Parseval's theorem: Worked examples — Hinglish deep dive

4.7.7 · D3 · Maths › Partial Differential Equations › Parseval's theorem

Intuition Yeh page kis liye hai

Parent note ne theorem prove kiya aur teen examples dikhaye. Yahan hum ise har us tarah ke input ke against stress-test karenge jo aap encounter kar sakte ho: odd functions, even functions, functions jo dono nahi hain, constants (degenerate case), ek signal jo finite sum of harmonics hai, ek limiting-behaviour check, aur ek exam twist. Agar aap yeh sab kar sakte ho, toh koi bhi Parseval question aapko surprise nahi kar sakta.

Ek rule jis par hum poori tarah rely karenge: $(- L, L)$ par $\frac{1}{L} \int_{- L}^{L} [f (x)]^{2} d x = \frac{a _{0}^{2}}{2} + \sum_{n = 1}^{\infty} (a_{n}^{2} + b_{n}^{2}) .$ Left side ek single integral hai jo aap hamesha directly compute kar sakte ho. Right side Fourier Series coefficients se banti hai. Parseval kehta hai yeh dono equal hain — isliye ek jaanna doosra de deta hai.

Scenario matrix

Har Parseval problem asal mein inhi cells mein se ek hoti hai. Is page ka poora point yeh hai ki har row ko walk karo taaki aap kabhi koi unseen case na mile.

#	Cell (case class)	Isme kya khaas hai	Covered by
A	Odd $f$ : sirf $b_{n}$ bachte hain	$a_{0} = a_{n} = 0$ , RHS $= \sum b_{n}^{2}$	Ex 1
B	Even $f$ : sirf $a_{0}, a_{n}$ bachte hain	$b_{n} = 0$ , RHS $= \frac{a _{0}^{2}}{2} + \sum a_{n}^{2}$	Ex 2
C	Na odd na even	teeno families present hain	Ex 3
D	Degenerate: constant $f = c$	sirf DC term $\frac{a _{0}^{2}}{2}$	Ex 4
E	Finite harmonic sum (signal)	koi infinite series nahi — direct power add	Ex 5
F	Alag interval / general $L$	$\frac{1}{L}$ factor kaafi important hai	Ex 6
G	Limiting / convergence sanity	partial sums ko LHS ki taraf approach karna chahiye, neeche se	Ex 7
H	Word problem: RMS & power	units, physical meaning	Ex 8
I	Exam twist: ek unknown $b_{n}$ solve karo	Parseval ko ulta use karo	Ex 9

Do zero-cases explicitly naam lene ke liye taaki aap kabhi trip na karo:

Zero function $f \equiv 0$ : dono sides $0$ hain. Trivially true; yeh neeche diye picture ka "origin" hai.
$a_{0}$ average hai, doubled: agar $f$ ka zero average hai toh $a_{0} = 0$ aur DC term nikal jaata hai (yahi cheez odd functions ko clean banati hai).

Example 1 — Cell A (odd function): Basel sum recover karo

Left side — integral directly compute karo. $\frac{1}{π} \int_{- π}^{π} x^{2} d x = \frac{1}{π} \cdot \frac{2 π ^{3}}{3} = \frac{2 π ^{2}}{3} .$ Yeh step kyun? Odd $f$ ke liye RHS pure $\sum b_{n}^{2}$ hai; ise use karne ke liye pehle LHS ke liye ek number chahiye, aur $\int x^{2}$ elementary hai.
Right side — coefficients ko square karke sum karo. $\sum_{n = 1}^{\infty} b_{n}^{2} = \sum_{n = 1}^{\infty} \frac{4 ( - 1 ) ^{2 (n + 1)}}{n ^{2}} = \sum_{n = 1}^{\infty} \frac{4}{n ^{2}} = 4 \sum_{n = 1}^{\infty} \frac{1}{n ^{2}} .$ Yeh step kyun? Square karne se sign $(- 1)^{n + 1}$ khatam ho jaata hai (yeh hamesha $+ 1$ ban jaata hai) — yeh signal hai ki Parseval mein phase information kho jaati hai; sirf magnitudes matter karti hain.
LHS $=$ RHS set karo aur solve karo. $\frac{2 π ^{2}}{3} = 4 \sum \frac{1}{n ^{2}} \Rightarrow \sum_{n = 1}^{\infty} \frac{1}{n ^{2}} = \frac{π ^{2}}{6} .$ Yeh step kyun? Yeh hai Parseval "ulta run" — ek integral jo hum kar sakte hain woh ek sum deliver karta hai jo hum directly nahi kar sakte. Dekho Basel Problem.

Verify: $π^{2} /6 \approx 1.6449$ — wakai $1.6$ se thoda zyada, forecast match karta hai. Numerically $\sum_{n = 1}^{5000} 1/ n^{2} \approx 1.6447$ , $1.6449$ ke paas aa raha hai. ✓

Example 2 — Cell B (even function): $\sum 1/ n^{4}$ recover karo

Left side. $\frac{1}{π} \int_{- π}^{π} x^{4} d x = \frac{1}{π} \cdot \frac{2 π ^{5}}{5} = \frac{2 π ^{4}}{5} .$ Yeh step kyun? $[f]^{2} = x^{4}$ ; interval symmetric hai isliye $\int x^{4} = 2 \int_{0}^{π} x^{4}$ .
Right side — DC term ko half weight ke saath include karo. $\frac{a _{0}^{2}}{2} + \sum a_{n}^{2} = \frac{1}{2} (\frac{2 π ^{2}}{3})^{2} + \sum \frac{16}{n ^{4}} = \frac{2 π ^{4}}{9} + 16 \sum \frac{1}{n ^{4}} .$ Yeh step kyun? Akela $a_{0}$ term factor $\frac{1}{2}$ carry karta hai — ise bhoolna classic error hai (parent note ka Mistake 2 dekho).
Equate karke sum isolate karo. $\frac{2 π ^{4}}{5} - \frac{2 π ^{4}}{9} = 2 π^{4} \cdot \frac{9 - 5}{45} = \frac{8 π ^{4}}{45} = 16 \sum \frac{1}{n ^{4}} \Rightarrow \sum \frac{1}{n ^{4}} = \frac{π ^{4}}{90} .$ Yeh step kyun? Ex 1 jaisi hi machinery, ek power higher — Parseval $x$ ke har even power tak scale hota hai.

Verify: $π^{4} /90 \approx 1.0823$ , forecast match karta hai. Partial sum $\sum_{n = 1}^{2000} 1/ n^{4} \approx 1.0823$ . ✓

Example 3 — Cell C (na odd na even): dono families present

Left side directly. $\frac{1}{π} \int_{- π}^{π} (1 + x)^{2} d x = \frac{1}{π} \int_{- π}^{π} (1 + 2 x + x^{2}) d x .$ Middle term $\int_{- π}^{π} 2 x d x = 0$ (symmetric interval par odd), isliye $= \frac{1}{π} (2 π + \frac{2 π ^{3}}{3}) = 2 + \frac{2 π ^{2}}{3} .$ Yeh step kyun? Yeh dikhata hai ki constant aur odd part ke beech cross-term integration par vanish ho jaata hai — orthogonality ki geometric echo: ek constant (DC) aur ek sine perpendicular hain.
Right side. $\frac{a _{0}^{2}}{2} + \sum b_{n}^{2} = \frac{2 ^{2}}{2} + 4 \sum \frac{1}{n ^{2}} = 2 + 4 \sum \frac{1}{n ^{2}} .$ Yeh step kyun? $a_{0} = 2$ contribute karta hai $\frac{a _{0}^{2}}{2} = 2$ , sines exactly Ex 1 ki tarah contribute karte hain.
Match karo. LHS $=$ RHS deta hai $2 + \frac{2 π ^{2}}{3} = 2 + 4 \sum \frac{1}{n ^{2}}$ , yaani phir $\sum \frac{1}{n ^{2}} = \frac{π ^{2}}{6}$ . Yeh step kyun? Constant piece dono sides se cancel ho jaata hai — Parseval energy ko frequency ke hisaab se separate karta hai, isliye DC energy ( $2$ ) aur oscillating energy apne-apne bins mein rehti hain.

Verify: LHS $= 2 + 2 π^{2} /3 \approx 8.5797$ ; RHS $= 2 + 4 (π^{2} /6) = 2 + 2 π^{2} /3 \approx 8.5797$ . ✓

Example 4 — Cell D (degenerate constant): sirf DC bin

Fourier coefficients. $a_{0} = \frac{1}{π} \int_{- π}^{π} 5 d x = \frac{1}{π} \cdot 10 π = 10$ ; saare $a_{n} = b_{n} = 0$ . Yeh step kyun? DC coefficient $a_{0} = \frac{1}{L} \int f$ hai; ek constant ka average $5$ hai, aur $a_{0} = 2 \times average = 10$ . Yeh "sirf constant term bachta hai" waala degenerate cell hai.
Left side. $\frac{1}{π} \int_{- π}^{π} 25 d x = \frac{1}{π} \cdot 50 π = 50.$
Right side. $\frac{a _{0}^{2}}{2} = \frac{100}{2} = 50.$ Yeh step kyun? Mysterious factor $\frac{1}{2}$ ki bilkul sahi confirmation hai — iske bina RHS $100 \neq = 50$ hota.

Verify: $50 = 50$ . ✓ $a_{0}$ par half-weight koi typo nahi hai; yeh constant case ko sahi banata hai.

Example 5 — Cell E (finite harmonic signal): energy harmonic by harmonic add hoti hai

Right side finite sum hai — koi infinite series nahi. $\frac{1}{π} \int_{- π}^{π} f^{2} d t = a_{1}^{2} + b_{2}^{2} = 9 + 16 = 25.$ Yeh step kyun? Finitely many harmonics ke saath, Parseval literally $3$ - $4$ - $5$ Pythagoras hai: perpendicular components quadrature mein add hote hain.
Raw integral paane ke liye un-average karo. $L = π$ se multiply karo: $\int_{- π}^{π} f^{2} d t = 25 π .$ Yeh step kyun? $\frac{1}{L}$ ek averaging factor hai; ise hatane se total energy restore hoti hai.
Mean-square value (full period $2 π$ par $f^{2}$ ka average): $\overline{f^{2}} = \frac{1}{2 π} \int_{- π}^{π} f^{2} d t = \frac{25 π}{2 π} = 12.5, RMS = 12.5 \approx 3.536.$ Yeh step kyun? Dhyan do do averaging windows ka: Parseval $\frac{1}{L}$ use karta hai (half-period), physics power $\frac{1}{2 π}$ use karta hai (full period). Inhe confuse mat karo.

Verify: $9 + 16 = 25$ ; $25 π \approx 78.54$ ; mean-square $= 12.5$ ; RMS $= 12.5 \approx 3.5355$ . ✓ Dekho RMS and Power Spectra.

Example 6 — Cell F (general $L$ ): $\frac{1}{L}$ factor kaata hai

$f = x$ ke liye $(- L, L)$ par coefficients. Odd function $x$ ka $b_{n} = \frac{2}{L} \int_{0}^{L} x sin \frac{nπ x}{L} d x = \frac{2 L ( - 1 ) ^{n + 1}}{nπ} .$ Yeh step kyun? Coefficient formula $L$ carry karta hai; dekho yeh kaise self-cancel karta hai.
Right side. $\sum b_{n}^{2} = \sum \frac{4 L ^{2}}{n ^{2} π ^{2}} = \frac{4 L ^{2}}{π ^{2}} \sum \frac{1}{n ^{2}} .$
Left side (apne $\frac{1}{L}$ ke saath). $\frac{1}{L} \int_{- L}^{L} x^{2} d x = \frac{1}{L} \cdot \frac{2 L ^{3}}{3} = \frac{2 L ^{2}}{3} .$ Yeh step kyun? Yahan $\frac{1}{L}$ apni zaroorat prove karta hai — iske bina dono sides mismatched powers of $L$ carry karti.
Equate karo. $\frac{2 L ^{2}}{3} = \frac{4 L ^{2}}{π ^{2}} \sum \frac{1}{n ^{2}} \Rightarrow \sum \frac{1}{n ^{2}} = \frac{2 L ^{2}}{3} \cdot \frac{π ^{2}}{4 L ^{2}} = \frac{π ^{2}}{6} .$ Yeh step kyun? Har $L^{2}$ cancel ho jaata hai — ek pure number aapki interval choice par depend nahi kar sakta. Explicitly $L = 2$ plug karo agar chaaho: LHS $= \frac{8}{3}$ , RHS $= \frac{16}{π ^{2}} \cdot \frac{π ^{2}}{6} = \frac{16}{6} = \frac{8}{3}$ . ✓

Verify: $L = 2$ ke saath: $\frac{2 L ^{2}}{3} = 8/3$ ; $\frac{4 L ^{2}}{π ^{2}} \cdot \frac{π ^{2}}{6} = 8/3$ . Equal, $L$ -free result $π^{2} /6$ . ✓

Example 7 — Cell G (limiting / convergence sanity)

Kuch partial sums compute karo. $S_{1} = 4, S_{2} = 4 (1 + \frac{1}{4}) = 5, S_{5} = 4 \sum_{1}^{5} \frac{1}{n ^{2}} \approx 5.854.$ Yeh step kyun? Staircase ko climb karte dekhna "convergence from below" ko abstract ki jagah concrete banata hai.
Bound. Kyunki har added term positive hai, $S_{N}$ increasing hai aur $\frac{2 π ^{2}}{3}$ se upar bounded hai. Yeh step kyun? Yeh exactly Bessel's inequality hai $\sum_{n = 1}^{N} (a_{n}^{2} + b_{n}^{2}) \leq \frac{1}{L} \int f^{2}$ ; Parseval complete basis ke liye $N \to \infty$ ka equality case hai.
Limit. $S_{N} \to 4 \cdot \frac{π ^{2}}{6} = \frac{2 π ^{2}}{3}$ .

Verify: $S_{1} = 4$ , $S_{2} = 5$ , $S_{5} \approx 5.8544$ , sab $< \frac{2 π ^{2}}{3} \approx 6.5797$ , aur $S_{N} \to 6.5797$ . ✓

Example 8 — Cell H (word problem: RMS & power)

Coefficients identify karo. $a_{1} = 2$ , $b_{3} = 6$ , $a_{0} = 0$ , baaki sab $0$ . Yeh step kyun? Amplitudes seedha padhlo — power $a_{n}^{2} + b_{n}^{2}$ mein rehti hai.
Total (Parseval). $\frac{1}{π} \int_{- π}^{π} v^{2} d t = a_{1}^{2} + b_{3}^{2} = 4 + 36 = 40.$ Isliye $\int_{- π}^{π} v^{2} d t = 40 π$ .
RMS. Full period par mean-square: $\overline{v^{2}} = \frac{1}{2 π} (40 π) = 20$ , isliye $V_{RMS} = 20 \approx 4.472 V .$ Yeh step kyun? RMS DC-equivalent heating voltage hai — units volts hain, aur $V^{2} = V$ check out karta hai.
Power fractions. Third harmonic weight $= \frac{b _{3}^{2}}{a _{1}^{2} + b _{3}^{2}} = \frac{36}{40} = 0.9 = 90%$ . Yeh step kyun? Forecast confirm karta hai: amplitude- $6$ harmonic $90%$ power hold karta hai. Fundamental $10%$ hold karta hai.

Verify: $4 + 36 = 40$ ; $\overline{v^{2}} = 20$ ; RMS $= 20 \approx 4.4721$ V; third-harmonic fraction $= 36/40 = 0.9$ . ✓

Example 9 — Cell I (exam twist: missing coefficient solve karo)

Parseval ko ek bookkeeping equation ki tarah likho. $10 = a_{1}^{2} + b_{2}^{2} = 1 + b_{2}^{2} .$ Yeh step kyun? Ulta use kiya gaya: total given hai, isliye Parseval ek solvable equation ban jaata hai ek unknown ke liye.
Solve karo. $b_{2}^{2} = 9 \Rightarrow b_{2} = \pm 3$ ; kyunki bataya gaya hai $b_{2} > 0$ , isliye $b_{2} = 3$ lo. Yeh step kyun? Parseval sirf magnitude (energy) fix karta hai — sign/phase extra information hai jo aapko diya jaana chahiye, yeh Ex 1 ki "phase is lost" wali note echo karta hai.

Verify: $1 + 3^{2} = 1 + 9 = 10$ . ✓ Given energy se exactly match karta hai.

Recall

Recall Kaun sa cell kaun sa hai? (click)

Odd $f$ ::: RHS pure $\sum b_{n}^{2}$ hai (Cell A). Even $f$ ::: RHS $\frac{a _{0}^{2}}{2} + \sum a_{n}^{2}$ hai (Cell B). Constant $f = c$ ::: sirf $\frac{a _{0}^{2}}{2}$ bachta hai, $a_{0} = 2 c$ (Cell D). Finite harmonics ::: Parseval exact Pythagoras hai, koi infinite sum nahi (Cell E). Partial sums vs total ::: woh neeche se total ki taraf badhte hain — Bessel's inequality (Cell G).

Connections

Parent: Parseval's theorem
Basel Problem — Ex 1 result $\sum 1/ n^{2} = π^{2} /6$ .
Bessel's Inequality — Ex 7 ka "approach from below".
RMS and Power Spectra — Ex 5 aur Ex 8.
Orthogonality of functions — cross-terms kyun khatam hote hain (Ex 3).
Fourier Series — coefficients kahan se aate hain.