Exercises — Half-range sine and cosine series

Answer: half-range sine series. The building blocks of a sine series are $\sin \frac{n\pi x}{L}$. Look at the figure below (Figure s01 — three sine curves $\sin x,\sin 2x,\sin 3x$ drawn in black over $[0,\pi ]$, with red dots marking the two endpoints where every curve crosses zero): each such curve is pinned to zero at $x=0$ (because $\sin 0=0$) and at $x=L$ (because $\sin n\pi =0$).

Since every term already vanishes at both ends, the whole sum automatically satisfies $u=0$ at the walls — exactly the Dirichlet condition. A cosine series would generally be non-zero there, so it cannot obey the boundary rule. See Heat Equation — separation of variables.

Answer: half-range cosine series. The blocks $\cos \frac{n\pi x}{L}$ have zero slope at both ends: their derivative $-\frac{n\pi }{L}\sin \frac{n\pi x}{L}$ is zero at $x=0$ and $x=L$ (same $\sin n\pi =0$ reason). Zero slope = no heat flux = insulated. This is the Neumann Boundary Conditions pairing. Sine blocks have maximum slope at the ends, so they'd force heat through — wrong.

False. On $[0,L]$ both reproduce $f$. But outside they are different inventions:

• The sine series is the odd extension (flip through the origin), period $2L$.
• The cosine series is the even extension (mirror across the $y$-axis), period $2L$. Same data on $[0,L]$, two different fantasies on $[-L,0]$. See Even and Odd Functions.

WHAT: compute $b_n=\frac{2}{\pi }{\int }_0^{\pi }1\cdot \sin (nx)dx$. WHY this integral: the sine formula with $L=\pi$ gives $\frac{n\pi x}{L}=nx$. ${\int }_0^{\pi }\sin (nx)dx={\left[-\frac{\cos nx}{n}\right]}_0^{\pi }=\frac{1-\cos n\pi }{n}=\frac{1-(-1{)}^n}{n}.$ So $b_n=\frac{2}{\pi }\cdot \frac{1-(-1{)}^n}{n}$. This is $0$ when $n$ is even and $\frac{4}{\pi n}$ when $n$ is odd. $\boxed{1=\frac{4}{\pi }\sum _{k=0}^{\infty }\frac{\sin ((2k+1)x)}{2k+1},0<x<\pi .}$ Meaning: the odd extension of the constant $1$ is a square wave (+1 then −1), and its Fourier series is exactly this classic odd-harmonics-only sum.

The shortcut: $f=1$ is already even. Its even extension is just the constant $1$ everywhere. $\frac{a_0}{2}=\frac{1}{\pi }{\int }_0^{\pi }1dx=1$, and for $n\ge 1$: $a_n=\frac{2}{\pi }{\int }_0^{\pi }\cos (nx)dx=\frac{2}{\pi }\cdot \frac{\sin n\pi }{n}=0.$ $\boxed{1=1.}$ A constant needs no wiggles — the whole function is its own average. This is the cleanest sanity check that the $\frac{a_0}{2}$ term really is the mean value.

$b_n=\frac{2}{\pi }{\int }_0^{\pi }\cos x\sin (nx)dx.$ WHY the product-to-sum identity? The integrand is a product of two different trig functions, $\cos x$ times $\sin (nx)$. We have no rule to integrate a product directly, and integration by parts here just bounces between sines and cosines forever. But a sum of single sines is trivial to integrate term by term. So we trade the product for a sum using $\sin A\cos B=\frac{1}{2}[\sin (A+B)+\sin (A-B)],$ which with $A=nx,B=x$ gives $\cos x\sin (nx)=\frac{1}{2}[\sin ((n+1)x)+\sin ((n-1)x)].$ Now every piece is a lone sine we already know how to integrate. For $n\ne 1$: ${\int }_0^{\pi }\sin (mx)dx=\frac{1-(-1{)}^m}{m}$. With $m=n\pm 1$ (opposite parity to $n$): $b_n=\frac{1}{\pi }\left[\frac{1-(-1{)}^{n+1}}{n+1}+\frac{1-(-1{)}^{n-1}}{n-1}\right].$ $(-1{)}^{n\pm 1}=-(-1{)}^n$, so $1-(-1{)}^{n\pm 1}=1+(-1{)}^n$, which is $0$ for odd $n$ and $2$ for even $n$. For even $n$: $b_n=\frac{1}{\pi }\left[\frac{2}{n+1}+\frac{2}{n-1}\right]=\frac{4n}{\pi (n^2-1)}.$ For odd $n\ge 3$: $b_n=0$. For $n=1$: ${\int }_0^{\pi }\cos x\sin xdx=\frac{1}{2}{\int }_0^{\pi }\sin 2xdx=0$, so $b_1=0$. $\boxed{\cos x=\frac{8}{\pi }\sum _{k=1}^{\infty }\frac{k}{4k^2-1}\sin (2kx),0<x<\pi .}$

Look at the two extensions (Figure s02 — over $[-\pi ,\pi ]$ the black "V" curve $|x|$ is the even extension, the red straight sawtooth line $y=x$ is the odd extension, with a red dot marking the jump at $x=\pi$):

• Odd extension of $x$ = a sawtooth: it jumps from $+\pi$ down to $-\pi$ at each period boundary. A jump (discontinuity) is "expensive" — you need lots of high harmonics to build a vertical cliff, so coefficients die only as $\frac{1}{n}$.
• Even extension of $x$ = $|x|$, a continuous "V" tent. No jumps, only a corner. Continuous functions need fewer high harmonics; coefficients die faster, as $\frac{1}{n^2}$. Rule of thumb: the smoother the periodic extension, the faster the decay. The sawtooth's cliffs make the sine partial sums spikier (Gibbs overshoot at the jump). See Dirichlet Conditions.

At $x=\pi$: every term has $\sin (n\pi )=0$, so the series sums to $0$. But $f(\pi )=\pi$. No contradiction: $x=\pi$ is a jump point of the odd (sawtooth) extension — it leaps from $+\pi$ (approaching from below) to $-\pi$ (the next period). By Dirichlet's theorem the series converges to the midpoint of the jump: $\frac{\pi +(-\pi )}{2}=0$. ✓ This is why we always write the equality for the open interval $0<x<\pi$.

The sine series converges to the odd extension, whose value at the ends is forced to $0$ (odd $\Rightarrow f(0)=0$; jump midpoint at $x=L$). So if $f(0)\ne 0$ or $f(L)\ne 0$, the series does not reproduce those endpoint heights — it snaps them to $0$. Physically that's fine for a Wave Equation on a finite string with fixed ends, because a real string is pinned to $0$ there. The claim is only "wrong" if you expected the series to honour a nonzero endpoint value it can never represent.

Rederive the coefficients (self-contained). Here $L=\pi$, so we use $a_n=\frac{2}{\pi }{\int }_0^{\pi }x\cos (nx)dx$ and $\frac{a_0}{2}=\frac{1}{\pi }{\int }_0^{\pi }xdx$.

• Constant term: $\frac{a_0}{2}=\frac{1}{\pi }\cdot \frac{{\pi }^2}{2}=\frac{\pi }{2}$ (the average height of $x$ on $[0,\pi ]$ — sanity check ✓).
• $a_n$ via integration by parts (why IBP? polynomial × trig; IBP lowers the polynomial power): with $u=x,dv=\cos (nx)dx$, ${\int }_0^{\pi }x\cos (nx)dx={\left[\frac{x\sin nx}{n}\right]}_0^{\pi }-\frac{1}{n}{\int }_0^{\pi }\sin (nx)dx.$ The bracket is $0$ because $\sin (n\pi )=0$. The remaining integral is $\frac{1}{n}\cdot \frac{1-\cos n\pi }{n}=\frac{1-(-1{)}^n}{n^2}$ with a minus sign in front, giving $\frac{(-1{)}^n-1}{n^2}$. Hence $a_n=\frac{2}{\pi }\cdot \frac{(-1{)}^n-1}{n^2},$ which is $0$ for even $n$ and $-\frac{4}{\pi n^2}$ for odd $n$. So $x=\frac{\pi }{2}-\frac{4}{\pi }{\sum }_{k=0}^{\infty }\frac{\cos ((2k+1)x)}{(2k+1{)}^2},0<x<\pi .$ Now evaluate at $x=0$ (a point of continuity of the even $|x|$ extension, so equality is exact): $0=\frac{\pi }{2}-\frac{4}{\pi }{\sum }_{k=0}^{\infty }\frac{1}{(2k+1{)}^2}⟹{\sum }_{k=0}^{\infty }\frac{1}{(2k+1{)}^2}=\frac{\pi }{2}\cdot \frac{\pi }{4}=\boxed{\frac{{\pi }^2}{8}}.$ Cross-check: this is the famous odd-square sum; combined with ${\sum }_{n\ge 1}\frac{1}{n^2}=\frac{{\pi }^2}{6}$ it gives the even part $\frac{1}{4}\cdot \frac{{\pi }^2}{6}=\frac{{\pi }^2}{24}$, and indeed $\frac{{\pi }^2}{8}+\frac{{\pi }^2}{24}=\frac{{\pi }^2}{6}$. ✓

We need blocks that are zero at $x=0$ (so use $\sin$, since $\sin 0=0$) and have zero slope at $x=\pi$. Slope of $\sin (\lambda x)$ is $\lambda \cos (\lambda x)$; it vanishes at $x=\pi$ when $\cos (\lambda \pi )=0$, i.e. $\lambda \pi =\frac{(2n-1)\pi }{2}$, giving $\lambda =\frac{2n-1}{2}$. $\boxed{\sin \left(\frac{(2n-1)x}{2}\right),n=1,2,3,\dots }$ These are half-integer frequencies. What "quarter-range" means: an ordinary half-range series knows $f$ on $[0,L]$ = half of one period $2L$, and both ends are the same type of boundary. When the two ends are different types (one fixed, one insulated), the natural period of the resulting waves is $4L$ instead of $2L$ — the known piece $[0,L]$ is now only a quarter of a full period, which is why we call it a quarter-range expansion. Concretely, $\sin \frac{(2n-1)x}{2}$ completes one full cycle over a length of $4\pi$, i.e. period $4L$ with $L=\pi$. Neither pure sine nor pure cosine of integer harmonics works; the mixed BC forces this new family. (Same separation-of-variables machinery as Heat Equation — separation of variables.)

Step 1 — separate variables (WHY: turns a PDE into two ODEs). Try $u=X(x)T(t)$. Then $\frac{T^{\prime }}{T}=\frac{X^{\prime \prime }}{X}=-\lambda$ (constant). The fixed-end BCs kill $X$ at both ends, so we need $X$ that vanishes there → sine blocks $X_n=\sin (nx)$ with $\lambda =n^2$ (period-$2\pi$ half-range, $L=\pi$). Step 2 — time factor. $T^{\prime }=-n^{2}T\Rightarrow T_n=e^{-n^{2}t}$. Step 3 — superpose and match the initial shape. $u(x,t)={\sum }_{n\ge 1}{b}_n{e}^{-n^{2}t}\sin (nx)$, and at $t=0$ this must equal $x$. That's exactly the half-range sine series of $x$ (parent Example 1): $b_n=\frac{2(-1{)}^{n+1}}{n}.$ Answer: $\boxed{u(x,t)=2\sum _{n=1}^{\infty }\frac{(-1{)}^{n+1}}{n}{e}^{-n^{2}t}\sin (nx).}$ Verify BCs: every term has $\sin (0)=\sin (n\pi )=0$ → $u(0,t)=u(\pi ,t)=0$. ✓ Verify IC: at $t=0$ the exponential is $1$ and the sum is the sine series of $x$, equal to $x$ on $0<x<\pi$. ✓ Physical picture: as $t$ grows, the $e^{-n^{2}t}$ factor crushes high harmonics fastest, so the sharp sawtooth smooths and the whole rod cools toward $0$.

WHY Parseval, and what it says. Parseval's identity is a statement of energy conservation: the "total squared size" of $f$ measured directly (by integrating $f^2$) must equal the total squared size measured harmonic-by-harmonic (by adding up $b_n^2$). Because the sine blocks $\sin (nx)$ are mutually orthogonal on $[0,\pi ]$, all cross terms integrate to zero and only the squares survive. For a half-range sine series on $[0,\pi ]$ the identity reads $\frac{2}{\pi }{\int }_0^{\pi }f(x{)}^{2}dx={\sum }_{n=1}^{\infty }{b}_n^2.$ This is the perfect tool here: it converts an integral of $f^2$ (easy) into an infinite sum of the $b_n^2$ (which contains the $\sum 1/n^2$ we want). Left side: $\frac{2}{\pi }{\int }_0^{\pi }{x}^{2}dx=\frac{2}{\pi }\cdot \frac{{\pi }^3}{3}=\frac{2{\pi }^2}{3}.$ Right side: $\sum b_n^2=\sum {\left(\frac{2(-1{)}^{n+1}}{n}\right)}^2=\sum \frac{4}{n^2}=4\sum \frac{1}{n^2}.$ Set them equal and solve: $4{\sum }_{n=1}^{\infty }\frac{1}{n^2}=\frac{2{\pi }^2}{3}⟹{\sum }_{n=1}^{\infty }\frac{1}{n^2}=\frac{1}{4}\cdot \frac{2{\pi }^2}{3}=\boxed{\frac{{\pi }^2}{6}}.$ ✓ The Basel sum, recovered from a rod-temperature expansion!

Fixed ends → sine; insulated ends → cosine; mixed ends → half-integer frequencies. Series equals $f$ on the open interval; at jumps it gives the midpoint.