu = e^{x^2/2}.

      Solve: z=−1+Ce−x2/2z = -1 + Ce^{-x^2/2}, so y=1−1+Ce−x2/2y = \frac{1}{-1 + Ce^{-x^2/2}}.


      Homogeneous Substitution

      Form: dydx=f(yx)\frac{dy}{dx} = f\left(\frac{y}{x}\right)

      Method: Let u=yxu = \frac{y}{x}, so y=uxy = ux and dydx=u+xdudx\frac{dy}{dx} = u + x\frac{du}{dx}.

      Substitute:

      u+xdudx=f(u)u + x\frac{du}{dx} = f(u)

      dudx=f(u)−ux\frac{du}{dx} = \frac{f(u) - u}{x}

      This is now separable.


      Exact Equations

      An equation dydx=−P(x,y)Q(x,y)\frac{dy}{dx} = -\frac{P(x,y)}{Q(x,y)} is exact if there exists F(x,y)F(x,y) such that:

      ∂F∂x=Pand∂F∂y=Q\frac{\partial F}{\partial x} = P \quad \text{and} \quad \frac{\partial F}{\partial y} = Q

      Test: ∂P∂y=∂Q∂x\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}

      Solution: The ODE becomes dF=0dF = 0, so F(x,y)=CF(x,y) = C.

      Find FF by integrating PP with respect to xx, then matching partial derivatives with QQ.


      Existence and Uniqueness

      For the initial value problem y′=f(x,y)y' = f(x,y), y(x0)=y0y(x_0) = y_0:

      If ff and ∂f∂y\frac{\partial f}{\partial y} are continuous near (x0,y0)(x_0, y_0), then a unique solution exists locally.

      Warning: Solutions may not exist for all xx (finite-time blow-up).


      Why These Methods Matter

      • Physical systems often yield ODEs
      • The classification tells you which tool to use
      • Numerical methods apply when analytical solutions don’t exist
      • Understanding behaviour (steady states, stability) comes from the solution structure
    • Second Order Linear ODEs

      Standard Form

      A second order linear ODE with constant coefficients:

      d2ydx2+adydx+by=f(x)\frac{d^2y}{dx^2} + a\frac{dy}{dx} + by = f(x)

      or equivalently:

      y′′+2γy′+ω02y=f(x)y'' + 2\gamma y' + \omega_0^2 y = f(x)


      Homogeneous Case (f(x)=0f(x) = 0)

      Try solution y=eλxy = e^{\lambda x}:

      λ2eλx+2γλeλx+ω02eλx=0\lambda^2 e^{\lambda x} + 2\gamma \lambda e^{\lambda x} + \omega_0^2 e^{\lambda x} = 0

      Dividing by eλxe^{\lambda x} gives the characteristic equation:

      λ2+2γλ+ω02=0\lambda^2 + 2\gamma\lambda + \omega_0^2 = 0

      Roots: λ=−γ±γ2−ω02\lambda = -\gamma \pm \sqrt{\gamma^2 - \omega_0^2}


      Three Cases for the Roots

      Three cases shown on complex plane: two real roots (overdamped), repeated root (critical), complex conjugate roots (underdamped)

      CaseConditionRootsGeneral Solution
      Two real distinctγ2>ω02\gamma^2 > \omega_0^2λ1,λ2\lambda_1, \lambda_2 (real)y=Aeλ1x+Beλ2xy = Ae^{\lambda_1 x} + Be^{\lambda_2 x}
      Repeated rootγ2=ω02\gamma^2 = \omega_0^2λ=−γ\lambda = -\gamma (double)y=(A+Bx)eλxy = (A + Bx)e^{\lambda x}
      Complex conjugateγ2<ω02\gamma^2 < \omega_0^2−γ±iΩ-\gamma \pm i\Omegay=e−γx(Acos⁡Ωx+Bsin⁡Ωx)y = e^{-\gamma x}(A\cos\Omega x + B\sin\Omega x)

      where Ω=ω02−γ2\Omega = \sqrt{\omega_0^2 - \gamma^2}.


      Derivation of the Repeated Root Solution

      When λ1=λ2\lambda_1 = \lambda_2, we only have one solution y1=eλxy_1 = e^{\lambda x}.

      Method of reduction of order: Seek second solution as y2=v(x)eλxy_2 = v(x)e^{\lambda x}.

      Substituting and solving gives v′=v' = constant, so v=Bxv = Bx.

      Hence y=(A+Bx)eλxy = (A + Bx)e^{\lambda x}.


      Inhomogeneous Case: Particular Integral

      Principle of Superposition

      General solution = Complementary Function + Particular Integral

      y=yc+ypy = y_c + y_p

      • ycy_c: general solution to homogeneous equation
      • ypy_p: any particular solution to inhomogeneous equation

      Finding Particular Integral

      Use trial solutions based on the form of f(x)f(x):

      f(x)f(x)Try ypy_p
      Polynomial of degree nnGeneral polynomial of degree nn
      FeKxFe^{Kx}AeKxAe^{Kx} (unless K=λ1K = \lambda_1 or λ2\lambda_2)
      Fsin⁡KxF\sin Kx or Fcos⁡KxF\cos KxAsin⁡Kx+Bcos⁡KxA\sin Kx + B\cos Kx
      Product/compoundCombine forms

      If trial form appears in ycy_c: Multiply by xx (or x2x^2 for double root).

      Examples

      Example 1: y′′+5y′+6y=12y'' + 5y' + 6y = 12

      Characteristic: λ2+5λ+6=0⇒λ=−2,−3\lambda^2 + 5\lambda + 6 = 0 \Rightarrow \lambda = -2, -3

      yc=Ae−2x+Be−3xy_c = Ae^{-2x} + Be^{-3x}

      Try yp=Cy_p = C. Then yp′′=0y_p'' = 0, yp′=0y_p' = 0, so 6C=126C = 12, giving C=2C = 2.

      y=Ae−2x+Be−3x+2y = Ae^{-2x} + Be^{-3x} + 2

      Example 2: y′′+4y=cos⁡xy'' + 4y = \cos x

      Characteristic: λ2+4=0⇒λ=±2i\lambda^2 + 4 = 0 \Rightarrow \lambda = \pm 2i

      yc=Acos⁡2x+Bsin⁡2xy_c = A\cos 2x + B\sin 2x

      Try yp=Ccos⁡x+Dsin⁡xy_p = C\cos x + D\sin x. Substituting:

      (−C+4C)cos⁡x+(−D+4D)sin⁡x=cos⁡x(-C + 4C)\cos x + (-D + 4D)\sin x = \cos x

      3Ccos⁡x+3Dsin⁡x=cos⁡x3C\cos x + 3D\sin x = \cos x

      C=13,D=0C = \frac{1}{3}, D = 0

      y=Acos⁡2x+Bsin⁡2x+13cos⁡xy = A\cos 2x + B\sin 2x + \frac{1}{3}\cos x


      Initial Value Problems

      Apply initial conditions y(x0)=y0y(x_0) = y_0, y′(x0)=y1y'(x_0) = y_1 to determine AA and BB.


      Physical Applications

      • Springs: mx¨+cx˙+kx=F(t)m\ddot{x} + c\dot{x} + kx = F(t)
      • Circuits: Lq¨+Rq˙+qC=V(t)L\ddot{q} + R\dot{q} + \frac{q}{C} = V(t)
      • Pendulum (small angle): θ¨+gLθ=0\ddot{\theta} + \frac{g}{L}\theta = 0

      The mathematical structure is universal; the physics provides the coefficients.

    • Damped Oscillations

      The Damped Harmonic Oscillator

      mx¨+cx˙+kx=0m\ddot{x} + c\dot{x} + kx = 0

      Or in standard form:

      x¨+2γx˙+ω02x=0\ddot{x} + 2\gamma\dot{x} + \omega_0^2 x = 0

      where:

      • γ=c2m\gamma = \frac{c}{2m} (damping coefficient, per unit mass)
      • ω0=km\omega_0 = \sqrt{\frac{k}{m}} (natural frequency, undamped)

      Three Damping Cases

      1. Overdamped (γ>ω0\gamma > \omega_0)

      Large damping, no oscillation.

      Overdamped response showing exponential decay without oscillation, returning slowly to equilibrium

      Characteristic equation roots are real, distinct, and negative:

      λ±=−γ±γ2−ω02<0\lambda_{\pm} = -\gamma \pm \sqrt{\gamma^2 - \omega_0^2} < 0

      Solution:

      x(t)=Aeλ+t+Beλ−tx(t) = Ae^{\lambda_+ t} + Be^{\lambda_- t}

      Decays monotonically to zero. The system returns to equilibrium without oscillating.


      2. Critically Damped (γ=ω0\gamma = \omega_0)

      Boundary case: fastest return to equilibrium without oscillation.

      Critically damped response showing fastest non-oscillatory decay to equilibrium

      Repeated root: λ=−γ\lambda = -\gamma

      Solution:

      x(t)=(A+Bt)e−γtx(t) = (A + Bt)e^{-\gamma t}

      Initially, the linear term may dominate if B≠0B \neq 0, but exponential decay eventually wins.

      Application: Car suspension systems are designed to be critically damped.


      3. Underdamped (γ<ω0\gamma < \omega_0)

      Small damping: oscillation with decaying amplitude.

      Underdamped response showing oscillatory decay with envelope e^(-γt)

      Complex conjugate roots: −γ±iΩ-\gamma \pm i\Omega where Ω=ω02−γ2\Omega = \sqrt{\omega_0^2 - \gamma^2}

      Solution:

      x(t)=e−γt(Acos⁡Ωt+Bsin⁡Ωt)x(t) = e^{-\gamma t}(A\cos\Omega t + B\sin\Omega t)

      Or equivalently:

      x(t)=Ce−γtcos⁡(Ωt−ϕ)x(t) = Ce^{-\gamma t}\cos(\Omega t - \phi)

      Properties:

      • Oscillates with frequency Ω\Omega (less than ω0\omega_0)
      • Amplitude decays as e−γte^{-\gamma t}
      • Period T=2πΩT = \frac{2\pi}{\Omega}

      Energy Dissipation

      For undamped oscillator: total energy E=12mx˙2+12kx2E = \frac{1}{2}m\dot{x}^2 + \frac{1}{2}kx^2 is conserved.

      With damping, multiply equation by x˙\dot{x}:

      mx¨x˙+cx˙2+kxx˙=0m\ddot{x}\dot{x} + c\dot{x}^2 + kx\dot{x} = 0

      ddt(12mx˙2+12kx2)=−cx˙2\frac{d}{dt}\left(\frac{1}{2}m\dot{x}^2 + \frac{1}{2}kx^2\right) = -c\dot{x}^2

      Energy decays at rate cx˙2c\dot{x}^2 (power dissipated by friction).


      Quality Factor (Q-factor)

      The Q-factor measures how underdamped the system is:

      Q=ω02γ=kmcQ = \frac{\omega_0}{2\gamma} = \frac{\sqrt{km}}{c}

      Interpretation:

      • Q≫1Q \gg 1: lightly damped, many oscillations
      • Q≈12Q \approx \frac{1}{2}: critically damped
      • QQ equals the number of oscillations for amplitude to decay by factor e−πe^{-\pi}

      Forced Oscillations

      Add external forcing:

      x¨+2γx˙+ω02x=f0cos⁡ωt\ddot{x} + 2\gamma\dot{x} + \omega_0^2 x = f_0\cos\omega t

      Particular Integral

      Try xp=acos⁡ωt+bsin⁡ωtx_p = a\cos\omega t + b\sin\omega t.

      Substituting and solving:

      a=f0(ω02−ω2)(ω02−ω2)2+4γ2ω2a = \frac{f_0(\omega_0^2 - \omega^2)}{(\omega_0^2 - \omega^2)^2 + 4\gamma^2\omega^2}

      b=2γωf0(ω02−ω2)2+4γ2ω2b = \frac{2\gamma\omega f_0}{(\omega_0^2 - \omega^2)^2 + 4\gamma^2\omega^2}

      Amplitude and Phase

      Write xp=μf0cos⁡(ωt−ϕ)x_p = \mu f_0\cos(\omega t - \phi) where:

      μ=1(ω02−ω2)2+4γ2ω2\mu = \frac{1}{\sqrt{(\omega_0^2 - \omega^2)^2 + 4\gamma^2\omega^2}}

      tan⁡ϕ=2γωω02−ω2\tan\phi = \frac{2\gamma\omega}{\omega_0^2 - \omega^2}


      Resonance

      Resonance occurs when the amplitude μ\mu is maximum.

      Resonance curve showing amplitude μ vs driving frequency ω, with peak near ω₀ and width controlled by γ

      For small γ\gamma: maximum near ω=ω0\omega = \omega_0.

      Resonant frequency (exact):

      ωres=ω02−2γ2\omega_{\text{res}} = \sqrt{\omega_0^2 - 2\gamma^2}

      At resonance:

      • Amplitude is large (inversely proportional to γ\gamma)
      • Phase shift is π/2\pi/2: response lags forcing by 90°90°

      Transients

      The complete solution is:

      x(t)=xc(t)+xp(t)x(t) = x_c(t) + x_p(t)

      where xcx_c (complementary function) decays as e−γte^{-\gamma t}.

      Transient behaviour: Initially dominated by xcx_c, which depends on initial conditions.

      Steady state: At large tt, only xpx_p remains (forced oscillation).

      Physical interpretation: The system “forgets” its initial state and responds only to the forcing.

    • Partial Derivatives

      Definition

      For f(x,y)f(x, y), the partial derivative with respect to xx is:

      ∂f∂x=lim⁡h→0f(x+h,y)−f(x,y)h\frac{\partial f}{\partial x} = \lim_{h \to 0} \frac{f(x+h, y) - f(x, y)}{h}

      Hold yy constant and differentiate with respect to xx.

      Alternative notation: fxf_x, ∂xf\partial_x f.


      Higher Derivatives

      Second partial derivatives:

      ∂2f∂x2=fxx,∂2f∂y2=fyy\frac{\partial^2 f}{\partial x^2} = f_{xx}, \quad \frac{\partial^2 f}{\partial y^2} = f_{yy}

      Mixed partials:

      ∂2f∂x∂y=fxy,∂2f∂y∂x=fyx\frac{\partial^2 f}{\partial x \partial y} = f_{xy}, \quad \frac{\partial^2 f}{\partial y \partial x} = f_{yx}

      Equality of mixed partials: For well-behaved functions (continuous second derivatives):

      fxy=fyxf_{xy} = f_{yx}


      Chain Rule

      One Independent Variable

      If f=f(x,y)f = f(x, y) and x=x(t)x = x(t), y=y(t)y = y(t):

      dfdt=∂f∂xdxdt+∂f∂ydydt\frac{df}{dt} = \frac{\partial f}{\partial x}\frac{dx}{dt} + \frac{\partial f}{\partial y}\frac{dy}{dt}

      Chain rule as flow diagram: t → (x, y) → f, showing partial derivatives along each path

      Two Independent Variables

      If f=f(x,y)f = f(x, y) and x=x(u,v)x = x(u, v), y=y(u,v)y = y(u, v):

      ∂f∂u=∂f∂x∂x∂u+∂f∂y∂y∂u\frac{\partial f}{\partial u} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial u} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial u}

      ∂f∂v=∂f∂x∂x∂v+∂f∂y∂y∂v\frac{\partial f}{\partial v} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial v} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial v}


      Implicit Differentiation

      For F(x,y)=0F(x, y) = 0:

      dydx=−∂F/∂x∂F/∂y\frac{dy}{dx} = -\frac{\partial F/\partial x}{\partial F/\partial y}

      Proof: Differentiate F(x,y(x))=0F(x, y(x)) = 0:

      ∂F∂x+∂F∂ydydx=0\frac{\partial F}{\partial x} + \frac{\partial F}{\partial y}\frac{dy}{dx} = 0


      Total Differential

      For f=f(x,y)f = f(x, y):

      df=∂f∂xdx+∂f∂ydydf = \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy

      This is the total differential: the linear approximation to the change in ff.


      Directional Derivative

      The rate of change of ff in direction n^=(nx,ny)\hat{\mathbf{n}} = (n_x, n_y):

      Dn^f=∇f⋅n^=∂f∂xnx+∂f∂ynyD_{\hat{\mathbf{n}}}f = \nabla f \cdot \hat{\mathbf{n}} = \frac{\partial f}{\partial x}n_x + \frac{\partial f}{\partial y}n_y

      The gradient ∇f\nabla f points in direction of steepest increase.


      Taylor Series for Two Variables

      f(a+h,b+k)=f(a,b)+hfx(a,b)+kfy(a,b)+12(h2fxx+2hkfxy+k2fyy)+⋯f(a+h, b+k) = f(a,b) + hf_x(a,b) + kf_y(a,b) + \frac{1}{2}(h^2f_{xx} + 2hkf_{xy} + k^2f_{yy}) + \cdots

      Compact form:

      f(a+h)=f(a)+∇f⋅h+12hTHh+⋯f(\mathbf{a} + \mathbf{h}) = f(\mathbf{a}) + \nabla f \cdot \mathbf{h} + \frac{1}{2}\mathbf{h}^T H \mathbf{h} + \cdots

      where HH is the Hessian matrix:

      H=(fxxfxyfyxfyy)H = \begin{pmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{pmatrix}


      Reciprocity and Cyclic Relations

      Reciprocity Relation

      For f(x,y,z)=0f(x, y, z) = 0:

      (∂x∂y)z=1(∂y∂x)z\left(\frac{\partial x}{\partial y}\right)_z = \frac{1}{\left(\frac{\partial y}{\partial x}\right)_z}

      The subscript indicates what variable is held constant.

      Cyclic Relation

      (∂x∂y)z⋅(∂y∂z)x⋅(∂z∂x)y=−1\left(\frac{\partial x}{\partial y}\right)_z \cdot \left(\frac{\partial y}{\partial z}\right)_x \cdot \left(\frac{\partial z}{\partial x}\right)_y = -1

      Application: Thermodynamics (relating partial derivatives of state functions).


      Why Partial Derivatives Matter

      • Optimisation: Find maxima and minima of multivariable functions
      • Thermodynamics: Relate state variables
      • PDEs: Model heat, waves, diffusion
      • Machine learning: Gradient descent optimises loss functions

      The key skill is determining which variables are independent vs dependent — this determines which derivatives make sense.

    • Exact Differentials

      The Total Differential

      For f=f(x,y)f = f(x, y):

      df=∂f∂xdx+∂f∂ydydf = \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy

      This describes an exact differential: the differential of a function.


      Definition

      An expression P(x,y)dx+Q(x,y)dyP(x,y)dx + Q(x,y)dy is exact if there exists f(x,y)f(x,y) such that:

      df=P dx+Q dydf = P\,dx + Q\,dy

      That is:

      • ∂f∂x=P\frac{\partial f}{\partial x} = P
      • ∂f∂y=Q\frac{\partial f}{\partial y} = Q

      Test for Exactness

      A differential P dx+Q dyP\,dx + Q\,dy is exact if and only if:

      ∂P∂y=∂Q∂x\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}

      Exact differential showing level curves of potential function f where df = P dx + Q dy

      Proof (if exact):

      • If df=P dx+Q dydf = P\,dx + Q\,dy, then P=fxP = f_x and Q=fyQ = f_y
      • Py=fxyP_y = f_{xy} and Qx=fyxQ_x = f_{yx}
      • By equality of mixed partials: Py=QxP_y = Q_x

      Finding the Potential Function

      If P dx+Q dyP\,dx + Q\,dy is exact, find ff by integration:

      Step 1: Integrate PP with respect to xx:

      f(x,y)=∫P(x,y) dx+g(y)f(x, y) = \int P(x, y)\,dx + g(y)

      where g(y)g(y) is an arbitrary function (the “constant” of integration).

      Step 2: Differentiate with respect to yy:

      ∂f∂y=∂∂y∫P dx+g′(y)\frac{\partial f}{\partial y} = \frac{\partial}{\partial y}\int P\,dx + g'(y)

      Step 3: Set equal to QQ and solve for g′(y)g'(y):

      Q(x,y)=∂∂y∫P dx+g′(y)Q(x, y) = \frac{\partial}{\partial y}\int P\,dx + g'(y)

      Step 4: Integrate g′(y)g'(y) to find g(y)g(y).


      Example

      Is (2xy+cos⁡y) dx+(x2−xsin⁡y) dy(2xy + \cos y)\,dx + (x^2 - x\sin y)\,dy exact?

      Check: Py=2x−sin⁡yP_y = 2x - \sin y, Qx=2x−sin⁡yQ_x = 2x - \sin y. Yes, exact.

      Find ff:

      Integrate P=2xy+cos⁡yP = 2xy + \cos y wrt xx:

      f=x2y+xcos⁡y+g(y)f = x^2y + x\cos y + g(y)

      Differentiate:

      fy=x2−xsin⁡y+g′(y)f_y = x^2 - x\sin y + g'(y)

      Set equal to Q=x2−xsin⁡yQ = x^2 - x\sin y:

      x2−xsin⁡y+g′(y)=x2−xsin⁡yx^2 - x\sin y + g'(y) = x^2 - x\sin y

      So g′(y)=0g'(y) = 0, hence g(y)=Cg(y) = C.

      f(x,y)=x2y+xcos⁡y+Cf(x, y) = x^2y + x\cos y + C


      Integrating Factors

      If P dx+Q dyP\,dx + Q\,dy is not exact, sometimes we can find μ(x,y)\mu(x, y) such that:

      μP dx+μQ dy\mu P\,dx + \mu Q\,dy

      is exact. Then μ\mu is an integrating factor.

      For the ODE P dx+Q dy=0P\,dx + Q\,dy = 0, we then solve:

      d(μf)=0⇒μf=constantd(\mu f) = 0 \Rightarrow \mu f = \text{constant}


      Physical Interpretation

      Thermodynamics: State functions (energy UU, entropy SS, etc.) have exact differentials.

      • dU=TdS−PdVdU = TdS - PdV is exact (energy is a state function)
      • Heat đQđQ and work đWđW are not exact (path-dependent)

      The notation đđ indicates inexact differentials.

      State functions: The change depends only on initial and final states, not the path.


      Exact vs Inexact: Summary

      PropertyExactInexact
      Definitiondfdf for some ffNot dfdf for any ff
      Test∂P/∂y=∂Q/∂x\partial P/\partial y = \partial Q/\partial x∂P/∂y≠∂Q/∂x\partial P/\partial y \neq \partial Q/\partial x
      Integral∫P dx+Q dy\int P\,dx + Q\,dy path-independentPath-dependent
      Example in physicsState functionsHeat, work

      Why Exact Differentials Matter

      • Identify when an ODE has a simple solution f(x,y)=Cf(x,y) = C
      • Connect to conservative fields in vector calculus
      • Fundamental to thermodynamics (state vs path functions)

      The exactness test quickly distinguishes solvable cases from ones needing integrating factors.

    • Stationary Points

      Definition

      A stationary point of f(x,y)f(x, y) occurs where:

      ∇f=0\nabla f = \mathbf{0}

      That is: ∂f∂x=0\frac{\partial f}{\partial x} = 0 and ∂f∂y=0\frac{\partial f}{\partial y} = 0.

      At a stationary point, the function is “flat” in all directions.


      Types of Stationary Points

      1. Local minimum: ff increases in all directions
      2. Local maximum: ff decreases in all directions
      3. Saddle point: ff increases in some directions, decreases in others

      3D surface showing three types: local minimum (bowl), local maximum (hill), and saddle point (hyperbolic paraboloid)


      The Hessian Matrix

      At a stationary point, the second-order behaviour is determined by the Hessian matrix:

      H=(fxxfxyfyxfyy)H = \begin{pmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{pmatrix}

      Since fxy=fyxf_{xy} = f_{yx} (for smooth functions), HH is symmetric.


      Classification Criterion

      Let D=det⁡(H)=fxxfyy−(fxy)2D = \det(H) = f_{xx}f_{yy} - (f_{xy})^2

      ConditionClassification
      D>0D > 0 and fxx>0f_{xx} > 0Local minimum
      D>0D > 0 and fxx<0f_{xx} < 0Local maximum
      D<0D < 0Saddle point
      D=0D = 0Inconclusive

      Why D>0D > 0 implies extremum: Both eigenvalues of HH have same sign.

      Why D<0D < 0 implies saddle: Eigenvalues have opposite signs.

      Why D=0D = 0 is inconclusive: One eigenvalue is zero; higher-order terms needed.


      Taylor Series Interpretation

      Near stationary point (a,b)(a, b):

      f(a+h,b+k)≈f(a,b)+12(h2fxx+2hkfxy+k2fyy)f(a+h, b+k) \approx f(a,b) + \frac{1}{2}(h^2 f_{xx} + 2hk f_{xy} + k^2 f_{yy})

      In matrix form: f≈f(a,b)+12hTHhf \approx f(a,b) + \frac{1}{2}\mathbf{h}^T H \mathbf{h}

      The quadratic form hTHh\mathbf{h}^T H \mathbf{h} determines whether we have:

      • Minimum: quadratic positive in all directions (HH positive definite)
      • Maximum: quadratic negative in all directions (HH negative definite)
      • Saddle: quadratic can be positive or negative depending on direction

      Example

      Find and classify stationary points of f(x,y)=x3−4x2+2xy−y2f(x, y) = x^3 - 4x^2 + 2xy - y^2.

      Find stationary points:

      • fx=3x2−8x+2y=0f_x = 3x^2 - 8x + 2y = 0
      • fy=2x−2y=0f_y = 2x - 2y = 0

      From fyf_y: y=xy = x.

      Substituting into fxf_x: 3x2−8x+2x=03x^2 - 8x + 2x = 0 3x2−6x=03x^2 - 6x = 0 x(3x−6)=0x(3x - 6) = 0

      So x=0x = 0 or x=2x = 2.

      Stationary points: (0,0)(0, 0) and (2,2)(2, 2).

      Compute Hessian:

      • fxx=6x−8f_{xx} = 6x - 8
      • fxy=2f_{xy} = 2
      • fyy=−2f_{yy} = -2

      At (0,0)(0, 0): H=(−822−2)H = \begin{pmatrix} -8 & 2 \\ 2 & -2 \end{pmatrix}

      D=(−8)(−2)−4=12>0D = (-8)(-2) - 4 = 12 > 0, and fxx=−8<0f_{xx} = -8 < 0.

      Local maximum.

      At (2,2)(2, 2): H=(422−2)H = \begin{pmatrix} 4 & 2 \\ 2 & -2 \end{pmatrix}

      D=(4)(−2)−4=−12<0D = (4)(-2) - 4 = -12 < 0.

      Saddle point.


      Global vs Local

      • Local minimum: f(x,y)≥f(a,b)f(x,y) \geq f(a,b) for all nearby (x,y)(x,y)
      • Global minimum: f(x,y)≥f(a,b)f(x,y) \geq f(a,b) for all (x,y)(x,y) in the domain

      Multiple stationary points may exist; the global minimum/maximum is found by comparing function values.


      Connection to Eigenvalues

      The Hessian is real symmetric, so has real eigenvalues λ1,λ2\lambda_1, \lambda_2.

      • Both positive → minimum (D>0,fxx>0D > 0, f_{xx} > 0)
      • Both negative → maximum (D>0,fxx<0D > 0, f_{xx} < 0)
      • One positive, one negative → saddle (D<0D < 0)

      D=λ1λ2D = \lambda_1\lambda_2 and fxxf_{xx} determines the sign of λ1\lambda_1 (for convention).


      Why Stationary Points Matter

      • Optimisation problems (minimise cost, maximise efficiency)
      • Machine learning (gradient descent finds minima of loss functions)
      • Physics (equilibrium states are stationary points of potential energy)
      • Economics (profit maximisation, cost minimisation)
    • Lagrange Multipliers

      Problem Statement

      Find the stationary points of f(x,y)f(x, y) subject to constraint g(x,y)=0g(x, y) = 0.

      The constraint restricts us to a curve (or surface), so we seek constrained extrema rather than global ones.


      Geometric Intuition

      At a constrained stationary point:

      • ∇f\nabla f is perpendicular to the level curve of ff passing through that point
      • ∇g\nabla g is perpendicular to the constraint curve g=0g = 0
      • For an extremum, these level curves must be tangent

      Level curves of f, constraint curve g=0, showing where ∇f and ∇g are parallel at optimum

      Therefore: ∇f\nabla f and ∇g\nabla g are parallel at the extremum.

      ∇f=λ∇g\nabla f = \lambda \nabla g

      where λ\lambda is called the Lagrange multiplier.


      Method

      To find stationary points of f(x,y)f(x, y) subject to g(x,y)=0g(x, y) = 0:

      1. Form the Lagrangian: L(x,y,λ)=f(x,y)−λg(x,y)\mathcal{L}(x, y, \lambda) = f(x, y) - \lambda g(x, y)

      2. Solve the system:

        • ∂L∂x=0\frac{\partial \mathcal{L}}{\partial x} = 0
        • ∂L∂y=0\frac{\partial \mathcal{L}}{\partial y} = 0
        • ∂L∂λ=0\frac{\partial \mathcal{L}}{\partial \lambda} = 0 (which is just g(x,y)=0g(x, y) = 0)
      3. Evaluate ff at each solution to classify.


      Example

      Maximise f(x,y)=xyf(x, y) = xy subject to x2+y2=1x^2 + y^2 = 1.

      Setup: L=xy−λ(x2+y2−1)\mathcal{L} = xy - \lambda(x^2 + y^2 - 1)

      Equations:

      • ∂L∂x=y−2λx=0\frac{\partial \mathcal{L}}{\partial x} = y - 2\lambda x = 0 …(i)
      • ∂L∂y=x−2λy=0\frac{\partial \mathcal{L}}{\partial y} = x - 2\lambda y = 0 …(ii)
      • ∂L∂λ=−(x2+y2−1)=0\frac{\partial \mathcal{L}}{\partial \lambda} = -(x^2 + y^2 - 1) = 0 …(iii)

      Solve: From (i): y=2λxy = 2\lambda x

      Substitute into (ii): x−2λ(2λx)=x(1−4λ2)=0x - 2\lambda(2\lambda x) = x(1 - 4\lambda^2) = 0

      So x=0x = 0 or λ=±12\lambda = \pm\frac{1}{2}.

      If x=0x = 0: Then y=±1y = \pm 1 from (iii). f=0f = 0.

      If λ=12\lambda = \frac{1}{2}: From (i), y=xy = x. From (iii), x2+x2=1x^2 + x^2 = 1, so x=±12x = \pm\frac{1}{\sqrt{2}}.

      Points: (12,12)(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}) and (−12,−12)(-\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}) with f=12f = \frac{1}{2}.

      If λ=−12\lambda = -\frac{1}{2}: y=−xy = -x. Points: (12,−12)(\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}) and (−12,12)(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}) with f=−12f = -\frac{1}{2}.

      Maximum occurs when x=y=±12x = y = \pm\frac{1}{\sqrt{2}}, giving f=12f = \frac{1}{2}.


      Multiple Constraints

      For constraints g1(x,y)=0g_1(x, y) = 0 and g2(x,y)=0g_2(x, y) = 0:

      ∇f=λ1∇g1+λ2∇g2\nabla f = \lambda_1 \nabla g_1 + \lambda_2 \nabla g_2

      The Lagrangian becomes:

      L=f−λ1g1−λ2g2\mathcal{L} = f - \lambda_1 g_1 - \lambda_2 g_2


      Interpretation of λ\lambda

      Adding a small change ε\varepsilon to the constraint:

      g(x,y)=εg(x, y) = \varepsilon

      The optimal value f∗(ε)f^*(\varepsilon) satisfies:

      df∗dε=λ\frac{df^*}{d\varepsilon} = \lambda

      λ\lambda measures the sensitivity: how much the optimal value changes per unit change in the constraint.

      Example: In economics, if ff is profit and gg is a budget constraint, λ\lambda is the marginal value of relaxing the budget.


      When Lagrange Multipliers are Needed

      • Constraints cannot be easily substituted into the objective
      • Multiple constraints interact
      • Need to understand the “cost” of constraints (via λ\lambda)
      • Generalising to higher dimensions (constrained on surfaces)

      When a single constraint can be substituted, that’s often simpler. Lagrange multipliers are a systematic method for complex situations.


      Why This Method Matters

      • Optimisation: Constraints arise naturally in real problems
      • Economics: Utility maximisation under budget constraints
      • Physics: Equilibrium under constraints (frictionless motion)
      • Machine learning: Regularisation as constrained optimisation
    • Scalar and Vector Fields

      Definitions

      Scalar Field

      A scalar field ϕ(r)\phi(\mathbf{r}) assigns a scalar (number) to each point in space.

      Examples:

      • Temperature in a room: T(x,y,z)T(x, y, z)
      • Pressure in the atmosphere: P(x,y,z)P(x, y, z)
      • Electric potential: V(r)V(\mathbf{r})

      Vector Field

      A vector field F(r)\mathbf{F}(\mathbf{r}) assigns a vector to each point in space.

      Examples:

      • Velocity of fluid flow: v(x,y,z)\mathbf{v}(x, y, z)
      • Electric field: E(r)\mathbf{E}(\mathbf{r})
      • Gravitational field: g(r)\mathbf{g}(\mathbf{r})

      The Gradient Operator

      The del or nabla operator ∇\nabla is:

      ∇=i^∂∂x+j^∂∂y+k^∂∂z\nabla = \hat{\mathbf{i}}\frac{\partial}{\partial x} + \hat{\mathbf{j}}\frac{\partial}{\partial y} + \hat{\mathbf{k}}\frac{\partial}{\partial z}

      Gradient of a Scalar Field

      ∇ϕ=∂ϕ∂xi^+∂ϕ∂yj^+∂ϕ∂zk^\nabla \phi = \frac{\partial \phi}{\partial x}\hat{\mathbf{i}} + \frac{\partial \phi}{\partial y}\hat{\mathbf{j}} + \frac{\partial \phi}{\partial z}\hat{\mathbf{k}}

      The gradient is a vector field pointing in the direction of steepest increase of ϕ\phi.


      Properties of the Gradient

      Direction

      ∇ϕ\nabla \phi points in the direction of maximum rate of increase of ϕ\phi.

      Magnitude

      ∣∇ϕ∣=rate of steepest increase|\nabla \phi| = \text{rate of steepest increase}

      Contour map with gradient vectors pointing uphill, perpendicular to level curves

      Directional Derivative

      The rate of change of ϕ\phi in direction n^\hat{\mathbf{n}}:

      dϕdn=∇ϕ⋅n^\frac{d\phi}{dn} = \nabla \phi \cdot \hat{\mathbf{n}}

      This is the component of ∇ϕ\nabla \phi in direction n^\hat{\mathbf{n}}.


      Level Surfaces

      A level surface of ϕ\phi is the set of points where ϕ\phi is constant:

      ϕ(x,y,z)=c\phi(x, y, z) = c

      Level surfaces of a scalar field, with gradient vector perpendicular to level surface

      Key theorem: ∇ϕ\nabla \phi is perpendicular to the level surface.

      Proof: If r(t)\mathbf{r}(t) lies on the surface, then ϕ(r(t))=c\phi(\mathbf{r}(t)) = c. Differentiating:

      ∇ϕ⋅drdt=0\nabla \phi \cdot \frac{d\mathbf{r}}{dt} = 0

      The tangent dr/dtd\mathbf{r}/dt to the surface is perpendicular to ∇ϕ\nabla \phi.


      Visualising Fields

      Scalar Fields

      • Contour plots: Level curves (2D) or level surfaces (3D)
      • Colour mapping: Assign colours to different values
      • Example: Weather maps with isobars (pressure contours)

      Contour plot of a scalar field f(x,y) showing level curves f = constant

      Vector Fields

      • Arrow plots: Draw vectors at grid points
      • Streamlines: Curves everywhere tangent to the field
      • Example: Weather maps showing wind velocity

      2D vector field visualisation showing arrows at grid points, demonstrating rotation or divergence patterns


      The Differential of a Scalar Field

      For a small displacement dr=dx i^+dy j^+dz k^d\mathbf{r} = dx\,\hat{\mathbf{i}} + dy\,\hat{\mathbf{j}} + dz\,\hat{\mathbf{k}}:

      dϕ=∇ϕ⋅dr=∂ϕ∂xdx+∂ϕ∂ydy+∂ϕ∂zdzd\phi = \nabla \phi \cdot d\mathbf{r} = \frac{\partial \phi}{\partial x}dx + \frac{\partial \phi}{\partial y}dy + \frac{\partial \phi}{\partial z}dz

      This is the total differential: the change in ϕ\phi due to a small displacement.


      Special Cases

      Radial Fields

      Distance from origin: r=x2+y2+z2r = \sqrt{x^2 + y^2 + z^2}

      ∇r=rr=r^\nabla r = \frac{\mathbf{r}}{r} = \hat{\mathbf{r}}

      For any function f(r)f(r) of distance:

      ∇f(r)=f′(r)r^\nabla f(r) = f'(r) \hat{\mathbf{r}}

      Example: Gravitational potential ϕ=−GMr\phi = -\frac{GM}{r}

      ∇ϕ=GMr2r^\nabla \phi = \frac{GM}{r^2}\hat{\mathbf{r}}

      The gravitational field (force per unit mass) is g=−∇ϕ\mathbf{g} = -\nabla \phi.


      Why Gradient Matters

      • Force fields: F=−∇V\mathbf{F} = -\nabla V (force from potential)
      • Heat flow: q=−k∇T\mathbf{q} = -k\nabla T (heat flows down temperature gradient)
      • Mass transport: J=−D∇C\mathbf{J} = -D\nabla C (diffusion down concentration gradient)

      The gradient encodes “downhill” directions for flow and equilibration processes.

    • Line Integrals

      Curves

      A curve in space is parameterised by:

      r(t)=(x(t),y(t),z(t)),t∈[t1,t2]\mathbf{r}(t) = (x(t), y(t), z(t)), \quad t \in [t_1, t_2]

      Orientation: Parameter tt gives direction along the curve.

      Closed curve: r(t1)=r(t2)\mathbf{r}(t_1) = \mathbf{r}(t_2)


      Definition of Line Integral

      Given vector field F(r)\mathbf{F}(\mathbf{r}) and curve Γ\Gamma:

      ∫ΓF⋅dr=∫t1t2F(r(t))⋅drdt dt\int_\Gamma \mathbf{F} \cdot d\mathbf{r} = \int_{t_1}^{t_2} \mathbf{F}(\mathbf{r}(t)) \cdot \frac{d\mathbf{r}}{dt}\,dt

      Curve Γ through a vector field F, showing small displacement dr and the dot product F·dr along the path

      Interpretation: The integral adds up the component of F\mathbf{F} along the curve at each point.


      Properties

      1. Parameterisation independent: Different parameterisations of the same curve give the same integral.

      2. Orientation matters: Reversing the curve multiplies by −1-1: ∫−ΓF⋅dr=−∫ΓF⋅dr\int_{-\Gamma} \mathbf{F} \cdot d\mathbf{r} = -\int_\Gamma \mathbf{F} \cdot d\mathbf{r}

      3. Additivity: For Γ=Γ1+Γ2\Gamma = \Gamma_1 + \Gamma_2: ∫Γ=∫Γ1+∫Γ2\int_\Gamma = \int_{\Gamma_1} + \int_{\Gamma_2}


      Work Done

      Physical interpretation: If F\mathbf{F} is force, the line integral gives work done:

      W=∫ΓF⋅drW = \int_\Gamma \mathbf{F} \cdot d\mathbf{r}

      Example: Work done by gravity moving an object along a path.


      Example

      Evaluate ∫ΓF⋅dr\int_\Gamma \mathbf{F} \cdot d\mathbf{r} for F=(2xy,x2)\mathbf{F} = (2xy, x^2) along:

      1. Straight line from (0,0)(0,0) to (1,1)(1,1)
      2. Parabola y=x2y = x^2 from (0,0)(0,0) to (1,1)(1,1)

      Line: Parameterise r(t)=(t,t)\mathbf{r}(t) = (t, t), t∈[0,1]t \in [0, 1].

      drdt=(1,1)\frac{d\mathbf{r}}{dt} = (1, 1)

      F=(2t2,t2)\mathbf{F} = (2t^2, t^2)

      F⋅drdt=2t2+t2=3t2\mathbf{F} \cdot \frac{d\mathbf{r}}{dt} = 2t^2 + t^2 = 3t^2

      ∫013t2 dt=[t3]01=1\int_0^1 3t^2\,dt = [t^3]_0^1 = 1

      Parabola: Parameterise r(t)=(t,t2)\mathbf{r}(t) = (t, t^2), t∈[0,1]t \in [0, 1].

      drdt=(1,2t)\frac{d\mathbf{r}}{dt} = (1, 2t)

      F=(2t3,t2)\mathbf{F} = (2t^3, t^2)

      F⋅drdt=2t3+2t3=4t3\mathbf{F} \cdot \frac{d\mathbf{r}}{dt} = 2t^3 + 2t^3 = 4t^3

      ∫014t3 dt=[t4]01=1\int_0^1 4t^3\,dt = [t^4]_0^1 = 1

      Both paths give the same result — F\mathbf{F} may be conservative.


      Conservative Fields

      A vector field is conservative if the line integral depends only on endpoints, not the path.

      Equivalent conditions:

      1. F=∇ϕ\mathbf{F} = \nabla \phi for some scalar field ϕ\phi (potential)
      2. F⋅dr\mathbf{F} \cdot d\mathbf{r} is an exact differential
      3. ∮CF⋅dr=0\oint_C \mathbf{F} \cdot d\mathbf{r} = 0 for all closed curves CC
      4. ∇×F=0\nabla \times \mathbf{F} = \mathbf{0} (curl-free)

      Path Independence

      If F=∇ϕ\mathbf{F} = \nabla \phi:

      ∫ΓF⋅dr=∫Γ∇ϕ⋅dr=ϕ(rend)−ϕ(rstart)\int_\Gamma \mathbf{F} \cdot d\mathbf{r} = \int_\Gamma \nabla \phi \cdot d\mathbf{r} = \phi(\mathbf{r}_\text{end}) - \phi(\mathbf{r}_\text{start})

      The line integral equals the potential difference.

      Proof:

      ∇ϕ⋅dr=∂ϕ∂xdx+∂ϕ∂ydy+∂ϕ∂zdz=dϕ\nabla \phi \cdot d\mathbf{r} = \frac{\partial \phi}{\partial x}dx + \frac{\partial \phi}{\partial y}dy + \frac{\partial \phi}{\partial z}dz = d\phi


      Finding the Potential

      If we know F=∇ϕ\mathbf{F} = \nabla \phi and want ϕ\phi:

      ϕ(r)=∫r0rF⋅dr+C\phi(\mathbf{r}) = \int_{\mathbf{r}_0}^{\mathbf{r}} \mathbf{F} \cdot d\mathbf{r} + C

      Choose any convenient path (e.g., along coordinate axes).


      Why Line Integrals Matter

      • Work done by forces: Fundamental in mechanics
      • Circulation: ∮v⋅dr\oint \mathbf{v} \cdot d\mathbf{r} measures rotation in fluids
      • Electromagnetic induction: ∮E⋅dr=−dΦdt\oint \mathbf{E} \cdot d\mathbf{r} = -\frac{d\Phi}{dt} (Faraday’s law)
      • Conservative systems: Potential energy reconstructs from force field
    • Flux Integrals

      Surfaces

      A surface SS in 3D is often parameterised by:

      r(u,v)=(x(u,v),y(u,v),z(u,v))\mathbf{r}(u, v) = (x(u,v), y(u,v), z(u,v))

      Orientation: A surface has two sides. The normal vector n\mathbf{n} points to one side.


      Surface Element

      For a small patch of surface:

      • Normal vector: n\mathbf{n}
      • Area: dSdS
      • Vector area element: dS=n dSd\mathbf{S} = \mathbf{n}\,dS

      Definition of Flux Integral

      The flux of vector field F\mathbf{F} through surface SS:

      Φ=∫SF⋅dS=∫SF⋅n dS\Phi = \int_S \mathbf{F} \cdot d\mathbf{S} = \int_S \mathbf{F} \cdot \mathbf{n}\,dS

      Surface S with normal vectors n, showing vector field F passing through and the flux integral interpretation

      Interpretation: F⋅n\mathbf{F} \cdot \mathbf{n} is the component of F\mathbf{F} through the surface. Added up over SS, this gives total flux.


      Physical Examples

      • Fluid flow: Flux = volume of fluid crossing surface per unit time
      • Electric field: Flux = ∫E⋅dS\int \mathbf{E} \cdot d\mathbf{S} (related to enclosed charge)
      • Heat flux: Rate of heat flow through surface

      Computing Flux Integrals

      Method 1: Parameterisation

      For r(u,v)\mathbf{r}(u, v):

      dS=∂r∂u×∂r∂v du dvd\mathbf{S} = \frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v}\,du\,dv

      Direction given by right-hand rule for (u,v)(u, v) coordinates.

      Method 2: Level Surface

      For surface g(x,y,z)=cg(x, y, z) = c:

      n=±∇g∣∇g∣\mathbf{n} = \pm \frac{\nabla g}{|\nabla g|}

      Choose sign for correct orientation.

      Method 3: Graph

      For z=f(x,y)z = f(x, y):

      n dS=(−∂f∂x,−∂f∂y,1)dx dy\mathbf{n}\,dS = \left(-\frac{\partial f}{\partial x}, -\frac{\partial f}{\partial y}, 1\right)dx\,dy


      Example: Flux Through a Hemisphere

      Find flux of F=(0,0,z)\mathbf{F} = (0, 0, z) through the hemisphere x2+y2+z2=a2x^2 + y^2 + z^2 = a^2, z≥0z \geq 0, with outward normal.

      On the hemisphere: The unit normal is n^=1a(x,y,z)\hat{\mathbf{n}} = \frac{1}{a}(x, y, z).

      F⋅n^=z2a\mathbf{F} \cdot \hat{\mathbf{n}} = \frac{z^2}{a}

      Flux = ∫Sz2a dS\int_S \frac{z^2}{a}\,dS

      By symmetry, ⟨x2⟩=⟨y2⟩=⟨z2⟩\langle x^2 \rangle = \langle y^2 \rangle = \langle z^2 \rangle on the sphere, so z2=a23z^2 = \frac{a^2}{3}.

      Surface area = 2πa22\pi a^2.

      Flux = a2/3a⋅2πa2=2πa33\frac{a^2/3}{a} \cdot 2\pi a^2 = \frac{2\pi a^3}{3}.


      Properties

      Orientation: Reversing normal multiplies flux by −1-1.

      Closed surfaces: For closed surface SS, by convention the normal points outward.


      Vector Area

      The vector area of a surface:

      S=∫SdS\mathbf{S} = \int_S d\mathbf{S}

      Closed surface: Vector area = 0\mathbf{0}

      (Each patch cancels with opposite contribution.)


      Example: Flat Surface

      Flux of F\mathbf{F} through flat surface SS in plane with normal n\mathbf{n}:

      Φ=n⋅∫SF dS\Phi = \mathbf{n} \cdot \int_S \mathbf{F}\,dS

      If F\mathbf{F} is constant: Φ=F⋅n⋅Area\Phi = \mathbf{F} \cdot \mathbf{n} \cdot \text{Area}


      Why Flux Matters

      • Gauss’s law: Flux through closed surface relates to enclosed source
      • Fluid dynamics: Conservation laws use flux integrals
      • Heat transfer: Thermal flux through boundaries
      • Electromagnetism: Flux of E\mathbf{E} and B\mathbf{B} through surfaces
    • Divergence and Gauss's Theorem

      The Divergence Operator

      For a vector field F=(Fx,Fy,Fz)\mathbf{F} = (F_x, F_y, F_z):

      ∇⋅F=∂Fx∂x+∂Fy∂y+∂Fz∂z\nabla \cdot \mathbf{F} = \frac{\partial F_x}{\partial x} + \frac{\partial F_y}{\partial y} + \frac{\partial F_z}{\partial z}

      The result is a scalar field.


      Physical Interpretation

      ∇⋅F\nabla \cdot \mathbf{F} measures expansion: the rate at which flow “spreads out” from each point.

      Vector field with positive divergence (source, flow outward) and negative divergence (sink, flow inward)

      • ∇⋅F>0\nabla \cdot \mathbf{F} > 0: Source (flow diverges)
      • ∇⋅F<0\nabla \cdot \mathbf{F} < 0: Sink (flow converges)
      • ∇⋅F=0\nabla \cdot \mathbf{F} = 0: Incompressible (volume preserved)

      The Laplacian

      Applying divergence to gradient of a scalar field:

      ∇⋅(∇ϕ)=∇2ϕ=∂2ϕ∂x2+∂2ϕ∂y2+∂2ϕ∂z2\nabla \cdot (\nabla \phi) = \nabla^2 \phi = \frac{\partial^2 \phi}{\partial x^2} + \frac{\partial^2 \phi}{\partial y^2} + \frac{\partial^2 \phi}{\partial z^2}

      The Laplacian ∇2\nabla^2 is fundamental in PDEs:

      PDEEquation
      Laplace’s equation∇2ϕ=0\nabla^2\phi = 0
      Heat equation∇2ϕ=1D∂ϕ∂t\nabla^2\phi = \frac{1}{D}\frac{\partial\phi}{\partial t}
      Wave equation∇2ϕ=1c2∂2ϕ∂t2\nabla^2\phi = \frac{1}{c^2}\frac{\partial^2\phi}{\partial t^2}

      Example

      For F=(x2,y2,z2)\mathbf{F} = (x^2, y^2, z^2):

      ∇⋅F=2x+2y+2z=2(x+y+z)\nabla \cdot \mathbf{F} = 2x + 2y + 2z = 2(x + y + z)

      At origin: ∇⋅F=0\nabla \cdot \mathbf{F} = 0.

      On axis (x,0,0)(x, 0, 0): ∇⋅F=2x\nabla \cdot \mathbf{F} = 2x, increases with distance.


      Gauss’s (Divergence) Theorem

      For a closed surface SS enclosing volume VV: ∫SF⋅dS=∫V(∇⋅F) dV\int_S \mathbf{F} \cdot d\mathbf{S} = \int_V (\nabla \cdot \mathbf{F})\,dV

      The flux through a closed surface equals the volume integral of divergence inside.

      Closed surface S enclosing volume V, with outward normal vectors and flux arrows through the surface


      Proof Sketch (Cuboid)

      Consider a cuboid VV: 0≤x≤a0 \leq x \leq a, 0≤y≤b0 \leq y \leq b, 0≤z≤c0 \leq z \leq c.

      Faces perpendicular to xx:

      • x=0x = 0: flux =−∫Fx(0,y,z) dy dz= -\int F_x(0, y, z)\,dy\,dz (normal points inward)
      • x=ax = a: flux =∫Fx(a,y,z) dy dz= \int F_x(a, y, z)\,dy\,dz

      Combined: ∫[Fx(a,y,z)−Fx(0,y,z)] dy dz\int [F_x(a, y, z) - F_x(0, y, z)]\,dy\,dz

      By fundamental theorem of calculus: =∫0a∂Fx∂xdx dy dz= \int_0^a \frac{\partial F_x}{\partial x} dx\,dy\,dz

      Sum over all six faces:

      Φ=∫V(∂Fx∂x+∂Fy∂y+∂Fz∂z)dV\Phi = \int_V \left(\frac{\partial F_x}{\partial x} + \frac{\partial F_y}{\partial y} + \frac{\partial F_z}{\partial z}\right)dV


      Proof Extension

      To generalise from cuboid to any shape:

      1. Subdivide volume into small cuboids
      2. Apply theorem to each cuboid
      3. Sum: internal surfaces cancel (opposite normals)
      4. Result: theorem holds for whole volume

      Requirement: Surface is smooth enough for subdivision.


      Applications

      Flux Calculation

      Often easier to compute volume integral than surface integral.

      Example: Flux of F=(x2,y2,z2)\mathbf{F} = (x^2, y^2, z^2) through unit sphere.

      ∇⋅F=2(x+y+z)\nabla \cdot \mathbf{F} = 2(x + y + z)

      Φ=∫V2(x+y+z) dV\Phi = \int_V 2(x + y + z)\,dV

      By symmetry: ∫x dV=∫y dV=∫z dV=0\int x\,dV = \int y\,dV = \int z\,dV = 0.

      So Φ=0\Phi = 0.

      Vector Area of Closed Surface

      ∮SdS=0\oint_S d\mathbf{S} = \mathbf{0}

      (Apply divergence theorem to constant vector field.)


      Solenoidal Fields

      A vector field with ∇⋅F=0\nabla \cdot \mathbf{F} = 0 everywhere is solenoidal.

      From divergence theorem: flux through any closed surface is zero.

      Examples:

      • Magnetic field B\mathbf{B}: ∇⋅B=0\nabla \cdot \mathbf{B} = 0 (no magnetic monopoles)
      • Incompressible fluid velocity

      Why These Theorems Matter

      Gauss’s theorem converts between:

      • Surface integrals (flux)
      • Volume integrals (divergence)

      This is powerful because:

      • Choose whichever is easier to evaluate
      • Derive conservation laws in fluid mechanics
      • Apply to Maxwell’s equations in electromagnetism
    • Curl and Stokes' Theorem

      The Curl Operator

      For a vector field F=(Fx,Fy,Fz)\mathbf{F} = (F_x, F_y, F_z), the curl is:

      ∇×F=∣i^j^k^∂∂x∂∂y∂∂zFxFyFz∣\nabla \times \mathbf{F} = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ F_x & F_y & F_z \end{vmatrix}

      Explicitly:

      ∇×F=(∂Fz∂y−∂Fy∂z,∂Fx∂z−∂Fz∂x,∂Fy∂x−∂Fx∂y)\nabla \times \mathbf{F} = \left(\frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z}, \frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x}, \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y}\right)

      The result is a vector field.


      Physical Interpretation

      ∇×F\nabla \times \mathbf{F} measures rotation: the tendency to cause circulation.

      Paddle wheel in rotating fluid, showing how curl measures local rotation rate

      • ∇×F≠0\nabla \times \mathbf{F} \neq \mathbf{0}: Field has rotational component
      • ∇×F=0\nabla \times \mathbf{F} = \mathbf{0}: Irrotational

      Example: Paddle wheel in a fluid placed at point r\mathbf{r}:

      • Angular velocity proportional to ∣∇×v∣|\nabla \times \mathbf{v}|
      • Axis of rotation aligned with ∇×v\nabla \times \mathbf{v}

      Example

      For F=(−y,x,0)\mathbf{F} = (-y, x, 0):

      ∇×F=(0−0,0−0,1−(−1))=(0,0,2)\nabla \times \mathbf{F} = \left(0 - 0, 0 - 0, 1 - (-1)\right) = (0, 0, 2)

      This field has uniform rotation about zz-axis.


      Stokes’ Theorem

      For an open surface SS with boundary curve CC: ∫S(∇×F)⋅dS=∮CF⋅dr\int_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S} = \oint_C \mathbf{F} \cdot d\mathbf{r}

      The flux of curl through a surface equals the circulation around its boundary.

      Open surface S with boundary curve C, showing circulation around the boundary and curl flux through the surface

      Orientation: Right-hand rule — fingers curl along CC, thumb points along n\mathbf{n}.


      Proof Sketch (Rectangle)

      Consider rectangle SS: 0≤x≤a0 \leq x \leq a, 0≤y≤b0 \leq y \leq b in z=0z = 0 plane.

      Boundary consists of four edges:

      • C1C_1: xx from 0 to aa at y=0y = 0
      • C2C_2: yy from 0 to bb at x=ax = a
      • C3C_3: xx from aa to 0 at y=by = b (reversed)
      • C4C_4: yy from bb to 0 at x=0x = 0 (reversed)

      Circulation:

      ∮CF⋅dr=∫C1Fx dx+∫C2Fy dy+∫C3Fx dx+∫C4Fy dy\oint_C \mathbf{F} \cdot d\mathbf{r} = \int_{C_1} F_x\,dx + \int_{C_2} F_y\,dy + \int_{C_3} F_x\,dx + \int_{C_4} F_y\,dy

      =∫0a[Fx(x,0)−Fx(x,b)] dx+∫0b[Fy(a,y)−Fy(0,y)] dy= \int_0^a [F_x(x, 0) - F_x(x, b)]\,dx + \int_0^b [F_y(a, y) - F_y(0, y)]\,dy

      =∫0a∫0b∂Fy∂x dy dx−∫0a∫0b∂Fx∂y dy dx= \int_0^a \int_0^b \frac{\partial F_y}{\partial x}\,dy\,dx - \int_0^a \int_0^b \frac{\partial F_x}{\partial y}\,dy\,dx

      =∫S(∂Fy∂x−∂Fx∂y)dS=∫S(∇×F)z dS= \int_S \left(\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y}\right)dS = \int_S (\nabla \times \mathbf{F})_z\,dS


      Path Independence

      Stokes’ theorem implies: for ∇×F=0\nabla \times \mathbf{F} = \mathbf{0}, ∮CF⋅dr=0\oint_C \mathbf{F} \cdot d\mathbf{r} = 0 for any CC.

      Consequence: The line integral from AA to BB is path-independent.

      Equivalent to F\mathbf{F} being conservative.


      Relations Between Operators

      Curl of Gradient = 0

      ∇×(∇ϕ)=0\nabla \times (\nabla \phi) = \mathbf{0}

      Proof: The xx-component is ∂2ϕ∂y∂z−∂2ϕ∂z∂y=0\frac{\partial^2\phi}{\partial y\partial z} - \frac{\partial^2\phi}{\partial z\partial y} = 0 by equality of mixed partials.

      Divergence of Curl = 0

      ∇⋅(∇×F)=0\nabla \cdot (\nabla \times \mathbf{F}) = 0

      Proof: Each term in the divergence pairs with another that cancels it.


      Summary of Relationships

      Operator combinationResultReason
      ∇×∇ϕ\nabla \times \nabla \phi0\mathbf{0}Mixed partials equal
      ∇⋅(∇×F)\nabla \cdot (\nabla \times \mathbf{F})0Terms pair and cancel
      ∇⋅∇ϕ\nabla \cdot \nabla \phi∇2ϕ\nabla^2\phiLaplacian (non-zero in general)

      Irrotational Fields

      ∇×F=0\nabla \times \mathbf{F} = \mathbf{0} everywhere ⇔\Leftrightarrow F\mathbf{F} is conservative ⇔\Leftrightarrow F=∇ϕ\mathbf{F} = \nabla \phi

      Example: Gravitational field g=−∇ϕG\mathbf{g} = -\nabla \phi_G is irrotational.


      Why Stokes’ Theorem Matters

      Stokes’ theorem converts between:

      • Surface integrals (flux of curl)
      • Line integrals (circulation)

      This connects:

      • Circulation to rotation in fluid mechanics
      • Electromagnetic induction (Faraday’s law)
      • Path independence of conservative fields
    • Fourier Series Definition

      Periodic Functions

      A function f(x)f(x) is periodic with period 2L2L if:

      f(x+2L)=f(x)f(x + 2L) = f(x)

      for all xx.


      Example: Temperature Variation

      Consider a function defined on −L≤x<L-L \leq x < L and extended periodically. Fourier series represent such functions as sums of sines and cosines.


      The Fourier Series

      If f(x)f(x) is periodic with period 2L2L and is “sufficiently nice” (piecewise continuous with finite discontinuities), then:

      f(x)=a02+∑n=1∞[ancos⁡(nπxL)+bnsin⁡(nπxL)]f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} \left[ a_n \cos\left(\frac{n\pi x}{L}\right) + b_n \sin\left(\frac{n\pi x}{L}\right) \right]


      Fourier Coefficients

      The coefficients are found by integration:

      an=1L∫−LLf(x)cos⁡(nπxL)dxa_n = \frac{1}{L} \int_{-L}^{L} f(x) \cos\left(\frac{n\pi x}{L}\right) dx

      bn=1L∫−LLf(x)sin⁡(nπxL)dxb_n = \frac{1}{L} \int_{-L}^{L} f(x) \sin\left(\frac{n\pi x}{L}\right) dx

      For n=0n = 0:

      a0=1L∫−LLf(x) dxa_0 = \frac{1}{L} \int_{-L}^{L} f(x)\,dx

      Interpretation: a0/2a_0/2 is the average value of ff over one period.


      Orthogonality of Trig Functions

      The Fourier coefficient formulas come from orthogonality:

      ∫−LLcos⁡(mπxL)cos⁡(nπxL)dx={0m≠nLm=n\int_{-L}^{L} \cos\left(\frac{m\pi x}{L}\right) \cos\left(\frac{n\pi x}{L}\right) dx = \begin{cases} 0 & m \neq n \\ L & m = n \end{cases}

      ∫−LLsin⁡(mπxL)sin⁡(nπxL)dx={0m≠nLm=n\int_{-L}^{L} \sin\left(\frac{m\pi x}{L}\right) \sin\left(\frac{n\pi x}{L}\right) dx = \begin{cases} 0 & m \neq n \\ L & m = n \end{cases}

      ∫−LLcos⁡(mπxL)sin⁡(nπxL)dx=0\int_{-L}^{L} \cos\left(\frac{m\pi x}{L}\right) \sin\left(\frac{n\pi x}{L}\right) dx = 0

      This is analogous to orthogonal vectors — the sines and cosines form an orthogonal “basis”.


      Even and Odd Functions

      Even Function: f(−x)=f(x)f(-x) = f(x)

      • bn=0b_n = 0 for all nn (no sine terms)
      • f(x)=a02+∑ancos⁡(nπxL)f(x) = \frac{a_0}{2} + \sum a_n \cos\left(\frac{n\pi x}{L}\right)

      Odd Function: f(−x)=−f(x)f(-x) = -f(x)

      • an=0a_n = 0 for all nn (no cosine terms)
      • f(x)=∑bnsin⁡(nπxL)f(x) = \sum b_n \sin\left(\frac{n\pi x}{L}\right)

      Example: Triangle Wave

      Find Fourier series for f(x)=∣x∣f(x) = |x| on −1≤x≤1-1 \leq x \leq 1, period 2.

      Triangle wave over two periods, showing the periodic extension of |x|

      Since ff is even: bn=0b_n = 0.

      a0=∫−11∣x∣ dx=2∫01x dx=1a_0 = \int_{-1}^{1} |x|\,dx = 2\int_0^1 x\,dx = 1

      an=∫−11∣x∣cos⁡(nπx) dx=2∫01xcos⁡(nπx) dxa_n = \int_{-1}^{1} |x|\cos(n\pi x)\,dx = 2\int_0^1 x\cos(n\pi x)\,dx

      Integration by parts: an=2n2π2[(−1)n−1]a_n = \frac{2}{n^2\pi^2}[(-1)^n - 1]

      For nn even: an=0a_n = 0. For n=2k+1n = 2k+1 odd: an=−4n2π2a_n = -\frac{4}{n^2\pi^2}.

      Result:

      ∣x∣=12−4π2∑k=0∞cos⁡((2k+1)πx)(2k+1)2|x| = \frac{1}{2} - \frac{4}{\pi^2}\sum_{k=0}^{\infty} \frac{\cos((2k+1)\pi x)}{(2k+1)^2}


      Example: Sawtooth Wave

      Find Fourier series for f(x)=xf(x) = x on −1<x<1-1 < x < 1, period 2.

      Sawtooth wave over two periods, showing the periodic extension of y = x

      Since ff is odd: an=0a_n = 0.

      bn=∫−11xsin⁡(nπx) dxb_n = \int_{-1}^{1} x\sin(n\pi x)\,dx

      Integration by parts: bn=2(−1)n+1nπb_n = \frac{2(-1)^{n+1}}{n\pi}

      Result:

      x=2π∑n=1∞(−1)n+1nsin⁡(nπx)x = \frac{2}{\pi}\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}\sin(n\pi x)

      =2π(sin⁡(πx)−12sin⁡(2πx)+13sin⁡(3πx)−⋯ )= \frac{2}{\pi}\left(\sin(\pi x) - \frac{1}{2}\sin(2\pi x) + \frac{1}{3}\sin(3\pi x) - \cdots\right)


      Complex Form

      Using Euler’s formula, Fourier series can be written:

      f(x)=∑n=−∞∞cneinπx/Lf(x) = \sum_{n=-\infty}^{\infty} c_n e^{in\pi x/L}

      where:

      cn=12L∫−LLf(x)e−inπx/L dxc_n = \frac{1}{2L}\int_{-L}^{L} f(x)e^{-in\pi x/L}\,dx

      Relations: c0=a0/2c_0 = a_0/2, cn=(an−ibn)/2c_n = (a_n - ib_n)/2 for n>0n > 0.

    • Fourier Series Properties

      Value at Discontinuities

      If ff has a jump discontinuity at x0x_0, the Fourier series converges to:

      Fourier value=f(x0−)+f(x0+)2\text{Fourier value} = \frac{f(x_0^-) + f(x_0^+)}{2}

      The average of left and right limits.

      Example: For sawtooth wave at x=1x = 1:

      • f(1−)=1f(1^-) = 1, f(1+)=−1f(1^+) = -1
      • Fourier series value = 1+(−1)2=0\frac{1 + (-1)}{2} = 0

      This matches the series: all sin⁡(nπ)=0\sin(n\pi) = 0.


      Parseval’s Theorem

      For periodic ff with period 2L2L:

      12L∫−LL[f(x)]2 dx=a024+12∑n=1∞(an2+bn2)\frac{1}{2L}\int_{-L}^{L} [f(x)]^2\,dx = \frac{a_0^2}{4} + \frac{1}{2}\sum_{n=1}^{\infty}(a_n^2 + b_n^2)

      Interpretation: The mean square value equals sum of squares of amplitudes.

      Application: Evaluate series.

      Example (sawtooth): f(x)=xf(x) = x on [−1,1][-1, 1].

      Left side: 12∫−11x2 dx=13\frac{1}{2}\int_{-1}^{1} x^2\,dx = \frac{1}{3}

      Right side: Using bn=2(−1)n+1nπb_n = \frac{2(-1)^{n+1}}{n\pi}, get 2π2∑1n2\frac{2}{\pi^2}\sum \frac{1}{n^2}

      Therefore: ∑n=1∞1n2=π26\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}


      Gibbs Phenomenon

      Near a jump discontinuity, partial Fourier series exhibit overshoot:

      • The overshoot approaches ~9% of the jump height
      • As more terms are added, overshoot stays constant in height
      • Overshoot moves closer to the discontinuity

      Square wave Fourier series partial sums showing Gibbs phenomenon - overshoot near discontinuities

      Example: Square wave f(x)={−1−1<x<010<x<1f(x) = \begin{cases} -1 & -1 < x < 0 \\ 1 & 0 < x < 1 \end{cases}

      Even with many terms, partial sums overshoot near x=0x = 0.


      Differentiation of Fourier Series

      If ff is continuous and f′f' is piecewise smooth, then:

      dfdx∼∑n=1∞[nπLbncos⁡(nπxL)−nπLansin⁡(nπxL)]\frac{df}{dx} \sim \sum_{n=1}^{\infty} \left[\frac{n\pi}{L}b_n \cos\left(\frac{n\pi x}{L}\right) - \frac{n\pi}{L}a_n \sin\left(\frac{n\pi x}{L}\right)\right]

      Warning: Differentiating term-by-term can fail if ff has discontinuities (Gibbs phenomenon).

      Example: Triangle wave differentiated gives square wave.


      Integration of Fourier Series

      If Fourier series for ff is known:

      ∫f(x) dx=a0x2+∑[anLnπsin⁡(nπxL)−bnLnπcos⁡(nπxL)]+C\int f(x)\,dx = \frac{a_0 x}{2} + \sum \left[\frac{a_n L}{n\pi}\sin\left(\frac{n\pi x}{L}\right) - \frac{b_n L}{n\pi}\cos\left(\frac{n\pi x}{L}\right)\right] + C

      Condition: Requires a0=0a_0 = 0 for the integral to be periodic.


      Decay of Coefficients

      Function smoothnessCoefficient decay
      DiscontinuousO(1/n)O(1/n)
      Continuous, f′f' discontinuousO(1/n2)O(1/n^2)
      ff and f′f' continuous, f′′f'' discontinuousO(1/n3)O(1/n^3)
      Analytic (smooth)Exponential

      Rule: Smoothness determines how quickly coefficients shrink.

      Intuition: High frequencies (large nn) are needed to capture sharp features.


      Half-Range Expansions

      Sine Series (odd extension)

      Extend f(x)f(x) for 0<x<L0 < x < L as an odd function of period 2L2L:

      f(x)=∑n=1∞bnsin⁡(nπxL)f(x) = \sum_{n=1}^{\infty} b_n \sin\left(\frac{n\pi x}{L}\right)

      bn=2L∫0Lf(x)sin⁡(nπxL) dxb_n = \frac{2}{L}\int_0^L f(x)\sin\left(\frac{n\pi x}{L}\right)\,dx

      Cosine Series (even extension)

      Extend f(x)f(x) for 0<x<L0 < x < L as an even function of period 2L2L:

      f(x)=a02+∑n=1∞ancos⁡(nπxL)f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} a_n \cos\left(\frac{n\pi x}{L}\right)

      an=2L∫0Lf(x)cos⁡(nπxL) dxa_n = \frac{2}{L}\int_0^L f(x)\cos\left(\frac{n\pi x}{L}\right)\,dx


      Convergence Conditions

      Dirichlet conditions: Fourier series converges to f(x)f(x) if ff is:

      1. Single-valued
      2. Finite number of maxima and minima in one period
      3. Finite number of discontinuities in one period
      4. Absolutely integrable over one period

      At discontinuities: converges to the average value.


      Why Fourier Analysis Matters

      • Signal processing: Decompose signals into frequencies
      • PDEs: Solve heat and wave equations (separation of variables)
      • Quantum mechanics: Superposition of wave functions
      • Audio/visual compression: Encode by frequency content

      Fourier series connect the “time domain” (function values) to “frequency domain” (coefficients).

    • Methods II Tripos Questions

      First Order ODEs

      Example from notes

      Solve: y′+3y=e−2xy' + 3y = e^{-2x}

      Solution:

      Integrating factor: μ=e∫3 dx=e3x\mu = e^{\int 3\,dx} = e^{3x}

      ddx(e3xy)=e3xe−2x=ex\frac{d}{dx}(e^{3x}y) = e^{3x}e^{-2x} = e^x

      e3xy=ex+Ce^{3x}y = e^x + C

      y=e−2x+Ce−3xy = e^{-2x} + Ce^{-3x}


      Second Order ODEs

      Example: Damped Oscillator

      Solve y′′+4y′+5y=0y'' + 4y' + 5y = 0 with y(0)=1y(0) = 1, y′(0)=0y'(0) = 0.

      Characteristic equation: λ2+4λ+5=0\lambda^2 + 4\lambda + 5 = 0

      λ=−4±16−202=−2±i\lambda = \frac{-4 \pm \sqrt{16 - 20}}{2} = -2 \pm i

      General solution: y=e−2x(Acos⁡x+Bsin⁡x)y = e^{-2x}(A\cos x + B\sin x)

      Apply initial conditions:

      y(0)=A=1y(0) = A = 1

      y′(x)=−2e−2x(cos⁡x+Bsin⁡x)+e−2x(−sin⁡x+Bcos⁡x)y'(x) = -2e^{-2x}(\cos x + B\sin x) + e^{-2x}(-\sin x + B\cos x)

      y′(0)=−2(1)+B=0⇒B=2y'(0) = -2(1) + B = 0 \Rightarrow B = 2

      Solution: y=e−2x(cos⁡x+2sin⁡x)y = e^{-2x}(\cos x + 2\sin x)


      Lagrange Multipliers

      Maximum of f(x,y)=xyf(x,y) = xy on x2+4y2=1x^2 + 4y^2 = 1

      Set up: L=xy−λ(x2+4y2−1)\mathcal{L} = xy - \lambda(x^2 + 4y^2 - 1)

      Equations:

      • y−2λx=0y - 2\lambda x = 0 … (i)
      • x−8λy=0x - 8\lambda y = 0 … (ii)
      • x2+4y2=1x^2 + 4y^2 = 1 … (iii)

      From (i): λ=y2x\lambda = \frac{y}{2x}

      Substitute into (ii): x−8y22x=x−4y2x=0x - \frac{8y^2}{2x} = x - \frac{4y^2}{x} = 0

      x2=4y2x^2 = 4y^2

      From (iii): 4y2+4y2=8y2=14y^2 + 4y^2 = 8y^2 = 1

      y=±122y = \pm\frac{1}{2\sqrt{2}}

      x=±12x = \pm\frac{1}{\sqrt{2}}

      Maximum: f=122⋅122=18f = \frac{1}{2\sqrt{2}} \cdot \frac{1}{2\sqrt{2}} = \frac{1}{8}

      (When signs match.)


      Vector Calculus

      Line Integral Example

      Compute ∫CF⋅dr\int_C \mathbf{F} \cdot d\mathbf{r} for F=(y,x)\mathbf{F} = (y, x) along CC:

      1. Straight line from (0,0)(0,0) to (1,1)(1,1)
      2. Curve y=x2y = x^2 from (0,0)(0,0) to (1,1)(1,1)

      Check if conservative:

      ∇×F=∂Fy∂x−∂Fx∂y=∂x∂x−∂y∂y=1−1=0\nabla \times \mathbf{F} = \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} = \frac{\partial x}{\partial x} - \frac{\partial y}{\partial y} = 1 - 1 = 0

      F\mathbf{F} is conservative. Find potential: ϕ=xy+C\phi = xy + C.

      Line integral = ϕ(1,1)−ϕ(0,0)=1\phi(1,1) - \phi(0,0) = 1

      Both paths give the same result: 1.


      Stokes’ Theorem Verification

      Verify Stokes’ theorem for F=(−y,x,0)\mathbf{F} = (-y, x, 0) on circle x2+y2≤a2x^2 + y^2 \leq a^2 in plane z=0z = 0.

      Curl: ∇×F=(0,0,2)\nabla \times \mathbf{F} = (0, 0, 2)

      Surface integral: ∫S(∇×F)⋅dS=∫S2 dS=2πa2\int_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S} = \int_S 2\,dS = 2\pi a^2

      Line integral: Boundary is circle x2+y2=a2x^2 + y^2 = a^2 at z=0z = 0.

      r=(acos⁡θ,asin⁡θ,0)\mathbf{r} = (a\cos\theta, a\sin\theta, 0)

      dr=(−asin⁡θ,acos⁡θ,0) dθd\mathbf{r} = (-a\sin\theta, a\cos\theta, 0)\,d\theta

      F=(−asin⁡θ,acos⁡θ,0)\mathbf{F} = (-a\sin\theta, a\cos\theta, 0)

      F⋅dr=a2sin⁡2θ+a2cos⁡2θ=a2\mathbf{F} \cdot d\mathbf{r} = a^2\sin^2\theta + a^2\cos^2\theta = a^2

      ∮F⋅dr=∫02πa2 dθ=2πa2\oint \mathbf{F} \cdot d\mathbf{r} = \int_0^{2\pi} a^2\,d\theta = 2\pi a^2

      Both equal: 2πa22\pi a^2


      Fourier Series

      Parseval’s Theorem Application

      Use Fourier series for f(x)=xf(x) = x on [−π,π][-\pi, \pi] to evaluate ∑n=1∞1n2\sum_{n=1}^{\infty} \frac{1}{n^2}.

      Fourier coefficients:

      Since ff is odd: an=0a_n = 0.

      bn=1π∫−ππxsin⁡(nx) dx=2(−1)n+1nb_n = \frac{1}{\pi}\int_{-\pi}^{\pi} x\sin(nx)\,dx = \frac{2(-1)^{n+1}}{n}

      Parseval:

      12π∫−ππx2 dx=12∑n=1∞bn2\frac{1}{2\pi}\int_{-\pi}^{\pi} x^2\,dx = \frac{1}{2}\sum_{n=1}^{\infty} b_n^2

      π23=12∑4n2\frac{\pi^2}{3} = \frac{1}{2}\sum \frac{4}{n^2}

      ∑n=1∞1n2=π26\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}


      Fourier Series of Square Wave

      f(x)={−1−π<x<010<x<πf(x) = \begin{cases} -1 & -\pi < x < 0 \\ 1 & 0 < x < \pi \end{cases}

      Even though ff is odd, this gives a series with only sine terms:

      f(x)=4π(sin⁡x+13sin⁡3x+15sin⁡5x+⋯ )f(x) = \frac{4}{\pi}\left(\sin x + \frac{1}{3}\sin 3x + \frac{1}{5}\sin 5x + \cdots\right)

      =4π∑k=0∞sin⁡((2k+1)x)2k+1= \frac{4}{\pi}\sum_{k=0}^{\infty} \frac{\sin((2k+1)x)}{2k+1}

      Note: This can also be obtained by differentiating the triangle wave series.

  • Methods III (Easter Term)

    Linear algebra (vector spaces, matrices, determinants, linear maps, eigenvalues), and partial differential equations (Laplace, wave, and diffusion equations)

    • Vector Spaces

      Definition

      A vector space VV over a field KK (usually R\mathbb{R} or C\mathbb{C}) is a set with two operations:

      • Addition: u+v∈V\mathbf{u} + \mathbf{v} \in V
      • Scalar multiplication: au∈Va\mathbf{u} \in V for a∈Ka \in K

      satisfying:

      PropertyCondition
      Closureu+v∈V\mathbf{u} + \mathbf{v} \in V, au∈Va\mathbf{u} \in V
      Commutativityu+v=v+u\mathbf{u} + \mathbf{v} = \mathbf{v} + \mathbf{u}
      Associativity(u+v)+w=u+(v+w)(\mathbf{u} + \mathbf{v}) + \mathbf{w} = \mathbf{u} + (\mathbf{v} + \mathbf{w})
      Zero vector∃ 0:u+0=u\exists\,\mathbf{0}: \mathbf{u} + \mathbf{0} = \mathbf{u}
      Additive inverse∃ (−u):u+(−u)=0\exists\,(-\mathbf{u}): \mathbf{u} + (-\mathbf{u}) = \mathbf{0}
      Distributivity 1a(u+v)=au+ava(\mathbf{u} + \mathbf{v}) = a\mathbf{u} + a\mathbf{v}
      Distributivity 2(a+b)u=au+bu(a + b)\mathbf{u} = a\mathbf{u} + b\mathbf{u}
      Compatibilitya(bu)=(ab)ua(b\mathbf{u}) = (ab)\mathbf{u}
      Identity1u=u1\mathbf{u} = \mathbf{u}

      Examples

      Vector SpaceElementsField
      Rn\mathbb{R}^nnn-tuples (x1,…,xn)(x_1, \ldots, x_n)R\mathbb{R}
      Matrices Mm,nM_{m,n}m×nm \times n matricesR\mathbb{R}
      Polynomials PnP_nDegree ≤n\leq n polynomialsR\mathbb{R}
      Functions C[a,b]C[a,b]Continuous functions on [a,b][a,b]R\mathbb{R}

      Linear Combinations

      A linear combination of vectors v1,…,vm\mathbf{v}_1, \ldots, \mathbf{v}_m is:

      a1v1+a2v2+⋯+amvma_1\mathbf{v}_1 + a_2\mathbf{v}_2 + \cdots + a_m\mathbf{v}_m

      where ai∈Ka_i \in K are scalars.


      Linear Independence

      Vectors v1,…,vm\mathbf{v}_1, \ldots, \mathbf{v}_m are linearly independent if:

      a1v1+⋯+amvm=0⇒a1=⋯=am=0a_1\mathbf{v}_1 + \cdots + a_m\mathbf{v}_m = \mathbf{0} \Rightarrow a_1 = \cdots = a_m = 0

      Linear independence in 2D - two non-collinear vectors are independent; three vectors in 2D are dependent

      Equivalently: no vector can be written as a combination of the others.

      Test: Set up the linear combination equal to zero, solve for coefficients.


      Span

      The span of vectors v1,…,vm\mathbf{v}_1, \ldots, \mathbf{v}_m is:

      span{v1,…,vm}={a1v1+⋯+amvm:ai∈K}\text{span}\{\mathbf{v}_1, \ldots, \mathbf{v}_m\} = \{a_1\mathbf{v}_1 + \cdots + a_m\mathbf{v}_m : a_i \in K\}

      All linear combinations of the vectors.


      Basis and Dimension

      A basis for VV is a set of vectors that:

      1. Is linearly independent
      2. Spans VV

      The dimension dim⁡(V)\dim(V) is the number of vectors in any basis.

      Key theorem: All bases have the same number of vectors.


      Example: R3\mathbb{R}^3

      Standard basis: e1=(1,0,0)\mathbf{e}_1 = (1,0,0), e2=(0,1,0)\mathbf{e}_2 = (0,1,0), e3=(0,0,1)\mathbf{e}_3 = (0,0,1)

      Any vector: v=v1e1+v2e2+v3e3\mathbf{v} = v_1\mathbf{e}_1 + v_2\mathbf{e}_2 + v_3\mathbf{e}_3

      Dimension: dim⁡(R3)=3\dim(\mathbb{R}^3) = 3


      Important Result

      If v1,…,vm\mathbf{v}_1, \ldots, \mathbf{v}_m are linearly independent in VV and dim⁡(V)=m\dim(V) = m, then they form a basis.

      In Rn\mathbb{R}^n:

      • More than nn vectors → linearly dependent
      • Fewer than nn vectors → cannot span

      Why Vector Spaces Matter

      • Unify disparate objects (vectors, matrices, functions) under one framework
      • Enable general results applicable to all linear problems
      • Foundation for quantum mechanics (Hilbert spaces)
      • Underpin linear algebra (matrices represent linear maps between vector spaces)

      The abstraction reveals common structure across seemingly different mathematical objects.

    • Matrices

      Definition

      An m×nm \times n matrix is a rectangular array:

      A=(a11a12⋯a1na21a22⋯a2n⋮⋮⋱⋮am1am2⋯amn)A = \begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \end{pmatrix}

      Element aija_{ij} is in row ii, column jj.


      Basic Operations

      Addition

      For m×nm \times n matrices AA and BB:

      (A+B)ij=aij+bij(A + B)_{ij} = a_{ij} + b_{ij}

      Scalar Multiplication

      (λA)ij=λaij(\lambda A)_{ij} = \lambda a_{ij}

      Matrix Multiplication

      For AA (m×pm \times p) and BB (p×np \times n):

      (AB)ij=∑k=1paikbkj(AB)_{ij} = \sum_{k=1}^{p} a_{ik}b_{kj}

      Matrix multiplication showing row i of A times column j of B gives element (i,j) of AB

      Condition: Number of columns of AA = number of rows of BB.

      Warning: Matrix multiplication is not commutative: AB≠BAAB \neq BA in general.


      Transpose

      AijT=AjiA^T_{ij} = A_{ji}

      Properties:

      • (AT)T=A(A^T)^T = A
      • (A+B)T=AT+BT(A + B)^T = A^T + B^T
      • (AB)T=BTAT(AB)^T = B^T A^T (reversed order)

      Trace

      The trace of a square n×nn \times n matrix:

      Tr(A)=∑i=1naii\text{Tr}(A) = \sum_{i=1}^{n} a_{ii}

      Properties:

      • Tr(A+B)=Tr(A)+Tr(B)\text{Tr}(A + B) = \text{Tr}(A) + \text{Tr}(B)
      • Tr(λA)=λ Tr(A)\text{Tr}(\lambda A) = \lambda\,\text{Tr}(A)
      • Tr(AB)=Tr(BA)\text{Tr}(AB) = \text{Tr}(BA) (cyclic property)

      Identity Matrix

      In=(10⋯001⋯0⋮⋮⋱⋮00⋯1)I_n = \begin{pmatrix} 1 & 0 & \cdots & 0 \\ 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 1 \end{pmatrix}

      AI=IA=AAI = IA = A for compatible matrices.


      Special Types

      TypeDefinition
      Diagonalaij=0a_{ij} = 0 for i≠ji \neq j
      Upper triangularaij=0a_{ij} = 0 for i>ji > j
      Lower triangularaij=0a_{ij} = 0 for i<ji < j
      SymmetricAT=AA^T = A
      AntisymmetricAT=−AA^T = -A
      OrthogonalATA=IA^T A = I

      Complex Matrices

      Complex Conjugate

      (A∗)ij=aˉij(A^*)_{ij} = \bar{a}_{ij}

      Hermitian Conjugate

      A†=(AT)∗A^\dagger = (A^T)^*

      Hermitian Matrix

      A†=AA^\dagger = A

      Hermitian matrices have real eigenvalues (important in quantum mechanics).


      Representing Linear Maps

      A matrix AA (m×nm \times n) represents a linear map Rn→Rm\mathbb{R}^n \to \mathbb{R}^m:

      y=Ax\mathbf{y} = A\mathbf{x}

      where x∈Rn\mathbf{x} \in \mathbb{R}^n (column vector) and y∈Rm\mathbf{y} \in \mathbb{R}^m.


      Example: Rotation Matrix

      2D rotation by angle θ\theta:

      R(θ)=(cos⁡θ−sin⁡θsin⁡θcos⁡θ)R(\theta) = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}

      Properties:

      • R(θ)R(ϕ)=R(θ+ϕ)R(\theta)R(\phi) = R(\theta + \phi)
      • R(θ)−1=R(−θ)R(\theta)^{-1} = R(-\theta)
      • det⁡(R)=1\det(R) = 1
      • Orthogonal: RTR=IR^T R = I

      Why Matrices Matter

      • Represent linear systems of equations compactly
      • Encode transformations (rotations, scaling, reflections)
      • Essential for computer graphics and physics simulations
      • Foundation for quantum mechanics (operators)
    • Determinants

      Definition for 2×22 \times 2

      det⁡(abcd)=ad−bc\det\begin{pmatrix} a & b \\ c & d \end{pmatrix} = ad - bc


      Definition for 3×33 \times 3

      Using cofactor expansion:

      det⁡(A)=a11M11−a12M12+a13M13\det(A) = a_{11}M_{11} - a_{12}M_{12} + a_{13}M_{13}

      where MijM_{ij} is the minor (determinant of 2×22 \times 2 matrix after removing row ii and column jj).


      Cofactor

      The cofactor CijC_{ij} is:

      Cij=(−1)i+jMijC_{ij} = (-1)^{i+j} M_{ij}

      Pattern of signs: (+−+−+−+−+)\begin{pmatrix} + & - & + \\ - & + & - \\ + & - & + \end{pmatrix}


      General Method

      The determinant can be computed by expansion along any row or column:

      det⁡(A)=∑j=1naijCij(expansion along row i)\det(A) = \sum_{j=1}^{n} a_{ij}C_{ij} \quad \text{(expansion along row } i)

      det⁡(A)=∑i=1naijCij(expansion along column j)\det(A) = \sum_{i=1}^{n} a_{ij}C_{ij} \quad \text{(expansion along column } j)

      Strategy: Expand along a row/column with many zeros.


      Levi-Civita Symbol

      For n=3n = 3, define ϵijk\epsilon_{ijk}:

      ϵijk={+1(i,j,k) even permutation of (1,2,3)−1(i,j,k) odd permutation of (1,2,3)0any index repeated\epsilon_{ijk} = \begin{cases} +1 & (i,j,k) \text{ even permutation of } (1,2,3) \\ -1 & (i,j,k) \text{ odd permutation of } (1,2,3) \\ 0 & \text{any index repeated} \end{cases}

      Determinant:

      det⁡(A)=∑i,j,k=13ϵijka1ia2ja3k\det(A) = \sum_{i,j,k=1}^{3} \epsilon_{ijk} a_{1i} a_{2j} a_{3k}


      Properties

      PropertyFormula
      Transposedet⁡(AT)=det⁡(A)\det(A^T) = \det(A)
      Productdet⁡(AB)=det⁡(A)det⁡(B)\det(AB) = \det(A)\det(B)
      Scalardet⁡(λA)=λndet⁡(A)\det(\lambda A) = \lambda^n\det(A) for n×nn \times n
      Inversedet⁡(A−1)=1/det⁡(A)\det(A^{-1}) = 1/\det(A)
      Triangulardet⁡=product of diagonal elements\det = \text{product of diagonal elements}

      Row Operations

      OperationEffect on determinant
      Swap two rowsMultiply by −1-1
      Multiply row by λ\lambdaMultiply by λ\lambda
      Add multiple of one row to anotherNo change

      Geometric Interpretation

      ∣det⁡(A)∣|\det(A)| = volume of parallelepiped formed by columns of AA.

      Parallelepiped formed by columns of a 3×3 matrix, with volume = |det(A)|

      2×22 \times 2: Area of parallelogram.

      3×33 \times 3: Volume of parallelepiped.

      det⁡(A)=0\det(A) = 0 means columns are linearly dependent (volume collapses).


      Example

      det⁡(1234567810)\det\begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 10 \end{pmatrix}

      Expand along first row:

      =1det⁡(56810)−2det⁡(46710)+3det⁡(4578)= 1\det\begin{pmatrix} 5 & 6 \\ 8 & 10 \end{pmatrix} - 2\det\begin{pmatrix} 4 & 6 \\ 7 & 10 \end{pmatrix} + 3\det\begin{pmatrix} 4 & 5 \\ 7 & 8 \end{pmatrix}

      =1(50−48)−2(40−42)+3(32−35)= 1(50 - 48) - 2(40 - 42) + 3(32 - 35)

      =2+4−9=−3= 2 + 4 - 9 = -3


      Determinant Test for Linear Independence

      nn vectors in Rn\mathbb{R}^n are linearly independent ⇔\Leftrightarrow determinant of the matrix formed by them is non-zero.


      Why Determinants Matter

      • Determine if linear system has unique solution
      • Compute matrix inverse
      • Calculate eigenvalues (characteristic polynomial)
      • Measure volume change under transformation
      • Test for singular matrices
    • Matrix Inverse

      Definition

      A−1A^{-1} is the inverse of AA if:

      A−1A=AA−1=IA^{-1}A = AA^{-1} = I

      Existence: Inverse exists if and only if det⁡(A)≠0\det(A) \neq 0.


      Methods for Finding Inverse

      1. Formula for 2×22 \times 2

      (abcd)−1=1ad−bc(d−b−ca)\begin{pmatrix} a & b \\ c & d \end{pmatrix}^{-1} = \frac{1}{ad - bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}

      2. Cofactor Method (Adjugate)

      A−1=1det⁡(A)(adj A)A^{-1} = \frac{1}{\det(A)}(\text{adj }A)

      where adj(A)=CT\text{adj}(A) = C^T is the transpose of the cofactor matrix.

      3. Row Reduction (Gaussian Elimination)

      Augment [A∣I][A|I] and perform row operations:

      [A∣I]→[I∣A−1][A|I] \to [I|A^{-1}]


      Example: 2×22 \times 2 Inverse

      A=(2153)A = \begin{pmatrix} 2 & 1 \\ 5 & 3 \end{pmatrix}

      det⁡(A)=6−5=1\det(A) = 6 - 5 = 1

      A−1=11(3−1−52)=(3−1−52)A^{-1} = \frac{1}{1}\begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix} = \begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix}

      Verify: AA−1=(6−5−2+215−15−5+6)=IAA^{-1} = \begin{pmatrix} 6-5 & -2+2 \\ 15-15 & -5+6 \end{pmatrix} = I


      Properties

      PropertyFormula
      UniquenessInverse is unique (if exists)
      Product(AB)−1=B−1A−1(AB)^{-1} = B^{-1}A^{-1} (reversed order)
      Transpose(AT)−1=(A−1)T(A^T)^{-1} = (A^{-1})^T
      Double inverse(A−1)−1=A(A^{-1})^{-1} = A
      Determinantdet⁡(A−1)=1/det⁡(A)\det(A^{-1}) = 1/\det(A)

      Kernel (Null Space)

      The kernel of AA is:

      ker⁡(A)={x:Ax=0}\ker(A) = \{\mathbf{x} : A\mathbf{x} = \mathbf{0}\}

      Kernel as the null space in R^n, showing vectors that map to zero

      Key results:

      • ker⁡(A)={0}\ker(A) = \{\mathbf{0}\} ⇔\Leftrightarrow det⁡(A)≠0\det(A) \neq 0 ⇔\Leftrightarrow A−1A^{-1} exists
      • Non-trivial kernel ⇔\Leftrightarrow columns are linearly dependent

      Solving Ax=yA\mathbf{x} = \mathbf{y}

      If det⁡(A)≠0\det(A) \neq 0:

      x=A−1y\mathbf{x} = A^{-1}\mathbf{y}

      Row reduction method: Augment [A∣y][A|\mathbf{y}] and solve.


      When Inverse Doesn’t Exist

      det⁡(A)=0\det(A) = 0 means:

      • Columns are linearly dependent
      • ker⁡(A)≠{0}\ker(A) \neq \{\mathbf{0}\}
      • Either no solution or infinitely many solutions to Ax=yA\mathbf{x} = \mathbf{y}

      Left and Right Inverses

      For m×nm \times n matrices:

      • Left inverse: BA=InBA = I_n (if columns of AA are independent)
      • Right inverse: AB=ImAB = I_m (if rows of AA are independent)

      For square matrices: left inverse = right inverse = unique inverse.


      Why Matrix Inverse Matters

      • Solve systems Ax=bA\mathbf{x} = \mathbf{b} directly: x=A−1b\mathbf{x} = A^{-1}\mathbf{b}
      • Undo transformations (inverse rotation, inverse scaling)
      • Essential for parameter estimation in statistics
      • Foundation for understanding linear maps

      Note: In practice, row reduction or decomposition methods are preferred for numerical stability. Direct inversion is rarely computationally optimal.

    • Linear Simultaneous Equations

      The Problem

      Solve the system:

      {a11x1+a12x2+⋯+a1nxn=b1a21x1+a22x2+⋯+a2nxn=b2⋮am1x1+am2x2+⋯+amnxn=bm\begin{cases} a_{11}x_1 + a_{12}x_2 + \cdots + a_{1n}x_n = b_1 \\ a_{21}x_1 + a_{22}x_2 + \cdots + a_{2n}x_n = b_2 \\ \vdots \\ a_{m1}x_1 + a_{m2}x_2 + \cdots + a_{mn}x_n = b_m \end{cases}

      In matrix form: Ax=bA\mathbf{x} = \mathbf{b}.


      Gaussian Elimination

      Transform to row echelon form:

      (a11a12⋯a1n∣b10a22′⋯a2n′∣b2′⋮⋱∣⋮00⋯ank′∣bm′)\begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n} & | & b_1 \\ 0 & a'_{22} & \cdots & a'_{2n} & | & b'_2 \\ \vdots & & \ddots & & | & \vdots \\ 0 & 0 & \cdots & a'_{nk} & | & b'_m \end{pmatrix}

      Row operations:

      1. Swap rows
      2. Multiply row by non-zero scalar
      3. Add multiple of one row to another

      Classification of Solutions

      Three cases for linear systems: unique solution (lines meet at point), no solution (parallel lines), infinitely many solutions (same line)

      CaseConditionSolutions
      Uniquedet⁡(A)≠0\det(A) \neq 0 (square)Exactly one
      UnderdeterminedRank < number of variablesInfinitely many
      OverdeterminedInconsistent equationsNo solutions
      Dependent equationsRows linearly dependentRedundant equations

      Rank

      The rank of AA is:

      • Number of non-zero rows in row echelon form
      • Number of linearly independent rows (or columns)
      • Dimension of the image: rank(A)=dim⁡(im(A))\text{rank}(A) = \dim(\text{im}(A))

      Rank theorem:

      rank(A)+dim⁡(ker⁡(A))=n\text{rank}(A) + \dim(\ker(A)) = n

      (for m×nm \times n matrix)


      Example: 2×22 \times 2 System

      Solve: {2x+y=53x+2y=8\begin{cases} 2x + y = 5 \\ 3x + 2y = 8 \end{cases}

      Matrix form: (2132)(xy)=(58)\begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 5 \\ 8 \end{pmatrix}

      Row operations: [2 1 ∣ 5][2\ 1\ |\ 5] [3 2 ∣ 8][3\ 2\ |\ 8] →\to subtract 32\frac{3}{2} of row 1 from row 2: [2 1 ∣ 5][2\ 1\ |\ 5] [0 12 ∣ 12][0\ \frac{1}{2}\ |\ \frac{1}{2}] →\to back-substitute: y=1y = 1, then 2x+1=52x + 1 = 5, so x=2x = 2.


      Cramer’s Rule

      For n×nn \times n system with det⁡(A)≠0\det(A) \neq 0:

      xi=det⁡(Ai)det⁡(A)x_i = \frac{\det(A_i)}{\det(A)}

      where AiA_i is AA with column ii replaced by b\mathbf{b}.

      Impractical for large nn — use Gaussian elimination instead.


      Homogeneous Systems

      Ax=0A\mathbf{x} = \mathbf{0}

      • Always has trivial solution x=0\mathbf{x} = \mathbf{0}
      • Non-trivial solutions exist ⇔\Leftrightarrow det⁡(A)=0\det(A) = 0

      Kernel and Image

      Kernel: ker⁡(A)={x:Ax=0}\ker(A) = \{\mathbf{x} : A\mathbf{x} = \mathbf{0}\}

      Image: im(A)={Ax:x∈Rn}\text{im}(A) = \{A\mathbf{x} : \mathbf{x} \in \mathbb{R}^n\}

      Fundamental theorem:

      • ker⁡(A)\ker(A) gives the “freedom” in solutions
      • b∈im(A)\mathbf{b} \in \text{im}(A) is required for solution to exist

      Why Linear Systems Matter

      • Engineering: structures, circuits, control systems
      • Economics: input-output models
      • Statistics: regression (least squares)
      • Physics: equilibrium problems, eigenvalue problems

      Almost every quantitative problem involves solving linear systems at some stage.

    • Linear Maps

      Definition

      A function f:V→Wf: V \to W between vector spaces is linear if:

      f(u+v)=f(u)+f(v)f(\mathbf{u} + \mathbf{v}) = f(\mathbf{u}) + f(\mathbf{v}) f(λu)=λf(u)f(\lambda\mathbf{u}) = \lambda f(\mathbf{u})

      for all u,v∈V\mathbf{u}, \mathbf{v} \in V and λ∈K\lambda \in K.


      Matrix Representation

      Every linear map f:Rn→Rmf: \mathbb{R}^n \to \mathbb{R}^m can be written:

      f(x)=Axf(\mathbf{x}) = A\mathbf{x}

      where AA is an m×nm \times n matrix.

      The jjth column of AA is the image of the jjth basis vector: AejA\mathbf{e}_j.


      Orthogonal Matrices

      A matrix OO is orthogonal if:

      OTO=IO^T O = I

      Properties:

      • O−1=OTO^{-1} = O^T
      • Columns form an orthonormal basis
      • det⁡(O)=±1\det(O) = \pm 1
      • Preserves length: ∣Ox∣=∣x∣|O\mathbf{x}| = |\mathbf{x}|
      • Preserves dot product: (Ou)⋅(Ov)=u⋅v(O\mathbf{u}) \cdot (O\mathbf{v}) = \mathbf{u} \cdot \mathbf{v}

      Rotation Matrices

      2D rotation by angle θ\theta:

      R(θ)=(cos⁡θ−sin⁡θsin⁡θcos⁡θ)R(\theta) = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}

      2D rotation showing point (x,y) rotated by angle θ to new position (x',y')

      Properties:

      • R(θ)R(ϕ)=R(θ+ϕ)R(\theta)R(\phi) = R(\theta + \phi)
      • R(θ)−1=R(−θ)R(\theta)^{-1} = R(-\theta)

      3D rotations about coordinate axes:

      Rx(θ)=(1000cos⁡θ−sin⁡θ0sin⁡θcos⁡θ)R_x(\theta) = \begin{pmatrix} 1 & 0 & 0 \\ 0 & \cos\theta & -\sin\theta \\ 0 & \sin\theta & \cos\theta \end{pmatrix}


      Reflection Matrices

      2D reflection across line at angle θ\theta to xx-axis:

      M(θ)=(cos⁡2θsin⁡2θsin⁡2θ−cos⁡2θ)M(\theta) = \begin{pmatrix} \cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{pmatrix}

      det⁡(M)=−1\det(M) = -1 (orientation-reversing)


      Geometric Transformations

      TransformationMatrix type
      RotationOrthogonal, det⁡=+1\det = +1
      ReflectionOrthogonal, det⁡=−1\det = -1
      ScalingDiagonal (or isotropic)
      ShearUpper/lower triangular with 1s on diagonal
      ProjectionA2=AA^2 = A (idempotent)

      Determinant and Volume

      For linear map with matrix AA:

      ∣det⁡(A)∣=volume of imagevolume of original|\det(A)| = \frac{\text{volume of image}}{\text{volume of original}}

      • det⁡>0\det > 0: orientation preserved
      • det⁡<0\det < 0: orientation reversed
      • det⁡=0\det = 0: dimension collapses (singular)

      Kernel and Image of Linear Map

      For f:V→Wf: V \to W:

      Kernel: ker⁡(f)={v∈V:f(v)=0}\ker(f) = \{\mathbf{v} \in V : f(\mathbf{v}) = \mathbf{0}\}

      Image: im(f)={f(v):v∈V}\text{im}(f) = \{f(\mathbf{v}) : \mathbf{v} \in V\}

      Rank-nullity theorem:

      dim⁡(V)=dim⁡(ker⁡(f))+dim⁡(im(f))\dim(V) = \dim(\ker(f)) + \dim(\text{im}(f))


      Change of Basis

      If AA represents a linear map in basis {ei}\{\mathbf{e}_i\}, then in basis {fi}\{\mathbf{f}_i\} where fj=∑Pijei\mathbf{f}_j = \sum P_{ij}\mathbf{e}_i:

      A′=P−1APA' = P^{-1}AP

      The same linear map has different matrix representations in different bases.


      Why Linear Maps Matter

      • Unified treatment of rotations, scaling, projections
      • Essential for computer graphics
      • Foundation for eigentheory (best basis for a linear map)
      • Quantum mechanics: observables are linear operators

      Linear maps preserve vector space structure — this is the essence of linearity.

    • Eigenvalues and Eigenvectors

      Definition

      For a square matrix AA, an eigenvector v≠0\mathbf{v} \neq \mathbf{0} satisfies:

      Av=λvA\mathbf{v} = \lambda\mathbf{v}

      where λ\lambda is the eigenvalue.

      Key: Applying AA to its eigenvector just scales it by λ\lambda.


      Characteristic Polynomial

      Rearrange: (A−λI)v=0(A - \lambda I)\mathbf{v} = \mathbf{0}

      Non-trivial solutions exist when det⁡(A−λI)=0\det(A - \lambda I) = 0.

      This gives the characteristic polynomial:

      PA(λ)=det⁡(A−λI)=0P_A(\lambda) = \det(A - \lambda I) = 0

      An n×nn \times n matrix has exactly nn eigenvalues (counting multiplicities).


      Finding Eigenvalues and Eigenvectors

      Step 1: Solve PA(λ)=0P_A(\lambda) = 0 for eigenvalues.

      Step 2: For each λ\lambda, solve (A−λI)v=0(A - \lambda I)\mathbf{v} = \mathbf{0} for v\mathbf{v}.


      Example

      Find eigenvalues and eigenvectors of A=(2112)A = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}.

      Eigenvectors of 2×2 symmetric matrix shown as vectors, demonstrating they are perpendicular

      Characteristic polynomial:

      det⁡(A−λI)=det⁡(2−λ112−λ)\det(A - \lambda I) = \det\begin{pmatrix} 2-\lambda & 1 \\ 1 & 2-\lambda \end{pmatrix}

      =(2−λ)2−1=λ2−4λ+3=(λ−1)(λ−3)= (2-\lambda)^2 - 1 = \lambda^2 - 4\lambda + 3 = (\lambda - 1)(\lambda - 3)

      Eigenvalues: λ1=1\lambda_1 = 1, λ2=3\lambda_2 = 3.

      Eigenvector for λ=1\lambda = 1:

      (A−I)v=(1111)v=0(A - I)\mathbf{v} = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}\mathbf{v} = \mathbf{0}

      ⇒v1+v2=0\Rightarrow v_1 + v_2 = 0

      v1=(1−1)\mathbf{v}_1 = \begin{pmatrix} 1 \\ -1 \end{pmatrix}

      Eigenvector for λ=3\lambda = 3:

      (A−3I)v=(−111−1)v=0(A - 3I)\mathbf{v} = \begin{pmatrix} -1 & 1 \\ 1 & -1 \end{pmatrix}\mathbf{v} = \mathbf{0}

      ⇒v1=v2\Rightarrow v_1 = v_2

      v2=(11)\mathbf{v}_2 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}


      Properties

      PropertyFormula
      Sum of eigenvaluesλ1+⋯+λn=Tr(A)\lambda_1 + \cdots + \lambda_n = \text{Tr}(A)
      Product of eigenvaluesλ1⋯λn=det⁡(A)\lambda_1 \cdots \lambda_n = \det(A)
      Inverse eigenvaluesIf Av=λvA\mathbf{v} = \lambda\mathbf{v}, then A−1v=λ−1vA^{-1}\mathbf{v} = \lambda^{-1}\mathbf{v}
      PolynomialAkv=λkvA^k\mathbf{v} = \lambda^k\mathbf{v}

      Real Symmetric Matrices

      For real symmetric AA (A=ATA = A^T):

      1. All eigenvalues are real
      2. Eigenvectors for different eigenvalues are orthogonal
      3. Can always find nn orthonormal eigenvectors

      Diagonalisation

      If AA has nn linearly independent eigenvectors v1,…,vn\mathbf{v}_1, \ldots, \mathbf{v}_n:

      A=SΛS−1A = S\Lambda S^{-1}

      where SS has eigenvectors as columns, and Λ=diag(λ1,…,λn)\Lambda = \text{diag}(\lambda_1, \ldots, \lambda_n).

      Diagonalisation A = SΛS⁻¹, showing transformation to eigenbasis, scaling, and transformation back

      For symmetric matrices: SS can be orthogonal, so S−1=STS^{-1} = S^T.


      Diagonalisation Example

      For A=(2112)A = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}:

      S=(11−11)S = \begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix}, Λ=(1003)\Lambda = \begin{pmatrix} 1 & 0 \\ 0 & 3 \end{pmatrix}

      Verify: SΛS−1=AS\Lambda S^{-1} = A.


      Application: Hessian Matrix

      For stationary point analysis, diagonalise the Hessian HH:

      • All λi>0\lambda_i > 0 → local minimum
      • All λi<0\lambda_i < 0 → local maximum
      • Mixed signs → saddle point

      Eigenvalues give the curvatures in principal directions.


      Similarity Transformation

      Matrices AA and BB are similar if B=P−1APB = P^{-1}AP for some invertible PP.

      Similar matrices represent the same linear map in different bases. They share:

      • Eigenvalues
      • Determinant
      • Trace

      Why Eigenvalues Matter

      • Dynamics: Akx=SΛkS−1xA^k\mathbf{x} = S\Lambda^k S^{-1}\mathbf{x} (long-term behaviour)
      • Stability: Eigenvalues of Jacobian determine stability of fixed points
      • Vibrations: Natural frequencies are eigenvalues
      • Quantum mechanics: Observable values are eigenvalues
      • Principal component analysis: Covariance matrix eigenvectors
    • Introduction to PDEs

      What is a PDE?

      A partial differential equation involves partial derivatives of an unknown function of several variables:

      F(x,y,…,u,∂u∂x,∂u∂y,…,∂2u∂x2,∂2u∂x∂y,∂2u∂y2,…)=0F\left(x, y, \ldots, u, \frac{\partial u}{\partial x}, \frac{\partial u}{\partial y}, \ldots, \frac{\partial^2 u}{\partial x^2}, \frac{\partial^2 u}{\partial x\partial y}, \frac{\partial^2 u}{\partial y^2}, \ldots\right) = 0


      Order and Linearity

      Order: Highest derivative appearing.

      Linear PDE: Linear in uu and its derivatives. Can be written:

      ∑∣α∣≤naα(x)∂∣α∣u∂x1α1⋯∂xmαm=f(x)\sum_{|\alpha| \leq n} a_\alpha(\mathbf{x})\frac{\partial^{|\alpha|}u}{\partial x_1^{\alpha_1}\cdots\partial x_m^{\alpha_m}} = f(\mathbf{x})


      Classification of Second Order Linear PDEs (2D)

      Consider:

      a∂2u∂x2+2b∂2u∂x∂y+c∂2u∂y2+lower order=0a\frac{\partial^2 u}{\partial x^2} + 2b\frac{\partial^2 u}{\partial x\partial y} + c\frac{\partial^2 u}{\partial y^2} + \text{lower order} = 0

      Visual comparison of elliptic (smooth elliptical contours), parabolic (spreading Gaussian), and hyperbolic (travelling waves) PDE behaviours

      Define discriminant: Δ=b2−ac\Delta = b^2 - ac.

      TypeConditionDiscriminantExample
      Ellipticb2<acb^2 < acΔ<0\Delta < 0Laplace: uxx+uyy=0u_{xx} + u_{yy} = 0
      Hyperbolicb2>acb^2 > acΔ>0\Delta > 0Wave: utt=c2uxxu_{tt} = c^2 u_{xx}
      Parabolicb2=acb^2 = acΔ=0\Delta = 0Heat: ut=Duxxu_t = Du_{xx}

      Boundary Conditions

      Dirichlet

      Specify the value of the function on the boundary:

      u(x)=f(x) on ∂Ωu(\mathbf{x}) = f(\mathbf{x}) \text{ on } \partial\Omega

      Neumann

      Specify the normal derivative on the boundary:

      ∂u∂n=n⋅∇u=g(x) on ∂Ω\frac{\partial u}{\partial n} = \mathbf{n} \cdot \nabla u = g(\mathbf{x}) \text{ on } \partial\Omega

      Mixed (Robin)

      Linear combination:

      αu+β∂u∂n=h(x)\alpha u + \beta\frac{\partial u}{\partial n} = h(\mathbf{x})

      Initial Conditions

      For time-dependent problems, specify initial state:

      u(x,0)=u0(x)u(\mathbf{x}, 0) = u_0(\mathbf{x})


      Well-Posed Problems

      A problem is well-posed (Hadamard) if:

      1. A solution exists
      2. The solution is unique
      3. The solution depends continuously on data

      Physical Interpretation

      PDEPhysical setting
      LaplaceSteady-state heat distribution, electrostatics
      Heat equationTemperature evolution in a conductor
      Wave equationVibrating string, electromagnetic waves
      DiffusionParticle transport, heat flow

      Superposition Principle

      For linear homogeneous PDEs: if u1u_1 and u2u_2 are solutions, so is u1+u2u_1 + u_2.

      This enables building complex solutions from simple ones.


      Why PDEs Matter

      • Model physical phenomena with spatial and temporal variation
      • Describe systems from quantum mechanics to fluid dynamics
      • Foundation for understanding continuum physics
      • Essential for finite element methods and numerical simulation
    • Laplace's Equation

      The Equation

      ∇2ϕ=∂2ϕ∂x2+∂2ϕ∂y2=0\nabla^2 \phi = \frac{\partial^2 \phi}{\partial x^2} + \frac{\partial^2 \phi}{\partial y^2} = 0

      In 3D: ∇2ϕ=∂2ϕ∂x2+∂2ϕ∂y2+∂2ϕ∂z2=0\nabla^2 \phi = \frac{\partial^2 \phi}{\partial x^2} + \frac{\partial^2 \phi}{\partial y^2} + \frac{\partial^2 \phi}{\partial z^2} = 0


      Physical Context

      • Steady-state temperature distribution
      • Electrostatic potential (no charges)
      • Gravitational potential (no masses)
      • Incompressible, irrotational fluid flow

      Temperature distribution in a rectangular plate with boundary conditions, showing isotherms (level curves) of potential

      Key property: Solutions are harmonic functions — smooth, with no local maxima/minima (maximum principle).


      Separation of Variables

      Problem: Find ϕ(x,y)\phi(x, y) in rectangle 0<x<a0 < x < a, 0<y<b0 < y < b with:

      Rectangle with boundary conditions on each edge - showing Dirichlet conditions

      • ϕ(0,y)=0\phi(0, y) = 0 (left edge)
      • ϕ(x,0)=0\phi(x, 0) = 0 (bottom edge)
      • ϕ(x,b)=0\phi(x, b) = 0 (top edge)
      • ϕ(a,y)=f(y)\phi(a, y) = f(y) (right edge)

      Ansatz: ϕ(x,y)=X(x)Y(y)\phi(x, y) = X(x)Y(y)

      Substituting into Laplace equation:

      X′′Y+XY′′=0X''Y + XY'' = 0

      X′′X=−Y′′Y=k\frac{X''}{X} = -\frac{Y''}{Y} = k


      Solution

      For homogeneous Dirichlet conditions on yy edges:

      The separation constant must be negative: k=−γ2<0k = -\gamma^2 < 0.

      YY equation: Y′′+γ2Y=0Y'' + \gamma^2 Y = 0

      Y(y)=Asin⁡(γy)+Bcos⁡(γy)Y(y) = A\sin(\gamma y) + B\cos(\gamma y)

      Boundary condition Y(0)=0⇒B=0Y(0) = 0 \Rightarrow B = 0.

      Y(b)=0⇒sin⁡(γb)=0Y(b) = 0 \Rightarrow \sin(\gamma b) = 0, so γ=nπb\gamma = \frac{n\pi}{b} for n=1,2,3,…n = 1, 2, 3, \ldots

      XX equation: X′′−γ2X=0X'' - \gamma^2 X = 0

      X(x)=Csinh⁡(γx)+Dcosh⁡(γx)X(x) = C\sinh(\gamma x) + D\cosh(\gamma x)

      Boundary condition X(0)=0⇒D=0X(0) = 0 \Rightarrow D = 0.

      Combined solution:

      ϕ(x,y)=∑n=1∞Cnsin⁡(nπyb)sinh⁡(nπxb)\phi(x, y) = \sum_{n=1}^{\infty} C_n \sin\left(\frac{n\pi y}{b}\right) \sinh\left(\frac{n\pi x}{b}\right)


      Applying the Inhomogeneous Boundary Condition

      At x=ax = a: ϕ(a,y)=f(y)\phi(a, y) = f(y)

      This is a Fourier sine series, so:

      f(y)=∑n=1∞Cnsin⁡(nπyb)sinh⁡(nπab)f(y) = \sum_{n=1}^{\infty} C_n \sin\left(\frac{n\pi y}{b}\right) \sinh\left(\frac{n\pi a}{b}\right)

      Fourier coefficients:

      Cn=2bsinh⁡(nπa/b)∫0bf(y)sin⁡(nπyb)dyC_n = \frac{2}{b\sinh(n\pi a/b)} \int_0^b f(y)\sin\left(\frac{n\pi y}{b}\right) dy


      Example: Constant f(y)=1f(y) = 1

      Cn=2bsinh⁡(nπa/b)∫0bsin⁡(nπyb)dyC_n = \frac{2}{b\sinh(n\pi a/b)} \int_0^b \sin\left(\frac{n\pi y}{b}\right) dy

      =2bsinh⁡(nπa/b)⋅bnπ[1−(−1)n]= \frac{2}{b\sinh(n\pi a/b)} \cdot \frac{b}{n\pi}[1 - (-1)^n]

      =2nπsinh⁡(nπa/b)×{2n odd0n even= \frac{2}{n\pi\sinh(n\pi a/b)} \times \begin{cases} 2 & n \text{ odd} \\ 0 & n \text{ even} \end{cases}


      Maximum Principle

      Statement: A harmonic function on a bounded domain attains its maximum and minimum on the boundary.

      Physical interpretation: Steady-state temperature cannot be hotter than the hottest boundary point.


      Uniqueness

      For given boundary conditions, Laplace’s equation has a unique solution.

      Proof: If ϕ1\phi_1 and ϕ2\phi_2 are solutions, then ψ=ϕ1−ϕ2\psi = \phi_1 - \phi_2 is harmonic with zero boundary conditions. By maximum principle, ψ=0\psi = 0.


      Why Laplace’s Equation Matters

      • Describes equilibrium in many physical systems
      • Solved analytically using separation of variables
      • Basis for understanding more complex PDEs
      • Important in electrostatics, fluid dynamics, and heat transfer
    • Wave Equation

      The Equation

      ∂2y∂t2=c02∂2y∂x2\frac{\partial^2 y}{\partial t^2} = c_0^2 \frac{\partial^2 y}{\partial x^2}

      Or: ytt=c02yxxy_{tt} = c_0^2 y_{xx}

      c0c_0 is the wave speed.


      Physical Derivation

      Consider a stretched string with:

      • Tension TT (constant for small displacements)
      • Mass per unit length μ\mu

      For small yy and small slope:

      Newton’s second law on element:

      μΔx∂2y∂t2≈T[(∂y∂x)x+Δx−(∂y∂x)x]\mu\Delta x \frac{\partial^2 y}{\partial t^2} \approx T\left[\left(\frac{\partial y}{\partial x}\right)_{x+\Delta x} - \left(\frac{\partial y}{\partial x}\right)_x\right]

      Dividing by Δx\Delta x and taking limit:

      μ∂2y∂t2=T∂2y∂x2\mu\frac{\partial^2 y}{\partial t^2} = T\frac{\partial^2 y}{\partial x^2}

      Wave speed: c02=T/μc_0^2 = T/\mu


      D’Alembert’s Solution

      The general solution is:

      y(x,t)=f(x−c0t)+g(x+c0t)y(x, t) = f(x - c_0 t) + g(x + c_0 t)

      Travelling waves - f(x-ct) moving right, g(x+ct) moving left, both at speed c

      where ff and gg are arbitrary functions.

      Interpretation:

      • f(x−c0t)f(x - c_0 t): wave travelling right at speed c0c_0
      • g(x+c0t)g(x + c_0 t): wave travelling left at speed c0c_0

      Standing Waves on a Finite String

      Consider string of length LL with fixed ends:

      Vibrating string of length L with fixed ends, showing standing wave pattern

      Boundary conditions:

      • y(0,t)=0y(0, t) = 0
      • y(L,t)=0y(L, t) = 0

      Separation of variables: Try y(x,t)=X(x)T(t)y(x, t) = X(x)T(t)

      Substituting: T′′c02T=X′′X=−ω2\frac{T''}{c_0^2 T} = \frac{X''}{X} = -\omega^2

      Spatial part: X(x)=Asin⁡(ωx/c0)+Bcos⁡(ωx/c0)X(x) = A\sin(\omega x/c_0) + B\cos(\omega x/c_0)

      X(0)=0⇒B=0X(0) = 0 \Rightarrow B = 0

      X(L)=0⇒sin⁡(ωL/c0)=0X(L) = 0 \Rightarrow \sin(\omega L/c_0) = 0, so ωL/c0=nπ\omega L/c_0 = n\pi

      Normal modes:

      yn(x,t)=sin⁡(nπxL)[Cncos⁡(nπc0tL)+Dnsin⁡(nπc0tL)]y_n(x, t) = \sin\left(\frac{n\pi x}{L}\right)\left[C_n\cos\left(\frac{n\pi c_0 t}{L}\right) + D_n\sin\left(\frac{n\pi c_0 t}{L}\right)\right]


      Initial Conditions

      Given y(x,0)=Y(x)y(x, 0) = Y(x) and y˙(x,0)=V(x)\dot{y}(x, 0) = V(x):

      Expand in Fourier sine series:

      Y(x)=∑n=1∞Ansin⁡(nπxL)Y(x) = \sum_{n=1}^{\infty} A_n \sin\left(\frac{n\pi x}{L}\right)

      V(x)=∑n=1∞Bn(nπc0L)sin⁡(nπxL)V(x) = \sum_{n=1}^{\infty} B_n \left(\frac{n\pi c_0}{L}\right) \sin\left(\frac{n\pi x}{L}\right)

      The solution combines position and velocity terms.


      Standing Waves

      The normal modes are standing waves with:

      • Wavelength: λn=2L/n\lambda_n = 2L/n
      • Frequency: fn=nc02Lf_n = \frac{nc_0}{2L}

      First three standing wave modes (n=1, 2, 3) on a string, showing nodes and antinodes

      n=1n = 1: fundamental mode
      n=2n = 2: first harmonic
      n=3n = 3: second harmonic

      The general solution is a superposition of all modes.


      Energy Conservation

      Total energy of the string:

      E=12∫0L[μ(∂y∂t)2+T(∂y∂x)2]dxE = \frac{1}{2}\int_0^L \left[\mu\left(\frac{\partial y}{\partial t}\right)^2 + T\left(\frac{\partial y}{\partial x}\right)^2\right] dx

      For free vibrations, energy is conserved.


      Why the Wave Equation Matters

      • Describes vibrations (strings, membranes, acoustics)
      • Electromagnetic waves (light, radio)
      • Seismic waves (earthquakes)
      • One of the fundamental equations of physics

      The separation of variables method generalises to many other PDEs.

    • Diffusion Equation

      The Equation

      ∂ϕ∂t=D∂2ϕ∂x2\frac{\partial \phi}{\partial t} = D\frac{\partial^2 \phi}{\partial x^2}

      Or: ϕt=Dϕxx\phi_t = D\phi_{xx}

      DD is the diffusion coefficient (or thermal diffusivity for heat).


      Physical Derivation

      Fick’s law: Diffusive flux J=−D∂C∂xJ = -D\frac{\partial C}{\partial x} (down concentration gradient).

      Conservation: Rate of change = -net flux out:

      ∂C∂t=D∂2C∂x2\frac{\partial C}{\partial t} = D\frac{\partial^2 C}{\partial x^2}

      Heat equation: From Fourier’s law (J=−k∇TJ = -k\nabla T):

      ∂T∂t=D∇2T\frac{\partial T}{\partial t} = D\nabla^2 T

      where D=k/(ρcp)D = k/(\rho c_p).


      Three Fundamental Solutions

      1. Diffusion from Step Function

      Initially: ϕ(x,0)={0x<01x>0\phi(x, 0) = \begin{cases} 0 & x < 0 \\ 1 & x > 0 \end{cases}

      Step function initial condition and its smooth diffusion over time

      Solution:

      ϕ(x,t)=12[1−erf(x2Dt)]\phi(x, t) = \frac{1}{2}\left[1 - \text{erf}\left(\frac{x}{2\sqrt{Dt}}\right)\right]

      where erf(z)=2π∫0ze−u2du\text{erf}(z) = \frac{2}{\sqrt{\pi}}\int_0^z e^{-u^2}du.

      Interpretation: Smooth spreading of the initial discontinuity.


      2. Point Source (Green’s Function)

      Initially: ϕ(x,0)=δ(x)\phi(x, 0) = \delta(x) (concentrated at origin)

      Solution:

      ϕ(x,t)=14πDtexp⁡(−x24Dt)\phi(x, t) = \frac{1}{\sqrt{4\pi Dt}}\exp\left(-\frac{x^2}{4Dt}\right)

      Gaussian spreading over time - narrow spike at t=0, wider and lower at later times

      Properties:

      • Gaussian shape at all times
      • Spreads: width ∼Dt\sim \sqrt{Dt}
      • Mass conserved: ∫−∞∞ϕ dx=1\int_{-\infty}^{\infty}\phi\,dx = 1

      3. Semi-Infinite System with Absorbing Wall

      Domain: 0<x<L0 < x < L with ϕ(0,t)=0\phi(0, t) = 0 (absorbing boundary).

      Solution:

      ϕ(x,t)=∑n oddAnsin⁡(nπx2L)exp⁡(−n2π2Dt4L2)\phi(x, t) = \sum_{n\text{ odd}} A_n \sin\left(\frac{n\pi x}{2L}\right) \exp\left(-\frac{n^2\pi^2 D t}{4L^2}\right)

      Long-time behaviour: Dominated by n=1n = 1 mode, decays as e−π2Dt/4L2e^{-\pi^2 Dt/4L^2}.


      Separation of Variables

      Try ϕ(x,t)=X(x)T(t)\phi(x, t) = X(x)T(t):

      XT′=DX′′TXT' = DX''T

      T′DT=X′′X=−k2\frac{T'}{DT} = \frac{X''}{X} = -k^2

      Time: T(t)=e−Dk2tT(t) = e^{-Dk^2 t}

      Space: X(x)=Asin⁡(kx)+Bcos⁡(kx)X(x) = A\sin(kx) + B\cos(kx)

      Apply boundary conditions to determine kk.


      Example: Finite Rod with Fixed Ends

      ϕ(0,t)=ϕ(L,t)=0\phi(0, t) = \phi(L, t) = 0

      X(0)=0⇒B=0X(0) = 0 \Rightarrow B = 0

      X(L)=0⇒sin⁡(kL)=0X(L) = 0 \Rightarrow \sin(kL) = 0, so k=nπ/Lk = n\pi/L

      Solution:

      ϕ(x,t)=∑n=1∞Ansin⁡(nπxL)exp⁡(−n2π2DtL2)\phi(x, t) = \sum_{n=1}^{\infty} A_n \sin\left(\frac{n\pi x}{L}\right) \exp\left(-\frac{n^2\pi^2 Dt}{L^2}\right)


      Fourier Coefficients

      For initial condition ϕ(x,0)=f(x)\phi(x, 0) = f(x):

      An=2L∫0Lf(x)sin⁡(nπxL)dxA_n = \frac{2}{L}\int_0^L f(x)\sin\left(\frac{n\pi x}{L}\right) dx

      Each mode decays exponentially; higher modes decay faster.


      Long-Time Behaviour

      At large tt, the n=1n = 1 mode dominates:

      ϕ∝sin⁡(πx/L)exp⁡(−π2Dt/L2)\phi \propto \sin(\pi x/L) \exp(-\pi^2 Dt/L^2)

      Characteristic decay time: τ=L2/(π2D)\tau = L^2/(\pi^2 D)


      Comparison with Wave Equation

      PropertyWaveDiffusion
      Equationytt=c2yxxy_{tt} = c^2 y_{xx}ϕt=Dϕxx\phi_t = D\phi_{xx}
      Order in ttSecondFirst
      EnergyConservedDissipated
      InformationPropagates at speed ccDiffuses, spreads

      Why the Diffusion Equation Matters

      • Heat conduction
      • Particle diffusion
      • Chemical transport
      • Financial mathematics (Black-Scholes)
      • Biological processes (morphogen gradients)

      The diffusion equation describes how things spread out and dissipate — a universal phenomenon in nature.

    • Methods III Tripos Questions

      Linear Algebra

      Eigenvalues and Eigenvectors

      Find eigenvalues and eigenvectors of A=(4213)A = \begin{pmatrix} 4 & 2 \\ 1 & 3 \end{pmatrix}.

      Characteristic polynomial:

      det⁡(A−λI)=(4−λ)(3−λ)−2=λ2−7λ+10\det(A - \lambda I) = (4-\lambda)(3-\lambda) - 2 = \lambda^2 - 7\lambda + 10

      =(λ−2)(λ−5)=0= (\lambda - 2)(\lambda - 5) = 0

      Eigenvalues: λ1=2\lambda_1 = 2, λ2=5\lambda_2 = 5.

      For λ1=2\lambda_1 = 2:

      (A−2I)v=(2211)v=0(A - 2I)\mathbf{v} = \begin{pmatrix} 2 & 2 \\ 1 & 1 \end{pmatrix}\mathbf{v} = \mathbf{0}

      v1+v2=0v_1 + v_2 = 0, so v1=(1−1)\mathbf{v}_1 = \begin{pmatrix} 1 \\ -1 \end{pmatrix}

      For λ2=5\lambda_2 = 5:

      (A−5I)v=(−121−2)v=0(A - 5I)\mathbf{v} = \begin{pmatrix} -1 & 2 \\ 1 & -2 \end{pmatrix}\mathbf{v} = \mathbf{0}

      −v1+2v2=0-v_1 + 2v_2 = 0, so v2=(21)\mathbf{v}_2 = \begin{pmatrix} 2 \\ 1 \end{pmatrix}


      Diagonalisation

      Diagonalise AA from the previous example.

      S=(12−11)S = \begin{pmatrix} 1 & 2 \\ -1 & 1 \end{pmatrix}

      S−1=13(1−211)S^{-1} = \frac{1}{3}\begin{pmatrix} 1 & -2 \\ 1 & 1 \end{pmatrix}

      Λ=(2005)\Lambda = \begin{pmatrix} 2 & 0 \\ 0 & 5 \end{pmatrix}

      Verify: SΛS−1=(4213)=AS\Lambda S^{-1} = \begin{pmatrix} 4 & 2 \\ 1 & 3 \end{pmatrix} = A


      Determinant Properties

      Show that if AA and BB are n×nn \times n, then det⁡(AB)=det⁡(A)det⁡(B)\det(AB) = \det(A)\det(B).

      For 2×22 \times 2 matrices:

      det⁡(AB)=(a11b11+a12b21)(a21b12+a22b22)−(a11b12+a12b22)(a21b11+a22b21)\det(AB) = (a_{11}b_{11} + a_{12}b_{21})(a_{21}b_{12} + a_{22}b_{22}) - (a_{11}b_{12} + a_{12}b_{22})(a_{21}b_{11} + a_{22}b_{21})

      Expanding and collecting terms gives (a11a22−a12a21)(b11b22−b12b21)=det⁡(A)det⁡(B)(a_{11}a_{22} - a_{12}a_{21})(b_{11}b_{22} - b_{12}b_{21}) = \det(A)\det(B).

      (General proof uses properties of multilinear forms.)


      Partial Differential Equations

      Laplace Equation

      Solve ∇2ϕ=0\nabla^2 \phi = 0 in 0<x<π0 < x < \pi, 0<y<π0 < y < \pi with:

      • ϕ(0,y)=0\phi(0, y) = 0
      • ϕ(π,y)=0\phi(\pi, y) = 0
      • ϕ(x,0)=0\phi(x, 0) = 0
      • ϕ(x,π)=sin⁡x\phi(x, \pi) = \sin x

      By separation: ϕ=X(x)Y(y)\phi = X(x)Y(y)

      From xx boundaries: X=sin⁡(nx)X = \sin(nx), n=1,2,3,...n = 1, 2, 3, ...

      From y=0y = 0 boundary: Y(0)=0Y(0) = 0, so Y=sinh⁡(ny)Y = \sinh(ny).

      Solution:

      ϕ=∑n=1∞Cnsin⁡(nx)sinh⁡(ny)\phi = \sum_{n=1}^{\infty} C_n \sin(nx)\sinh(ny)

      At y=πy = \pi: ϕ(x,π)=sin⁡x\phi(x, \pi) = \sin x

      This matches only n=1n = 1: C1sinh⁡(π)=1C_1\sinh(\pi) = 1, so C1=1/sinh⁡(π)C_1 = 1/\sinh(\pi).

      ϕ=sin⁡(x)sinh⁡(y)sinh⁡(π)\phi = \frac{\sin(x)\sinh(y)}{\sinh(\pi)}


      Heat Equation

      Solve ut=uxxu_t = u_{xx} on 0<x<10 < x < 1, t>0t > 0 with u(0,t)=u(1,t)=0u(0, t) = u(1, t) = 0 and u(x,0)=sin⁡(2πx)u(x, 0) = \sin(2\pi x).

      By separation: u=X(x)T(t)u = X(x)T(t)

      From boundaries: Xn=sin⁡(nπx)X_n = \sin(n\pi x), n=1,2,3,...n = 1, 2, 3, ...

      Tn(t)=e−n2π2tT_n(t) = e^{-n^2\pi^2 t}

      All Cn=0C_n = 0 except C2C_2 (matching initial condition).

      u(x,t)=sin⁡(2πx)e−4π2tu(x, t) = \sin(2\pi x)e^{-4\pi^2 t}


      Wave Equation

      Solve ytt=yxxy_{tt} = y_{xx} on 0<x<10 < x < 1 with:

      • y(0,t)=y(1,t)=0y(0, t) = y(1, t) = 0 (fixed ends)
      • y(x,0)=sin⁡(πx)y(x, 0) = \sin(\pi x)
      • y˙(x,0)=0\dot{y}(x, 0) = 0

      Solution:

      y(x,t)=sin⁡(πx)cos⁡(πt)y(x, t) = \sin(\pi x)\cos(\pi t)

      (Standing wave with n=1n = 1.)

  • Other Resources

    • Complex Logarithms and Loci

      Complex Logarithm

      For z=x+iy=∣z∣eiθz = x + iy = |z|e^{i\theta}, the logarithm is:

      ln⁡z=ln⁡∣z∣+iθ\ln z = \ln|z| + i\theta

      Since θ\theta is defined up to multiples of 2π2\pi:

      ln⁡z=ln⁡∣z∣+i(θ+2nπ)\ln z = \ln|z| + i(\theta + 2n\pi)

      for integer nn.

      Principal value: Choose −π<θ≤π-\pi < \theta \leq \pi, giving:

      Ln z=ln⁡∣z∣+iArg(z)\text{Ln } z = \ln|z| + i\text{Arg}(z)


      Example: Evaluating Powers

      Evaluate 2i2^i and iii^i in form x+iyx + iy.

      For zw=ewln⁡zz^w = e^{w\ln z}, using principal values:

      2i2^i:

      • ln⁡2=ln⁡2+i⋅0\ln 2 = \ln 2 + i \cdot 0 (real, so θ=0\theta = 0)
      • 2i=eiln⁡2=cos⁡(ln⁡2)+isin⁡(ln⁡2)2^i = e^{i\ln 2} = \cos(\ln 2) + i\sin(\ln 2)
      • =0.769+0.639i= 0.769 + 0.639i (approximately)

      iii^i:

      • i=eiπ/2i = e^{i\pi/2}, so ln⁡i=iπ/2\ln i = i\pi/2
      • ii=ei⋅iπ/2=e−π/2i^i = e^{i \cdot i\pi/2} = e^{-\pi/2}
      • =0.208= 0.208 (real!)

      Multi-Valued Nature

      Unlike real logarithms, ln⁡z\ln z is multi-valued:

      • Different branches correspond to different choices of nn
      • The principal branch has −π<θ≤π-\pi < \theta \leq \pi
      • Moving around the origin by 2π2\pi returns to the same point but adds 2πi2\pi i to the logarithm

      Branch cut: To make ln⁡z\ln z single-valued, we cut the complex plane (usually along negative real axis).


      Loci in the Argand Diagram

      A locus is a set of points satisfying a geometric condition.

      Circle

      ∣z−a∣=r|z - a| = r describes a circle:

      • Centre: aa (complex number)
      • Radius: rr

      Circle in complex plane centred at point a with radius r, showing distance |z-a| always equals r

      Example: ∣z−1−2i∣=3|z - 1 - 2i| = 3 is circle centred at (1,2)(1, 2) with radius 3.


      Perpendicular Bisector

      ∣z−z1∣=∣z−z2∣|z - z_1| = |z - z_2| describes the perpendicular bisector of the line segment joining z1z_1 and z2z_2.

      Two points z₁ and z₂ in complex plane, with the perpendicular bisector shown as the locus where |z-z₁| = |z-z₂|

      Example: ∣z∣=∣z−i∣|z| = |z - i| describes the line y=12y = \frac{1}{2}.

      Derivation: Let z=x+iyz = x + iy: ∣z∣2=∣z−i∣2|z|^2 = |z - i|^2 x2+y2=x2+(y−1)2x^2 + y^2 = x^2 + (y-1)^2 y2=y2−2y+1y^2 = y^2 - 2y + 1 y=12y = \frac{1}{2}


      Half-Line (Ray)

      arg⁡(z−a)=θ\arg(z - a) = \theta describes a half-line from point aa making angle θ\theta with positive real axis.


      Region Inside a Circle

      ∣z−a∣<r|z - a| < r describes the interior of the circle.


      Example: Combined Loci

      Question: What shape is described by ∣z∣=3∣z−i∣|z| = 3|z - i|?

      Let z=x+iyz = x + iy:

      x2+y2=9(x2+(y−1)2)x^2 + y^2 = 9(x^2 + (y-1)^2)

      x2+y2=9x2+9y2−18y+9x^2 + y^2 = 9x^2 + 9y^2 - 18y + 9

      8x2+8y2−18y+9=08x^2 + 8y^2 - 18y + 9 = 0

      Dividing by 8 and completing the square:

      x2+y2−9y4+98=0x^2 + y^2 - \frac{9y}{4} + \frac{9}{8} = 0

      x2+(y−98)2=8164−98=−964x^2 + \left(y - \frac{9}{8}\right)^2 = \frac{81}{64} - \frac{9}{8} = -\frac{9}{64}

      This would be impossible (negative radius squared), so let me recalculate:

      x2+y2=9(x2+y2−2y+1)x^2 + y^2 = 9(x^2 + y^2 - 2y + 1)

      x2+y2=9x2+9y2−18y+9x^2 + y^2 = 9x^2 + 9y^2 - 18y + 9

      0=8x2+8y2−18y+90 = 8x^2 + 8y^2 - 18y + 9

      x2+y2−9y4+98=0x^2 + y^2 - \frac{9y}{4} + \frac{9}{8} = 0

      x2+(y−98)2=8164−7264=964x^2 + \left(y - \frac{9}{8}\right)^2 = \frac{81}{64} - \frac{72}{64} = \frac{9}{64}

      Circle centred at (0,98)(0, \frac{9}{8}) with radius 38\frac{3}{8}.


      General Powers

      For complex zz and ww:

      zw=ewln⁡zz^w = e^{w\ln z}

      Since ln⁡z\ln z is multi-valued, so is zwz^w (unless ww is an integer).

    • Kronecker Delta and Levi-Civita Symbol

      Kronecker Delta

      δij={1i=j0i≠j\delta_{ij} = \begin{cases} 1 & i = j \\ 0 & i \neq j \end{cases}

      This is the identity matrix in index notation: Iij=δijI_{ij} = \delta_{ij}.


      Properties

      Contraction:

      δikAkj=Aij\delta_{ik} A_{kj} = A_{ij}

      δikδkj=δij\delta_{ik} \delta_{kj} = \delta_{ij}

      Trace:

      δii=n\delta_{ii} = n

      (sum over repeated index ii in nn dimensions)


      Levi-Civita Symbol (3D)

      ϵijk={+1(i,j,k) even permutation of (1,2,3)−1(i,j,k) odd permutation of (1,2,3)0any index repeated\epsilon_{ijk} = \begin{cases} +1 & (i,j,k) \text{ even permutation of } (1,2,3) \\ -1 & (i,j,k) \text{ odd permutation of } (1,2,3) \\ 0 & \text{any index repeated} \end{cases}

      Even permutations: (1,2,3)(1,2,3), (2,3,1)(2,3,1), (3,1,2)(3,1,2)

      Odd permutations: (1,3,2)(1,3,2), (3,2,1)(3,2,1), (2,1,3)(2,1,3))


      Determinant Formula

      For a 3×33 \times 3 matrix AA:

      det⁡(A)=∑i,j,k=13ϵijkA1iA2jA3k\det(A) = \sum_{i,j,k=1}^{3} \epsilon_{ijk} A_{1i} A_{2j} A_{3k}

      Expanding: this is a11a22a33+a12a23a31+a13a21a32−a11a23a32−a12a21a33−a13a22a31a_{11}a_{22}a_{33} + a_{12}a_{23}a_{31} + a_{13}a_{21}a_{32} - a_{11}a_{23}a_{32} - a_{12}a_{21}a_{33} - a_{13}a_{22}a_{31}.


      Cross Product

      Using Levi-Civita:

      (a×b)i=∑j,k=13ϵijkajbk(\mathbf{a} \times \mathbf{b})_i = \sum_{j,k=1}^{3} \epsilon_{ijk} a_j b_k

      Or with Einstein summation convention (implied sum over repeated indices):

      (a×b)i=ϵijkajbk(\mathbf{a} \times \mathbf{b})_i = \epsilon_{ijk} a_j b_k


      Relationship Between δ\delta and ϵ\epsilon

      Contracted identity:

      ϵijkϵimn=δjmδkn−δjnδkm\epsilon_{ijk} \epsilon_{imn} = \delta_{jm}\delta_{kn} - \delta_{jn}\delta_{km}

      This can be verified by checking all cases.

      Special cases:

      ϵijkϵijn=2δkn\epsilon_{ijk} \epsilon_{ijn} = 2\delta_{kn}

      ϵijkϵijk=6\epsilon_{ijk} \epsilon_{ijk} = 6


      Curl as Levi-Civita Expression

      (∇×F)i=ϵijk∂Fk∂xj(\nabla \times \mathbf{F})_i = \epsilon_{ijk} \frac{\partial F_k}{\partial x_j}


      Why These Symbols Matter

      • Compact notation for determinants and cross products
      • Essential in tensor calculus and general relativity
      • Express vector identities elegantly (e.g., a×(b×c)=b(a⋅c)−c(a⋅b)\mathbf{a} \times (\mathbf{b} \times \mathbf{c}) = \mathbf{b}(\mathbf{a} \cdot \mathbf{c}) - \mathbf{c}(\mathbf{a} \cdot \mathbf{b}))
      • Basis for understanding antisymmetric tensors

      The Levi-Civita symbol encodes the “orientation” of a coordinate system — it changes sign under any swap of two indices, just like the determinant changes sign under row swaps.