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Part IA Michaelmas, Lent, Easter Term

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:

12LLL[f(x)]2dx=a024+12n=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: 1211x2dx=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 2π21n2\frac{2}{\pi^2}\sum \frac{1}{n^2}

Therefore: n=11n2=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)={11<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 ff' is piecewise smooth, then:

dfdxn=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, ff' discontinuousO(1/n2)O(1/n^2)
ff and ff' continuous, ff'' 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=1bnsin(nπxL)f(x) = \sum_{n=1}^{\infty} b_n \sin\left(\frac{n\pi x}{L}\right)

bn=2L0Lf(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=1ancos(nπxL)f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} a_n \cos\left(\frac{n\pi x}{L}\right)

an=2L0Lf(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).