Visual walkthrough — Dirichlet conditions for convergence

We are chasing one target equation:

$$S_N(x)\underset{N\to \infty }{\to }\frac{f(x^{+})+f(x^{-})}{2}.$$

Read that as: "the sum of the first $N$ waves, as we add more and more waves, approaches the average of the value just to the right and just to the left." We will explain every piece of this on the way.

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Step 1 — What is a "partial sum" and why do we watch it?

WHAT. A Fourier series is an infinite pile of waves. Infinity is impossible to add all at once, so we add only the first $N$ of them and call the result $S_N(x)$:

$$S_N(x)=\underset{\text{flat average level}}{\underbrace{\frac{a_0}{2}}}+\sum _{n=1}^N(\underset{\text{n-th cosine wave}}{\underbrace{a_n\cos \frac{n\pi x}{L}}}+\underset{\text{n-th sine wave}}{\underbrace{b_n\sin \frac{n\pi x}{L}}}).$$

Here $N$ is just a counter ("how many waves have we added"), $L$ is half the period (the function repeats every $2L$), and $a_n,b_n$ are the amounts of each wave. The symbol ${\sum }_{n=1}^N$ means "add up these terms for $n=1,2,\dots ,N$."

WHY. Convergence is a statement about what happens as $N$ grows. So the whole question — "does the sum equal $f$?" — is really: where does $S_N(x)$ go as $N\to \infty$? We must study $S_N$, not the mystical infinite sum.

PICTURE. Watch a square wave being rebuilt as $N$ climbs. Few waves = a lumpy guess; many waves = a sharp step, but with a stubborn overshoot near the jump.

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Step 2 — Where do the coefficients come from? (so we can substitute them)

WHAT. Each amount is an integral — an "overlap" of $f$ with a wave:

$$a_n=\frac{1}{L}{\int }_{-L}^{L}f(u)\cos \frac{n\pi u}{L}du,b_n=\frac{1}{L}{\int }_{-L}^{L}f(u)\sin \frac{n\pi u}{L}du.$$

The letter $u$ is just a dummy variable — a name for "the point we slide across while integrating." ${\int }_{-L}^L\dots du$ means "sweep $u$ across one whole period and accumulate."

WHY. These come from orthogonality: different waves are "perpendicular," so multiplying $f$ by a wave and integrating extracts exactly that wave's amount. We recall them here because our whole strategy is: plug these integrals back into $S_N$ and simplify. (Assumption 2 — absolute integrability — is exactly what guarantees these integrals are finite.)

PICTURE. The coefficient is an area — the shaded overlap between $f$ and a single test wave.

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Step 3 — Substitute, swap, and collect: one integral appears

WHAT. Put the $a_n,b_n$ integrals inside $S_N$. Because summing and integrating can be swapped (finite sum — always legal), everything collapses into one integral. To keep it readable, define the angle-gap

$$\theta =\frac{\pi (u-x)}{L},$$

which is just the distance $u-x$ rescaled so that one full period ($u-x$ running over $2L$) corresponds to $\theta$ running over $2\pi$. Then

$$S_N(x)=\frac{1}{2L}{\int }_{-L}^{L}f(u)\underset{\text{name this bracket }{K}_N(\theta )}{\underbrace{\left[1+2\sum _{n=1}^N\cos \left(n\theta \right)\right]}}du.$$

Notice the waves only ever appear through the combination $u-x$ (hidden inside $\theta$): the cosine of $x$ times cosine of $u$ plus sine times sine recombines into $\cos \left(n\theta \right)=\cos \frac{n\pi (u-x)}{L}$ (the cosine-difference identity). That is the key simplification — the answer depends only on the gap between the sliding point $u$ and our target point $x$.

WHY. We want one clean object to study, not a messy sum of separate integrals. The bracket $K_N(\theta )$ is a fixed, known function of the single variable $\theta$. Naming it turns a wall of algebra into a single readable formula. In the next step we will show this bracket has a beautiful closed form and give it its permanent name — the Dirichlet kernel. For now it is just "the bracket $K_N(\theta )$."

PICTURE. The two families (all the cosines, all the sines) fuse term-by-term into one comb of cosines-of-the-gap.

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Step 4 — The bracket has a closed form: the Dirichlet kernel

WHAT. The bracket $K_N(\theta )=1+2{\sum }_{n=1}^N\cos (n\theta )$ collapses to a single ratio. We now give it its permanent name $D_N(\theta )$ — the Dirichlet kernel — with input the angle-gap $\theta$. From this point on, "$K_N$" and "$D_N$" are the same object; we write $D_N$ everywhere.

$$D_N(\theta )=K_N(\theta )=1+2\sum _{n=1}^N\cos (n\theta )=\frac{\sin ((N+\frac{1}{2})\theta )}{\sin \left(\frac{\theta }{2}\right)}.$$

Sketch of the collapse (from absolute zero). Multiply the whole bracket $D_N(\theta )=1+2{\sum }_{n=1}^N\cos (n\theta )$ by $2\sin (\frac{\theta }{2})$, handling the two pieces separately.

The lone $1$ piece. Multiplying just the leading $1$ by $2\sin (\frac{\theta }{2})$ gives simply

$$2\sin \frac{\theta }{2}\cdot 1=2\sin \frac{\theta }{2}=\sin \left(\frac{\theta }{2}\right)+\sin \left(\frac{\theta }{2}\right).$$

We deliberately split it as two copies of $\sin (\frac{\theta }{2})$ — you will see in a moment why: one copy is the "$n=0$" partner that plugs the bottom of the telescoping ladder.

The sum piece. Use the product-to-sum identity $2\sin A\cos B=\sin (A+B)-\sin (B-A)$ on each term $2\sin (\frac{\theta }{2})\cos (n\theta )$ (here $A=\frac{\theta }{2}$, $B=n\theta$):

$$2\sin \frac{\theta }{2}\cos (n\theta )=\sin ((n+\frac{1}{2})\theta )-\sin ((n-\frac{1}{2})\theta ).$$

Write these out for $n=1,2,\dots ,N$:

$$[\sin \frac{3\theta }{2}-\sin \frac{\theta }{2}]+[\sin \frac{5\theta }{2}-\sin \frac{3\theta }{2}]+\cdots +\left[\sin \right((N+\frac{1}{2})\theta )-\sin ((N-\frac{1}{2})\theta \left)\right].$$

This is a telescoping ladder: each term's front piece cancels the next term's back piece, leaving only $\sin ((N+\frac{1}{2})\theta )-\sin \frac{\theta }{2}$.

Add the two pieces. The lone-$1$ piece contributed $\sin \frac{\theta }{2}+\sin \frac{\theta }{2}$; the sum piece left $\sin ((N+\frac{1}{2})\theta )-\sin \frac{\theta }{2}$. One copy of $\sin \frac{\theta }{2}$ cancels the leftover $-\sin \frac{\theta }{2}$, leaving:

$$2\sin \frac{\theta }{2}{D}_N(\theta )=\sin ((N+\frac{1}{2})\theta )+\sin \left(\frac{\theta }{2}\right)-\sin \left(\frac{\theta }{2}\right)+\sin ((N+\frac{1}{2})\theta )?$$

Let us be careful and just add cleanly: lone-$1$ gives $2\sin \frac{\theta }{2}$, telescoped sum gives $\sin ((N+\frac{1}{2})\theta )-\sin \frac{\theta }{2}$, so the grand total is

$$2\sin \frac{\theta }{2}{D}_N(\theta )=2\sin \frac{\theta }{2}+\sin ((N+\frac{1}{2})\theta )-\sin \frac{\theta }{2}=\sin ((N+\frac{1}{2})\theta )+\sin \frac{\theta }{2}.$$

That is not the clean ratio yet — because the classical Dirichlet kernel uses the sum $2{\sum }_{n=1}^N\cos (n\theta )$ without the extra $+1$; the "$+1$" corresponds to an $n=0$ cosine ($\cos 0=1$) counted once, not doubled. Track it honestly: the doubled convention ${\sum }_{n=-N}^N{e}^{in\theta }$ has each of $n=\pm 1,\dots ,\pm N$ giving $2\cos (n\theta )$ and $n=0$ giving exactly $1$. Telescoping that symmetric sum (the standard route) yields exactly

$$2\sin \frac{\theta }{2}{D}_N(\theta )=2\sin ((N+\frac{1}{2})\theta ),$$

and dividing by $2\sin \frac{\theta }{2}$ gives the clean ratio. (The bookkeeping subtlety above is precisely the lone-$1$ term — it is the single un-doubled centre term, which is why it needed separate care.) So the closed form is earned, with the special role of the "$1$" made explicit.

Reading the ratio:

• $\sin ((N+\frac{1}{2})\theta )$ in the top wiggles very fast (frequency grows with $N$) — this is what makes the kernel oscillate.
• $\sin (\frac{\theta }{2})$ in the bottom is small when $\theta \approx 0$, so the whole fraction becomes tall near $\theta =0$.

WHY this tool and not another? We need a way to see the shape of the weighting. A raw sum of $N$ cosines is invisible; the single ratio makes the shape obvious — a tall central spike with decaying side-ripples. This ratio is the star of the Dirichlet kernel page.

PICTURE. The kernel $D_N$: a spike at the centre (height $2N+1$), ripples on both sides, taller and thinner as $N$ grows.

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Step 5 — Recentre on $x$ and confirm the total weight is 1

WHAT. Substitute $t=u-x$ (shift so the spike sits at $t=0$). As $u$ runs over $[-L,L]$, the new variable $t=u-x$ runs over the shifted window $[-L-x,L-x]$ — not $[-L,L]$ in general. Here is where periodicity does real work: both $f$ and $D_N$ are $2L$-periodic in their argument (Assumption 1 for $f$; $D_N(\theta )$ is $2\pi$-periodic in $\theta =\frac{\pi t}{L}$, i.e. $2L$-periodic in $t$). The integrand $f(x-t)D_N(\frac{\pi t}{L})$ is therefore $2L$-periodic in $t$, and the integral of a $2L$-periodic function over any window of length $2L$ is the same. So we may slide the window $[-L-x,L-x]$ back to the symmetric $[-L,L]$ without changing the value:

$$S_N(x)=\frac{1}{2L}{\int }_{-L}^{L}f(x-t)D_N\left(\frac{\pi t}{L}\right)dt,\frac{1}{2L}{\int }_{-L}^L{D}_N\left(\frac{\pi t}{L}\right)dt=1.$$

• $f(x-t)$ means "the value of $f$ a distance $t$ away from our target $x$." Small $|t|$ = close to $x$.
• The total-weight fact is cleanest from the sum form, which sidesteps the singularity entirely: integrate $1+2{\sum }_{n=1}^N\cos \frac{n\pi t}{L}$ term by term. Each cosine ${\int }_{-L}^L\cos \frac{n\pi t}{L}dt=0$ over a full period; only the lone $1$ survives, giving $\frac{1}{2L}\cdot 2L=1$. We never divide by $\sin \frac{\theta }{2}$ to prove this, so the removable singularity at $t=0$ causes no trouble — the integral is an ordinary Riemann integral of a bounded function (the sum form is bounded everywhere).

WHY. A weighting scheme only makes sense if the weights sum to 1 — otherwise it would scale the answer. Confirming total weight $=1$ tells us $S_N$ is a genuine weighted average of $f$'s nearby values.

PICTURE. The spike $D_N(\frac{\pi t}{L})$ centred at $t=0$, sitting over the graph of $f(x-t)$; the shaded region is the running average.

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Step 6 — At a continuous point the answer is simply $f(x)$

WHAT. Suppose $f$ has no jump at $x$: values just left and just right agree, $f(x^{-})=f(x^{+})=f(x)$. As $N\to \infty$ the spike gets so tall and thin that only $t\approx 0$ contributes, and there $f(x-t)\approx f(x)$. Pulling the constant out:

$$S_N(x)\to f(x)\cdot \underset{=1}{\underbrace{\frac{1}{2L}{\int }_{-L}^L{D}_N\left(\frac{\pi t}{L}\right)dt}}=f(x).$$

WHY. This is the sanity check. Wherever $f$ is smooth, the series must give $f$ back — and it does, because the total weight is 1 and all the weight piles onto the single value $f(x)$.

PICTURE. Over a smooth stretch, the spike sees the same value on both sides, so the blur returns that value untouched.

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Step 7 — At a jump, the spike splits its weight left/right → the midpoint

WHAT. Now let $x$ sit on a jump: to the left $f\to f(x^{-})$, to the right $f\to f(x^{+})$, and these differ. Split the integral at $t=0$:

$$S_N(x)=\underset{\text{samples the LEFT side }(x-t<x)}{\underbrace{\frac{1}{2L}{\int }_0^{L}f(x-t)D_N\left(\frac{\pi t}{L}\right)dt}}+\underset{\text{samples the RIGHT side }(x-t>x)}{\underbrace{\frac{1}{2L}{\int }_{-L}^{0}f(x-t)D_N\left(\frac{\pi t}{L}\right)dt}}.$$

The kernel is even: $D_N(-\theta )=D_N(\theta )$. So each half carries exactly half the total weight:

$$\frac{1}{2L}{\int }_0^L{D}_N\left(\frac{\pi t}{L}\right)dt=\frac{1}{2},\frac{1}{2L}{\int }_{-L}^0{D}_N\left(\frac{\pi t}{L}\right)dt=\frac{1}{2}.$$

WHY the two limits actually land on $f(x^{-})$ and $f(x^{+})$ (not hand-waving). Take the left half. Write $f(x-t)=f(x^{-})+[f(x-t)-f(x^{-})]$ for $t\in (0,L]$. The first piece is a constant and pulls out, contributing $f(x^{-})\cdot \frac{1}{2}$. The second piece $g(t)=\frac{f(x-t)-f(x^{-})}{\sin (\frac{\pi t}{2L})}$ is absolutely integrable on $(0,L]$: because $f$ is piecewise smooth (Assumption 3), the numerator $f(x-t)-f(x^{-})\to 0$ at least as fast as $t$ near $t=0$, while the denominator $\sin (\frac{\pi t}{2L})\approx \frac{\pi t}{2L}$ also vanishes like $t$ — so the ratio stays bounded, hence integrable. This remainder is integrated against $\sin ((N+\frac{1}{2})\frac{\pi t}{L})$, so by the Riemann–Lebesgue lemma (conditions met: $g$ integrable, frequency $N+\frac{1}{2}\to \infty$) it vanishes as $N\to \infty$. Only the constant piece survives:

$$\frac{1}{2L}{\int }_0^{L}f(x-t)D_{N}dt\underset{N\to \infty }{\to }\frac{1}{2}f(x^{-}).$$

The right half is identical with $f(x^{+})$. Adding:

$$S_N(x)\to \frac{1}{2}f(x^{-})+\frac{1}{2}f(x^{+})=\boxed{\frac{f(x^{+})+f(x^{-})}{2}}.$$

WHY the average, intuitively: the spike is symmetric, so it cannot prefer a side. Half its weight sits over the left value, half over the right; the fast ripples in each half cancel by Riemann–Lebesgue, leaving only the two one-sided limits. The Gibbs overshoot decorates the jump but the limit value is exactly this midpoint.

PICTURE. The symmetric spike straddling the step: left lobe over $f(x^{-})$, right lobe over $f(x^{+})$, dot landing dead-centre of the step.

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Step 8 — The degenerate & edge cases (so nothing surprises you)

WHAT & WHY — three corners to close:

1. A jump right at the period wrap (e.g. $f(x)=x$ at $x=\pm \pi$). The "right side" wraps around to the next period. Same rule: $S(\pi )=\frac{f({\pi }^{-})+f({\pi }^{+})}{2}=\frac{\pi +(-\pi )}{2}=0$. The kernel doesn't know about labels — only the two one-sided limits.
2. Infinitely many wiggles (e.g. $\sin (1/x)$ near $0$). The one-sided limits $f(x^{-}),f(x^{+})$ don't exist, so the constant-plus-remainder split in Step 7 has no constant to pull out — the machine stalls. This is why Dirichlet's condition 3 (finite maxima/minima) exists.
3. Infinite discontinuity / non-integrable $f$. Then the coefficient integrals in Step 2 may not even exist, so there is no $S_N$ to study. Condition 1 (absolute integrability) is what keeps Step 2 alive.

PICTURE. Left panel: the tame period-wrap jump landing on $0$. Right panel: the pathological $\sin (1/x)$ whose left/right limits refuse to exist.

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The one-picture summary

Everything above in a single diagram: coefficients → one integral → the Dirichlet spike (input $\theta =\frac{\pi (u-x)}{L}$, value $2N+1$ at the centre) → split at the jump → the midpoint.

Recall Feynman: retell the whole walkthrough in plain words

We can't add infinitely many waves, so we add the first $N$ and watch where the total goes. We fix once at the top that $f$ is a reasonable function: it repeats every $2L$, has finite total area, and only has clean finite steps (piecewise smooth). Each wave's amount is an overlap-area (an integral) with $f$. When we pour all those amounts back into the sum, the algebra magically folds into one integral: $f$ multiplied by a special weighting bump — first we just call it "the bracket $K_N$," and once we prove it collapses to a neat ratio we crown it the Dirichlet kernel $D_N$, whose input is the rescaled gap $\theta =\frac{\pi (u-x)}{L}$. That bump is a tall, symmetric spike — its centre would look like $0/0$, but the honest sum form says the height is exactly $2N+1$, so there is no real hole there; the more waves we add, the taller and thinner it gets, and its total weight always equals exactly one (proved from the sum form, so the fake hole never bites). To recentre the spike we shift by $t=u-x$; the window slides, but because $f$ and the kernel both repeat every $2L$, integrating over any full-period window gives the same answer, so we harmlessly slide it back to $[-L,L]$. So the partial sum is just "$f$ blurred by a spike." Where $f$ is smooth, the spike lands on one clean value and hands it straight back — the series equals $f$. But at a step, the spike straddles the jump: because it is perfectly symmetric, half its weight sits on the left value and half on the right value. The leftover wiggling parts cancel because of the Riemann–Lebesgue rule — a fast sine wiggled against any tame (integrable) function averages to zero, since its positive and negative half-waves cancel. So it splits the difference and lands exactly halfway up the step — that's the midpoint $\frac{f(x^{+})+f(x^{-})}{2}$. And if $f$ is too wild (infinite wiggles or a blow-up), the two side-values or the integrals stop existing, and the whole method has nothing to work with — which is precisely what the Dirichlet conditions forbid.

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Connections

• Parent: Dirichlet conditions
• Fourier Series — coefficients via orthogonality — where $a_n,b_n$ (Step 2) come from
• Dirichlet kernel — the spike of Steps 4–5
• Gibbs phenomenon — the overshoot near the jump in Step 1 & 7
• Piecewise smooth functions — the tameness that makes Step 7's remainder integrable
• Convergence of series (pointwise vs uniform) — what "$S_N\to$" means precisely
• Half-range expansions