Parseval's theorem

4.7.7 · Maths › Partial Differential Equations

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

Poora proof sines aur cosines ki orthogonality par tika hai. Chaliye ise step by step build karte hain.

Ab derivation. $[f (x)]^{2}$ lo — yaani series ko khud se multiply karo:

$[f (x)]^{2} = (\frac{a _{0}}{2} + \sum_{n} a_{n} cos \frac{nπ x}{L} + \sum_{n} b_{n} sin \frac{nπ x}{L})^{2} .$

Ab $(- L, L)$ par integrate karo. Ek sum ko square karne se diagonal terms (har piece squared) aur cross terms (alag pieces ke products) milte hain.

Yeh step kyun? Integration linear hai, isliye $\int (sum)^{2} = \sum \int (products)$ . Orthogonality relations decide karte hain ki kaun se survive karte hain.

Cross terms (jaise $\int cos \frac{mπ x}{L} cos \frac{nπ x}{L}$ jab $m \neq = n$ , ya koi bhi cos·sin) sab 0 mein integrate ho jaate hain. ✗ gone.
Constant-squared term: $\int_{- L}^{L} (\frac{a _{0}}{2})^{2} d x = \frac{a _{0}^{2}}{4} \cdot 2 L = \frac{a _{0}^{2} L}{2}$ .
Har $\int_{- L}^{L} a_{n}^{2} cos^{2} \frac{nπ x}{L} d x = a_{n}^{2} \cdot L$ .
Har $\int_{- L}^{L} b_{n}^{2} sin^{2} \frac{nπ x}{L} d x = b_{n}^{2} \cdot L$ .

Bachne waale terms add karo: $\int_{- L}^{L} [f (x)]^{2} d x = \frac{a _{0}^{2} L}{2} + L \sum_{n = 1}^{\infty} (a_{n}^{2} + b_{n}^{2}) .$

Dono sides ko $L$ se divide karo: $\frac{1}{L} \int_{- L}^{L} [f (x)]^{2} d x = \frac{a _{0}^{2}}{2} + \sum_{n = 1}^{\infty} (a_{n}^{2} + b_{n}^{2}) . ■$

Yeh step kyun? $L$ se divide karne par constant term $\frac{a _{0}^{2}}{2}$ pattern se match karta hai aur formula symmetric ban jaata hai.

3. HOW to use it (Worked Examples)

5. Active Recall

Recall Kya tum ise reconstruct kar sakte ho? (click to check)

$(- L, L)$ ke liye Parseval state karo. → $\frac{1}{L} \int_{- L}^{L} f^{2} = \frac{a _{0}^{2}}{2} + \sum (a_{n}^{2} + b_{n}^{2})$ .
Proof ko kya kaam ka banata hai? → Sines/cosines ki orthogonality cross terms ko khatam kar deti hai.
Iska ek famous result? → $\sum 1/ n^{2} = π^{2} /6$ .

Recall Feynman: ek 12-saal ke bacche ko samjhao

Ek gaana imagine karo. Tum use pure tones mein tod sakte ho — kuch bass, kuch treble. Har pure tone kuch "loudness" carry karta hai. Parseval's theorem kehta hai: poore gaane ki total loudness bass ki loudness plus treble plus har doosre tone ki loudness ke barabar hai, sab add karo. Kuch bhi bahar nahi jaata. Toh messy poore gaane ko measure karne ki jagah, tum sirf har simple tone ki loudness add kar sakte ho. Aur chalaaki se, yeh humein $1 + \frac{1}{4} + \frac{1}{9} + \dots$ jaise weird kabhi-na-khatam hone waale number lists add karne aur yeh discover karne deta hai ki woh $π^{2} /6$ ke barabar hain!

6. Flashcards

Parseval's theorem kya equate karta hai?

f

ki mean-square value (energy) ko uske Fourier coefficients ke squares ke sum se.

(- L, L)

par Parseval's theorem state karo.

\frac{1}{L} \int_{- L}^{L} f^{2} d x = \frac{a _{0}^{2}}{2} + \sum_{n = 1}^{\infty} (a_{n}^{2} + b_{n}^{2})

Sines/cosines ki kaun si property proof ko kaam ka banati hai?

Orthogonality — cross-term integrals

m \neq = n

ke liye vanish ho jaate hain.

Constant term

a_{0}^{2} /2

kyun hai,

a_{0}^{2}

kyun nahi?

(a_{0} /2)^{2}

times

\int 1 d x = 2 L

se,

L

se divide karne par ⇒

a_{0}^{2} /2

Parseval ko

f (x) = x

par

(- π, π)

apply karne se kaun sa famous sum aata hai?

\sum 1/ n^{2} = π^{2} /6

f (x) = x^{2}

(- π, π)

se kaun sa sum aata hai?

\sum 1/ n^{4} = π^{4} /90

\frac{a _{n}^{2} + b _{n}^{2}}{2}

ka physical meaning?

n

-th harmonic ki power.

\int_{- L}^{L} cos \frac{mπ x}{L} sin \frac{nπ x}{L} d x

ki value kya hai?

Hamesha

0

7. Connections

Fourier Series — Parseval iska energy/completeness statement hai.
Orthogonality of functions — proof ka engine.
Basel Problem — $\sum 1/ n^{2} = π^{2} /6$ ek corollary ke roop mein.
Bessel's Inequality — Parseval equality case hai (complete basis).
Plancherel Theorem — continuous Fourier transform analogue.
RMS and Power Spectra — engineering application.

1. WHAT is it?