Exercises — Full Fourier series — coefficients derivation

Formulas: $a_0=\frac{1}{L}{\int }_{-L}^{L}fdx$, $a_m=\frac{1}{L}{\int }_{-L}^{L}f\cos \frac{m\pi x}{L}dx$, $b_m=\frac{1}{L}{\int }_{-L}^{L}f\sin \frac{m\pi x}{L}dx$.

First integral: $m=2\ne n=5$, so by orthogonality it is $\boxed{0}$. Second integral: this is $m=n=3$ sine–sine, which equals $\boxed{L}$.

True. $x^2$ is even; $\sin \frac{m\pi x}{L}$ is odd; even $\times$ odd $=$ odd; the integral of an odd function over the symmetric interval $[-L,L]$ is $0$. Hence every $b_m=0$.

No symmetry here, so compute all three.

$a_0$: $a_0=\frac{1}{\pi }{\int }_{-\pi }^{\pi }fdx=\frac{1}{\pi }\left({\int }_{-\pi }^{0}1dx+{\int }_0^{\pi }0dx\right)=\frac{1}{\pi }(\pi )=1.$ Constant term $=a_0/2=\frac{1}{2}$ (the average of $f$).

$a_m$: $a_m=\frac{1}{\pi }{\int }_{-\pi }^0\cos (mx)dx=\frac{1}{\pi }{\left[\frac{\sin mx}{m}\right]}_{-\pi }^0=\frac{1}{\pi }\left(0-\frac{\sin (-m\pi )}{m}\right)=0$ (since $\sin m\pi =0$). So all $a_m=0$ for $m\ge 1$.

$b_m$: $b_m=\frac{1}{\pi }{\int }_{-\pi }^0\sin (mx)dx=\frac{1}{\pi }{\left[-\frac{\cos mx}{m}\right]}_{-\pi }^0=\frac{1}{\pi }\left(-\frac{1}{m}+\frac{\cos (-m\pi )}{m}\right)=\frac{(-1{)}^m-1}{m\pi }.$ This is $0$ for even $m$ and $-\frac{2}{m\pi }$ for odd $m$.

$f(x)=\frac{1}{2}-\frac{2}{\pi }{\sum }_{k=0}^{\infty }\frac{1}{2k+1}\sin ((2k+1)x).$

Value at $x=0$. Here $f$ jumps: the left limit is $f(0^{-})=1$, the right limit is $f(0^{+})=0$. By the Dirichlet convergence theorem the series converges to the midpoint $\frac{1}{2}(f(0^{-})+f(0^{+}))=\frac{1}{2}(1+0)=\frac{1}{2}$. Check: at $x=0$ every $\sin ((2k+1)\cdot 0)=0$, so the series literally sums to the constant $\frac{1}{2}$. This is why the constant term ($a_0/2=\frac{1}{2}$) already lands on the midpoint — reassuring, not coincidence.

$a_0$: $a_0=\frac{1}{\pi }{\int }_{-\pi }^{\pi }{x}^{2}dx=\frac{1}{\pi }\cdot \frac{2{\pi }^3}{3}=\frac{2{\pi }^2}{3}.$ Constant term $=a_0/2=\frac{{\pi }^2}{3}$.

$a_m$: since $x^2\cos mx$ is even, $a_m=\frac{2}{\pi }{\int }_0^{\pi }{x}^2\cos mxdx$. Why integration by parts, and why twice? We have a polynomial times a trig function. There is no product rule for integrals, but integration by parts lets us trade a derivative for an integral: each time we choose $u=$ the polynomial part, differentiating $x^2\to 2x\to 2$ lowers its degree by one. After two rounds the polynomial is a constant and the integral becomes a plain trig integral we can finish. (Choosing $u=$ the trig part instead would raise complexity — always differentiate the polynomial.) With $u=x^2,dv=\cos mxdx$: ${\int }_0^{\pi }{x}^2\cos mxdx={\left[\frac{x^2\sin mx}{m}\right]}_0^{\pi }-\frac{2}{m}{\int }_0^{\pi }x\sin mxdx.$ The bracket is $0$ ($\sin m\pi =0$). Then ${\int }_0^{\pi }x\sin mxdx=-\frac{\pi (-1{)}^m}{m}$ (the second round of parts, as in the parent's Example 2). So ${\int }_0^{\pi }{x}^2\cos mxdx=-\frac{2}{m}\left(-\frac{\pi (-1{)}^m}{m}\right)=\frac{2\pi (-1{)}^m}{m^2}.$ Thus $a_m=\frac{2}{\pi }\cdot \frac{2\pi (-1{)}^m}{m^2}=\frac{4(-1{)}^m}{m^2}.$ $x^2=\frac{{\pi }^2}{3}+4{\sum }_{m=1}^{\infty }\frac{(-1{)}^m}{m^2}\cos mx.$

Put $x=\pi$. By the convergence theorem the series equals $f(\pi )={\pi }^2$. Why is this point legal? The $2\pi$-periodic extension of $x^2$ is continuous at $x=\pi$ (both sides approach ${\pi }^2$), so there is no jump and the series hits the true value — not a midpoint. Since $\cos m\pi =(-1{)}^m$: ${\pi }^2=\frac{{\pi }^2}{3}+4{\sum }_{m=1}^{\infty }\frac{(-1{)}^m}{m^2}(-1{)}^m=\frac{{\pi }^2}{3}+4{\sum }_{m=1}^{\infty }\frac{1}{m^2}.$ Rearrange: $4\sum \frac{1}{m^2}={\pi }^2-\frac{{\pi }^2}{3}=\frac{2{\pi }^2}{3}$, so ${\sum }_{m=1}^{\infty }\frac{1}{m^2}=\frac{{\pi }^2}{6}.$ $✓$

Left side: $f^2=1$ everywhere, so $\frac{1}{\pi }{\int }_{-\pi }^{\pi }1dx=\frac{1}{\pi }(2\pi )=2$. Right side: $a_0=0$, all $a_n=0$, and $\sum b_m^2={\sum }_{\text{odd }m}{\left(\frac{4}{m\pi }\right)}^2=\frac{16}{{\pi }^2}{\sum }_{k=0}^{\infty }\frac{1}{(2k+1{)}^2}.$ Set equal: $2=\frac{16}{{\pi }^2}{\sum }_{k=0}^{\infty }\frac{1}{(2k+1{)}^2}$, giving ${\sum }_{k=0}^{\infty }\frac{1}{(2k+1{)}^2}=\frac{{\pi }^2}{8}.$

No symmetry — compute all. First, why the two helper integrals below have that shape. When you integrate $e^x\cos mx$ by parts twice, the integral reappears on the right, and you solve for it algebraically (like solving $I=(\text{stuff})-kI$). The clean results are $\int e^x\cos mxdx=\frac{e^x(\cos mx+m\sin mx)}{1+m^2},\int e^x\sin mxdx=\frac{e^x(\sin mx-m\cos mx)}{1+m^2}.$ You can verify either by differentiating the right side (product rule) and watching it collapse back to the integrand — the $1+m^2$ in the denominator is exactly what makes the cross-terms cancel. Write $\sinh \pi =\frac{e^{\pi }-e^{-\pi }}{2}$.

$a_0$: $a_0=\frac{1}{\pi }{\int }_{-\pi }^{\pi }{e}^{x}dx=\frac{1}{\pi }(e^{\pi }-e^{-\pi })=\frac{2\sinh \pi }{\pi }.$

$a_m$: $a_m=\frac{1}{\pi }{\left[\frac{e^x(\cos mx+m\sin mx)}{1+m^2}\right]}_{-\pi }^{\pi }$. At $x=\pm \pi$: $\sin m\pi =0,\cos m\pi =(-1{)}^m$, so the bracket is $\frac{(-1{)}^m{e}^x}{1+m^2}$ evaluated at the ends: $a_m=\frac{1}{\pi }\cdot \frac{(-1{)}^m(e^{\pi }-e^{-\pi })}{1+m^2}=\frac{2(-1{)}^m\sinh \pi }{\pi (1+m^2)}.$

$b_m$: $b_m=\frac{1}{\pi }{\left[\frac{e^x(\sin mx-m\cos mx)}{1+m^2}\right]}_{-\pi }^{\pi }$. At $x=\pm \pi$ the bracket is $\frac{-m(-1{)}^m{e}^x}{1+m^2}$, so $b_m=\frac{1}{\pi }\cdot \frac{-m(-1{)}^m(e^{\pi }-e^{-\pi })}{1+m^2}=\frac{-2m(-1{)}^m\sinh \pi }{\pi (1+m^2)}.$

$e^x=\frac{2\sinh \pi }{\pi }\left[\frac{1}{2}+{\sum }_{m=1}^{\infty }\frac{(-1{)}^m}{1+m^2}(\cos mx-m\sin mx)\right].$

Quick refresher on the complex framework. Using Euler's formula $e^{\pm inx}=\cos nx\pm i\sin nx$, the real series can be repackaged as $f(x)={\sum }_{n=-\infty }^{\infty }{c}_n{e}^{inx}$ where each complex coefficient is $c_n=\frac{1}{2\pi }{\int }_{-\pi }^{\pi }f(x)e^{-inx}dx$ (the $e^{-inx}$ is the "matching fork" being projected out, exactly as $\cos ,\sin$ were in the real case). The bridge back to real coefficients: grouping the $+n$ and $-n$ terms and using $\cos nx=\frac{e^{inx}+e^{-inx}}{2}$ gives $a_n=c_n+c_{-n}$ and $b_n=i(c_n-c_{-n})$. So checking $c_n+c_{-n}$ against $a_n$ tests both frameworks at once.

$c_n=\frac{1}{2\pi }{\int }_{-\pi }^{\pi }{e}^{(1-in)x}dx=\frac{1}{2\pi }\cdot \frac{e^{(1-in)\pi }-e^{-(1-in)\pi }}{1-in}$. Since $e^{\mp in\pi }=(-1{)}^n$ (integer $n$: $e^{-in\pi }=\cos n\pi -i\sin n\pi =(-1{)}^n$, purely real), the numerator is $(-1{)}^n(e^{\pi }-e^{-\pi })=2(-1{)}^n\sinh \pi$. So $c_n=\frac{(-1{)}^n\sinh \pi }{\pi (1-in)}$.

Then $a_n=c_n+c_{-n}=\frac{(-1{)}^n\sinh \pi }{\pi }\left(\frac{1}{1-in}+\frac{1}{1+in}\right)=\frac{(-1{)}^n\sinh \pi }{\pi }\cdot \frac{2}{1+n^2}=\frac{2(-1{)}^n\sinh \pi }{\pi (1+n^2)}.$ For $n=0$: $a_0=\frac{2\sinh \pi }{\pi }$; for $n=1$: $a_1=\frac{-2\sinh \pi }{2\pi }=\frac{-\sinh \pi }{\pi }$. Both match Ex 4.1. $✓$

Expand $x(\pi -x)=\pi x-x^2$. We need ${\int }_0^{\pi }x\sin mxdx$ and ${\int }_0^{\pi }{x}^2\sin mxdx$. Why parts again? Same reason as Ex 2.2 — polynomial $\times$ trig — so differentiate the polynomial to grind its degree down.

From the parent, ${\int }_0^{\pi }x\sin mxdx=-\frac{\pi (-1{)}^m}{m}$. For the second, integrate by parts twice: ${\int }_0^{\pi }{x}^2\sin mxdx={\left[-\frac{x^2\cos mx}{m}\right]}_0^{\pi }+\frac{2}{m}{\int }_0^{\pi }x\cos mxdx.$ The bracket $=-\frac{{\pi }^2(-1{)}^m}{m}$. And ${\int }_0^{\pi }x\cos mxdx={\left[\frac{x\sin mx}{m}\right]}_0^{\pi }-\frac{1}{m}{\int }_0^{\pi }\sin mxdx=0-\frac{1}{m}\cdot \frac{1-(-1{)}^m}{m}=\frac{(-1{)}^m-1}{m^2}.$ So ${\int }_0^{\pi }{x}^2\sin mxdx=-\frac{{\pi }^2(-1{)}^m}{m}+\frac{2}{m}\cdot \frac{(-1{)}^m-1}{m^2}=-\frac{{\pi }^2(-1{)}^m}{m}+\frac{2((-1{)}^m-1)}{m^3}.$

Now combine ($\pi \cdot$ first $-$ second): ${\int }_0^{\pi }(\pi x-x^2)\sin mxdx=\pi \left(-\frac{\pi (-1{)}^m}{m}\right)-\left(-\frac{{\pi }^2(-1{)}^m}{m}+\frac{2((-1{)}^m-1)}{m^3}\right)=-\frac{2((-1{)}^m-1)}{m^3}.$ The ${\pi }^2(-1{)}^m/m$ terms cancel exactly. Therefore $B_m=\frac{2}{\pi }\cdot \left(-\frac{2((-1{)}^m-1)}{m^3}\right)=\frac{4(1-(-1{)}^m)}{\pi m^3}=\{\begin{array}{ll}\frac{8}{\pi m^3}& m\text{ odd}\\ 0& m\text{ even}.\end{array}$ Only the odd harmonics appear — consistent with the profile being symmetric about the rod's midpoint $x=\pi /2$.

Left side: ${\int }_0^{\pi }{x}^2(\pi -x{)}^{2}dx$. Let $I={\int }_0^{\pi }({\pi }^2{x}^2-2\pi x^3+x^4)dx={\pi }^2\frac{{\pi }^3}{3}-2\pi \frac{{\pi }^4}{4}+\frac{{\pi }^5}{5}={\pi }^5\left(\frac{1}{3}-\frac{1}{2}+\frac{1}{5}\right)=\frac{{\pi }^5}{30}.$ So LHS $=\frac{2}{\pi }\cdot \frac{{\pi }^5}{30}=\frac{{\pi }^4}{15}.$ Right side: $\sum B_m^2={\sum }_{\text{odd }m}{\left(\frac{8}{\pi m^3}\right)}^2=\frac{64}{{\pi }^2}{\sum }_{k=0}^{\infty }\frac{1}{(2k+1{)}^6}.$ Equate: $\frac{{\pi }^4}{15}=\frac{64}{{\pi }^2}\sum \frac{1}{(2k+1{)}^6}$, so ${\sum }_{k=0}^{\infty }\frac{1}{(2k+1{)}^6}=\frac{{\pi }^6}{15\cdot 64}=\frac{{\pi }^6}{960}.$

Read each line as "Prompt ::: Answer" — cover the text after the :::, say your answer aloud, then reveal to check. Interval $[-L,L]$ uses which prefactor for every coefficient? ::: $\frac{1}{L}$ Odd $f$ has only which coefficients nonzero? ::: only $b_m$ (all $a_m=0$) Even $f$ has only which coefficients nonzero? ::: only $a_m$ (all $b_m=0$) At a jump discontinuity the series converges to what value? ::: the midpoint of the two one-sided limits Does a pure sine (half-range) series have a constant $B_0$ term? ::: No — $\sin 0=0$, so there is no constant Parseval links $\int f^2$ to what? ::: the sum of the squared coefficients