Parseval's theorem: Exercises — Hinglish deep dive

4.7.7 · D4 · Maths › Partial Differential Equations › Parseval's theorem

Level 1 — Recognition

Goal: series padho aur coefficients sahi-sahi identify karo.

Exercise 1.1

$(- π, π)$ pe ek function ka Fourier series hai $f (x) = 2 + 3 cos x - cos 2 x + 4 sin x$ . $a_{0}, a_{1}, a_{2}, b_{1}$ , aur baaki sab likhke dikhao.

Recall Solution

General series hai $f (x) = \frac{a _{0}}{2} + \sum (a_{n} cos n x + b_{n} sin n x)$ (yahan $L = π$ ). Constant ka matlab: akela number jo dikhta hai woh $\frac{a _{0}}{2}$ hai, $a_{0}$ nahi. Isliye $\frac{a _{0}}{2} = 2 \Rightarrow a_{0} = 4.$ Term by term match karo:

$cos x$ ka coefficient $3 \Rightarrow a_{1} = 3$ .
$cos 2 x$ ka coefficient $- 1 \Rightarrow a_{2} = - 1$ .
$sin x$ ka coefficient $4 \Rightarrow b_{1} = 4$ .
Baaki sab $a_{n} = b_{n} = 0$ .

Factor-of-2 trap kyun important hai: constant term ek built-in $\frac{1}{2}$ ke saath likha jaata hai taaki coefficient formulas uniform rahein. Yahan $a_{0} = 2$ padhna aage ke har Parseval calculation ko kharaab kar dega.

Exercise 1.2

Usi $f$ ke liye, Parseval ka left-hand side kaunsi ek quantity compute karta hai? Use words mein batao.

Recall Solution

Left side hai $\frac{1}{L} \int_{- L}^{L} [f (x)]^{2} d x = \frac{1}{π} \int_{- π}^{π} f^{2} d x$ . Words mein: $f$ ka twice the mean-square value ek period ke upar. (Isse 2 se divide karo toh mean square milta hai; uska square root hai RMS value.) Yeh signal ki total "energy per unit length" measure karta hai, frequencies mein split karne se pehle.

Level 2 — Application

Goal: Parseval mein numbers daalo aur value nikalo.

Exercise 2.1

$f (x) = 2 + 3 cos x - cos 2 x + 4 sin x$ on $(- π, π)$ ke liye, Parseval use karke $\frac{1}{π} \int_{- π}^{π} f^{2} d x$ compute karo.

Recall Solution

Coefficients (Ex 1.1 se): $a_{0} = 4, a_{1} = 3, a_{2} = - 1, b_{1} = 4$ . $\frac{1}{π} \int_{- π}^{π} f^{2} d x = \frac{a _{0}^{2}}{2} + (a_{1}^{2} + b_{1}^{2}) + (a_{2}^{2} + b_{2}^{2})$ $= \frac{16}{2} + (9 + 16) + (1 + 0) = 8 + 25 + 1 = 34.$ Toh $\int_{- π}^{π} f^{2} d x = 34 π .$ Yeh integrate karne se kyun aasaan hai: $f$ ko directly square karo toh 16 cross terms bante hain; orthogonality guarantee karta hai ki saare cross terms integrate ho ke zero ho jaate hain, toh unhe kabhi chhuna nahi padta.

Exercise 2.2

$(- π, π)$ pe ek signal mein sirf $a_{0} = 0, a_{3} = 6, b_{5} = 8$ hain. Uski RMS value nikalo.

Recall Solution

Parseval: $\frac{1}{π} \int_{- π}^{π} f^{2} d t = a_{3}^{2} + b_{5}^{2} = 36 + 64 = 100.$ Toh $\int_{- π}^{π} f^{2} d t = 100 π$ . Mean square = $\frac{1}{2 π} \int f^{2} = \frac{100 π}{2 π} = 50$ . RMS = $50 = 52 \approx 7.07$ . Mean ke liye $2 π$ se kyun divide karte hain: full period ki length $2 π$ hai; poore period pe squares ka average karo toh mean square milta hai, jiska square root root-mean-square hota hai.

Exercise 2.3

$f (x) = x$ on $(- π, π)$ ke liye, jahan $b_{n} = \frac{2 ( - 1 ) ^{n + 1}}{n}$ hai, verify karo ki Parseval se $n = 1 \sum \infty \frac{1}{n ^{2}} = \frac{π ^{2}}{6}$ milta hai.

Recall Solution

Left: $\frac{1}{π} \int_{- π}^{π} x^{2} d x = \frac{1}{π} \cdot \frac{2 π ^{3}}{3} = \frac{2 π ^{2}}{3} .$ Right: $\sum b_{n}^{2} = \sum \frac{4}{n ^{2}} = 4 \sum \frac{1}{n ^{2}}$ (sign $(- 1)^{n + 1}$ square hoke $1$ ban jaata hai). Equate karo: $\frac{2 π ^{2}}{3} = 4 \sum \frac{1}{n ^{2}} \Rightarrow \sum \frac{1}{n ^{2}} = \frac{π ^{2}}{6} .$ Yeh Basel Problem hai.

Level 3 — Analysis

Goal: reason karo ki kaunse terms survive karte hain aur kyun.

Exercise 3.1

Bina koi integral compute kiye, explain karo ki $\int_{- π}^{π} (3 cos x) (4 sin x) d x = 0$ kyun hai, aur yeh fact Parseval ko possible kyun banata hai.

Recall Solution

$cos x$ aur $sin x$ ek full period pe orthogonal hain: $\int_{- π}^{π} cos \frac{mπ x}{L} sin \frac{nπ x}{L} d x = 0 (always) .$ Geometrically iska matlab: har basis function ko ek infinite-dimensional space mein perpendicular axis ki tarah socho (figure dekho). Do alag axes ka product integrate ho ke zero hota hai, bilkul jaise perpendicular vectors ka dot product: $\overset{x}{^} \cdot \overset{y}{^} = 0$ . Parseval ko iska kyun zaroorat hai: $(f)^{2}$ expand karne pe har pairwise product aata hai; orthogonality saare cross products ko zero kar deta hai, sirf diagonal squared terms bache rehte hain. Parseval literally is space mein Pythagoras hai.

Exercise 3.2

Jab hum $\int_{- π}^{π} cos^{2} n x d x$ integrate karte hain toh $π$ milta hai, lekin $\int_{- π}^{π} 1 d x = 2 π$ . Explain karo ki yeh do alag values $\frac{a _{0}^{2}}{2}$ vs $a_{n}^{2}$ wali asymmetry kaise produce karti hain.

Recall Solution

Series ka DC piece hai $\frac{a _{0}}{2}$ . Uska squared integral: $\int_{- π}^{π} (\frac{a _{0}}{2})^{2} d x = \frac{a _{0}^{2}}{4} \cdot 2 π = \frac{a _{0}^{2} π}{2} .$ Ek cosine piece: $\int_{- π}^{π} (a_{n} cos n x)^{2} d x = a_{n}^{2} \cdot π .$ Sab ko $L = π$ se divide karo: $DC \to \frac{a _{0}^{2}}{2}, harmonic \to a_{n}^{2} .$ DC term alag kyun hai: constant "1" ka period-length norm $2 π$ hai (double), lekin $\frac{a _{0}}{2}$ mein $\frac{1}{2}$ square ho ke $\frac{1}{4}$ ban jaata hai (quarter). $2 π \times \frac{1}{4} = \frac{π}{2}$ per unit $a_{0}^{2}$ , jabki harmonics ke liye $π \times 1$ — yahi woh akela factor $\frac{1}{2}$ chhod jaata hai.

Exercise 3.3

Ek student claim karta hai: "Agar do functions ke Fourier coefficients ki magnitudes same hain lekin signs alag hain, toh unki energy alag hogi." Sach hai ya jhooth — justify karo.

Recall Solution

Jhooth. Parseval sirf $a_{n}^{2}$ aur $b_{n}^{2}$ pe depend karta hai. Kyunki $(+ c)^{2} = (- c)^{2}$ , koi bhi sign flip karne se har squared term unchanged rehta hai, toh total energy $\frac{1}{L} \int f^{2}$ same hi rahega. Kya badalta hai aur kya nahi: shape (waveform) badal jaata hai kyunki relative phases shift hote hain, lekin energy phase-blind hai. Yahi wajah hai ki Ex 2.3 mein $b_{n} = \frac{2 ( - 1 ) ^{n + 1}}{n}$ se $\sum 4/ n^{2}$ mila jismein alternating sign khatam ho gayi.

Level 4 — Synthesis

Goal: coefficient-finding ko Parseval ke saath combine karo aur ek naya result derive karo.

Exercise 4.1

$f (x) = x^{2}$ on $(- π, π)$ use karo (jahan $a_{0} = \frac{2 π ^{2}}{3}$ , $a_{n} = \frac{4 ( - 1 ) ^{n}}{n ^{2}}$ , $b_{n} = 0$ ) aur prove karo ki $n = 1 \sum \infty \frac{1}{n ^{4}} = \frac{π ^{4}}{90}$ .

Recall Solution

Left: $\frac{1}{π} \int_{- π}^{π} x^{4} d x = \frac{1}{π} \cdot \frac{2 π ^{5}}{5} = \frac{2 π ^{4}}{5} .$ Right: $\frac{a _{0}^{2}}{2} + \sum a_{n}^{2} = \frac{1}{2} (\frac{2 π ^{2}}{3})^{2} + \sum \frac{16}{n ^{4}} = \frac{2 π ^{4}}{9} + 16 \sum \frac{1}{n ^{4}} .$ Equate karo aur isolate karo: $\frac{2 π ^{4}}{5} - \frac{2 π ^{4}}{9} = 2 π^{4} \cdot \frac{9 - 5}{45} = \frac{8 π ^{4}}{45} = 16 \sum \frac{1}{n ^{4}} .$ $\sum \frac{1}{n ^{4}} = \frac{8 π ^{4}}{45 \cdot 16} = \frac{π ^{4}}{90} . ■$ Higher power → higher zeta kyun: $x$ ki har power coefficients ke denominators ko upar push karti hai (yahan $1/ n^{2}$ ), aur Parseval mein squaring $1/ n^{2} \to 1/ n^{4}$ kar deta hai. Yeh $ζ (4)$ hai.

Exercise 4.2

$f (x) = x$ on $(- π, π)$ lo lekin ab sum $n = 1 \sum \infty \frac{1}{( 2 n - 1 ) ^{2}}$ (sirf odd terms) nikalna hai. $\sum \frac{1}{n ^{2}} = \frac{π ^{2}}{6}$ se uski value derive karo.

Recall Solution

Full sum ko odd aur even indices mein split karo: $\sum_{n = 1}^{\infty} \frac{1}{n ^{2}} = S_{odd} odd \sum \frac{1}{n ^{2}} + S_{even} even \sum \frac{1}{n ^{2}} .$ Even terms: $n = 2 k$ , toh $S_{even} = k = 1 \sum \infty \frac{1}{( 2 k ) ^{2}} = \frac{1}{4} \sum \frac{1}{k ^{2}} = \frac{1}{4} \cdot \frac{π ^{2}}{6} = \frac{π ^{2}}{24} .$ Isliye $S_{odd} = \frac{π ^{2}}{6} - \frac{π ^{2}}{24} = \frac{4 π ^{2} - π ^{2}}{24} = \frac{3 π ^{2}}{24} = \frac{π ^{2}}{8} .$ Yeh trick kyun kaam karti hai: even-indexed terms poore sum ki ek scaled copy hain (factor $\frac{1}{4}$ ), toh subtract karne se bina koi naya integral kiye odd part mil jaata hai.

Exercise 4.3

$(- π, π)$ pe ek aisa signal design karo jiska total Parseval energy $\frac{1}{π} \int f^{2}$ exactly $10$ ho, sirf $a_{0}$ aur $b_{1}$ use karke. Ek valid choice do.

Recall Solution

Constraint: $\frac{a _{0}^{2}}{2} + b_{1}^{2} = 10.$ Ek clean choice: maano $a_{0} = 2 \Rightarrow \frac{4}{2} = 2$ , phir chahiye $b_{1}^{2} = 8 \Rightarrow b_{1} = 22 .$ Check: $\frac{2 ^{2}}{2} + (22)^{2} = 2 + 8 = 10.$ ✓ Toh $f (x) = 1 + 22 sin x$ kaam karta hai (yaad raho constant $\frac{a _{0}}{2} = 1$ hai). Infinitely many answers kyun exist karte hain: hamare paas do free knobs hain aur ek equation — $(a_{0}, b_{1})$ space mein solutions ka ek circle. Uska koi bhi point valid hai; energy ek level curve hai.

Level 5 — Mastery

Goal: limits, degenerate cases, aur cross-theorem links mein jaao.

Exercise 5.1

Dikhao ki kisi bhi valid Fourier series wale $f$ ke liye, partial-energy sum $E_{N} = \frac{a _{0}^{2}}{2} + \sum_{n = 1}^{N} (a_{n}^{2} + b_{n}^{2})$ har finite $N$ ke liye $E_{N} \leq \frac{1}{L} \int_{- L}^{L} f^{2} d x$ satisfy karta hai, aur equality sirf $N \to \infty$ pe hoti hai. Is mein involved do theorems ke naam batao.

Recall Solution

Maano $S_{N}$ series ka $N$ -term partial sum hai aur $r_{N} = f - S_{N}$ remainder hai. Kyunki basis orthogonal hai, $\int f^{2} = \int S_{N}^{2} + \int r_{N}^{2}$ (cross term zero ho jaata hai — $r_{N}$ har rakhe gaye mode ke liye orthogonal hai). $L$ se divide karo: $\frac{1}{L} \int f^{2} = E_{N} + \geq 0 \frac{1}{L} \int r_{N}^{2} .$ Kyunki $\frac{1}{L} \int r_{N}^{2} \geq 0$ hamesha, $E_{N} \leq \frac{1}{L} \int_{- L}^{L} f^{2} d x .$ Yeh inequality Bessel's Inequality hai. Jab $N \to \infty$ tab remainder energy $\to 0$ (sine–cosine basis ki completeness), aur equality mil jaati hai — woh equality hi Parseval's theorem khud hai. Yeh kaisa dikhta hai: har naya harmonic total energy bar ka ek slice bharta hai; bar sirf true total ki taraf badh sakta hai, kabhi overshoot nahi kar sakta (figure dekho).

Exercise 5.2 (Degenerate case)

Zero function $f \equiv 0$ ke liye aur pure constant $f \equiv c$ ke liye Parseval kya kehta hai? Dono consistent hain yeh confirm karo.

Recall Solution

Zero function: saare coefficients $0$ . Left $= \frac{1}{L} \int 0 d x = 0$ ; right $= 0$ . Consistent — na energy andar, na bahar. Constant $f = c$ : sirf DC term, $\frac{a _{0}}{2} = c \Rightarrow a_{0} = 2 c$ ; saare $a_{n}, b_{n} = 0$ . Left: $\frac{1}{L} \int_{- L}^{L} c^{2} d x = \frac{1}{L} \cdot c^{2} \cdot 2 L = 2 c^{2} .$ Right: $\frac{a _{0}^{2}}{2} = \frac{( 2 c ) ^{2}}{2} = \frac{4 c ^{2}}{2} = 2 c^{2} .$ ✓ Degenerates kyun test karein: yeh sabse sasta sanity check hai — agar $a_{0} /2$ ki bookkeeping galat hoti, toh constant case immediately disagree kar deta.

Exercise 5.3 (Synthesis with power spectra)

$(- π, π)$ pe ek current $i (t)$ ke harmonics hain $a_{1} = 5, b_{1} = 0, a_{3} = 0, b_{3} = 12$ . Nikalo (a) RMS current aur (b) total power mein 3rd harmonic ka fraction.

Recall Solution

Total: $\frac{1}{π} \int i^{2} d t = a_{1}^{2} + b_{3}^{2} = 25 + 144 = 169.$ Toh $\int i^{2} d t = 169 π .$ (a) RMS: mean square $= \frac{1}{2 π} \int i^{2} = \frac{169 π}{2 π} = 84.5$ ; $RMS = 84.5 \approx 9.192.$ (b) Power fractions: $n$ -th harmonic mein power $\propto a_{n}^{2} + b_{n}^{2}$ . $3rd harmonic fraction = \frac{b _{3}^{2}}{a _{1}^{2} + b _{3}^{2}} = \frac{144}{169} \approx 0.852 (85.2%) .$ Yeh RMS and Power Spectra ka poora point kyun hai: Parseval ek engineer ko total power ko frequency bins pe sum ke roop mein padhne deta hai, aur instantly dikhata hai kaunsa harmonic dominate kar raha hai — yahan 3rd $85%$ power carry karta hai.

Recall Ladder

Recall Kaun se level ne kya test kiya? (click karo)

L1 ::: coefficients padhna, $\frac{a _{0}}{2}$ identify karna.
L2 ::: Parseval mein plug karna, RMS values.
L3 ::: cross terms kyun khatam hote hain, $\frac{1}{2}$ asymmetry, phase-blindness.
L4 ::: $ζ (4) = \frac{π ^{4}}{90}$ derive karna, sums split karna, energy design karna.
L5 ::: Bessel→Parseval limit, degenerate cases, power-spectrum fractions.

Connections

Parseval's theorem — woh parent jo yeh drill karta hai.
Fourier Series · Orthogonality of functions — machinery.
Basel Problem · Bessel's Inequality · Plancherel Theorem · RMS and Power Spectra — jahan yeh results aage jaate hain.