Question bank — Dirichlet conditions for convergence

$a_n,b_n$ all exist, the series must equal $f(x)$ at every point. False ::: Existence of the coefficients only needs the integrals $\int f\cos$, $\int f\sin$ to be finite (absolute integrability). That is a much weaker demand than convergence — the "$\sim$" sign is a warning that the sum still has to be tested against the full I-J-W checklist, and even then it lands on the midpoint at jumps.

False ::: Dirichlet allows finitely many finite jumps. At each such jump the series simply converges to the midpoint $\frac{f(x^{+})+f(x^{-})}{2}$ instead of to $f$ — convergence is never lost, only re-aimed.

False ::: They are sufficient, not necessary. Some functions that violate a condition still happen to converge; Dirichlet just draws a clean fence around functions where convergence is guaranteed without further work.

False ::: Sines and cosines are symmetric ingredients with no left/right preference, so the sum splits the difference: it converges to the average of both one-sided limits, not to either side alone.

$f$ is continuous everywhere on one period, the series converges to $f(x)$ at every point. True ::: At a point of continuity $f(x^{+})=f(x^{-})=f(x)$, so the midpoint formula collapses to $f(x)$ itself — provided the wiggle and integrability conditions also hold (continuity alone is not the whole checklist).

False ::: The overshoot is a feature of the partial sums $S_N$ near the jump; its height never shrinks, but it slides ever closer to the jump as $N\to \infty$. At the jump point itself the limiting value is still exactly the midpoint. See Gibbs phenomenon.

$[-L,L]$ satisfies the Dirichlet conditions. False ::: Boundedness handles blow-ups but not wiggles. $f(x)=\sin (1/x)$ is bounded by $1$ yet has infinitely many maxima and minima near $x=0$, so it fails condition 3.

False ::: On a finite period, boundedness ($|f|\le M$) forces ${\int }_{-L}^L|f|dx\le 2LM<\infty$. Absolute integrability can only fail through an unbounded spike (e.g. $1/x$ near $0$).

$f(x)=\tan x$ on $(-\pi /2,\pi /2)$ is fine — it's smooth and has no wiggles, so its Fourier series converges to $f$ everywhere." The error is ignoring the infinite discontinuity. $\tan x\to \pm \infty$ at $x=\pm \pi /2$, so $\int |\tan x|dx$ diverges — condition 1 (absolute integrability) and condition 2 (no infinite discontinuities) both fail.

$f(x)=x$ on $(-\pi ,\pi )$ the series at $x=\pi$ should give $\pi$, because that's where the graph ends." The error is forgetting periodicity creates a jump at the wrap point: $f({\pi }^{-})=\pi$ but $f({\pi }^{+})$ wraps to $-\pi$. So the series gives $\frac{\pi +(-\pi )}{2}=0$, not $\pi$.

$\sin (1/x)$ fails Dirichlet because it blows up near $x=0$." The error is naming the wrong condition. $\sin (1/x)$ stays inside $[-1,1]$ — it never blows up. It fails because of infinitely many maxima and minima (condition 3), not an infinite discontinuity.

The error is reading "sufficient" as "necessary." Violating the conditions only means the guarantee is void — some such functions still converge; you just have to check by other means. See Convergence of series (pointwise vs uniform).

The error is calling it an approximation. The midpoint is the exact limit of the partial sums at a jump — derived from the symmetric Dirichlet kernel sampling both sides equally, not a fudge or rounding.

$[0,1]$ has only finite jumps, so it satisfies Dirichlet." The error is missing that "finite jumps" must also be finitely many. Infinitely many jumps (one at every rational) violates condition 2 even though each individual jump is finite.

average at a jump, rather than to one side? Because the partial sum is a convolution with the Dirichlet kernel $D_N(t)$, which is symmetric about $t=0$. As $N\to \infty$ it becomes a spike with equal mass on the left ($t>0$) and right ($t<0$), so it samples $f(x^{-})$ and $f(x^{+})$ with equal weight $\frac{1}{2}$ each.

absolutely integrable, not merely integrable? Absolute integrability, $\int |f|dx<\infty$, guarantees the coefficient integrals $\int f\cos$ and $\int f\sin$ are finite regardless of sign cancellation — it prevents the very definition of $a_n,b_n$ from breaking down before any convergence question is even asked.

Each smooth wave in the series has a fixed maximum frequency at any partial-sum stage, so no finite collection of waves can track infinitely rapid oscillation near a point — the series simply cannot keep up, and convergence is not guaranteed.

$\sim$" symbol used instead of "$=$" when first writing a Fourier series? Because writing the coefficients only asserts they exist; it says nothing about whether the infinite sum converges or to what value. The "$\sim$" reserves judgement until the I-J-W checklist is verified, at which point $=$ holds (with midpoints at jumps).

$\mathbb{R}$? A periodic function repeats, so anything true over one period automatically holds everywhere. Checking one period is both necessary and enough — and it keeps the integrals finite (over all of $\mathbb{R}$, $\int |f|$ would diverge for any nonzero periodic $f$).

derivative at some points and still converge? Convergence needs piecewise smoothness, not differentiability everywhere. Corners (finite jumps in the slope) are perfectly allowed — the value stays continuous, so the series converges to $f$ there; only jumps in $f$ itself trigger the midpoint rule.

continuous but has a sharp corner (like $|x|$ at $x=0$)? To $f(x)$ itself. A corner is a jump in the slope, not in the value, so $f(0^{+})=f(0^{-})=0$ and the midpoint rule returns $f(0)=0$ — no surprise there.

$f$ is redefined at a single jump point to equal, say, the left limit, does the series's value there change? No. The series value depends only on the one-sided limits $f(x^{\pm })$, which are unaffected by the value assigned exactly at the point. The sum still gives the midpoint regardless of how you define $f$ at that single point.

$x=\pm L$ for a general period-$2L$ function? By periodicity $f(-L)$ and $f(L)$ are the same point on the circle. If the left limit approaching $L$ and the right limit (wrapping to $-L$) differ, that is a jump, and the series converges to their midpoint — exactly as with any interior jump.

$f=-1$ on $(-\pi ,0)$, $f=+1$ on $(0,\pi )$, what value does the series give at $x=0$, where $f$ is not even defined? $\frac{f(0^{+})+f(0^{-})}{2}=\frac{1+(-1)}{2}=0$. The series happily returns $0$ — a value the function never actually takes — because it averages the two sides of the jump.

Yes. A half-range expansion is just the full Fourier series of the odd/even periodic extension of $f$. The same I-J-W checklist applies to that extension, and jumps introduced by the extension (often at $x=0$ or $x=L$) are resolved by the same midpoint rule.

$f(x)=x$ at $x=0$? $0$. The function is continuous at $x=0$ with $f(0)=0$, so no midpoint averaging is needed — the series converges straight to $f(0)=0$. Contrast this with $x=\pi$, where the periodic wrap forces a midpoint of $0$ from $\pi$ and $-\pi$.