Exercises — Legendre's equation and Legendre polynomials (intro)

WHAT/WHY: Hum Rodrigues' formula use karte hain, kyunki yeh $P_n$ ko explicitly express karta hai aur sign-flip $x\to -x$ ko track karna aasaan banata hai. Maano $g(x)=(x^2-1{)}^n$. Kyunki $(-x{)}^2=x^2$, hame $g(-x)=g(x)$ milta hai — $g$ ek even function hai. Rodrigues deta hai $P_n(x)=\frac{1}{2^{n}n!}{g}^{(n)}(x)$, $g$ ki $n$-th derivative. Derivatives ke baare mein key fact: har derivative parity flip karti hai. Agar $g$ even hai, toh $g^{\prime }$ odd hai, $g^{\prime \prime }$ even hai, aur generally $g^{(n)}(-x)=(-1{)}^n{g}^{(n)}(x)$. (Chain rule: $g(-x)$ differentiate karne par har derivative ke saath ek factor $-1$ aata hai.) Isliye $P_n(-x)=\frac{1}{2^{n}n!}{g}^{(n)}(-x)=(-1{)}^n\frac{1}{2^{n}n!}{g}^{(n)}(x)=(-1{)}^n{P}_n(x).■$ Consequence: even $n$ $\Rightarrow$ $P_n$ even (sirf even powers); odd $n$ $\Rightarrow$ $P_n$ odd (sirf odd powers). Ab hum ise freely use karte hain.

WHAT/WHY: Normalization convention hai $P_n(1)=1$ har $n$ ke liye, toh koi computation nahi chahiye. Parity fact $P_n(-x)=(-1{)}^n{P}_n(x)$ (Problem 0) kehta hai $P_n(-1)=(-1{)}^n$.

• $P_5(1)=1$.
• $P_5(-1)=(-1{)}^5=-1$.
• $n=5$ odd hai $\Rightarrow$ $P_5$ ek odd function hai. Sanity check explicit form $P_5=\frac{1}{8}(63x^5-70x^3+15x)$ ke against: $x=1$ pe, $\frac{1}{8}(63-70+15)=\frac{1}{8}(8)=1$. ✔ Sirf odd powers appear karte hain, oddness confirm karti hai.

WHAT: Equation ko $y^{\prime \prime }$ coefficient expose karke likho: $(1-x^2)y^{\prime \prime }-2x{y}^{\prime }+n(n+1)y=0$. Wo point singular hota hai jahan leading coefficient $(1-x^2)$ vanish karta hai. WHY: $1-x^2=0\Rightarrow x=\pm 1$. Un points pe ODE apna "highest-derivative dominance" kho deta hai, toh power-series behaviour break ho sakti hai.

• Singular points: $x=+1$ aur $x=-1$ (ye regular singular points hain).
• Kyunki $x=\cos \theta$, ye $\theta =0$ aur $\theta =\pi$ hain: sphere ke north aur south poles.

WHY $a_1=0$ (parity argument): recurrence $a_{m+2}=(\cdots )a_m$ coefficients ko do-do alag link karta hai, isliye solution do alag chains mein split hoti hai — even chain $a_0\to a_2\to a_4\to \cdots$ aur ek bilkul alag odd chain $a_1\to a_3\to a_5\to \cdots$. Ye dono chains kabhi mix nahi hote. Problem 0 se, $P_4$ (even $n$) ek even function hai, jisme sirf $x$ ki even powers hain. Even powers saari even chain mein hain, isliye poori odd chain absent honi chahiye — yaani hum iska seed $a_1=0$ set karte hain. Sirf even chain reh jaati hai. WHAT/WHY ratios: $n=4$ ke saath recurrence ka numerator $m(m+1)-20$ hai. $m=0,2,4$ par chalte hain:

• $m=0$: $\frac{0-20}{2\cdot 1}=\frac{-20}{2}=-10$, toh $a_2/a_0=-10$.
• $m=2$: $\frac{6-20}{4\cdot 3}=\frac{-14}{12}=-\frac{7}{6}$, toh $a_4/a_2=-\frac{7}{6}$.
• $m=4$: $\frac{20-20}{6\cdot 5}=0$, toh $a_6/a_4=0$. Termination: $m=n=4$ pe numerator $0$ ho jaata hai, $a_6$ aur uske baad ke saare even coefficients ko khatam kar deta hai. Bachne wali powers hain $x^0,x^2,x^4$ $\Rightarrow$ degree 4. ✔ $P_4$ se match karta hai.

WHAT: $a_2=-10a_0$ aur $a_4=-\frac{7}{6}{a}_2=-\frac{7}{6}(-10a_0)=\frac{70}{6}{a}_0=\frac{35}{3}{a}_0$ ke saath, $y=a_0(1-10x^2+\frac{35}{3}{x}^4).$ WHY normalize: ODE shape fix karta hai lekin ek free constant $a_0$ chhod deta hai; convention $P_4(1)=1$ use pin karta hai. $x=1$ pe: $y(1)=a_0(1-10+\frac{35}{3})=a_0\cdot \frac{3-30+35}{3}=a_0\cdot \frac{8}{3}$. Ise $=1$ set karne pe $a_0=\frac{3}{8}$ milta hai. $P_4(x)=\frac{3}{8}(1-10x^2+\frac{35}{3}{x}^4)=\frac{1}{8}(35x^4-30x^2+3).✔$

WHAT/WHY Rodrigues: yeh $P_n$ ko ek known polynomial $(x^2-1{)}^n$ ko exactly $n$ baar differentiate karke produce karta hai, recurrence bookkeeping se bachata hai. $n=3$ ke saath: $P_3=\frac{1}{2^3\cdot 3!}\frac{d^3}{d{x}^3}(x^2-1{)}^3=\frac{1}{48}\frac{d^3}{d{x}^3}(x^2-1{)}^3$. $(x^2-1{)}^3=x^6-3x^4+3x^2-1$ expand karo. Teen baar differentiate karo:

• $\frac{d}{dx}:6x^5-12x^3+6x$
• $\frac{d^2}{d{x}^2}:30x^4-36x^2+6$
• $\frac{d^3}{d{x}^3}:120x^3-72x$ Toh $P_3=\frac{1}{48}(120x^3-72x)=\frac{120}{48}{x}^3-\frac{72}{48}x=\frac{5}{2}{x}^3-\frac{3}{2}x=\frac{1}{2}(5x^3-3x).✔$

WHY parity pehle: $P_1=x$ odd hai, $P_2=\frac{1}{2}(3x^2-1)$ even hai. Product = odd $\times$ even = odd function. Kisi bhi odd function ka integral symmetric interval $[-1,1]$ pe $0$ hota hai — left half right half ko cancel karta hai (figure dekho). Direct check: ${\int }_{-1}^{1}x\cdot \frac{1}{2}(3x^2-1)dx=\frac{1}{2}{\int }_{-1}^1(3x^3-x)dx.$ $3x^3$ aur $x$ dono odd hain $\Rightarrow$ dono $[-1,1]$ pe $0$ integrate ho jaate hain. Isliye integral $0$ hai. ✔ Yeh concrete banaya gaya orthogonality hai.

WHAT: $P_2^2=\frac{1}{4}(3x^2-1{)}^2=\frac{1}{4}(9x^4-6x^2+1)$. ${\int }_{-1}^{1}9x^{4}dx=9\cdot \frac{2}{5}=\frac{18}{5},{\int }_{-1}^{1}6x^{2}dx=6\cdot \frac{2}{3}=4,{\int }_{-1}^{1}1dx=2.$ Toh ${\int }_{-1}^1{P}_2^2=\frac{1}{4}(\frac{18}{5}-4+2)=\frac{1}{4}(\frac{18}{5}-2)=\frac{1}{4}\cdot \frac{8}{5}=\frac{2}{5}.$ ✔ Matlab: $\frac{2}{2n+1}$ $P_n$ ka $L^2[-1,1]$ inner product mein "squared length" hai. Yeh woh divisor hai jo tumhe chahiye jab tum kisi function ko Legendre polynomials mein expand karte ho taaki coefficient $c_n$ nikaal sako — bilkul Fourier cosine series mein $\pi$ se divide karne ki tarah. Note karo ki yeh exactly $m=n$ (diagonal, ${\delta }_{mn}=1$) case hai orthogonality tool ka.

WHAT: $y=\frac{1}{2}(3x^2-1)$, toh $y^{\prime }=3x$, $y^{\prime \prime }=3$. $(1-x^2)y^{\prime \prime }-2x{y}^{\prime }+n(n+1)y$ mein $n(n+1)=6$ ke saath plug karo: $(1-x^2)(3)-2x(3x)+6\cdot \frac{1}{2}(3x^2-1).$ $=3-3x^2-6x^2+3(3x^2-1)=3-3x^2-6x^2+9x^2-3.$ Collect karo: constants $3-3=0$; $x^2$ terms $-3-6+9=0$. Total $=0$. ✔ Toh $P_2$ genuinely ek solution hai, sirf fit nahi.

(a) Matching. Kyunki $P_2=\frac{1}{2}(3x^2-1)$, $x^2$ ke liye solve karo: $x^2=\frac{1}{3}(2P_2+1)=\frac{2}{3}{P}_2+\frac{1}{3}$. Aur $P_0=1$, toh $x^2=\frac{1}{3}{P}_0+\frac{2}{3}{P}_2\Rightarrow c_0=\frac{1}{3},c_2=\frac{2}{3}.$ (b) Orthogonality formula $c_n=\frac{2n+1}{2}{\int }_{-1}^{1}f{P}_{n}dx$.

• $c_0=\frac{1}{2}{\int }_{-1}^1{x}^2\cdot 1dx=\frac{1}{2}\cdot \frac{2}{3}=\frac{1}{3}.$ ✔
• $c_2=\frac{5}{2}{\int }_{-1}^1{x}^2\cdot \frac{1}{2}(3x^2-1)dx=\frac{5}{4}{\int }_{-1}^1(3x^4-x^2)dx=\frac{5}{4}(3\cdot \frac{2}{5}-\frac{2}{3})=\frac{5}{4}(\frac{6}{5}-\frac{2}{3}).$ $\frac{6}{5}-\frac{2}{3}=\frac{18-10}{15}=\frac{8}{15}$, toh $c_2=\frac{5}{4}\cdot \frac{8}{15}=\frac{40}{60}=\frac{2}{3}.$ ✔ Dono methods dete hain $c_0=\frac{1}{3},c_2=\frac{2}{3}$ — "Fourier-in-polynomials" idea action mein.

WHAT: $x^3{P}_1=x^3\cdot x=x^4$. Hum ${\int }_{-1}^1{x}^{4}dx$ chahte hain, lekin aao "structural" tarika karein machinery dikhane ke liye. $x^4$ ko Legendre form mein likho. $P_4=\frac{1}{8}(35x^4-30x^2+3)$ se milta hai $35x^4=8P_4+30x^2-3$. $x^2=\frac{1}{3}{P}_0+\frac{2}{3}{P}_2$ (Problem 9) use karke: $30x^2=10P_0+20P_2$, aur $3=3P_0$. Toh $35x^4=8P_4+10P_0+20P_2-3P_0=8P_4+20P_2+7P_0.$ $x^4=\frac{8}{35}{P}_4+\frac{4}{7}{P}_2+\frac{1}{5}{P}_0.$ WHY yeh help karta hai: ${\int }_{-1}^1{x}^{4}dx={\int }_{-1}^1{x}^4{P}_{0}dx$, aur orthogonality se sirf $P_0$-component bachta hai: ${\int }_{-1}^1{x}^{4}dx=\frac{1}{5}{\int }_{-1}^1{P}_0^{2}dx=\frac{1}{5}\cdot 2=\frac{2}{5}.$ Direct check: ${\int }_{-1}^1{x}^{4}dx=\frac{2}{5}$. ✔ Same answer, koi brute force nahi — orthogonality ne projection kar di.

WHAT: Standard form ko singular point $x=1$ pe evaluate karo. Wahan $(1-x^2)=0$, toh $y^{\prime \prime }$ term drop ho jaata hai: $0\cdot y^{\prime \prime }(1)-2(1)y^{\prime }(1)+n(n+1)y(1)=0\Rightarrow -2y^{\prime }(1)+n(n+1)y(1)=0.$ Yeh boundary relation deta hai $y^{\prime }(1)=\frac{1}{2}n(n+1)y(1)$. WHY yeh nonzero value force karta hai: maano $y(1)=0$. Toh $y^{\prime }(1)=0$ bhi. Lekin ek 2nd-order linear ODE ka bounded polynomial solution regular singular point ke paas uski value aur derivative se determine hota hai; dono zero hone se $y\equiv 0$ (trivial solution) force hoti hai. Kyunki $P_n$ ek nontrivial degree-$n$ polynomial hai, $y(1)\ne 0$. Isliye hum $y(1)$ se divide kar sakte hain $P_n(1)=1$ normalize karne ke liye, aur yeh choice consistent hai. $■$ Numeric sanity ($n=2$): $P_2^{\prime }(1)=3\cdot 1=3$ aur $\frac{1}{2}\cdot 2\cdot 3\cdot P_2(1)=3\cdot 1=3$. ✔ boundary relation hold karti hai.

WHAT (self-adjoint form): equation hai $\frac{d}{dx}[(1-x^2)y^{\prime }]+n(n+1)y=0$. Maano $L[y]=\frac{d}{dx}[(1-x^2)y^{\prime }]$. Maano $P_m,P_n$ ise eigenvalues ${\lambda }_m=m(m+1),{\lambda }_n=n(n+1)$ ke saath solve karte hain: $L[P_m]=-{\lambda }_m{P}_m,L[P_n]=-{\lambda }_n{P}_n.$ WHY multiply-and-subtract: $({\lambda }_n-{\lambda }_m)$ times overlap integral isolate karne ke liye. Pehle ko $P_n$ se multiply karo, doosre ko $P_m$ se, subtract karo, $[-1,1]$ pe integrate karo: ${\int }_{-1}^1(P_{n}L[P_m]-P_{m}L[P_n])dx=({\lambda }_n-{\lambda }_m){\int }_{-1}^1{P}_m{P}_{n}dx.$ Left side ek boundary term hai (Lagrange's identity): $P_{n}L[P_m]-P_{m}L[P_n]=\frac{d}{dx}[(1-x^2)(P_n{P}_m^{\prime }-P_m{P}_n^{\prime }\left)\right].$ (Derivative expand karke verify karo: cross terms $(1-x^2)(P_n{P}_m^{\prime \prime }-P_m{P}_n^{\prime \prime })$ aur $-2x(\dots )$ $P_{n}L[P_m]-P_{m}L[P_n]$ reconstruct karte hain.) Total derivative integrate karne par: $[(1-x^2)(P_n{P}_m^{\prime }-P_m{P}_n^{\prime }){]}_{-1}^1.$ KAHAN yeh khatam hoti hai: factor $(1-x^2)$ dono endpoints $x=\pm 1$ pe zero hai. Toh poora boundary term vanish ho jaata hai — yahi wajah hai ki ODE ke singular points exactly woh hain jo ise $[-1,1]$ pe self-adjoint banate hain. Isliye $({\lambda }_n-{\lambda }_m){\int }_{-1}^1{P}_m{P}_{n}dx=0.$ $m\ne n$ ke liye, ${\lambda }_n\ne {\lambda }_m$, toh ${\int }_{-1}^1{P}_m{P}_{n}dx=0$. $■$ Is instance ka general theorem Sturm-Liouville Theory mein dekho.