Fourier series — motivation from periodic functions: Exercises

4.7.3 · D4 · Maths › Partial Differential Equations › Fourier series — motivation from periodic functions

Period $2 L$ wale function ke liye do machines yaad rakho, jo baar baar kaam aayengi:

Level 1 — Recognition

L1.1

$g (x) = cos (3 x) + sin (5 x)$ ka fundamental period $T$ batao, aur usme maujood sabse slow wave ki angular frequency batao.

Recall Solution

Hum kya dhundh rahe hain: ek aisa period $T$ jisme har piece repeat ho. $cos (3 x)$ tab repeat hota hai jab $3 x$ mein $2 π$ ka advancement ho, yaani har $2 π /3$ par. $sin (5 x)$ har $2 π /5$ par repeat hota hai. Poora sum uss sabse chhote common length par repeat hoga jo dono ka whole-number multiple ho.

$\frac{2 π}{3}$ aur $\frac{2 π}{5}$ dono $2 π$ ko divide karte hain (kyunki $3$ aur $5$ integers hain). To $== T = 2 π ==$ .

Sabse slow wave $cos (3 x)$ hai; iski angular frequency $ω = 3$ (radians per unit $x$ ) hai. Sabse slow wave ki frequency number sabse chhoti hoti hai.

L1.2

Kya $h (x) = x^{2} cos x$ even hai, odd hai, ya kuch bhi nahi? Iske aadhar par, symmetric interval par kaun si family ke Fourier coefficients zero ho jaate hain?

Recall Solution

Test: $x$ ki jagah $- x$ rakh do. $h (- x) = (- x)^{2} cos (- x) = x^{2} cos x = h (x)$ . Even ka even se product even hota hai, aur cosine bhi even hai, isliye $h$ even hai.

Ek even function ke saare $b_{n} = 0$ hote hain — sirf cosines bachte hain. Kyun: $b_{n}$ mein $h (x) sin \frac{nπ x}{L}$ integrate hota hai, jo even × odd = odd hai, aur symmetric interval $[- L, L]$ par odd integrand ka integral $0$ hota hai.

Level 2 — Application

L2.1

$f (x) = x$ ka, $(- π, π)$ par, period $2 π$ ke saath extend karke, full Fourier series nikalo.

Recall Solution

Pehle symmetry (kyun: aadha kaam bachta hai). $f (- x) = - x = - f (x)$ , to $f$ odd hai ⟹ saare $a_{n} = 0$ . Sirf sines.

Yahan $L = π$ hai, to machine hai $b_{n} = \frac{1}{π} \int_{- π}^{π} x sin (n x) d x$ . Odd × odd = even integrand, to right half double karo: $b_{n} = \frac{2}{π} \int_{0}^{π} x sin (n x) d x .$

Integration by parts (kyun: polynomial ka wave ke saath product — parts $x$ ko peel kar deta hai). $u = x, d v = sin (n x) d x$ leke: $\int_{0}^{π} x sin (n x) d x = [- \frac{x c o s ( n x )}{n}]_{0}^{π} + \frac{1}{n} \int_{0}^{π} cos (n x) d x = - \frac{π c o s ( nπ )}{n} + \frac{1}{n} \cdot = 0 [\frac{s i n ( n x )}{n}]_{0}^{π} .$ Kyunki $cos (nπ) = (- 1)^{n}$ : $b_{n} = \frac{2}{π} \cdot (- \frac{π ( - 1 ) ^{n}}{n}) = - \frac{2 ( - 1 ) ^{n}}{n} = \frac{2 ( - 1 ) ^{n + 1}}{n} .$

Result: $x = 2 (sin x - \frac{s i n 2 x}{2} + \frac{s i n 3 x}{3} - \dots) = 2 \sum_{n = 1}^{\infty} \frac{( - 1 ) ^{n + 1}}{n} sin (n x) .$

L2.2

$f (x) = x^{2}$ ke liye, $(- π, π)$ par period $2 π$ ke saath, $a_{0}$ aur $a_{1}$ nikalo.

Recall Solution

$f$ even hai, to $b_{n} = 0$ ; hum sirf $a_{n}$ nikalenge. $L = π$ .

$a_{0} = \frac{1}{π} \int_{- π}^{π} x^{2} d x = \frac{1}{π} \cdot \frac{2 π ^{3}}{3} = \frac{2 π ^{2}}{3} .$ To constant term $a_{0} /2 = π^{2} /3$ = $x^{2}$ ka average over $(- π, π)$ . Check karo: yeh positive hai, jo $x^{2}$ ke liye hona hi chahiye.

$a_{1} = \frac{1}{π} \int_{- π}^{π} x^{2} cos x d x = \frac{2}{π} \int_{0}^{π} x^{2} cos x d x .$ Parts do baar (kyun: power $x^{2}$ ko constant tak laana hai). Standard result $\int_{0}^{π} x^{2} cos x d x = 2 π cos π = - 2 π$ hai, to $a_{1} = \frac{2}{π} \cdot (- 2 π) = - 4.$

Level 3 — Analysis

L3.1

Parent note ka square wave $\frac{4}{π} \sum_{odd n} \frac{s i n n x}{n}$ hai. Predict karo ki series $x = 0$ par kis value par converge karegi, aur jump rule se explain karo.

Recall Solution

$x = 0$ par square wave jump karta hai $- 1$ (just left) se $+ 1$ (just right) tak. Har term $sin (n \cdot 0) = 0$ hai, to series literally $0$ pe sum hoti hai.

Jump rule kehta hai ki Fourier series discontinuity par midpoint par converge karti hai: $\frac{1}{2} [f (x^{+}) + f (x^{-})] = \frac{1}{2} [(+ 1) + (- 1)] = 0$ . ✔ Dono predictions agree karte hain: series sirf dono sides ka average de sakti hai, koi ek side nahi. Jump ke paas (par nahi) overshoot ke liye Gibbs Phenomenon dekho.

L3.2

$f (x) = x^{2}$ ki series (jo $\frac{π ^{2}}{3} + \sum_{n \geq 1} \frac{4 ( - 1 ) ^{n}}{n ^{2}} cos n x$ hai) ko $x = π$ par evaluate karke ek famous sum derive karo.

Recall Solution

$x = π$ kyun: $x^{2}$ wahan continuous hai (period- $2 π$ extension match karti hai: left se $π^{2}$ , right se $π^{2}$ ), to series exactly $f (π) = π^{2}$ ke barabar hai.

$x = π$ par: $cos (nπ) = (- 1)^{n}$ , to $(- 1)^{n} cos (nπ) = (- 1)^{2 n} = 1$ . Isliye $π^{2} = \frac{π ^{2}}{3} + \sum_{n = 1}^{\infty} \frac{4}{n ^{2}} .$ Rearrange karo: $\sum \frac{4}{n ^{2}} = π^{2} - \frac{π ^{2}}{3} = \frac{2 π ^{2}}{3}$ , jisse Basel sum milta hai: $\sum_{n = 1}^{\infty} \frac{1}{n ^{2}} = \frac{π ^{2}}{6} .$

Level 4 — Synthesis

L4.1

Ek function $(0, π)$ par $f (x) = x$ define hai. Tumhe sirf cosines ki series chahiye jo $(0, π)$ par valid ho. $f$ ka kaun sa extension use karoge, $L$ kya hoga, aur $a_{0}$ nikalo.

Recall Solution

Idea (half-range expansion). $(0, π)$ par tumhare paas doosra half invent karne ki freedom hai. Saare sines ko khatam karke cosines rakhne ke liye, $f$ ko even extend karo: $x$ ko $y$ -axis ke across reflect karo taaki $f (- x) = f (x) = ∣ x ∣$ on $(- π, π)$ ho jaye. Ab interval $(- π, π)$ hai, to $L = π$ , aur even extension guarantee karta hai $b_{n} = 0$ .

$a_{0} = \frac{1}{π} \int_{- π}^{π} ∣ x ∣ d x = \frac{2}{π} \int_{0}^{π} x d x = \frac{2}{π} \cdot \frac{π ^{2}}{2} = π .$ Constant term $a_{0} /2 = π /2$ = $∣ x ∣$ ka average on $(- π, π)$ . ✔ Figure mein even reflection dikhaya gaya hai.

Figure — Fourier series — motivation from periodic functions

L4.2

Usi $f (x) = x$ ke liye, $(0, π)$ par even extension ke saath, $n \geq 1$ ke liye $a_{n}$ nikalo.

Recall Solution

$a_{n} = \frac{2}{π} \int_{0}^{π} x cos (n x) d x .$ Parts ( $u = x, d v = cos n x d x$ ): $\int_{0}^{π} x cos n x d x = [\frac{x s i n n x}{n}]_{0}^{π} - \frac{1}{n} \int_{0}^{π} sin n x d x = 0 - \frac{1}{n} [- \frac{c o s n x}{n}]_{0}^{π} = \frac{c o s nπ - 1}{n ^{2}} = \frac{( - 1 ) ^{n} - 1}{n ^{2}} .$ To $a_{n} = \frac{2}{π} \cdot \frac{( - 1 ) ^{n} - 1}{n ^{2}}$ , jo even $n$ ke liye $0$ hai aur odd $n$ ke liye $- \frac{4}{π n ^{2}}$ hai. Final cosine series: $x = \frac{π}{2} - \frac{4}{π} (cos x + \frac{c o s 3 x}{9} + \frac{c o s 5 x}{25} + \dots), 0 < x < π .$

Level 5 — Mastery

L5.1

Orthogonality integral $\int_{- L}^{L} sin \frac{nπ x}{L} cos \frac{mπ x}{L} d x = 0$ ko saare integers $n, m \geq 0$ ke liye prove karo — degenerate case $n = m$ bhi include karo.

Recall Solution

Tool choice: product-to-sum identity $sin A cos B = \frac{1}{2} [sin (A + B) + sin (A - B)]$ . Yeh tool kyun: yeh product (integrate karna mushkil) ko single sines ke sum (har ek trivially integrable) mein convert kar deta hai.

$A = \frac{nπ x}{L}, B = \frac{mπ x}{L}$ leke: $\int_{- L}^{L} sin \frac{nπ x}{L} cos \frac{mπ x}{L} d x = \frac{1}{2} \int_{- L}^{L} [sin \frac{( n + m ) π x}{L} + sin \frac{( n - m ) π x}{L}] d x .$ Yeh zero kyun hota hai: har $sin (k π x / L)$ , $x$ ki odd function hai, aur interval $[- L, L]$ symmetric hai, to har ek ka integral $0$ hoga — chahe $n = m$ ho, $n \neq = m$ ho, ya dono mein se koi zero ho. ( $n = m$ case mein bhi $sin (2 nπ x / L) + sin 0$ milta hai, dono ka integral $0$ .) Isliye mixed integral hamesha $0$ hai. Yahi woh pillar hai jo sines aur cosines ko ek doosre mein interfere karne se rokta hai — Orthogonality of Functions ki machinery.

L5.2

Parent note ka plug-in result tha $1 - \frac{1}{3} + \frac{1}{5} - \dots = \frac{π}{4}$ . Square-wave series ko $x = π /2$ par evaluate karke isko independently re-derive karo, har step dikhao.

Recall Solution

Square wave: $f (x) = \frac{4}{π} (sin x + \frac{s i n 3 x}{3} + \frac{s i n 5 x}{5} + \dots)$ , aur $f (π /2) = + 1$ (wahan continuous hai — $π /2$ " $+ 1$ " block ke andar hai, kisi jump par nahi, to series exactly $f$ ke barabar hai).

Har sine ko $x = π /2$ par evaluate karo: $sin \frac{π}{2} = 1, sin \frac{3 π}{2} = - 1, sin \frac{5 π}{2} = 1, sin \frac{7 π}{2} = - 1, \dots$ To odd terms $+ 1, - 1, + 1, \dots$ alternate karte hain denominators $1, 3, 5, 7, \dots$ ke saath: $1 = \frac{4}{π} (1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \dots) ⟹ 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \dots = \frac{π}{4} .$ Yeh Leibniz series for $π$ hai, jo ek square wave se naturally nikal aaya.

L5.3

$2 π$ -periodic odd function ke liye Parseval's identity kehti hai $\frac{1}{π} \int_{- π}^{π} f^{2} d x = \sum_{n \geq 1} b_{n}^{2}$ . Ise square wave par apply karo ( $f^{2} = 1$ har jagah) aur $\sum_{odd n} \frac{1}{n ^{2}} = \frac{π ^{2}}{8}$ derive karo.

Recall Solution

Parseval kyun: yeh kehta hai total "energy" $\int f^{2}$ , squared amplitudes ke sum ke barabar hai — wahi orthogonality idea, ab norm ke liye. Left side: $f^{2} = 1$ hai, to $\frac{1}{π} \int_{- π}^{π} 1 d x = \frac{1}{π} \cdot 2 π = 2$ .

Right side: odd $n$ ke liye $b_{n} = \frac{4}{π n}$ , baaki $0$ , to $b_{n}^{2} = \frac{16}{π ^{2} n ^{2}}$ (odd $n$ ). Isliye $2 = \sum_{odd n} \frac{16}{π ^{2} n ^{2}} = \frac{16}{π ^{2}} \sum_{odd n} \frac{1}{n ^{2}} .$ Solve karo: $\sum_{odd n} \frac{1}{n ^{2}} = \frac{2 π ^{2}}{16} = \frac{π ^{2}}{8} .$ ✔ (Aur kyunki $\sum_{all} \frac{1}{n ^{2}} - \sum_{even} \frac{1}{n ^{2}} = \frac{π ^{2}}{6} - \frac{1}{4} \cdot \frac{π ^{2}}{6} = \frac{π ^{2}}{8}$ hai, yeh L3.2 ke Basel result se consistent hai.)

Recall Big-picture recall

Inhe exercises ne chaar moves rehearse kiye: (1) symmetry padho taaki half coefficients delete ho jaayein, (2) integration by parts se calculate karo, (3) smart points par evaluate karo (continuous vs jump) taaki number identities niklen, (4) orthogonality ko reuse karo — coefficients isolate karne ke liye bhi aur Parseval prove karne ke liye bhi. Yahi modes Heat Equation aur Wave Equation ko Separation of Variables se solve karte waqt wapas aate hain.