Full Fourier series — coefficients derivation

4.7.5 · Maths › Partial Differential Equations

Hum kya karne ki koshish kar rahe hain

KYA chahiye: $a_0,a_n,b_n$ ke liye formulas. KYUN yeh basis functions: $\cos \frac{n\pi x}{L}$ aur $\sin \frac{n\pi x}{L}$ sabke period $2L$ ko divide karte hain, isliye unka koi bhi sum $2L$-periodic hoga — $f$ se match karta hua.

Key tool: Orthogonality (pehle derive karo)

Sab kuch in integrals par depend karta hai ek full period $[-L,L]$ par. Maano $m,n$ positive integers hain.

Ek ko derive kaise karein (cosine–cosine case). Product-to-sum identity use karo $\cos A\cos B=\frac{1}{2}[\cos (A-B)+\cos (A+B)].$ Toh $A=\frac{m\pi x}{L},B=\frac{n\pi x}{L}$ ke saath: ${\int }_{-L}^L\cos \frac{m\pi x}{L}\cos \frac{n\pi x}{L}dx=\frac{1}{2}{\int }_{-L}^L\cos \frac{(m-n)\pi x}{L}dx+\frac{1}{2}{\int }_{-L}^L\cos \frac{(m+n)\pi x}{L}dx.$

Yeh step kyun? Product ko do simple cosines mein split karne se har integral trivial ho jaata hai.

• Agar $m\ne n$: dono $m-n$ aur $m+n$ nonzero integers $k$ hain, aur ${\int }_{-L}^L\cos \frac{k\pi x}{L}dx=\frac{L}{k\pi }\sin \frac{k\pi x}{L}{|}_{-L}^L=\frac{L}{k\pi }[\sin k\pi -\sin (-k\pi )]=0.$ Toh poori cheez $0$ hai.
• Agar $m=n$: pehla integrand $\cos 0=1$ hai jo $\frac{1}{2}\cdot 2L=L$ deta hai, aur doosra ($k=2n\ne 0$) $0$ deta hai. Total $=L$. ✔

Sine–sine aur sine–cosine cases bilkul usi tarah follow karte hain using $\sin A\sin B=\frac{1}{2}[\cos (A-B)-\cos (A+B)]$ aur $\sin A\cos B=\frac{1}{2}[\sin (A+B)+\sin (A-B)]$.

Coefficient formulas derive karna (scratch se)

Shuru karo yeh assume karke ki series $f$ ki taraf converge karti hai aur hum term by term integrate kar sakte hain: $f(x)=\frac{a_0}{2}+{\sum }_{n=1}^{\infty }\left[a_n\cos \frac{n\pi x}{L}+b_n\sin \frac{n\pi x}{L}\right].$

Step 1 — $a_0$ nikalo

Dono sides ko $[-L,L]$ par integrate karo: ${\int }_{-L}^{L}fdx=\frac{a_0}{2}\underset{2L}{\underbrace{{\int }_{-L}^{L}1dx}}+{\sum }_n{a}_n\underset{0}{\underbrace{{\int }_{-L}^L\cos \frac{n\pi x}{L}dx}}+{\sum }_n{b}_n\underset{0}{\underbrace{{\int }_{-L}^L\sin \frac{n\pi x}{L}dx}}.$ Yeh step kyun? Integrate karne se har oscillating term khatam ho jaata hai (unka ek full period par zero average hota hai), sirf constant bachta hai. $\Rightarrow {\int }_{-L}^{L}fdx=a_{0}L⟹\boxed{a_0=\frac{1}{L}{\int }_{-L}^{L}f(x)dx.}$

Step 2 — $a_m$ nikalo ($m\ge 1$ ke liye)

Dono sides ko $\cos \frac{m\pi x}{L}$ se multiply karo, phir integrate karo: ${\int }_{-L}^{L}f\cos \frac{m\pi x}{L}dx=\frac{a_0}{2}\underset{0}{\underbrace{\int \cos \frac{m\pi x}{L}dx}}+{\sum }_n{a}_n\int \cos \frac{n\pi x}{L}\cos \frac{m\pi x}{L}dx+{\sum }_n{b}_n\underset{0}{\underbrace{\int \sin \frac{n\pi x}{L}\cos \frac{m\pi x}{L}dx}}.$ Yeh step kyun? Orthogonality se, $a_n$ sum mein sirf $n=m$ wala term bachta hai (equals $L$); baaqi sab zero ho jaate hain, aur har $b_n$ term bhi vanish ho jaata hai. $\Rightarrow {\int }_{-L}^{L}f\cos \frac{m\pi x}{L}dx=a_{m}L⟹\boxed{a_m=\frac{1}{L}{\int }_{-L}^{L}f(x)\cos \frac{m\pi x}{L}dx.}$

Step 3 — $b_m$ nikalo ($m\ge 1$ ke liye)

$\sin \frac{m\pi x}{L}$ se multiply karo aur integrate karo. Same reasoning: sirf $n=m$ wala sine–sine term bachta hai ($=L$): $\boxed{b_m=\frac{1}{L}{\int }_{-L}^{L}f(x)\sin \frac{m\pi x}{L}dx.}$

Worked examples

Common mistakes

Active recall

Connections

• Orthogonality of functions — woh inner-product viewpoint jo projection ko kaam karaata hai.
• Half-range Fourier series (sine and cosine) — $[0,L]$ par $\frac{2}{L}$ factor ke saath special cases.
• Even and odd functions — coefficients ke liye symmetry shortcuts.
• Convergence of Fourier series (Dirichlet conditions) — kab series actually $f$ ke barabar hoti hai.
• Complex (exponential) Fourier series — $a_n,b_n$ ko $c_n=\frac{1}{2L}\int f{e}^{-in\pi x/L}dx$ ke roop mein repackage karna.
• Separation of variables for the heat equation — jahan Fourier coefficients initial condition fix karte hain.
• Parseval's theorem — energy relation $\frac{1}{L}\int f^2=\frac{a_0^2}{2}+\sum (a_n^2+b_n^2)$.

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