Fourier series — motivation from periodic functions

4.7.3 · Maths › Partial Differential Equations

HUM YEH CHAHTE HI KYUN HAIN?

Periodic function KIYA HOTA HAI?

Sabse simple periodic functions $\cos (n\omega x)$ aur $\sin (n\omega x)$ hain. Ye sab period $T$ share karte hain (ye ek period ke andar pure $n$ cycles fit karte hain). Isliye ye period $T$ ki kisi bhi cheez ke liye perfect Lego bricks hain.

SERIES KAISE BANATE HAIN? (Derivation from scratch)

Hum claim karte hain ki period $2L$ ki ek periodic function $f$ (to $\omega =\pi /L$) ko likha ja sakta hai:

$f(x)=\frac{a_0}{2}+{\sum }_{n=1}^{\infty }(a_n\cos \frac{n\pi x}{L}+b_n\sin \frac{n\pi x}{L})$

Poora problem yeh hai: $a_n,b_n$ numbers kya hain? Hum inhe derive karte hain, yaad nahi karte.

Step 1 — Orthogonality trick (WHY it works)

Jinke integrals chahiye (sab $-L$ se $L$ tak):

${\int }_{-L}^L\cos \frac{n\pi x}{L}\cos \frac{m\pi x}{L}dx=\{\begin{array}{ll}0& n\ne m\\ L& n=m\ne 0\\ 2L& n=m=0\end{array}$

${\int }_{-L}^L\sin \frac{n\pi x}{L}\sin \frac{m\pi x}{L}dx=\{\begin{array}{ll}0& n\ne m\\ L& n=m\end{array}$

${\int }_{-L}^L\sin \frac{n\pi x}{L}\cos \frac{m\pi x}{L}dx=0\text{(always)}$

Ye sach kyun hain? Product-to-sum use karo, jaise $\cos A\cos B=\frac{1}{2}[\cos (A-B)+\cos (A+B)]$. Har piece $\cos (k\pi x/L)$ hota hai integer $k$ ke saath; $[-L,L]$ par ek poori number of cycles integrate karne se $0$ milta hai, siwaaye jab $k=0$ (ek constant) ho, jo survive karta hai.

Step 2 — $a_m$ nikaalte hain (HOW)

Series ko $\cos (m\pi x/L)$ se multiply karo aur $-L$ se $L$ tak integrate karo:

${\int }_{-L}^{L}f(x)\cos \frac{m\pi x}{L}dx=\underset{=0(m\ge 1)}{\underbrace{\frac{a_0}{2}\int \cos \frac{m\pi x}{L}dx}}+{\sum }_n{a}_n\underset{=L\text{only if }n=m}{\underbrace{\int \cos \frac{n\pi x}{L}\cos \frac{m\pi x}{L}dx}}+{\sum }_n{b}_n\underset{=0}{\underbrace{(\dots )}}$

Yeh step kyun? Orthogonality har term ko khatam kar deti hai siwaaye us ek $n=m$ cosine term ke, jo $a_m\cdot L$ bachata hai. Isliye:

$\boxed{a_m=\frac{1}{L}{\int }_{-L}^{L}f(x)\cos \frac{m\pi x}{L}dx}$

$a_0/2$ ki choice se $a_0=\frac{1}{L}{\int }_{-L}^{L}fdx$ = average se double, isliye $a_0/2$ = average value. Neat.

Step 3 — $b_m$ nikaalte hain

Wahi trick $\sin (m\pi x/L)$ ke saath:

$\boxed{b_m=\frac{1}{L}{\int }_{-L}^{L}f(x)\sin \frac{m\pi x}{L}dx}$

Symmetry shortcut (80/20 ka fayda)

Worked Examples

Forecast-then-Verify

Common Mistakes (Steel-manned)

Flashcards

Connections

• Separation of Variables — woh sine/cosine modes produce karta hai jo Fourier series assemble karti hai.
• Heat Equation — initial temperature profile ko Fourier (sine) series ke roop mein expand kiya jaata hai.
• Wave Equation — plucked-string ki shape harmonics mein decompose hoti hai.
• Orthogonality of Functions — inner-product structure $⟨f,g⟩=\int fg$.
• Even and Odd Functions — coefficients ke liye symmetry shortcuts.
• Fourier Transform — Fourier series ki limit jab period $\to \infty$ ho.
• Gibbs Phenomenon — jumps par convergence behaviour.

Concept Map