Parseval's theorem

1. WHAT is it?

2. WHY is it true? (Derivation from scratch)

The whole proof rests on orthogonality of sines and cosines. Let me build it.

Formula Step 0 — Orthogonality relations (the engine)

Over $(-L,L)$ , with $m,n\ge 1$ : $\int_{-L}^{L}\cos\frac{m\pi x}{L}\cos\frac{n\pi x}{L}\,dx=\begin{cases}0 & m\ne n\\ L & m=n\end{cases}$ $\int_{-L}^{L}\sin\frac{m\pi x}{L}\sin\frac{n\pi x}{L}\,dx=\begin{cases}0 & m\ne n\\ L & m=n\end{cases}$ $\int_{-L}^{L}\cos\frac{m\pi x}{L}\sin\frac{n\pi x}{L}\,dx=0\quad(\text{always}),\qquad \int_{-L}^{L}1\,dx=2L.$ Why this step? Different frequency components don't "overlap" when integrated — they are perpendicular vectors. Cross-terms vanish, leaving only the squared terms.

Now the derivation. Take $[f(x)]^2$ — i.e. multiply the series by itself:

$[f(x)]^2=\left(\frac{a_0}{2}+\sum_{n}a_n\cos\frac{n\pi x}{L}+\sum_{n}b_n\sin\frac{n\pi x}{L}\right)^2.$

Now integrate over $(-L,L)$ . Squaring a sum gives diagonal terms (each piece squared) and cross terms (products of different pieces).

Why this step? Integration is linear, so $\int(\text{sum})^2 = \sum\int(\text{products})$ . The orthogonality relations decide which survive.

The cross terms (e.g. $\int \cos\frac{m\pi x}{L}\cos\frac{n\pi x}{L}$ with $m\ne n$ , or any cos·sin) all integrate to 0. ✗ gone.
The constant-squared term: $\int_{-L}^{L}\left(\frac{a_0}{2}\right)^2 dx=\frac{a_0^2}{4}\cdot 2L=\frac{a_0^2 L}{2}$ .
Each $\int_{-L}^{L} a_n^2\cos^2\frac{n\pi x}{L}\,dx=a_n^2\cdot L$ .
Each $\int_{-L}^{L} b_n^2\sin^2\frac{n\pi x}{L}\,dx=b_n^2\cdot L$ .

Add the survivors: $\int_{-L}^{L}[f(x)]^2\,dx=\frac{a_0^2 L}{2}+L\sum_{n=1}^{\infty}(a_n^2+b_n^2).$

Divide both sides by $L$ : $\frac{1}{L}\int_{-L}^{L}[f(x)]^2\,dx=\frac{a_0^2}{2}+\sum_{n=1}^{\infty}(a_n^2+b_n^2).\qquad\blacksquare$

Why this step? Dividing by $L$ packages it so the constant term matches the $\frac{a_0^2}{2}$ pattern and the formula is symmetric.

3. HOW to use it (Worked Examples)

Worked example Example 1 — Summing

$\sum 1/n^2$ from $f(x)=x$ Take $f(x)=x$ on $(-\pi,\pi)$ (so $L=\pi$ ). It's odd ⇒ $a_n=0$ . Its sine coefficients are $b_n=\frac{2(-1)^{n+1}}{n}.$ Why this step? $x$ is odd, only sines appear; the $b_n$ come from the standard integral.

Left side: $\dfrac{1}{\pi}\int_{-\pi}^{\pi}x^2\,dx=\dfrac{1}{\pi}\cdot\dfrac{2\pi^3}{3}=\dfrac{2\pi^2}{3}.$ Why? $\int_{-\pi}^{\pi}x^2dx = \frac{2\pi^3}{3}$ , then divide by $L=\pi$ .

Right side: $\sum b_n^2=\sum_{n=1}^{\infty}\dfrac{4}{n^2}=4\sum\dfrac{1}{n^2}.$

Parseval: $\dfrac{2\pi^2}{3}=4\sum\dfrac{1}{n^2}\Rightarrow \boxed{\sum_{n=1}^{\infty}\dfrac{1}{n^2}=\dfrac{\pi^2}{6}}.$ Why this matters: Parseval turns a Fourier series into a way to sum infinite series that look impossible directly. This is the famous Basel problem!

Worked example Example 2 —

$f(x)=x^2$ on $(-\pi,\pi)$ giving $\sum 1/n^4$ Here $a_0=\frac{2\pi^2}{3}$ , $a_n=\frac{4(-1)^n}{n^2}$ , $b_n=0$ (even function). Why? $x^2$ is even ⇒ cosine series only.

Left: $\dfrac{1}{\pi}\int_{-\pi}^{\pi}x^4dx=\dfrac{1}{\pi}\cdot\dfrac{2\pi^5}{5}=\dfrac{2\pi^4}{5}.$

Right: $\dfrac{a_0^2}{2}+\sum a_n^2=\dfrac{1}{2}\left(\dfrac{2\pi^2}{3}\right)^2+\sum\dfrac{16}{n^4}=\dfrac{2\pi^4}{9}+16\sum\dfrac{1}{n^4}.$

Set equal: $\dfrac{2\pi^4}{5}=\dfrac{2\pi^4}{9}+16\sum\dfrac{1}{n^4}.$ $\dfrac{2\pi^4}{5}-\dfrac{2\pi^4}{9}=2\pi^4\cdot\dfrac{9-5}{45}=\dfrac{8\pi^4}{45}.$ So $16\sum\dfrac{1}{n^4}=\dfrac{8\pi^4}{45}\Rightarrow \boxed{\sum\dfrac{1}{n^4}=\dfrac{\pi^4}{90}}.$ Why this matters: Same trick, deeper result. Higher powers of $x$ give higher zeta values.

4. Steel-manned Mistakes

5. Active Recall

Recall Can you reconstruct it? (click to check)

State Parseval for $(-L,L)$ . → $\frac{1}{L}\int_{-L}^{L}f^2=\frac{a_0^2}{2}+\sum(a_n^2+b_n^2)$ .
What makes the proof work? → Orthogonality of sines/cosines kills cross terms.
One famous result it gives? → $\sum 1/n^2=\pi^2/6$ .

Recall Feynman: explain to a 12-year-old

Imagine a song. You can break it into pure tones — some bass, some treble. Each pure tone carries some "loudness." Parseval's theorem says: the total loudness of the whole song equals the loudness of the bass plus the treble plus every other tone, added up. Nothing leaks away. So instead of measuring the messy whole song, you can just add up the loudness of each simple tone. And cleverly, this lets us add up weird never-ending number lists like $1+\frac14+\frac19+\dots$ and discover they equal $\pi^2/6$ !

6. Flashcards

What does Parseval's theorem equate?

The mean-square value of

f

(energy) with the sum of squares of its Fourier coefficients.

State Parseval's theorem on

(-L,L)

\frac{1}{L}\int_{-L}^{L}f^2\,dx=\frac{a_0^2}{2}+\sum_{n=1}^\infty(a_n^2+b_n^2)

What property of sines/cosines makes the proof work?

Orthogonality — cross-term integrals vanish for

m\ne n

Why is the constant term

a_0^2/2

not

a_0^2

From

(a_0/2)^2

times

\int 1\,dx=2L

, divided by

L

⇒

a_0^2/2

Which famous sum comes from applying Parseval to

f(x)=x

(-\pi,\pi)

\sum 1/n^2=\pi^2/6

Which sum comes from

f(x)=x^2

(-\pi,\pi)

\sum 1/n^4=\pi^4/90

Physical meaning of

\frac{a_n^2+b_n^2}{2}

Power carried by the

n

-th harmonic.

What is the value of

\int_{-L}^L\cos\frac{m\pi x}{L}\sin\frac{n\pi x}{L}dx

Always

0

7. Connections

Fourier Series — Parseval is its energy/completeness statement.
Orthogonality of functions — the engine of the proof.
Basel Problem — $\sum 1/n^2=\pi^2/6$ as a corollary.
Bessel's Inequality — Parseval is the equality case (complete basis).
Plancherel Theorem — continuous Fourier transform analogue.
RMS and Power Spectra — engineering application.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, Parseval ka theorem bilkul Pythagoras theorem jaisa hai, bas functions ke liye. Jaise vector ka length-square uske components ke squares ka sum hota hai ( $|v|^2 = v_x^2+v_y^2+v_z^2$ ), waise hi ek function ki total "energy" uske Fourier coefficients ke squares ke sum ke barabar hoti hai. Matlab agar tum function ko sine aur cosine ke chote chote tukdo me tod do, to har tukde ki energy alag-alag hoti hai, aur sab ko jod do to poora function ki energy mil jaati hai. Kuch bhi gum nahi hota.

Formula yaad rakhna: $\frac{1}{L}\int_{-L}^{L}f^2\,dx=\frac{a_0^2}{2}+\sum(a_n^2+b_n^2)$ . Left side me $1/L$ aur $f^2$ ka integral — yeh "average energy" hai. Right side me sirf coefficients. Proof ka magic hai orthogonality — jab tum $(\text{series})^2$ ko integrate karte ho, to alag-alag frequency wale cross-terms zero ho jaate hain, sirf squared terms bachte hain.

Iska sabse mast use: hum impossible-lagne wale infinite sums nikaal sakte hain. Jaise $f(x)=x$ lo, Parseval lagao, aur seedha $\sum 1/n^2 = \pi^2/6$ aa jaata hai (famous Basel problem!). $f(x)=x^2$ se $\sum 1/n^4=\pi^4/90$ . Engineering me bhi yeh signal ki total power = har harmonic ki power ka sum batata hai.

Common galti: $1/L$ bhool jaana, ya constant term ko $a_0^2$ likh dena (sahi hai $a_0^2/2$ — kyunki DC term $a_0/2$ likha jaata hai). Yeh do cheezein exam me sambhal lo, baaki sab seedha hai.