Visual walkthrough — Half-range sine and cosine series

We will derive, from absolute zero, this one result:

$$b_n=\frac{2}{L}{\int }_0^{L}f(x)\sin \frac{n\pi x}{L}dx$$

and understand every symbol in it.

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Step 1 — We only know half the story

WHAT. We are handed a function $f(x)$ that is defined only for $x$ between $0$ and $L$. Think of a temperature reading along a metal rod that starts at the left wall ($x=0$) and ends at the right wall ($x=L$). To the left of $0$, there is nothing — no rule, no values.

WHY it matters. The Fourier machine (the tool that writes any repeating wiggle as a sum of sines and cosines) needs a function on a full symmetric interval $[-L,L]$ so it can compare the left and right sides. We only have the right side $[0,L]$. So we are missing exactly half the input the machine wants.

PICTURE. The solid burnt-orange curve is all we were given; the grey shaded region on the left is the "unknown zone" — completely blank.

Symbols so far:

• $x$ ::: position along the rod, a number from $0$ to $L$.
• $L$ ::: the length of the interval we know (here also the right wall).
• $f(x)$ ::: the given values, drawn only on the right.

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Step 2 — We are FREE to invent the left half

WHAT. Because nobody told us what happens for $x<0$, we get to choose the missing values. Two natural choices exist. In this walkthrough we pick the odd choice: reflect the curve through the origin — flip it left–right and upside-down at the same time.

WHY this choice. An odd function is one where $f(-x)=-f(x)$: whatever height you have on the right at distance $x$, you put the negative of it on the left at the same distance. This particular flip has a magic property (revealed in Step 4) that annihilates every cosine. It also forces the curve to pass through zero at the origin — handy for problems with fixed walls.

PICTURE. The original (burnt orange) stays; the new plum curve on the left is the point-flipped copy. Notice the little rotation arrow at the origin — an odd extension is a $180^{\circ }$ spin about the point $(0,0)$.

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Step 3 — Now the full Fourier machine applies

WHAT. With values on all of $[-L,L]$, we may write the standard full Fourier series. It says: any well-behaved periodic wiggle is a constant plus a pile of cosines plus a pile of sines.

$$f(x)=\underset{\text{average height}}{\underbrace{\frac{a_0}{2}}}+\sum _{n=1}^{\infty }[\underset{\begin{array}{c}\text{how much}\\ \text{of }n\text{-th cosine}\end{array}}{\underbrace{a_n}}\underset{\begin{array}{c}\text{even wave,}\\ n\text{ humps}\end{array}}{\underbrace{\cos \frac{n\pi x}{L}}}+\underset{\begin{array}{c}\text{how much}\\ \text{of }n\text{-th sine}\end{array}}{\underbrace{b_n}}\underset{\begin{array}{c}\text{odd wave,}\\ n\text{ humps}\end{array}}{\underbrace{\sin \frac{n\pi x}{L}}}]$$

WHY these pieces. $\cos$ and $\sin$ of $\frac{n\pi x}{L}$ are the only pure waves that fit a whole number $n$ of humps into the interval and repeat with period $2L$. The numbers $a_n,b_n$ are "recipe amounts" — how loud each wave plays. The coefficients come from integrals (the tool that measures overlap):

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

WHY an integral here? An integral is a continuous sum — it adds up the product $f\times$(a test wave) over the whole interval. If $f$ leans the same way as that test wave, the products pile up positive and the coefficient is big; if they disagree, positives and negatives cancel. It is a similarity meter. See Fourier Series — full range.

PICTURE. The bar chart shows the "recipe": some cosine bars (teal) and some sine bars (plum), each of height $a_n$ or $b_n$. Right now we do not know which are zero — that is Step 4's job.

Symbols introduced:

• $n$ ::: a counting number $1,2,3,\dots$ — how many humps the wave has.
• $\frac{n\pi x}{L}$ ::: the angle fed into $\sin /\cos$, scaled so that $n$ full humps fit in $[0,L]$.
• $a_0/2$ ::: the average height of $f$ over the interval.

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Step 4 — Watch the cosines die (odd × even = odd)

WHAT. Take the cosine coefficient integrand: $f(x)\cos \frac{n\pi x}{L}$. Our $f$ is odd (we made it so). Cosine is even ($\cos (-\theta )=\cos \theta$ — its graph is a mirror across the vertical axis). The product of an odd thing and an even thing is odd.

WHY that kills it. Integrate an odd function over a symmetric interval $[-L,L]$ and you get exactly zero: every positive bump on the right is matched by an equal negative bump on the left. They cancel in pairs. So

$$a_n=\frac{1}{L}{\int }_{-L}^L\underset{\text{odd}}{\underbrace{f(x)}}\underset{\text{even}}{\underbrace{\cos \frac{n\pi x}{L}}}dx=\frac{1}{L}{\int }_{-L}^L(\text{odd})dx=0.$$

Every cosine amount is zero. The teal bars from Step 3 all vanish.

PICTURE. The left half of the shaded area (plum, negative) is the exact mirror-and-flip of the right half (orange, positive). The two signed areas are equal in size, opposite in sign — total area $=0$.

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Step 5 — Watch the sines double (odd × odd = even)

WHAT. Now the sine coefficient integrand: $f(x)\sin \frac{n\pi x}{L}$. Sine is odd ($\sin (-\theta )=-\sin \theta$ — flip left gives flip down). Odd $\times$ odd $=$ even.

WHY it doubles. An even function ($g(-x)=g(x)$) has a left half that is a perfect mirror copy of the right half — same sign, same size. So the area from $-L$ to $0$ equals the area from $0$ to $L$. Instead of integrating over the whole symmetric interval, we integrate over the right half and multiply by two:

$$b_n=\frac{1}{L}{\int }_{-L}^L\underset{\text{even}}{\underbrace{f(x)\sin \frac{n\pi x}{L}}}dx=\frac{1}{L}\cdot 2{\int }_0^{L}f(x)\sin \frac{n\pi x}{L}dx=\frac{2}{L}{\int }_0^{L}f(x)\sin \frac{n\pi x}{L}dx.$$

PICTURE. Both shaded halves now have the same sign (both orange, both above the axis in this example). Their areas are identical — the left is a mirror of the right — so the total is twice one half. That factor of $2$ is literally the second copy of the area.

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Step 6 — Where the average went (the missing $a_0$)

WHAT. In the full series there was a constant term $a_0/2$. What happened to it? The constant is just the $n=0$ cosine (since $\cos 0=1$). Its integrand is $f(x)\cdot 1=f(x)$, which is odd. Over $[-L,L]$ that integrates to zero as well.

WHY it must vanish. An odd extension has as much curve below the axis on the left as above on the right; its overall average is zero. A sine series therefore has no constant term — it cannot represent a nonzero background level. (If your problem needs that background, you use the cosine series instead — see the cosine branch in the parent note.)

PICTURE. The full odd curve over $[-L,L]$: the orange area above the axis exactly matches the plum area below. Net signed area (the average) is zero.

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Step 7 — The edge cases: endpoints and jumps

WHAT. Two degenerate spots deserve their own look: the origin $x=0$ and the wall $x=L$.

At $x=0$: every $\sin \frac{n\pi \cdot 0}{L}=\sin 0=0$. So the sine series always returns $0$ at the left end, no matter what $f(0)$ was. If $f(0)\ne 0$, the series simply cannot match it — it splits the difference (the midpoint rule) between the value coming from the right ($f(0^{+})$) and the flipped value from the left ($-f(0^{+})$), whose average is $0$.

At $x=L$: the periodic extension may have a jump. If $f(L)\ne 0$, the flipped copy just past $L$ drops to $-f(L)$, a sudden cliff. There the series converges to the midpoint $\frac{1}{2}(f(L^{-})+(-f(L^{-})))=0$, again not to $f(L)$.

WHY care. These are exactly the points where writing "$f(x)=\sum \dots$" is a lie. The honest statement holds on the open interval $0<x<L$.

PICTURE. Zoom on the sawtooth jump at $x=L$: the two branches (orange coming up to $f(L^{-})$, plum starting at $-f(L^{-})$) with a plum dot marking the midpoint the series actually hits.

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The one-picture summary

Everything at once. Left panel: the given right half → odd-reflected to fill the left. Middle: pairing with even cosine gives cancelling areas (⇒ $a_n=0$). Right: pairing with odd sine gives matching areas (⇒ double the half-integral). The three moves are the whole derivation.

Recall Feynman: retell the whole walkthrough in plain words

I was given a wiggle drawn only on the right side of the page, from $0$ to $L$. To use the Fourier machine — which needs both sides — I drew the missing left side myself by spinning the picture $180^{\circ }$ around the centre dot; that makes it "odd." Then I asked the machine for its recipe of waves. When it tried to measure how much cosine was inside, the left area (now upside-down) cancelled the right area perfectly, so the answer was zero — no cosines at all. When it measured how much sine was inside, the left area matched the right area sign-for-sign, so I just measured the right half and doubled it. That doubling is the little "$2$" out front, and the "only measure the right half" is the ${\int }_0^L$. The constant background also cancelled to zero, because an upside-down mirror has as much curve below the line as above. The only place I have to be careful is at the two ends: sines are always zero at $x=0$, and if my wiggle didn't already touch zero at $x=L$, the mirror makes a cliff there, and the series lands on the middle of that cliff, not the top. So I write the equality only for $0<x<L$. That is the entire half-range sine story.

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Connections

• Parent topic — the full statement and the cosine twin of this walkthrough.
• Even and Odd Functions — the reflection symmetry that powers Steps 4–6.
• Fourier Series — full range — the machine we specialised in Step 3.
• Dirichlet Conditions — the midpoint rule used in Step 7.
• Heat Equation — separation of variables — where these sine series get used.
• Wave Equation on a finite string — fixed ends → this exact sine expansion.
• Neumann Boundary Conditions — the even/cosine alternative to Step 2's choice.