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Part IA Michaelmas, Lent, Easter Term

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.