Full Fourier series — coefficients derivation: Question bank

4.7.5 · D5 · Maths › Partial Differential Equations › Full Fourier series — coefficients derivation

Shuru karne se pehle, teen quick reminders taaki sawaal sahi jagah land karein.

Recall Teen formulas (taaki tum inke khilaf argue kar sako)

$[- L, L]$ par (period $2 L$ ), ek single summation index use karte hue — hum use neeche har jagah $n$ kehte hain confusion se bachne ke liye: $a_{0} = \frac{1}{L} \int_{- L}^{L} f d x$ , aur $n \geq 1$ ke liye, $a_{n} = \frac{1}{L} \int_{- L}^{L} f cos \frac{nπ x}{L} d x$ , $b_{n} = \frac{1}{L} \int_{- L}^{L} f sin \frac{nπ x}{L} d x$ . Series hai $\frac{a _{0}}{2} + n \geq 1 \sum [a_{n} cos \frac{nπ x}{L} + b_{n} sin \frac{nπ x}{L}]$ . (Jab do alag indices ek orthogonality integral ke andar saath aate hain toh hum unhe $m$ aur $n$ kehte hain — lekin jis coefficient ko hum solve karte hain woh hamesha $a_{n}$ ya $b_{n}$ hota hai.)

True ya false — justify karo

Neeche har prompt ek claim hai. True/false decide karo, phir reason do.

Series ka constant term,

f

ka ek period par average value ke barabar hota hai.

True — average hai

\frac{1}{2 L} \int_{- L}^{L} f d x

, aur constant term hai

\frac{a _{0}}{2} = \frac{1}{2 L} \int_{- L}^{L} f d x

; ye literally same number hain.

Agar

f

odd hai, toh har

a_{n}

(including

a_{0}

) zero hota hai.

True —

f

odd hone se

f cos \frac{nπ x}{L}

odd ho jaata hai (even

\times

odd), isliye

0

ke daayein wala graph, left ka negated mirror hota hai, aur dono halve

[- L, L]

par cancel ho jaate hain.

Agar

f

even hai, toh har

b_{n}

zero hota hai.

True —

f

even hone se

f sin \frac{nπ x}{L}

odd ho jaata hai (even

\times

odd), isliye left aur right halve cancel ho jaate hain aur

[- L, L]

par har

b_{n}

integral zero ho jaata hai.

Integers

m, n \geq 1

ke liye,

\int_{- L}^{L} cos \frac{mπ x}{L} cos \frac{nπ x}{L} d x = 0

sabhi aisi pairs ke liye hold karta hai.

False — ye

0

sirf tab hota hai jab

m \neq = n

; jab

m = n (\geq 1)

tab integral

L

ke barabar hota hai, aur special case

m = n = 0

mein

\int_{- L}^{L} 1 \cdot 1 d x = 2 L

milta hai. Wahi surviving value hai jo exactly hume

a_{n}

solve karne deti hai.

Sine–cosine integral

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

sirf tab zero hota hai jab

m \neq = n

False — ye har

m, n

ke liye zero hota hai, including

m = n

, kyunki

sin (\cdot) cos (\cdot)

ek odd function hai jiske dono halve cancel ho jaate hain; ek sine aur ek cosine kabhi overlap nahi karte.

Ek full Fourier series mein dono sine aur cosine terms zaroor honi chahiye.

False — "full" ka matlab hai hum dono ko allow karte hain; agar

f

mein koi symmetry hai, toh ek poori family drop ho jaati hai (ek odd

f

pure sine series ban jaata hai), phir bhi woh full-interval series hi kehlaata hai.

Formula

a_{n} = \frac{1}{L} \int_{- L}^{L} f cos \frac{nπ x}{L} d x

mein

n = 0

plug karne par

a_{0}

sahi milta hai.

True — exactly isliye constant ko

\frac{a _{0}}{2}

likha jaata hai:

n = 0

par,

cos 0 = 1

aur formula

\frac{1}{L} \int_{- L}^{L} f d x = a_{0}

return karta hai.

Do functions ka orthogonal hona matlab unke graphs kabhi cross nahi karte.

False — orthogonal ka matlab hai

\int_{- L}^{L} f g d x = 0

, yaani product ke neeche signed area zero hai; graphs kai baar cross kar sakte hain aur karte bhi hain jabki axis ke upar ka area neeche wale ko cancel karta hai.

f

cos \frac{nπ x}{L}

se multiply karke integrate karna

a_{n}

ko isolate karta hai kyunki yeh

f

ko us ek wave par "project" karta hai.

True —

\int f g

ko dot product ki tarah sochte hue, yeh ek perpendicular axis par projection hai; har doosri basis wave zero signed area contribute karti hai, sirf

a_{n} L

term bachta hai.

Basis function

f (x) = cos \frac{3 π x}{L}

ki Fourier series mein infinitely many nonzero coefficients hote hain.

False — orthogonality

a_{3} = 1

ke alaawa sab kuch khatam kar deta hai; ek basis function ki apni series bas woh khud hi hoti hai.

Error dhundo

Har line mein ek mathematical ya factual mistake hai. Batao kya galat hai aur kya hona chahiye.

[- L, L]

par,

b_{n} = \frac{2}{L} \int_{- L}^{L} f sin \frac{nπ x}{L} d x

Galat prefactor: full-interval coefficient mein

\frac{1}{L}

hota hai,

\frac{2}{L}

nahi.

\frac{2}{L}

sirf

[0, L]

par half-range series mein aata hai, jahan tum half period par integrate karte ho.

"Kyunki

f

ka average

\frac{1}{2 L} \int f

hai, isliye

a_{0} = \frac{1}{2 L} \int_{- L}^{L} f d x

\frac{1}{2}

double count ho gaya: average

\frac{a _{0}}{2}

ke barabar hota hai, isliye

a_{0} = \frac{1}{L} \int_{- L}^{L} f d x

. Woh

\frac{1}{2}

pehle se constant term ke andar hai.

\int_{- L}^{L} cos \frac{nπ x}{L} d x = 2 L

har

n

ke liye."

Sirf

n = 0

ke liye true hai;

n \geq 1

ke liye integral

0

hai kyunki ek cosine ka poora period axis ke upar aur neeche equal area rakhta hai. Nonzero case constant mein hai, oscillating mein nahi.

a_{n}

derive karne ke liye, series ko term by term integrate karo — yeh hamesha allowed hai."

Is step mein assume kiya gaya hai ki term-by-term integration valid hai, jiske liye

f

ka kaafi accha hona zaroori hai (piecewise smooth); yeh ek assumption hai, free lunch nahi, aur pathological functions ke liye fail ho sakta hai.

cos mπ = 1

sabhi integers

m

ke liye."

Galat:

cos mπ = (- 1)^{m}

, isliye yeh

- 1, 1, - 1, \dots

alternate karta hai. Yahi sign exactly woh hai jo square wave ke even-indexed

b_{n}

ko vanish karata hai.

"Square wave ke liye,

a_{0} = \frac{1}{L} [(- 1) L + (1) L]

se nonzero constant milta hai."

Bracket

- L + L = 0

hai, isliye

a_{0} = 0

; arithmetic (aur odd symmetry) dono ise zero force karte hain.

"Kyunki

sin \frac{mπ x}{L} cos \frac{nπ x}{L}

positive ho sakta hai, isliye

[- L, L]

par iska integral positive hai."

Nahi — graph ke kuch hisse ka pointwise positive hona signed area ke baare mein kuch nahi kehta; product odd hai, uske dono halve exactly cancel ho jaate hain, isliye integral

0

hai.

Why wale sawaal

Sirf "kya" nahi, "kyun" answer karo.

Hum

[- L, L]

par poore period par integrate kyun karte hain, iske kuch hisse par nahi?

Ek poore period mein hi oscillating waves ka axis ke upar aur neeche equal area hota hai (isliye unka average zero hota hai) aur orthogonality cancellations hold hoti hain; partial interval par woh cancellations fail ho jaati hain aur projection trick toot jaati hai.

Orthogonality prove karne mein product-to-sum identity kyun use hoti hai?

Yeh do cosines ke product ko single cosines ki sum mein convert kar deti hai, jinka ek full period par signed area trivially

0

(ya matched case mein

L

) hota hai — raw product seedha integrate karna mushkil hai.

Jab hum

cos \frac{nπ x}{L}

se multiply karke integrate karte hain toh exactly ek term kyun bachti hai?

Kyunki

\int cos \frac{mπ x}{L} cos \frac{nπ x}{L} d x

m \neq = n

ke liye

0

aur

m = n

ke liye

L

hota hai; poori sum sirf

a_{n} L

term tak collapse ho jaati hai.

Constant ko sirf

a_{0}

ki jagah

\frac{a _{0}}{2}

kyun likhte hain?

Pure bookkeeping: isse single formula

a_{n} = \frac{1}{L} \int f cos \frac{nπ x}{L} d x

sabhi

n \geq 0

ke liye valid ho jaata hai, constant ko cosine coefficients ke saath unify karta hai.

Do alag functions ke

[- L, L]

par same Fourier coefficients kyun ho sakte hain?

Agar woh sirf isolated points par alag hain (jaise kisi jump par) toh coefficients define karne wale integrals unchanged rehte hain, isliye series unhe alag nahi kar sakti — woh values ko averaged (mean-square) sense mein match karti hai, yahi Parseval's theorem ke peeche ka idea hai.

Symmetry (

f

even ya odd) kaam kyun aadha kar deti hai?

Ek odd

f

har

a_{n}

ko aur ek even

f

har

b_{n}

ko kisi bhi integration se pehle khatam kar deta hai, kyunki relevant product odd hoti hai aur uske dono halve cancel ho jaate hain — isliye tum sirf ek family of coefficients compute karte ho.

Period

2 L

ke liye

cos \frac{nπ x}{L}

aur

sin \frac{nπ x}{L}

sahi building blocks kyun hain?

Har ek

2 L

length par poore number of cycles complete karta hai, isliye har term — aur unka koi bhi sum — har

2 L

pe repeat karta hai,

f

ki periodicity se match karta hai.

Complex form

\sum c_{n} e^{inπ x / L}

, real sine–cosine form ke equivalent kyun hai?

Euler's formula har exponential ko cosine plus

i

times sine mein split karta hai, isliye complex coefficients sirf

a_{n}, b_{n}

ko repackage karte hain; dono same function describe karte hain (dekho Complex (exponential) Fourier series).

Edge cases

Boundary aur degenerate situations jinhe formulas chupke handle karte hain.

Jump discontinuity par, series kis value par converge karti hai?

Left aur right limits ka midpoint,

\frac{f ( x ^{-} ) + f ( x ^{+} )}{2}

, difference ko split karta hai, chahe jump par

f

ki koi bhi value defined ho.

Jump ke paas, partial sums ek fixed amount se overshoot karte hain jo kabhi nahi shrinkta — ise kya kehte hain?

Gibbs phenomenon: zyada terms add karne se overshoot ki width kam hoti hai lekin uski height jump ka lagbhag

9%

rehti hai, isliye "ringing" spike kabhi disappear nahi hoti — yeh classic trap hai jab students har jagah perfect convergence expect karte hain.

Zero function

f (x) = 0

ki Fourier series kya hai?

Har coefficient zero hai, isliye series bas

0

hai; project karne ke liye kuch nahi hai aur koi wave present nahi hai.

Constant function

f (x) = c

ke liye, kaun se coefficients bachte hain?

Sirf

a_{0} = \frac{1}{L} \int_{- L}^{L} c d x = 2 c

, jisse constant term

\frac{a _{0}}{2} = c

milta hai; saare

a_{n}

(

n \geq 1

) aur

b_{n}

vanish ho jaate hain kyunki

c \times

(oscillation) ka axis ke upar aur neeche equal area hota hai.

n = 0

plug karne par

b_{n}

formula ke saath kya hota hai?

sin 0 = 0

identically hota hai, isliye

b_{0} = 0

trivially ho jaata hai — koi sine constant term nahi hoti, isliye sine sum

n = 1

se shuru hoti hai.

Endpoints

x = \pm L

par series ki value

f (\pm L)

ke barabar hoti hai?

Generally nahi — periodic extension

x = L

aur

x = - L

ko tiled line par same point bana deta hai, isliye series

f (- L^{+})

aur

f (L^{-})

ke midpoint par converge karti hai, yani "wrap-around" jump.

Agar

f

na even hai na odd, toh kya sine aur cosine dono families guaranteed nonzero hain?

Nonzero guaranteed nahi, lekin koi bhi family vanish hone par forced nahi hai; tumhe dono compute karne padenge kyunki koi symmetry tumhare liye kaam nahi karti.

Agar

f

continuous aur

2 L

-periodic hai aur uski continuous derivative hai, toh kya hoga?

Tab series har jagah

f

par converge karti hai (uniformly), yeh sabse clean case hai — koi jump-midpoint nahi aur koi Gibbs overshoot nahi, kyunki koi jump hai hi nahi.