Exercises — Half-range sine and cosine series

Answer: half-range sine series. Sine series ke building blocks hain $\sin \frac{n\pi x}{L}$. Neeche wali figure dekho (Figure s01 — teen sine curves $\sin x,\sin 2x,\sin 3x$ black mein $[0,\pi ]$ par draw ki gayi hain, red dots se dono endpoints mark kiye gaye hain jahan har curve zero cross karti hai): har aisi curve $x=0$ par zero par pin hoti hai (kyunki $\sin 0=0$) aur $x=L$ par (kyunki $\sin n\pi =0$).

Kyunki har term dono ends par automatically zero ho jaati hai, poora sum automatically $u=0$ dono walls par satisfy karta hai — exactly Dirichlet condition. Cosine series generally wahan non-zero hoti hai, isliye woh boundary rule nahi maan sakti. Dekho Heat Equation — separation of variables.

Answer: half-range cosine series. Blocks $\cos \frac{n\pi x}{L}$ ke dono ends par zero slope hota hai: unka derivative $-\frac{n\pi }{L}\sin \frac{n\pi x}{L}$ $x=0$ aur $x=L$ par zero hota hai (wahi $\sin n\pi =0$ wali wajah). Zero slope = koi heat flux nahi = insulated. Yeh Neumann Boundary Conditions pairing hai. Sine blocks ke ends par maximum slope hota hai, isliye woh heat ko force karenge bahar — galat hai.

False. $[0,L]$ par dono $f$ ko reproduce karti hain. Lekin bahar yeh alag-alag inventions hain:

• Sine series odd extension hai (origin se flip karo), period $2L$.
• Cosine series even extension hai ($y$-axis ke across mirror), period $2L$. $[0,L]$ par same data, $[-L,0]$ par do alag-alag fantasies. Dekho Even and Odd Functions.

KYA: $b_n=\frac{2}{\pi }{\int }_0^{\pi }1\cdot \sin (nx)dx$ compute karo. YEH INTEGRAL KYUN: $L=\pi$ ke saath sine formula $\frac{n\pi x}{L}=nx$ deta hai. ${\int }_0^{\pi }\sin (nx)dx={\left[-\frac{\cos nx}{n}\right]}_0^{\pi }=\frac{1-\cos n\pi }{n}=\frac{1-(-1{)}^n}{n}.$ Toh $b_n=\frac{2}{\pi }\cdot \frac{1-(-1{)}^n}{n}$. Yeh $0$ hota hai jab $n$ even ho aur $\frac{4}{\pi n}$ jab $n$ odd ho. $\boxed{1=\frac{4}{\pi }\sum _{k=0}^{\infty }\frac{\sin ((2k+1)x)}{2k+1},0<x<\pi .}$ Matlab: constant $1$ ka odd extension ek square wave hai (+1 phir −1), aur uski Fourier series exactly yeh classic odd-harmonics-only sum hai.

Shortcut: $f=1$ pehle se hi even hai. Uska even extension bas har jagah constant $1$ hai. $\frac{a_0}{2}=\frac{1}{\pi }{\int }_0^{\pi }1dx=1$, aur $n\ge 1$ ke liye: $a_n=\frac{2}{\pi }{\int }_0^{\pi }\cos (nx)dx=\frac{2}{\pi }\cdot \frac{\sin n\pi }{n}=0.$ $\boxed{1=1.}$ Ek constant ko koi wiggle nahi chahiye — poori function apna average khud hai. Yeh sabse clean sanity check hai ki $\frac{a_0}{2}$ term sach mein mean value hai.

$b_n=\frac{2}{\pi }{\int }_0^{\pi }\cos x\sin (nx)dx.$ Product-to-sum identity kyun? Integrand do alag trig functions ka product hai, $\cos x$ aur $\sin (nx)$ ka. Hmare paas koi rule nahi hai product ko directly integrate karne ka, aur yahan integration by parts sines aur cosines ke beech bounce karta rehta hai. Lekin sum of single sines ko term by term integrate karna trivial hai. Toh hum product ko sum se replace karte hain using $\sin A\cos B=\frac{1}{2}[\sin (A+B)+\sin (A-B)],$ jo $A=nx,B=x$ ke saath $\cos x\sin (nx)=\frac{1}{2}[\sin ((n+1)x)+\sin ((n-1)x)$ deta hai.$Abharpieceekakelasinehaijisehumpehleseintegratekarnajaantehain.$n\ne1$keliye:$\int_0^\pi\sin(mx),dx=\frac{1-(-1)^m}{m}$.$m=n\pm1$kesaath($nke opposite parity ke): $$b_n=\frac1\pi\left[\frac{1-(-1)^{n+1}}{n+1}+\frac{1-(-1)^{n-1}}{n-1}\right].$$(-1)^{n\pm1}=-(-1)^n$,toh$1-(-1)^{n\pm1}=1+(-1)^n$,joodd$n$keliye$0$haiaureven$n$keliye$2$hai.\ast \ast Even\ast \ast$n$keliye:$b_n=\frac1\pi\left[\frac{2}{n+1}+\frac{2}{n-1}\right]=\frac{4n}{\pi(n^2-1)}.$\ast \ast Odd\ast \ast$n\ge3$keliye:$b_n=0$.$n=1$keliye:$\int_0^\pi\cos x\sin x,dx=\frac12\int_0^\pi\sin 2x,dx=0$,toh$b_1=0$. $\boxed{\cos x=\frac{8}{\pi }\sum _{k=1}^{\infty }\frac{k}{4k^2-1}\sin (2kx),0<x<\pi .}$

Do extensions dekho (Figure s02 — $[-\pi ,\pi ]$ par black "V" curve $|x|$ even extension hai, red straight sawtooth line $y=x$ odd extension hai, red dot $x=\pi$ par jump mark karta hai):

• $x$ ka odd extension = ek sawtooth: yeh har period boundary par $+\pi$ se $-\pi$ tak jump karta hai. Ek jump (discontinuity) "expensive" hai — ek vertical cliff banane ke liye tumhe bahut saare high harmonics chahiye, isliye coefficients sirf $\frac{1}{n}$ ki tarah die karte hain.
• $x$ ka even extension = $|x|$, ek continuous "V" tent. Koi jump nahi, sirf ek corner. Continuous functions ko kam high harmonics chahiye; coefficients zyada tez die karte hain, $\frac{1}{n^2}$ ki tarah. Rule of thumb: periodic extension jitni smooth hogi, decay utni tez hogi. Sawtooth ke cliffs sine partial sums ko spikier banate hain (Gibbs overshoot jump par). Dekho Dirichlet Conditions.

$x=\pi$ par: har term mein $\sin (n\pi )=0$ hai, toh series $0$ sum karta hai. Lekin $f(\pi )=\pi$. Koi contradiction nahi: $x=\pi$ odd (sawtooth) extension ka ek jump point hai — yeh $+\pi$ (neeche se approach karte hue) se $-\pi$ (agla period) tak leap karta hai. Dirichlet's theorem ke anusaar series jump ke midpoint par converge karti hai: $\frac{\pi +(-\pi )}{2}=0$. ✓ Isliye hum equality hamesha open interval $0<x<\pi$ ke liye likhte hain.

Sine series odd extension par converge karti hai, jiska value ends par $0$ forced hoti hai (odd $\Rightarrow f(0)=0$; $x=L$ par jump midpoint). Toh agar $f(0)\ne 0$ ya $f(L)\ne 0$, toh series un endpoint heights ko reproduce nahi karti — woh unhe $0$ par snap karti hai. Physically yeh ek Wave Equation on a finite string ke liye theek hai jiske fixed ends hain, kyunki ek real string wahan $0$ par pin hoti hai. Claim sirf "galat" hai agar tumne expect kiya tha ki series ek nonzero endpoint value honour karegi jo woh kabhi represent nahi kar sakti.

Coefficients rederive karo (self-contained). Yahan $L=\pi$, toh hum $a_n=\frac{2}{\pi }{\int }_0^{\pi }x\cos (nx)dx$ aur $\frac{a_0}{2}=\frac{1}{\pi }{\int }_0^{\pi }xdx$ use karte hain.

• Constant term: $\frac{a_0}{2}=\frac{1}{\pi }\cdot \frac{{\pi }^2}{2}=\frac{\pi }{2}$ ($[0,\pi ]$ par $x$ ki average height — sanity check ✓).
• $a_n$ integration by parts se (IBP kyun? polynomial × trig; IBP polynomial power ghata deta hai): $u=x,dv=\cos (nx)dx$ ke saath, ${\int }_0^{\pi }x\cos (nx)dx={\left[\frac{x\sin nx}{n}\right]}_0^{\pi }-\frac{1}{n}{\int }_0^{\pi }\sin (nx)dx.$ Bracket $0$ hai kyunki $\sin (n\pi )=0$. Remaining integral $\frac{1}{n}\cdot \frac{1-\cos n\pi }{n}=\frac{1-(-1{)}^n}{n^2}$ hai jiske aage minus sign hai, giving $\frac{(-1{)}^n-1}{n^2}$. Hence $a_n=\frac{2}{\pi }\cdot \frac{(-1{)}^n-1}{n^2},$ jo even $n$ ke liye $0$ aur odd $n$ ke liye $-\frac{4}{\pi n^2}$ hai. Toh $x=\frac{\pi }{2}-\frac{4}{\pi }{\sum }_{k=0}^{\infty }\frac{\cos ((2k+1)x)}{(2k+1{)}^2},0<x<\pi .$ Ab $x=0$ par evaluate karo (even $|x|$ extension ka continuity point, toh equality exact hai): $0=\frac{\pi }{2}-\frac{4}{\pi }{\sum }_{k=0}^{\infty }\frac{1}{(2k+1{)}^2}⟹{\sum }_{k=0}^{\infty }\frac{1}{(2k+1{)}^2}=\frac{\pi }{2}\cdot \frac{\pi }{4}=\boxed{\frac{{\pi }^2}{8}}.$ Cross-check: yeh famous odd-square sum hai; ${\sum }_{n\ge 1}\frac{1}{n^2}=\frac{{\pi }^2}{6}$ ke saath combine karne par even part $\frac{1}{4}\cdot \frac{{\pi }^2}{6}=\frac{{\pi }^2}{24}$ deta hai, aur sach mein $\frac{{\pi }^2}{8}+\frac{{\pi }^2}{24}=\frac{{\pi }^2}{6}$. ✓

Humein aisi blocks chahiye jo $x=0$ par zero hon (toh $\sin$ use karo, kyunki $\sin 0=0$) aur $x=\pi$ par zero slope hon. $\sin (\lambda x)$ ka slope $\lambda \cos (\lambda x)$ hai; yeh $x=\pi$ par zero hota hai jab $\cos (\lambda \pi )=0$, yani $\lambda \pi =\frac{(2n-1)\pi }{2}$, giving $\lambda =\frac{2n-1}{2}$. $\boxed{\sin \left(\frac{(2n-1)x}{2}\right),n=1,2,3,\dots }$ Yeh half-integer frequencies hain. "Quarter-range" ka matlab: ek ordinary half-range series $f$ ko $[0,L]$ par jaanti hai = ek period $2L$ ka aadha hissa, aur dono ends ek hi type ki boundary hoti hain. Jab do ends alag type ki hoti hain (ek fixed, ek insulated), resulting waves ki natural period $2L$ ki jagah $4L$ hoti hai — known piece $[0,L]$ ab ek full period ka sirf quarter hai, isliye ise quarter-range expansion kehte hain. Concretely, $\sin \frac{(2n-1)x}{2}$ $4\pi$ length par ek full cycle complete karta hai, yani $L=\pi$ ke saath period $4L$. Na pure sine na pure cosine of integer harmonics kaam karta hai; mixed BC yeh nayi family force karta hai. (Wahi separation-of-variables machinery jaise Heat Equation — separation of variables.)

Step 1 — variables separate karo (KYU: ek PDE ko do ODEs mein convert karta hai). $u=X(x)T(t)$ try karo. Tab $\frac{T^{\prime }}{T}=\frac{X^{\prime \prime }}{X}=-\lambda$ (constant). Fixed-end BCs $X$ ko dono ends par kill karte hain, toh humein aisi $X$ chahiye jo wahan vanish kare → sine blocks $X_n=\sin (nx)$ with $\lambda =n^2$ (period-$2\pi$ half-range, $L=\pi$). Step 2 — time factor. $T^{\prime }=-n^{2}T\Rightarrow T_n=e^{-n^{2}t}$. Step 3 — superpose aur initial shape match karo. $u(x,t)={\sum }_{n\ge 1}{b}_n{e}^{-n^{2}t}\sin (nx)$, aur $t=0$ par yeh $x$ ke barabar hona chahiye. Yeh exactly $x$ ki half-range sine series hai (parent Example 1): $b_n=\frac{2(-1{)}^{n+1}}{n}.$ Answer: $\boxed{u(x,t)=2\sum _{n=1}^{\infty }\frac{(-1{)}^{n+1}}{n}{e}^{-n^{2}t}\sin (nx).}$ BCs verify karo: har term mein $\sin (0)=\sin (n\pi )=0$ hai → $u(0,t)=u(\pi ,t)=0$. ✓ IC verify karo: $t=0$ par exponential $1$ hai aur sum $x$ ki sine series hai, $0<x<\pi$ par $x$ ke barabar. ✓ Physical picture: jaise-jaise $t$ badhta hai, $e^{-n^{2}t}$ factor high harmonics ko sabse tez crush karta hai, toh sharp sawtooth smooth hota jaata hai aur poora rod $0$ ki taraf cool hota hai.

Parseval kyun, aur yeh kya kehta hai. Parseval's identity energy conservation ka statement hai: $f$ ki "total squared size" seedhe measure ki jaaye (by integrating $f^2$) toh woh harmonic-by-harmonic measured total squared size ke barabar honi chahiye (by adding up $b_n^2$). Kyunki sine blocks $\sin (nx)$ $[0,\pi ]$ par mutually orthogonal hain, saare cross terms integrate hokar zero ho jaate hain aur sirf squares bachte hain. $[0,\pi ]$ par half-range sine series ke liye identity yeh hai: $\frac{2}{\pi }{\int }_0^{\pi }f(x{)}^{2}dx={\sum }_{n=1}^{\infty }{b}_n^2.$ Yeh yahan perfect tool hai: yeh $f^2$ ka ek integral (aasaan) ko $b_n^2$ ki infinite sum mein convert karta hai (jisme woh $\sum 1/n^2$ hai jo hum chahte hain). Left side: $\frac{2}{\pi }{\int }_0^{\pi }{x}^{2}dx=\frac{2}{\pi }\cdot \frac{{\pi }^3}{3}=\frac{2{\pi }^2}{3}.$ Right side: $\sum b_n^2=\sum {\left(\frac{2(-1{)}^{n+1}}{n}\right)}^2=\sum \frac{4}{n^2}=4\sum \frac{1}{n^2}.$ Dono equal set karo aur solve karo: $4{\sum }_{n=1}^{\infty }\frac{1}{n^2}=\frac{2{\pi }^2}{3}⟹{\sum }_{n=1}^{\infty }\frac{1}{n^2}=\frac{1}{4}\cdot \frac{2{\pi }^2}{3}=\boxed{\frac{{\pi }^2}{6}}.$ ✓ Basel sum, ek rod-temperature expansion se recover kiya!

Fixed ends → sine; insulated ends → cosine; mixed ends → half-integer frequencies. Series $f$ ke barabar hai open interval par; jumps par yeh midpoint deti hai.