Fourier Series Expansions

If every reasonable $2\\pi$-periodic function is a sum of sines and cosines, how do you find the coefficients? The answer is one of the most elegant constructions in mathematics: the trigonometric system is orthogonal in the integral inner product, so each coefficient is recovered by a single integration. The Fourier-coefficient formula is just the inner product of $f$ with the basis function, divided by the basis function's norm. Once you internalize that one idea, every special-symmetry rule (odd functions have no cosines, even functions have no sines) becomes obvious.

The Fourier series

Let $f$ be a $2\\pi$-periodic function on \[-\\pi, \\pi\]. Its Fourier series is the formal sum

where the Fourier coefficients are computed by:

The factor of $1/\\pi$ comes from the inner-product norm, and the constant term gets a factor of $a\_0 / 2$ for tidiness. The $\\sim$ symbol (instead of $=$) is a reminder that convergence of the right side to $f$ is a separate question, taken up in section 13.3.

Why the formula works: orthogonality

Equip the space of square-integrable functions on \[-\\pi, \\pi\] with the inner product $\\langle f, g \\rangle = \\int\_{-\\pi}^{\\pi} f(x) g(x) \\, dx$. The system $\\{1, \\cos(x), \\sin(x), \\cos(2x), \\sin(2x), \\ldots\\}$ is orthogonal: any two distinct basis functions integrate to zero against each other. The product-to-sum identities give

and analogous formulas for sine-sine and sine-cosine products. To find $a\_m$, multiply the Fourier series by $\\cos(mx)$ and integrate term by term: all terms vanish except the $m$-th cosine term, which contributes $a\_m \\cdot \\pi$. Solving gives the formula $a\_m = (1/\\pi) \\int f \\cos(mx) \\, dx$.

The series summer lets you stack partial sums of a Fourier series. Try the square wave $f(x) = \\text{sign}(x)$ on \[-\\pi, \\pi\], the first few odd harmonics with coefficients $4/(\\pi n)$ already give a recognisable square shape. The amplitude of each new term decays as $1/n$, so adding terms rapidly diminishes the visible error away from the jumps.

Even/odd shortcuts

The integrand $f(x) \\cos(nx)$ is even if $f$ is even (even times even). Its integral over \[-\\pi, \\pi\] equals twice the integral over \[0, \\pi\], you save half the work. The integrand $f(x) \\sin(nx)$ is odd if $f$ is even (even times odd), so it integrates to zero: every $b\_n = 0$ for an even $f$. Symmetrically, odd $f$ has every $a\_n = 0$ and only sine terms appear. Always check the symmetry of $f$ before computing any integral.

• MP3, AAC, Opus audio: Modern audio codecs apply a windowed Fourier-like transform (modified DCT) to short frames of a recording, quantise the coefficients aggressively, and store only the perceptually significant ones. The "frequency-domain compression" the standards talk about is exactly Fourier coefficient quantisation.
• JPEG image compression: JPEG splits an image into 8x8 pixel blocks and computes a 2D discrete cosine transform of each block. Cosine basis functions are chosen because images tend to be smooth (the highest-frequency cosine coefficient is usually tiny), and the few large coefficients suffice to reconstruct the block.
• Heat equation solutions: When you separate variables on the heat equation $u\_t = u\_{xx}$ on \[0, L\] with zero boundary, the solution is a Fourier sine series in $x$ with time-dependent coefficients. Section 14.4 makes this explicit.
• Music synthesis: Additive synthesisers build a tone by summing sinusoids at integer multiples of the fundamental, each with its own amplitude envelope. This is literally constructing a sound from its Fourier coefficients in real time.

Pause and think: Suppose you compute $a\_n = 0$ for every $n$. Does that force $f \\equiv 0$? (Hint: think about what happens if $f$ is odd. Cosine coefficients vanish, but the function is non-zero.)

Try it

• Compute $a\_0$ for $f(x) = x^2$ on \[-\\pi, \\pi\]. Use the formula $a\_0 = (1/\\pi) \\int\_{-\\pi}^{\\pi} x^2 \\, dx$ and the symmetry of $x^2$.
• Before integrating: which Fourier coefficients of $f(x) = x^3$ on \[-\\pi, \\pi\] vanish, and why?
• Verify directly: $\\int\_{-\\pi}^{\\pi} \\sin(2x) \\cos(3x) \\, dx = 0$. (Use the product-to-sum identity \\sin a \\cos b = \\tfrac{1}{2}\[\\sin(a+b) + \\sin(a-b)\].)
• On the series-summer widget, build $\\sum\_{n=1,3,5,\\ldots}^{N} (4/(\\pi n)) \\sin(nx)$ for $N = 1, 3, 5, 9$. Each partial sum is the Fourier series of the square wave truncated at $N$ terms. Sketch how the graph evolves.
• Find the Fourier coefficients of the constant function $f(x) = 7$ on \[-\\pi, \\pi\].

A trap to watch for

The constant term $a\_0/2$ (not $a\_0$) is the average value of $f$ over \[-\\pi, \\pi\]. Students routinely write the Fourier series as $a\_0 + \\sum (a\_n \\cos nx + b\_n \\sin nx)$, dropping the factor of $1/2$. That extra factor exists because the cosine of $0x$ is the constant $1$, and the inner-product norm of $1$ is $2\\pi$ rather than $\\pi$, without the $1/2$, the formula for $a\_0$ would double-count the constant. The convention is annoying but universal; always include the $1/2$.

What you now know

You can compute Fourier coefficients of a $2\\pi$-periodic function, use even/odd symmetry to predict which integrals vanish, and explain why the formulas come from inner-product projection onto an orthogonal basis. The next section asks the convergence question: does the Fourier series actually reconstruct $f$, and in what sense?

Mark section complete →

References

• Garrity, T. (2002). All the Mathematics You Missed: But Need to Know for Graduate School. Cambridge University Press, ch. 13.
• Stein, E. M., Shakarchi, R. (2003). Fourier Analysis: An Introduction. Princeton University Press, ch. 2.
• Folland, G. B. (1992). Fourier Analysis and Its Applications. Wadsworth & Brooks/Cole, ch. 2.
• Bracewell, R. N. (1999). The Fourier Transform and Its Applications (3rd ed.). McGraw-Hill, ch. 2.
• Korner, T. W. (1989). Fourier Analysis. Cambridge University Press, ch. 2 (orthogonality and Fourier coefficients).