Recursive Relations of Legendre Polynomials
📂Functions Recursive Relations of Legendre Polynomials Theorem P l ( x ) = 1 2 l l ! d l d x l ( x 2 − 1 ) l (1)
P_{l}(x)=\dfrac{1}{2^l l!} \dfrac{d^l}{dx^l}(x^2-1)^l \tag{1}
P l ( x ) = 2 l l ! 1 d x l d l ( x 2 − 1 ) l ( 1 )
Such a Legendre Polynomial P l P_{l} P l
P l + 1 ′ ( x ) − P l − 1 ′ ( x ) = ( 2 l + 1 ) P l ( x ) (a)
P^{\prime}_{l+1}(x)-P^{\prime}_{l-1}(x)=(2l+1)P_{l}(x) \tag{a}
P l + 1 ′ ( x ) − P l − 1 ′ ( x ) = ( 2 l + 1 ) P l ( x ) ( a )
l P l ( x ) = ( 2 l − 1 ) x P l − 1 ( x ) − ( l − 1 ) P l − 2 ( x ) (b)
lP_{l}(x)=(2l-1)xP_{l-1}(x)-(l-1)P_{l-2}(x) \tag{b}
l P l ( x ) = ( 2 l − 1 ) x P l − 1 ( x ) − ( l − 1 ) P l − 2 ( x ) ( b )
x P l ′ ( x ) − P l − 1 ′ ( x ) = l P l ( x ) (c)
xP^{\prime}_{l}(x)-P^{\prime}_{l-1}(x)=lP_{l}(x)\tag{c}
x P l ′ ( x ) − P l − 1 ′ ( x ) = l P l ( x ) ( c )
Proof ( a ) (a) ( a ) First, if we calculate the derivative of P l ( x ) P_{l}(x) P l ( x )
d d x P l ( x ) = 1 2 l l ! d d x d l d x l ( x 2 − 1 ) l = 1 2 l l ! d l d x l d d x ( x 2 − 1 ) l = 2 l 2 l l ! d l d x l [ x ( x 2 − 1 ) l − 1 ] = 1 2 l − 1 ( l − 1 ) ! d l − 1 d x l − 1 [ ( x 2 − 1 ) l − 1 + 2 ( l − 1 ) x 2 ( x 2 − 1 ) l − 2 ] = 1 2 l − 1 ( l − 1 ) ! d l − 1 d x l − 1 [ ( x 2 − 1 ) + 2 l x 2 − 2 x 2 ] ( x 2 − 1 ) l − 2 = 1 2 l − 1 ( l − 1 ) ! d l − 1 d x l − 1 [ ( 2 l − 1 ) x 2 − 1 ] ( x 2 − 1 ) l − 2
\begin{align*}
\frac{d}{dx}P_{l}(x)
&= \frac{1}{2^l l!}\frac{d}{dx} \dfrac{d^l}{dx^l}(x^2-1)^l \\
&= \frac{1}{ 2^{l}l! }\frac{ d ^{l} }{ dx^{l} }\frac{d}{dx}(x^{2}-1)^{l} \\
&= \frac{2l}{ 2^{l}l! }\frac{ d ^{l} }{ dx^{l} }\left[ x(x^{2}-1)^{l-1} \right] \\
&= \frac{1}{ 2^{l-1}(l-1)! }\frac{ d ^{l-1} }{ dx^{l-1} }\left[ (x^{2}-1)^{l-1}+2(l-1)x^{2}(x^{2}-1)^{l-2} \right] \\
&= \frac{1}{ 2^{l-1}(l-1)! }\frac{ d ^{l-1} }{ dx^{l-1} }[(x^{2}-1)+2lx^{2}-2x^{2}] (x^{2}-1)^{l-2} \\
&= \frac{1}{ 2^{l-1}(l-1)! }\frac{ d ^{l-1} }{ dx^{l-1} }[(2l-1)x^{2}-1] (x^{2}-1)^{l-2}
\end{align*}
d x d P l ( x ) = 2 l l ! 1 d x d d x l d l ( x 2 − 1 ) l = 2 l l ! 1 d x l d l d x d ( x 2 − 1 ) l = 2 l l ! 2 l d x l d l [ x ( x 2 − 1 ) l − 1 ] = 2 l − 1 ( l − 1 )! 1 d x l − 1 d l − 1 [ ( x 2 − 1 ) l − 1 + 2 ( l − 1 ) x 2 ( x 2 − 1 ) l − 2 ] = 2 l − 1 ( l − 1 )! 1 d x l − 1 d l − 1 [( x 2 − 1 ) + 2 l x 2 − 2 x 2 ] ( x 2 − 1 ) l − 2 = 2 l − 1 ( l − 1 )! 1 d x l − 1 d l − 1 [( 2 l − 1 ) x 2 − 1 ] ( x 2 − 1 ) l − 2
and substitute l + 1 l+1 l + 1 l l l
P l + 1 ′ ( x ) = 1 2 l ! d l d x l [ ( 2 l + 1 ) x 2 − 1 ] ( x 2 − 1 ) l − 1 (2)
P^{\prime}_{l+1}(x)=\frac{1}{2l!}\frac{ d ^{l} }{ dx^{l} }[(2l+1)x^{2}-1] (x^{2} -1 )^{l-1} \tag{2}
P l + 1 ′ ( x ) = 2 l ! 1 d x l d l [( 2 l + 1 ) x 2 − 1 ] ( x 2 − 1 ) l − 1 ( 2 )
Then, substitute l l l l − 1 l-1 l − 1 ( 1 ) (1) ( 1 )
P l − 1 ′ ( x ) = 1 2 l − 1 ( l − 1 ) ! d d x d l − 1 d x l − 1 ( x 2 − 1 ) l − 1 = 2 l 2 l l ! d l d x l ( x 2 − 1 ) l − 1
\begin{align*}
P^{\prime}_{l-1}(x)
&= \frac{1}{2^{l-1}(l-1)!}\frac{d}{dx} \frac{ d^{l-1} }{ d x^{l-1} }(x^{2}-1)^{l-1} \\
&= \frac{2l}{2^{l}l!}\frac{ d^{l} }{ dx^{l} }(x^{2}-1)^{l-1} \tag{3}
\end{align*}
P l − 1 ′ ( x ) = 2 l − 1 ( l − 1 )! 1 d x d d x l − 1 d l − 1 ( x 2 − 1 ) l − 1 = 2 l l ! 2 l d x l d l ( x 2 − 1 ) l − 1 ( 3 )
Now, calculating ( 2 ) − ( 3 ) (2)-(3) ( 2 ) − ( 3 )
P l + 1 ′ ( x ) − P l − 1 ′ ( x ) = 1 2 l ! d l d x l [ ( 2 l + 1 ) x 2 − 1 ] ( x 2 − 1 ) l − 1 − 2 l 2 l l ! d l d x l ( x 2 − 1 ) l − 1 = 1 2 l l ! d l d x l [ ( 2 l + 1 ) x 2 − ( 2 l + 1 ) ] ( x 2 − 1 ) l − 1 = 1 2 l l ! d l d x l ( 2 l + 1 ) ( x 2 − 1 ) l = 2 l + 1 2 l l ! d l d x l ( x 2 − 1 ) l
\begin{align*}
P^{\prime}_{l+1}(x)-P^{\prime}_{l-1}(x)
&= \frac{1}{2l!}\frac{ d ^{l} }{ dx^{l} }[(2l+1)x^{2}-1] (x^{2}-1)^{l-1} -\frac{2l}{2^{l}l!}\frac{ d^{l} }{ dx^{l} }(x^{2}-1)^{l-1} \\
&= \frac{1}{2^{l}l!}\frac{ d ^{l} }{ d x^{l} }[(2l+1)x^{2}-(2l+1)] (x^{2}-1)^{l-1} \\
&= \frac{1}{2^{l}l!}\frac{ d ^{l} }{ d x^{l} }(2l+1)(x^{2}-1)^{l} \\
&= \frac{2l+1}{2^{l}l!}\frac{ d ^{l} }{ d x^{l} }(x^{2}-1)^{l}
\end{align*}
P l + 1 ′ ( x ) − P l − 1 ′ ( x ) = 2 l ! 1 d x l d l [( 2 l + 1 ) x 2 − 1 ] ( x 2 − 1 ) l − 1 − 2 l l ! 2 l d x l d l ( x 2 − 1 ) l − 1 = 2 l l ! 1 d x l d l [( 2 l + 1 ) x 2 − ( 2 l + 1 )] ( x 2 − 1 ) l − 1 = 2 l l ! 1 d x l d l ( 2 l + 1 ) ( x 2 − 1 ) l = 2 l l ! 2 l + 1 d x l d l ( x 2 − 1 ) l
Due to ( 1 ) (1) ( 1 ) ( 2 l + 1 ) P l ( x ) (2l+1)P_{l}(x) ( 2 l + 1 ) P l ( x )
P l + 1 ′ ( x ) − P l − 1 ′ ( x ) = ( 2 l + 1 ) P l ( x )
P^{\prime}_{l+1}(x)-P^{\prime}_{l-1}(x)=(2l+1)P_{l}(x)
P l + 1 ′ ( x ) − P l − 1 ′ ( x ) = ( 2 l + 1 ) P l ( x )
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( b ) (b) ( b ) Generating Function of Legendre Polynomials
The function Φ ( x , h ) \Phi (x,h) Φ ( x , h )
Φ ( x , h ) = 1 1 − 2 x h + h 2 , ∣ h ∣ < 1
\Phi (x,h)=\frac{1}{\sqrt{1-2xh+h^{2}}},\quad |h|<1
Φ ( x , h ) = 1 − 2 x h + h 2 1 , ∣ h ∣ < 1
The generating function satisfies the following equation:
Φ ( x , h ) = P 0 ( x ) + h P 1 ( x ) + h 2 P 2 ( x ) + ⋯ = ∑ l = 0 ∞ h l P l ( x )
\Phi (x,h)=P_{0}(x)+hP_{1}(x)+h^{2}P_{2}(x)+\cdots =\sum \limits_{l=0}^{\infty}h^{l}P_{l}(x)
Φ ( x , h ) = P 0 ( x ) + h P 1 ( x ) + h 2 P 2 ( x ) + ⋯ = l = 0 ∑ ∞ h l P l ( x )
Part 2. Proof of Recursive Relation
First, differentiate Φ ( x , t ) \Phi (x,t) Φ ( x , t ) t t t ∂ Φ ∂ t = − 1 2 ( − 2 x + 2 t ) ( 1 − 2 x t + t 2 ) − 3 2 ⟹ ( 1 − 2 x t + t 2 ) ∂ Φ ∂ t = ( x − t ) Φ
\begin{align*}
&& \frac{ \partial \Phi}{ \partial t } &= -\frac{1}{2}(-2x+2t)(1-2xt+t^{2})^{-\frac{3}{2}} \\ \implies && (1-2xt+t^{2})\frac{ \partial \Phi}{ \partial t} &= (x-t)\Phi
\end{align*}
⟹ ∂ t ∂ Φ ( 1 − 2 x t + t 2 ) ∂ t ∂ Φ = − 2 1 ( − 2 x + 2 t ) ( 1 − 2 x t + t 2 ) − 2 3 = ( x − t ) Φ
Then, substitute Φ ( x , t ) = ∑ l = 0 ∞ t l P l ( x ) \Phi (x,t)=\sum \limits_{l=0}^{\infty}t^{l}P_{l}(x) Φ ( x , t ) = l = 0 ∑ ∞ t l P l ( x )
( 1 − 2 x t + t 2 ) ∑ l = 1 ∞ l t l − 1 P l ( x ) = ( x − t ) ∑ l = 0 ∞ t l P l ( x )
(1-2xt+t^{2})\sum \limits _{l=1}^{\infty}lt^{l-1}P_{l}(x)=(x-t)\sum\limits_{l=0}^{\infty}t^{l}P_{l}(x)
( 1 − 2 x t + t 2 ) l = 1 ∑ ∞ l t l − 1 P l ( x ) = ( x − t ) l = 0 ∑ ∞ t l P l ( x )
Aligning the indices,
( 1 − 2 x t + t 2 ) ∑ l = 0 ∞ ( l + 1 ) t l P l + 1 ( x ) = ( x − t ) ∑ l = 0 ∞ t l P l ( x )
(1-2xt+t^{2})\sum \limits _{l=0}^{\infty}(l+1)t^{l}P_{l+1}(x)=(x-t)\sum\limits_{l=0}^{\infty}t^{l}P_{l}(x)
( 1 − 2 x t + t 2 ) l = 0 ∑ ∞ ( l + 1 ) t l P l + 1 ( x ) = ( x − t ) l = 0 ∑ ∞ t l P l ( x )
This equation is an identity with respect to t t t t l t^{l} t l t l t^{l} t l
1 [ ( l + 1 ) t l P l + 1 ( x ) ] − 2 x t [ ( l ) t l − 1 P l ( x ) ] + t 2 [ ( l − 1 ) t l − 2 P l − 1 ( x ) ] = x [ t l P l ( x ) ] − t [ t l − 1 P l − 1 ( x ) ]
1 [(l+1)t^{l}P_{l+1}(x)] -2xt[(l)t^{l-1}P_{l}(x)]+t^{2}[(l-1)t^{l-2}P_{l-1}(x)]=x [t^{l}P_{l}(x)]-t[t^{l-1}P_{l-1}(x)]
1 [( l + 1 ) t l P l + 1 ( x )] − 2 x t [( l ) t l − 1 P l ( x )] + t 2 [( l − 1 ) t l − 2 P l − 1 ( x )] = x [ t l P l ( x )] − t [ t l − 1 P l − 1 ( x )]
⟹ ( l + 1 ) t l P l + 1 ( x ) − 2 x l t l P l ( x ) + ( l − 1 ) t l P l − 1 ( x ) = x t l P l ( x ) − t l P l − 1 ( x )
\implies (l+1)t^{l}P_{l+1}(x) -2xlt^{l}P_{l}(x)+(l-1)t^{l}P_{l-1}(x)=x t^{l}P_{l}(x)-t^{l}P_{l-1}(x)
⟹ ( l + 1 ) t l P l + 1 ( x ) − 2 x l t l P l ( x ) + ( l − 1 ) t l P l − 1 ( x ) = x t l P l ( x ) − t l P l − 1 ( x )
⟹ ( l + 1 ) P l + 1 ( x ) − 2 x l P l ( x ) + ( l − 1 ) P l − 1 ( x ) = x P l ( x ) − P l − 1 ( x )
\implies (l+1)P_{l+1}(x) -2xlP_{l}(x)+(l-1)P_{l-1}(x)=x P_{l}(x)-P_{l-1}(x)
⟹ ( l + 1 ) P l + 1 ( x ) − 2 x l P l ( x ) + ( l − 1 ) P l − 1 ( x ) = x P l ( x ) − P l − 1 ( x )
Here, if we substitute l − 1 l-1 l − 1 l l l
l P l ( x ) − 2 x ( l − 1 ) P l − 1 ( x ) + ( l − 2 ) P l − 2 ( x ) = x P l − 1 ( x ) − P l − 2 ( x )
lP_{l}(x) -2x(l-1)P_{l-1}(x)+(l-2)P_{l-2}(x)=x P_{l-1}(x)-P_{l-2}(x)
l P l ( x ) − 2 x ( l − 1 ) P l − 1 ( x ) + ( l − 2 ) P l − 2 ( x ) = x P l − 1 ( x ) − P l − 2 ( x )
And rearrange the left side with respect to P l ( x ) P_{l}(x) P l ( x )
l P l ( x ) = ( 2 l − 1 ) x P l − 1 ( x ) − ( l − 1 ) P l − 2 ( x )
lP_{l}(x) =(2l-1)x P_{l-1}(x)-(l-1)P_{l-2}(x)
l P l ( x ) = ( 2 l − 1 ) x P l − 1 ( x ) − ( l − 1 ) P l − 2 ( x )
■
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( c ) (c) ( c ) It is easy to verify that the generating function satisfies the following:
( x − h ) ∂ Φ ∂ x = h ∂ Φ ∂ h
(x-h)\frac{ \partial \Phi}{ \partial x }=h\frac{ \partial \Phi}{ \partial h }
( x − h ) ∂ x ∂ Φ = h ∂ h ∂ Φ
Now, substituting the series form Φ ( x , h ) = ∑ l = 0 ∞ h l P l ( x ) \Phi (x,h)=\sum\limits_{l=0}^{\infty}h^{l}P_{l}(x) Φ ( x , h ) = l = 0 ∑ ∞ h l P l ( x )
( x − h ) ∑ l = 0 ∞ h l P l ′ ( x ) = h ∑ l = 1 ∞ l h l − 1 P l ( x )
(x-h)\sum\limits_{l=0}^{\infty}h^{l}P^{\prime}_{l}(x)=h\sum\limits_{l=1}^{\infty}lh^{l-1}P_{l}(x)
( x − h ) l = 0 ∑ ∞ h l P l ′ ( x ) = h l = 1 ∑ ∞ l h l − 1 P l ( x )
The above equation is an identity with respect to h h h h l h^{l} h l h l h^{l} h l
x [ h l P l ′ ( x ) ] − h [ h l − 1 P l − 1 ′ ( x ) ] = h [ l h l − 1 P l ( x ) ] ⟹ h l [ x P l ′ ( x ) − P l − 1 ′ ( x ) ] = h l l P l ( x ) ⟹ x P l ′ ( x ) − P l − 1 ′ ( x ) = l P l ( x )
\begin{align*}
&& x[h^{l}P^{\prime}_{l}(x)]-h[h^{l-1}P^{\prime}_{l-1}(x)] = h[lh^{l-1}P_{l}(x)] \\
\implies && h^{l}[xP^{\prime}_{l}(x)-P^{\prime}_{l-1}(x)] = h^{l}lP_{l}(x) \\
\implies && xP^{\prime}_{l}(x)-P^{\prime}_{l-1}(x) = lP_{l}(x)
\end{align*}
⟹ ⟹ x [ h l P l ′ ( x )] − h [ h l − 1 P l − 1 ′ ( x )] = h [ l h l − 1 P l ( x )] h l [ x P l ′ ( x ) − P l − 1 ′ ( x )] = h l l P l ( x ) x P l ′ ( x ) − P l − 1 ′ ( x ) = l P l ( x )
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