Exercises — Full Fourier series — coefficients derivation

Formulas: $a_0=\frac{1}{L}{\int }_{-L}^{L}fdx$, $a_m=\frac{1}{L}{\int }_{-L}^{L}f\cos \frac{m\pi x}{L}dx$, $b_m=\frac{1}{L}{\int }_{-L}^{L}f\sin \frac{m\pi x}{L}dx$.

Pehla integral: $m=2\ne n=5$, isliye orthogonality se yeh $\boxed{0}$ hai. Doosra integral: yeh $m=n=3$ sine–sine hai, jo $\boxed{L}$ ke barabar hai.

True. $x^2$ even hai; $\sin \frac{m\pi x}{L}$ odd hai; even $\times$ odd $=$ odd; symmetric interval $[-L,L]$ par ek odd function ka integral $0$ hota hai. Isliye har $b_m=0$.

Yahan koi symmetry nahi hai, to teeno compute karo.

$a_0$: $a_0=\frac{1}{\pi }{\int }_{-\pi }^{\pi }fdx=\frac{1}{\pi }\left({\int }_{-\pi }^{0}1dx+{\int }_0^{\pi }0dx\right)=\frac{1}{\pi }(\pi )=1.$ Constant term $=a_0/2=\frac{1}{2}$ ($f$ ka average).

$a_m$: $a_m=\frac{1}{\pi }{\int }_{-\pi }^0\cos (mx)dx=\frac{1}{\pi }{\left[\frac{\sin mx}{m}\right]}_{-\pi }^0=\frac{1}{\pi }\left(0-\frac{\sin (-m\pi )}{m}\right)=0$ (kyunki $\sin m\pi =0$). To $m\ge 1$ ke liye sab $a_m=0$.

$b_m$: $b_m=\frac{1}{\pi }{\int }_{-\pi }^0\sin (mx)dx=\frac{1}{\pi }{\left[-\frac{\cos mx}{m}\right]}_{-\pi }^0=\frac{1}{\pi }\left(-\frac{1}{m}+\frac{\cos (-m\pi )}{m}\right)=\frac{(-1{)}^m-1}{m\pi }.$ Yeh even $m$ ke liye $0$ hai aur odd $m$ ke liye $-\frac{2}{m\pi }$ hai.

$f(x)=\frac{1}{2}-\frac{2}{\pi }{\sum }_{k=0}^{\infty }\frac{1}{2k+1}\sin ((2k+1)x).$

$x=0$ par value. Yahan $f$ jump karti hai: left limit $f(0^{-})=1$ hai, right limit $f(0^{+})=0$ hai. Dirichlet convergence theorem se series midpoint $\frac{1}{2}(f(0^{-})+f(0^{+}))=\frac{1}{2}(1+0)=\frac{1}{2}$ par converge karti hai. Check karo: $x=0$ par har $\sin ((2k+1)\cdot 0)=0$ hota hai, to series literally constant $\frac{1}{2}$ tak sum hoti hai. Isliye constant term ($a_0/2=\frac{1}{2}$) pehle se midpoint par land karta hai — yeh reassuring hai, coincidence nahi.

$a_0$: $a_0=\frac{1}{\pi }{\int }_{-\pi }^{\pi }{x}^{2}dx=\frac{1}{\pi }\cdot \frac{2{\pi }^3}{3}=\frac{2{\pi }^2}{3}.$ Constant term $=a_0/2=\frac{{\pi }^2}{3}$.

$a_m$: kyunki $x^2\cos mx$ even hai, $a_m=\frac{2}{\pi }{\int }_0^{\pi }{x}^2\cos mxdx$. Integration by parts kyun, aur kyun do baar? Humare paas polynomial times trig function hai. Integrals ke liye koi product rule nahi hota, lekin integration by parts ek derivative ko ek integral se trade karne deta hai: har baar jab hum $u=$ polynomial part choose karte hain, $x^2\to 2x\to 2$ differentiate karna uski degree ek se ghatata hai. Do rounds ke baad polynomial ek constant ban jaata hai aur integral ek plain trig integral ban jaata hai jo hum finish kar sakte hain. (Trig part ko $u$ choose karna complexity badhayega — polynomial ko hamesha differentiate karo.) $u=x^2,dv=\cos mxdx$ lekar: ${\int }_0^{\pi }{x}^2\cos mxdx={\left[\frac{x^2\sin mx}{m}\right]}_0^{\pi }-\frac{2}{m}{\int }_0^{\pi }x\sin mxdx.$ Bracket $0$ hai ($\sin m\pi =0$). Phir ${\int }_0^{\pi }x\sin mxdx=-\frac{\pi (-1{)}^m}{m}$ (parts ka doosra round, jaise parent ke Example 2 mein). To ${\int }_0^{\pi }{x}^2\cos mxdx=-\frac{2}{m}\left(-\frac{\pi (-1{)}^m}{m}\right)=\frac{2\pi (-1{)}^m}{m^2}.$ Isliye $a_m=\frac{2}{\pi }\cdot \frac{2\pi (-1{)}^m}{m^2}=\frac{4(-1{)}^m}{m^2}.$ $x^2=\frac{{\pi }^2}{3}+4{\sum }_{m=1}^{\infty }\frac{(-1{)}^m}{m^2}\cos mx.$

$x=\pi$ rakho. Convergence theorem se series $f(\pi )={\pi }^2$ ke barabar hai. Yeh point legal kyun hai? $x^2$ ka $2\pi$-periodic extension $x=\pi$ par continuous hai (dono sides ${\pi }^2$ approach karti hain), to koi jump nahi hai aur series true value hit karti hai — midpoint nahi. Kyunki $\cos m\pi =(-1{)}^m$: ${\pi }^2=\frac{{\pi }^2}{3}+4{\sum }_{m=1}^{\infty }\frac{(-1{)}^m}{m^2}(-1{)}^m=\frac{{\pi }^2}{3}+4{\sum }_{m=1}^{\infty }\frac{1}{m^2}.$ Rearrange karo: $4\sum \frac{1}{m^2}={\pi }^2-\frac{{\pi }^2}{3}=\frac{2{\pi }^2}{3}$, to ${\sum }_{m=1}^{\infty }\frac{1}{m^2}=\frac{{\pi }^2}{6}.$ $✓$

Left side: $f^2=1$ har jagah, to $\frac{1}{\pi }{\int }_{-\pi }^{\pi }1dx=\frac{1}{\pi }(2\pi )=2$. Right side: $a_0=0$, sab $a_n=0$, aur $\sum b_m^2={\sum }_{\text{odd }m}{\left(\frac{4}{m\pi }\right)}^2=\frac{16}{{\pi }^2}{\sum }_{k=0}^{\infty }\frac{1}{(2k+1{)}^2}.$ Equal set karo: $2=\frac{16}{{\pi }^2}{\sum }_{k=0}^{\infty }\frac{1}{(2k+1{)}^2}$, jisse ${\sum }_{k=0}^{\infty }\frac{1}{(2k+1{)}^2}=\frac{{\pi }^2}{8}$ milta hai.

Koi symmetry nahi — sab compute karo. Pehle, neeche diye gaye do helper integrals ki shape aisi kyun hai. Jab tum $e^x\cos mx$ ko do baar parts se integrate karte ho, integral right side par wapas aata hai, aur tum algebraically uske liye solve karte ho (jaise $I=(\text{stuff})-kI$ solve karna). Clean results hain: $\int e^x\cos mxdx=\frac{e^x(\cos mx+m\sin mx)}{1+m^2},\int e^x\sin mxdx=\frac{e^x(\sin mx-m\cos mx)}{1+m^2}.$ Tum kisi bhi ek ko verify kar sakte ho right side differentiate karke (product rule) aur dekhke ki woh integrand par wapas collapse ho jaata hai — denominator mein $1+m^2$ exactly wahi hai jo cross-terms ko cancel karta hai. $\sinh \pi =\frac{e^{\pi }-e^{-\pi }}{2}$ likho.

$a_0$: $a_0=\frac{1}{\pi }{\int }_{-\pi }^{\pi }{e}^{x}dx=\frac{1}{\pi }(e^{\pi }-e^{-\pi })=\frac{2\sinh \pi }{\pi }.$

$a_m$: $a_m=\frac{1}{\pi }{\left[\frac{e^x(\cos mx+m\sin mx)}{1+m^2}\right]}_{-\pi }^{\pi }$. $x=\pm \pi$ par: $\sin m\pi =0,\cos m\pi =(-1{)}^m$, to bracket $\frac{(-1{)}^m{e}^x}{1+m^2}$ ban jaata hai ends par evaluate kiya hua: $a_m=\frac{1}{\pi }\cdot \frac{(-1{)}^m(e^{\pi }-e^{-\pi })}{1+m^2}=\frac{2(-1{)}^m\sinh \pi }{\pi (1+m^2)}.$

$b_m$: $b_m=\frac{1}{\pi }{\left[\frac{e^x(\sin mx-m\cos mx)}{1+m^2}\right]}_{-\pi }^{\pi }$. $x=\pm \pi$ par bracket $\frac{-m(-1{)}^m{e}^x}{1+m^2}$ ban jaata hai, to $b_m=\frac{1}{\pi }\cdot \frac{-m(-1{)}^m(e^{\pi }-e^{-\pi })}{1+m^2}=\frac{-2m(-1{)}^m\sinh \pi }{\pi (1+m^2)}.$

$e^x=\frac{2\sinh \pi }{\pi }\left[\frac{1}{2}+{\sum }_{m=1}^{\infty }\frac{(-1{)}^m}{1+m^2}(\cos mx-m\sin mx)\right].$

Complex framework ka quick refresher. Euler's formula $e^{\pm inx}=\cos nx\pm i\sin nx$ use karke, real series ko $f(x)={\sum }_{n=-\infty }^{\infty }{c}_n{e}^{inx}$ ke roop mein repackage kiya ja sakta hai jahan har complex coefficient $c_n=\frac{1}{2\pi }{\int }_{-\pi }^{\pi }f(x)e^{-inx}dx$ hota hai ($e^{-inx}$ woh "matching fork" hai jo project out ho rahi hai, exactly jaise real case mein $\cos ,\sin$ the). Real coefficients par wapas aane ka bridge: $+n$ aur $-n$ terms group karke aur $\cos nx=\frac{e^{inx}+e^{-inx}}{2}$ use karke milta hai $a_n=c_n+c_{-n}$ aur $b_n=i(c_n-c_{-n})$. To $c_n+c_{-n}$ ko $a_n$ se check karna dono frameworks ko ek saath test karta hai.

$c_n=\frac{1}{2\pi }{\int }_{-\pi }^{\pi }{e}^{(1-in)x}dx=\frac{1}{2\pi }\cdot \frac{e^{(1-in)\pi }-e^{-(1-in)\pi }}{1-in}$. Kyunki $e^{\mp in\pi }=(-1{)}^n$ (integer $n$: $e^{-in\pi }=\cos n\pi -i\sin n\pi =(-1{)}^n$, purely real), numerator $(-1{)}^n(e^{\pi }-e^{-\pi })=2(-1{)}^n\sinh \pi$ hai. To $c_n=\frac{(-1{)}^n\sinh \pi }{\pi (1-in)}$.

Phir $a_n=c_n+c_{-n}=\frac{(-1{)}^n\sinh \pi }{\pi }\left(\frac{1}{1-in}+\frac{1}{1+in}\right)=\frac{(-1{)}^n\sinh \pi }{\pi }\cdot \frac{2}{1+n^2}=\frac{2(-1{)}^n\sinh \pi }{\pi (1+n^2)}.$ $n=0$ ke liye: $a_0=\frac{2\sinh \pi }{\pi }$; $n=1$ ke liye: $a_1=\frac{-2\sinh \pi }{2\pi }=\frac{-\sinh \pi }{\pi }$. Dono Ex 4.1 se match karte hain. $✓$

$x(\pi -x)=\pi x-x^2$ expand karo. Humhe ${\int }_0^{\pi }x\sin mxdx$ aur ${\int }_0^{\pi }{x}^2\sin mxdx$ chahiye. Parts phir kyun? Same reason jaise Ex 2.2 — polynomial $\times$ trig — to polynomial differentiate karo uski degree grind karne ke liye.

Parent se, ${\int }_0^{\pi }x\sin mxdx=-\frac{\pi (-1{)}^m}{m}$. Doosre ke liye, do baar parts integrate karo: ${\int }_0^{\pi }{x}^2\sin mxdx={\left[-\frac{x^2\cos mx}{m}\right]}_0^{\pi }+\frac{2}{m}{\int }_0^{\pi }x\cos mxdx.$ Bracket $=-\frac{{\pi }^2(-1{)}^m}{m}$ hai. Aur ${\int }_0^{\pi }x\cos mxdx={\left[\frac{x\sin mx}{m}\right]}_0^{\pi }-\frac{1}{m}{\int }_0^{\pi }\sin mxdx=0-\frac{1}{m}\cdot \frac{1-(-1{)}^m}{m}=\frac{(-1{)}^m-1}{m^2}.$ To ${\int }_0^{\pi }{x}^2\sin mxdx=-\frac{{\pi }^2(-1{)}^m}{m}+\frac{2}{m}\cdot \frac{(-1{)}^m-1}{m^2}=-\frac{{\pi }^2(-1{)}^m}{m}+\frac{2((-1{)}^m-1)}{m^3}.$

Ab combine karo ($\pi \cdot$ pehla $-$ doosra): ${\int }_0^{\pi }(\pi x-x^2)\sin mxdx=\pi \left(-\frac{\pi (-1{)}^m}{m}\right)-\left(-\frac{{\pi }^2(-1{)}^m}{m}+\frac{2((-1{)}^m-1)}{m^3}\right)=-\frac{2((-1{)}^m-1)}{m^3}.$ ${\pi }^2(-1{)}^m/m$ terms exactly cancel ho jaate hain. Isliye $B_m=\frac{2}{\pi }\cdot \left(-\frac{2((-1{)}^m-1)}{m^3}\right)=\frac{4(1-(-1{)}^m)}{\pi m^3}=\{\begin{array}{ll}\frac{8}{\pi m^3}& m\text{ odd}\\ 0& m\text{ even}.\end{array}$ Sirf odd harmonics appear karte hain — consistent hai is baat se ki profile rod ke midpoint $x=\pi /2$ ke baare mein symmetric hai.

Left side: ${\int }_0^{\pi }{x}^2(\pi -x{)}^{2}dx$. $I={\int }_0^{\pi }({\pi }^2{x}^2-2\pi x^3+x^4)dx={\pi }^2\frac{{\pi }^3}{3}-2\pi \frac{{\pi }^4}{4}+\frac{{\pi }^5}{5}={\pi }^5\left(\frac{1}{3}-\frac{1}{2}+\frac{1}{5}\right)=\frac{{\pi }^5}{30}$ lete hain. To LHS $=\frac{2}{\pi }\cdot \frac{{\pi }^5}{30}=\frac{{\pi }^4}{15}.$ Right side: $\sum B_m^2={\sum }_{\text{odd }m}{\left(\frac{8}{\pi m^3}\right)}^2=\frac{64}{{\pi }^2}{\sum }_{k=0}^{\infty }\frac{1}{(2k+1{)}^6}.$ Equal karo: $\frac{{\pi }^4}{15}=\frac{64}{{\pi }^2}\sum \frac{1}{(2k+1{)}^6}$, to ${\sum }_{k=0}^{\infty }\frac{1}{(2k+1{)}^6}=\frac{{\pi }^6}{15\cdot 64}=\frac{{\pi }^6}{960}.$

Har line ko "Prompt ::: Answer" ke roop mein padho — ::: ke baad ke text ko cover karo, apna answer zor se bolo, phir check karne ke liye reveal karo. Interval $[-L,L]$ har coefficient ke liye kaun sa prefactor use karta hai? ::: $\frac{1}{L}$ Odd $f$ mein sirf kaun se coefficients nonzero hote hain? ::: sirf $b_m$ (sab $a_m=0$) Even $f$ mein sirf kaun se coefficients nonzero hote hain? ::: sirf $a_m$ (sab $b_m=0$) Jump discontinuity par series kis value par converge karti hai? ::: do one-sided limits ka midpoint Kya pure sine (half-range) series mein constant $B_0$ term hota hai? ::: Nahi — $\sin 0=0$, isliye koi constant nahi Parseval $\int f^2$ ko kis se link karta hai? ::: squared coefficients ke sum se