1^()circ Show that the Fourier coefficients of a 2l
$1^\circ$ Show that the Fourier coefficients of a $2l$-periodic function can be computed using $a_n=\frac1l\int_\lambda^{\lambda+2l}f(x)\cos\frac{n\pi x}l\,dx$, $b_n=\frac1l\int_\lambda^{\lambda+2l}f(x)\sin\frac{n\pi x}l\,dx$ for any $\lambda$. $2^\circ$ Expand $f(x)=\frac{\pi-x}2$ on $[0,2\pi]$, period $2\pi$, in a Fourier series.
Explanation of Method
Since $f(x)$ and the trigonometric kernel $\cos(n\pi x/l)$ (or $\sin$) both have period $2l$, their product is $2l$-periodic, and the integral of any $2l$-periodic function over one full period is independent of the starting point — a standard shift argument proves this. Applying it with $\lambda=0$ then lets us compute the Fourier series of $(\pi-x)/2$ directly over $[0,2\pi]$.
- Prove the shift-invariance of the integral of a periodic function over one period.
- Apply it to justify integrating over $[0,2\pi]$ instead of $[-\pi,\pi]$.
- Compute $a_n$ and $b_n$ for $f(x)=(\pi-x)/2$ directly.
Part 1° · Arbitrary integration interval
Let $g(x)=f(x)\cos\frac{n\pi x}l$, which has period $2l$ since both factors do (as $\cos\frac{n\pi(x+2l)}l=\cos\left(\frac{n\pi x}l+2n\pi\right)=\cos\frac{n\pi x}l$). Split the integral: $$ \int_\lambda^{\lambda+2l}g(x)\,dx = \int_\lambda^{-l}g+\int_{-l}^lg+\int_l^{\lambda+2l}g. $$ In the first integral substitute $u=x+2l$ (so $x=\lambda\to u=\lambda+2l$, $x=-l\to u=l$); since $g$ has period $2l$, $g(u-2l)=g(u)$, giving $$ \int_\lambda^{-l}g(x)\,dx = \int_l^{\lambda+2l}g(u)\,du. $$ Substituting back: $$ \int_\lambda^{\lambda+2l}g(x)\,dx = \int_{-l}^lg(x)\,dx, $$ independent of $\lambda$. The same argument applies with $\sin$ in place of $\cos$. $$ \boxed{\text{The Fourier coefficients may be computed over any interval of length } 2l.} $$
Part 2° · Fourier series of $(\pi-x)/2$
Here $l=\pi$; take $\lambda=0$, integrating over $[0,2\pi]$. $$ a_0 = \frac1\pi\int_0^{2\pi}\frac{\pi-x}2\,dx = \frac1{2\pi}\left[\pi x-\frac{x^2}2\right]_0^{2\pi} = \frac1{2\pi}(2\pi^2-2\pi^2)=0. $$ For $n\ge1$, both $\int_0^{2\pi}\cos(nx)\,dx=0$ and $\int_0^{2\pi}x\cos(nx)\,dx=0$ (each a full period of an integer-frequency trig function, or a boundary term that vanishes since $\sin(2n\pi)=0$ and $\cos(2n\pi)=1=\cos(0)$), so $a_n=0$ for all $n$. For $b_n$: $$ b_n = \frac1{2\pi}\int_0^{2\pi}(\pi-x)\sin(nx)\,dx = \frac1{2\pi}\left[0-\left(-\frac{2\pi}n\right)\right] = \frac1n $$ (using $\int_0^{2\pi}x\sin(nx)\,dx=-\frac{2\pi}n$, from integration by parts with $\cos(2n\pi)=1$, $\sin(2n\pi)=0$). $$ \boxed{\frac{\pi-x}2 = \sum_{n=1}^\infty\frac{\sin(nx)}n = \sin x+\frac{\sin2x}2+\frac{\sin3x}3+\cdots} $$ (Verified numerically at $x=1$ and $x=4$: partial sums to $200{,}000$ terms match $(\pi-x)/2$ to 4+ decimal places.)
Summary Table
| Property | Value |
|---|---|
| Shift-invariance | integral over any length-$2l$ interval is the same |
| Fourier series | $\frac{\pi-x}2=\sum\frac{\sin(nx)}n$ on $(0,2\pi)$ |
Additional Notes
- This series is the $2\pi$-shifted twin of Problem 324's $x=\sum\frac{2(-1)^{n+1}}n\sin(nx)$ on $(-\pi,\pi)$ — both represent the same underlying sawtooth wave, just centered on different intervals.
- The shift-invariance proven in Part $1^\circ$ is what justifies freely choosing $\lambda=0$ (rather than $\lambda=-\pi$) whenever a function is more naturally described on $[0,2\pi]$, as used throughout Problems 328–329, 338.
- Because the coefficients decay only like $1/n$, this series (like the Leibniz series in Problem 324) converges conditionally, not absolutely.