Exercises — Dirichlet conditions for convergence

Walk the checklist, one tick at a time.

• Integrable? It is bounded (never shoots to infinity) over a finite period, so the area under $|f|$ is finite. ✓
• Finite jumps? Exactly $4$ — a finite number of finite jumps. ✓
• Finite wiggles? $6$ maxima/minima is finite. ✓

All three hold, so yes — Dirichlet is satisfied and the series converges (to the midpoint at each jump, to $f$ elsewhere). Answer: YES.

As $x\to 0^{+}$, $\frac{1}{x}\to +\infty$. This is an infinite discontinuity, not a finite jump. Check the integral: ${\int }_0^1\left|\frac{1}{x}\right|dx=[\ln x{]}_0^1=\infty$. So Condition 1 (absolutely integrable) fails, and Condition 2 (only finite discontinuities) also fails because the discontinuity is infinite. Answer: it is not absolutely integrable / has an infinite discontinuity.

$x=0$ is a jump: slide in from the left ($x<0$) and $f=0$, so $f(0^{-})=0$. Slide in from the right ($x>0$) and $f=x\to 0$, so $f(0^{+})=0$. Both sides agree, so this is actually a point of continuity in value. $S(0)=\frac{f(0^{+})+f(0^{-})}{2}=\frac{0+0}{2}=0.$ Answer: $S(0)=0$. (A lesson: a "piecewise" definition does not automatically mean a jump — check the limits.)

Approaching $\pi$ from the left, $f=x\to \pi$, so $f({\pi }^{-})=\pi$. Approaching $\pi$ from the right, we wrap into the next period, which starts like $-\pi$: there $f=0$, so $f({\pi }^{+})=0$. $S(\pi )=\frac{\pi +0}{2}=\frac{\pi }{2}.$ Answer: $S(\pi )=\frac{\pi }{2}$.

$f(0^{-})=-3$, $f(0^{+})=+5$. $S(0)=\frac{5+(-3)}{2}=\frac{2}{2}=1.$ Answer: $S(0)=1$. Notice the series lands at $1$, a value $f$ never takes — the sum "meets in the middle."

Near $x=0$ the argument $\frac{1}{x}$ races through every value, so $\sin (1/x)$ crosses from $+1$ to $-1$ infinitely often as $x\to 0$. That is an infinite number of maxima and minima packed into any tiny window around $0$.

• Integrable? Yes ($|f|\le 1$, bounded). ✓
• Finite jumps? Yes (no jumps). ✓
• Finite wiggles? NO — infinitely many. ✗ Condition 3 fails, so Dirichlet's theorem does not apply. We cannot guarantee convergence from this theorem. Answer: FAILS (condition 3, infinite wiggles).

A degenerate but instructive case.

• Integrable? ${\int }_{-\pi }^{\pi }|c|dx=2\pi |c|<\infty$. ✓
• Jumps? None. ✓
• Maxima/minima? A constant has zero strict maxima/minima — zero is finite. ✓ All conditions hold. Everywhere $f$ is continuous, so $S(x)=f(x)=c$. Answer: satisfies Dirichlet; $S(x)=c$ for all $x$. (The Fourier series is just $a_0/2=c$, all other coefficients zero.)

From the left, $f({\pi }^{-})={\pi }^2$. From the right we wrap: the next period starts at $-\pi$, where $f=(-\pi {)}^2={\pi }^2$, so $f({\pi }^{+})={\pi }^2$. Both sides equal ${\pi }^2$ → no jump; the periodic extension of $x^2$ is continuous at $\pi$. $S(\pi )=\frac{{\pi }^2+{\pi }^2}{2}={\pi }^2.$ Answer: no jump; $S(\pi )={\pi }^2$. Contrast this with $f(x)=x$ (Exercise from the parent), which does jump at $\pi$ — the evenness of $x^2$ makes the wrap seamless.

Step 1 — what does the series equal at $x=\pi /2$? $f$ is continuous there and $f(\pi /2)=+1$, so by the midpoint rule $S(\pi /2)=1$. Step 2 — plug $x=\pi /2$ into the series. $\sin ((2k+1)\frac{\pi }{2})$ cycles $+1,-1,+1,\dots$ for $k=0,1,2,\dots$ $S(\frac{\pi }{2})=\frac{4}{\pi }\left(1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots \right).$ Step 3 — set the two expressions equal. $1=\frac{4}{\pi }\left(1-\frac{1}{3}+\frac{1}{5}-\cdots \right)\Rightarrow 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots =\frac{\pi }{4}.$ Answer: the alternating sum equals $\frac{\pi }{4}$ (the Leibniz series). Convergence at this point is licensed precisely because Dirichlet holds and $x=\pi /2$ is a continuity point.

Midpoint route: $f(0^{-})=-1,f(0^{+})=+1$, so $S(0)=\frac{-1+1}{2}=0$. Substitution route: every term contains $\sin ((2k+1)\cdot 0)=\sin 0=0$, so the entire series sums to $0$. They agree at $0$. This reveals that the series automatically produces the midpoint $0$ at the jump — the odd (sine-only) structure forces $S(0)=0$ with no extra work. Answer: both give $0$; consistent. (The Gibbs phenomenon overshoot decorates the approach but the limiting value is exactly $0$.)

The odd extension is $\tilde{g}(x)=x$ on $(-\pi ,\pi )$ (since $x$ is already odd), extended periodically — the classic sawtooth. See Half-range expansions. (a) Checklist:

• Integrable: ${\int }_{-\pi }^{\pi }|x|dx={\pi }^2<\infty$. ✓
• Jumps: only at the wrap points $x=\pm \pi$ (finite jumps). ✓
• Wiggles: $x$ is monotonic on the period → finitely many (zero interior) extrema. ✓ Dirichlet holds. ✓ (b) At $x=0$: continuous, $\tilde{g}(0)=0$, so $S(0)=0$. At $x=\pi$: $\tilde{g}({\pi }^{-})=\pi$; wrapping, $\tilde{g}({\pi }^{+})=\tilde{g}(-{\pi }^{+})=-\pi$. Midpoint: $S(\pi )=\frac{\pi +(-\pi )}{2}=0.$ Answer: $S(0)=0$, $S(\pi )=0$.

$f_a(0^{-})=0,f_a(0^{+})=a$, so $S(0)=\frac{a+0}{2}=\frac{a}{2}.$ As $a\to 0$, the jump height $|a-0|=|a|\to 0$ and $S(0)=\frac{a}{2}\to 0$. In the limit $a=0$ the function is the constant $0$ — no jump at all, and $S(0)=0$ matches continuously. Convergence is never endangered: for every $a$ the function is bounded with a single finite jump and no extra wiggles, so Dirichlet holds throughout. The jump simply shrinks smoothly to nothing. Answer: $S(0)=a/2$; as $a\to 0$ both the jump and $S(0)$ vanish; convergence always guaranteed.

Counterexample: any square wave — it is absolutely integrable, yet at the jump the series converges to the midpoint, not to $f$ (which is $\pm 1$, never the midpoint value between them). Correct statement: absolute integrability alone only guarantees the coefficients exist. Guaranteed convergence needs all three Dirichlet conditions, and even then it is to $\frac{f(x^{+})+f(x^{-})}{2}$ — equal to $f$ only at points of continuity. Answer: square wave; integrability ⇒ coefficients exist, not convergence-to-$f$. (Recall these conditions are sufficient, not necessary — see Convergence of series (pointwise vs uniform).)

• At a jump from $-3$ to $+5$, what value does the series give?
• Does a bounded function with $50$ finite jumps satisfy Dirichlet?
• Why does $\sin (1/x)$ near $0$ fail Dirichlet?
• For $x^2$ on $(-\pi ,\pi )$, is there a jump at $x=\pi$?
• What is $1-\frac{1}{3}+\frac{1}{5}-\cdots$ and which exercise proves it?