Legendre Polynomial
Show that the coefficient of \( x^n \) in \( P_n(x) \) is \( \frac{(2n)!}{2^n(n!)^2} \)
Solution: The n-th Legendre Polynomial is given by
\[ P_n(x)=\frac{1}{2^nn!}\frac{d^n}{dx^n}\left( x^2-1\right)^n \]
Expanding \( \left( x^2-1 \right) \) using Binomial theorem, we get
\begin{align*}
P_n(x) &=\frac{1}{2^nn!}\frac{d^n}{dx^n}\left[ x^{2n}-nx^{2(n-1)}+\cdots \right]\\[0.2cm]
P_n(x) &=\frac{1}{2^nn!}\left[ \frac{d^n}{dx^n} \left( x^{2n}\right) -\cdots \right]\\[0.2cm]
P_n(x) &=\frac{1}{2^nn!}\left[ (2n)(2n-1)\ldots(2n-(n-1))x^{2n-n} -\cdots \right]\\[0.2cm]
P_n(x) &=\frac{1}{2^nn!}\left[ (2n)(2n-1)\ldots(n+1)x^{n} -\cdots \right]\\[0.2cm]
P_n(x) &=\frac{1}{2^nn!}\left[ \frac{(2n)(2n-1)\ldots(n+1)\cdot n(n-1)\ldots 3\cdot 2\cdot 1}{n(n-1)\ldots 3\cdot 2\cdot 1}x^{n} -\cdots \right]\\[0.2cm]
P_n(x) &=\frac{1}{2^nn!}\left[ \frac{(2n)!}{n!}x^{n} -\cdots \right]\\[0.2cm]
P_n(x) &=\frac{(2n)!}{2^n(n!)^2}x^n -\cdots
\end{align*}
Therefore the coefficient of \( x^n \) in \( P_n(x) \) is \( \frac{(2n)!}{2^n(n!)^2} \)
Show that \( \int_{-1}^1P_n(x)P_m(x)\, dx =0 , \quad (n\neq m )\)
Solution: Consider the Legendre equation
\begin{align*}
\left( 1-x^2 \right)y''-2xy'&+n(n+1)y=0 \quad \ldots (1) \\[0.2cm]
\Rightarrow \left[ \left( 1-x^2 \right) y' \right]'&+n(n+1)y=0 \\[0.2cm]
\Rightarrow \left[ \left( 1-x^2 \right) y' \right]'&=-n(n+1)y
\end{align*}
Since \( P_n(x) \) is the solution of the equation \( (1) \) we then have,
\begin{align*}
\left[ \left( 1-x^2 \right) P'_n(x) \right]'&=-n(n+1)P_n(x)\quad \ldots (2)
\end{align*}
Similarly for the solution \( P_m(x) \) of the Legendre equation \( \left( 1-x^2 \right)y''-2xy'+m(m+1)y=0 \), we have,
\begin{align*}
\left[ \left( 1-x^2 \right) P_m'(x) \right]'&=-m(m+1)P_m(x)\quad \ldots (3)
\end{align*}
Multiplying the equation \( (2) \) by \( P_m(x) \) and the equation \( (3) \) by \( P_n(x) \), we have,
\begin{align*}
P_m(x)\left[ \left( 1-x^2 \right) P'_n(x) \right]'&=-n(n+1)P_n(x)P_m(x)\quad \ldots (4)\\[0.2cm]
P_n(x)\left[ \left( 1-x^2 \right) P_m'(x) \right]'&=-m(m+1)P_n(x)P_m(x)\quad \ldots (5)
\end{align*}
Subtracting equation \((5)\) from equation \((4)\), we get,
\begin{align*}
P_m(x)\left[ \left( 1-x^2 \right) P'_n(x) \right]'- P_n(x)\left[ \left( 1-x^2 \right) P_m'(x) \right]'&=\left[ m(m+1)-n(n+1)\right]P_n(x)P_m(x)\\[0.2cm]
\left\{ \left(1-x^2\right) \left[ P_m(x)P_n'(x)-P_m'(x)P_n(x) \right] \right\}'&=\left[ m(m+1)-n(n+1)\right]P_n(x)P_m(x)
\end{align*}
Integrating both sides from \( -1 \) to \( 1 \), we get
\begin{align*}
\left\{ \left(1-x^2\right) \left[ P_m(x)P_n'(x)-P_m'(x)P_n(x) \right] \right\}_{-1}^1&=\left[ m(m+1)-n(n+1)\right]\int_{-1}^1P_n(x)P_m(x)\,dx \\[0.2cm]
0&=\left[ m(m+1)-n(n+1)\right]\int_{-1}^1P_n(x)P_m(x)\,dx \\[0.2cm]
0&=\int_{-1}^1P_n(x)P_m(x)\,dx \\[0.2cm]
\Rightarrow \int_{-1}^1P_n(x)P_m(x)\,dx &=0
\end{align*}
Show that \( P_n(-x)=(-1)^nP_n(x) \) and hence show that \( P_n(-1)=(-1)^n \)
Solution: The n-th Legendre Polynomial is given by
\[ P_n(x)=\frac{1}{2^nn!}\frac{d^n}{dx^n}\left( x^2-1\right)^n \]
Replacing \( x\) by \( -x \), we have
\begin{align*}
P_n(-x)&=\frac{1}{2^nn!}\frac{\mathrm{d}^n}{\mathrm{d}(-x)^n}\left( (-x)^2-1 \right)^n \\[0.3cm]
&=\frac{1}{2^nn!}\frac{\mathrm{d}^n}{\mathrm{d}(-x)^n}\left( x^2-1 \right)^n \\[0.3cm]
&=\frac{1}{2^nn!}\frac{\mathrm{d}^{n-1}}{\mathrm{d}(-x)^{n-1}}\left\{ \frac{\mathrm{d}}{\mathrm{d}(-x)} \left( x^2-1 \right)^n \right\} \\[0.3cm]
&=\frac{1}{2^nn!}\frac{\mathrm{d}^{n-1}}{\mathrm{d}(-x)^{n-1}}\left\{ \frac{\frac{\mathrm{d}}{\mathrm{d}x}\left(x^2-1 \right)^n}{\frac{\mathrm{d}}{\mathrm{d}x}(-x)} \right\} \\[0.3cm]
&=\frac{1}{2^nn!}\frac{\mathrm{d}^{n-1}}{\mathrm{d}(-x)^{n-1}}\left\{ \frac{\frac{\mathrm{d}}{\mathrm{d}x}\left(x^2-1 \right)^n}{(-1)} \right\} \\[0.3cm]
&=(-1)^1\frac{1}{2^nn!}\frac{\mathrm{d}^{n-2}}{\mathrm{d}(-x)^{n-2}}\left\{ \frac{\mathrm{d}}{\mathrm{d}(-x)} \left[ \frac{\mathrm{d}}{\mathrm{d}x}\left(x^2-1 \right)^n \right] \right\} \\[0.3cm]
&=(-1)^1\frac{1}{2^nn!}\frac{\mathrm{d}^{n-2}}{\mathrm{d}(-x)^{n-2}}\left\{ \frac{\frac{\mathrm{d}}{\mathrm{d}x} \left[ \frac{\mathrm{d}}{\mathrm{d}x}\left(x^2-1 \right)^n \right]}{\frac{\mathrm{d}}{\mathrm{d}x}(-x)} \right\} \\[0.3cm]
&=(-1)^1\frac{1}{2^nn!}\frac{\mathrm{d}^{n-2}}{\mathrm{d}(-x)^{n-2}}\left\{ \frac{\frac{\mathrm{d}^2}{\mathrm{d}x^2}\left( x^2-1 \right)^n}{-1} \right\} \\[0.3cm]
&=(-1)^2\frac{1}{2^nn!}\frac{\mathrm{d}^{n-3}}{\mathrm{d}(-x)^{n-3}}\left\{ \frac{\mathrm{d}}{\mathrm{d}(-x)} \left[ \frac{\mathrm{d}^2}{\mathrm{d}x^2}\left(x^2-1 \right)^n \right] \right\} \\[0.3cm]
&=(-1)^2\frac{1}{2^nn!}\frac{\mathrm{d}^{n-3}}{\mathrm{d}(-x)^{n-3}}\left\{ \frac{\frac{\mathrm{d}^3}{\mathrm{d}x^3}\left( x^2-1 \right)^n}{-1} \right\} \\[0.3cm]
&=(-1)^3\frac{1}{2^nn!}\frac{\mathrm{d}^{n-3}}{\mathrm{d}(-x)^{n-3}}\left\{ {\frac{\mathrm{d}^3}{\mathrm{d}x^3}\left( x^2-1 \right)^n} \right\} \\[0.3cm]
&\vdots \\[0.3cm]
P_n(-x)&=(-1)^{n-1}\frac{1}{2^nn!}\frac{\mathrm{d}}{\mathrm{d}(-x)}\left\{ {\frac{\mathrm{d}^{n-1}}{\mathrm{d}x^{n-1}}\left( x^2-1 \right)^n} \right\} \\[0.3cm]
&= (-1)^{n}\frac{1}{2^nn!}\frac{\mathrm{d}^n}{\mathrm{d}x^n}\left( x^2-1\right)^n\\[0.3cm]
\Rightarrow P_n(-x)&= (-1)^n P_n(x)
\end{align*}
Now, put \( x=1 \)
\begin{align*}
P_n(-1)&=(-1)^n P_n(1) \\[0.3cm]
\Rightarrow P_n(-1)&=(-1)^n \quad \quad [ \text{since}\, P_n(1)=1 ]
\end{align*}
Show that \( \int_{-1}^1 P_n^2(x) \, \mathrm{d}x = \frac{2}{2n+1} \)
Solution: The n-th Legendre Polynomial is given by
\[ P_n(x)=\frac{1}{2^nn!}\frac{d^n}{dx^n}\left( x^2-1\right)^n \\[0.3cm] \]
\begin{align*}
\Rightarrow \int_{-1}^1P_n^2(x)\, \mathrm{d}x &= \int_{-1}^1 \frac{1}{2^nn!}\frac{d^n}{dx^n}\left( x^2-1\right)^n \cdot \frac{1}{2^nn!}\frac{d^n}{dx^n}\left( x^2-1\right)^n \, \mathrm{d}x\\[0.3cm]
&= \frac{1}{\left( 2^n n! \right)^2}\int_{-1}^1 \underbrace{ \frac{d^n}{dx^n}\left( x^2-1\right)^n}_u \cdot \underbrace{\frac{d^n}{dx^n}\left( x^2-1\right)^n}_v \, \mathrm{d}x\\[0.3cm]
&= \frac{1}{\left( 2^n n! \right)^2}\left[ \underbrace{ \frac{d^n}{dx^n}\left( x^2-1\right)^n}_u \cdot \underbrace{\frac{d^{n-1}}{dx^{n-1}}\left( x^2-1\right)^n}_{\int v} \right]_{-1}^1-\frac{1}{\left( 2^n n! \right)^2}\int_{-1}^1 \underbrace{ \frac{d^{n+1}}{dx^{n+1}}\left( x^2-1\right)^n}_{\frac{du}{dx}} \cdot \underbrace{\frac{d^{n-1}}{dx^{n-1}}\left( x^2-1\right)^n}_{\int v} \, \mathrm{d}x\\[0.3cm]
&= 0-\frac{1}{\left( 2^n n! \right)^2}\int_{-1}^1 \frac{d^{n+1}}{dx^{n+1}}\left( x^2-1\right)^n \cdot \frac{d^{n-1}}{dx^{n-1}}\left( x^2-1\right)^n \, \mathrm{d}x\\[0.3cm]
&= \frac{(-1)^1}{\left( 2^n n! \right)^2}\int_{-1}^1 \underbrace{ \frac{d^{n+1}}{dx^{n+1}}\left( x^2-1\right)^n}_{u} \cdot \underbrace{\frac{d^{n-1}}{dx^{n-1}}\left( x^2-1\right)^n}_{v} \, \mathrm{d}x\\[0.3cm]
&= \frac{(-1)^1}{\left( 2^n n! \right)^2}\left[ \underbrace{ \frac{d^{n+1}}{dx^{n+1}}\left( x^2-1\right)^n}_u \cdot \underbrace{\frac{d^{n-2}}{dx^{n-2}}\left( x^2-1\right)^n}_{\int v} \right]_{-1}^1-\frac{(-1)^1}{\left( 2^n n! \right)^2}\int_{-1}^1 \underbrace{ \frac{d^{n+2}}{dx^{n+2}}\left( x^2-1\right)^n}_{\frac{du}{dx}} \cdot \underbrace{\frac{d^{n-2}}{dx^{n-2}}\left( x^2-1\right)^n}_{\int v} \, \mathrm{d}x\\[0.3cm]
&= 0-\frac{(-1)^1}{\left( 2^n n! \right)^2}\int_{-1}^1 \frac{d^{n+2}}{dx^{n+2}}\left( x^2-1\right)^n \cdot \frac{d^{n-2}}{dx^{n-2}}\left( x^2-1\right)^n \, \mathrm{d}x\\[0.3cm]
&= \frac{(-1)^2}{\left( 2^n n! \right)^2}\int_{-1}^1 \frac{d^{n+2}}{dx^{n+2}}\left( x^2-1\right)^n \cdot \frac{d^{n-2}}{dx^{n-2}}\left( x^2-1\right)^n \, \mathrm{d}x\\[0.3cm]
&= \frac{(-1)^3}{\left( 2^n n! \right)^2}\int_{-1}^1 \frac{d^{n+3}}{dx^{n+3}}\left( x^2-1\right)^n \cdot \frac{d^{n-3}}{dx^{n-3}}\left( x^2-1\right)^n \, \mathrm{d}x\\[0.3cm]
&= \frac{(-1)^4}{\left( 2^n n! \right)^2}\int_{-1}^1 \frac{d^{n+4}}{dx^{n+4}}\left( x^2-1\right)^n \cdot \frac{d^{n-4}}{dx^{n-4}}\left( x^2-1\right)^n \, \mathrm{d}x\\[0.3cm]
&\vdots \\[0.3cm]
&= \frac{(-1)^n}{\left( 2^n n! \right)^2}\int_{-1}^1 \frac{d^{2n}}{dx^{2n}}\left( x^2-1\right)^n \cdot \left( x^2-1\right)^n \, \mathrm{d}x\\[0.3cm]
&= \frac{(-1)^n}{\left( 2^n n! \right)^2}\int_{-1}^1 (2n)! \cdot \left( x^2-1\right)^n \, \mathrm{d}x\\[0.3cm]
&= \frac{(-1)^n(2n)!}{\left( 2^n n! \right)^2}\int_{-1}^1 \left( x^2-1\right)^n \, \mathrm{d}x\\[0.3cm]
&= \frac{(-1)^n(2n)!}{\left( 2^n n! \right)^2}\int_{-1}^1 2^{2n+1}(-1)^nt^n(1-t)^n \, \mathrm{d}t,\quad \quad ( \, \text{by setting} \, x =2t-1 )\\[0.3cm]
&= \frac{(-1)^{2n}(2n)!2^{2n+1}}{\left( 2^n n! \right)^2}\int_{-1}^1 t^n(1-t)^n \, \mathrm{d}t\\[0.3cm]
&= \frac{(2n)!\cdot 2}{(n!)^2} \,B[n+1,n+1], \quad \quad \text{where}\, B[m,n] \, \text{ is Beta Function} \\[0.3cm]
&= \frac{(2n)!\cdot 2}{(n!)^2} \, \frac{\Gamma{(n+1)}\cdot\Gamma{(n+1)}}{\Gamma{(2n+2)}}, \quad \quad \text{where}\, B[m,n]=\frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)}\\[0.3cm]
&= \frac{(2n)!\cdot 2}{(n!)^2} \, \frac{n!\cdot n!}{(2n+1)!}, \quad \quad \text{where}\, \Gamma(n+1)=n!\\[0.3cm]
&= \frac{(2n)!\cdot 2}{(2n+1)(2n)!}\\[0.3cm]
\Rightarrow \int_{-1}^1P_n^2(x)\, \mathrm{d}x&= \frac{2}{2n+1}
\end{align*}
