Expand f(x)=cos(3x) in a Fourier series on (- pi,pi)
Expand $f(x)=\cos(3x)$ in a Fourier series on $(-\pi,\pi)$.
Explanation of Method
Since $\{1,\cos x,\sin x,\cos2x,\sin2x,\ldots\}$ is an orthogonal basis on $(-\pi,\pi)$, and $\cos(3x)$ is itself one of these basis functions, its Fourier expansion is trivial by orthogonality: every coefficient except $a_3$ integrates to zero.
- Note that $\cos(3x)$ is even, so $b_n=0$ automatically.
- Compute $a_n$ via the product-to-sum identity, showing it vanishes except at $n=3$.
- Conclude the series equals $\cos(3x)$ exactly.
Step 1 · Symmetry
$\cos(3x)$ is even, so $b_n=0$ for all $n\ge1$, and $a_0=\frac2\pi\int_0^\pi\cos(3x)\,dx=\frac2{3\pi}[\sin(3\pi)-0]=0$.
Step 2 · Computing $a_n$ via orthogonality
$$ a_n = \frac1\pi\int_{-\pi}^\pi\cos(3x)\cos(nx)\,dx = \frac1\pi\int_{-\pi}^\pi\frac12\bigl[\cos((3-n)x)+\cos((3+n)x)\bigr]dx. $$ For $n\ne3$, both $(3-n)$ and $(3+n)$ are nonzero integers, and $\int_{-\pi}^\pi\cos(kx)\,dx=0$ for any nonzero integer $k$, so $a_n=0$. For $n=3$: $$ a_3 = \frac1\pi\int_{-\pi}^\pi\frac12[1+\cos(6x)]\,dx = \frac1\pi\cdot\pi = 1. $$
Step 3 · Conclusion
$$ \boxed{f(x) = \cos(3x)} $$ with $a_3=1$ and every other coefficient equal to zero — the Fourier series of a basis function is the function itself.
Summary Table
| Property | Value |
|---|---|
| $a_3$ | $1$ |
| All other $a_n,b_n$ | $0$ |
| Fourier series | $\cos(3x)$ (itself) |
Additional Notes
- This is the $n=3$ special case of the general fact proven in Problem 335: any finite trigonometric polynomial is its own Fourier series, with coefficients equal to zero beyond its highest frequency.
- Contrast sharply with Problem 332's $\sin(ax)$ for non-integer $a$: there, since the frequency isn't an integer multiple of the fundamental, an infinite (non-terminating) expansion is genuinely required.
- The vanishing of $a_n$ for $n\ne3$ follows purely from orthogonality of $\{\cos(kx)\}$ on $(-\pi,\pi)$ — no cancellation "coincidence" is involved, just the defining property of the basis.