Recurrence relation and Rodrigue's formula of Legendre differential equation

Recurrence relation:

We know the generating function of Legendre polynomial is

$$(1-2xt+t^2)^{-\frac12}= \sum_{n=0}^\infty t^n P_n(x)$$

Differentiating on both side with respect to 't', we get,

$$-\frac12 (1-2xt+t^2)^{-\frac32}(-2x+2t)=\sum_{n=0}^\infty nt^{n-1} p_n(x)$$

$$or,\;\; (1-2xt+t^2)^{-\frac12}\times \frac{x-t}{(1-2xt+t^2)}=\sum_{n=0}^\infty nt^{n-1} p_n(x)$$

$$or,\;\; (x-t)\sum_{n=0}^\infty t^n p_n(x)=\sum_{n=0}^\infty nt^{n-1}p_n(x)-2x\sum_{n=0}^\infty n^{tn} \times p_n(x)+\sum_{n=0}^\infty nt^{n+1}p_n(x)$$

$$or,\;\; x\sum_{n=0}^\infty t^n P_n(x)-\sum_{n=0}^\infty t^{n+1}P_n(x)=\sum_{n=0}^\infty nt^{n-1} P_n(x)- 2x\sum_{n=0}^\infty nt^n P_n(x)+\sum_{n=0}^\infty nt^{n-1}P_n(x)$$

Now, comparing the coefficient of \(t^n\) on both sides, we get,

$$or,\;\; xP_n(x)-P_{n-1}(x) = (n+1) P_{n+1}(x)- 2x_n P_n(x)+ (n-1) P_{n-1} (x)$$

$$or,\;\; P_{n+1}=\frac{1}{n+1} \biggl[ xP_n(x) - P_{n-1}(x) + 2x_nP_n(x)-nP_{n-1}(x)+P_{n-1}(x)\biggr]$$

$$or,\;\; P_{n+1} = \frac{1}{n+1} \biggl[ P_n(x) x (1+2n) - nP_{n-1} (x)\biggr]$$

This is the first recurrence relation.

Again, we have

$$(1-2xt+t^2)^{-\frac12}=\sum_{n=0}^\infty t^n P_n(x)$$

Now , differentiating both sides with respect to x, we get,

$$-\frac12 (1-2xt+t^2)^{-\frac32}\times (-2t) = \sum_{n=0}^\infty t^n P_n'(x)$$

$$or,\;\; (1-2xt+t^2)^{-\frac12} \times \frac{t}{(1-2xt+t^2)}=\sum_{n=0}^\infty t^n P_n'(x)$$

$$or,\;\;\sum_{n=0}^\infty t^n P_n(x) t=\sum_{n=0}^\infty t^n P_n'(x) -2x\sum_{n=0}^\infty t^{n+1} P_n'(x)+\sum_{n=0}^\infty t^{n+2} P_n'(x)$$

$$or\;\;\sum_{n=0}^\infty t^{n+1} P_n(x)=\sum_{n=0}^\infty t^n P_n'(x)- 2x\sum_{n=0}^\infty t^{n+1} P_n'(x)+\sum_{n=0}^\infty t^{n+2} P_n'(x)$$

Now, collecting the coefficient of \(t^n\) on both sides, we get

$$P_{n-1}(x)= P_n'(x) -2xP_{n-1}'(x)+ P_{n-2}'(x)$$

This is the 2nd recurrence relation of Legendre polynomial.

Note: We know the generating function of Legendre polynomial is,

$$(1-2xt+t^2)^{-\frac12}=\sum_{n=0}^\infty t^n P_n(x)\dotsm(1)$$

Put \(x=\frac{e^{i\theta}+ e^{-i\theta}}{2}+cos\theta\)

$$or,\;\; \sum_{n=0}^\infty t^n P_n(cos\theta)=\biggl[1-\frac{2(e^{i\theta}+e^{-i\theta}}{2}t+t^2-1\biggr]^{-\frac12}$$

$$ = [1- te^{i\theta}-te^{-i\theta}+t^2\cdot e^{i\theta}\cdot e^{-i\theta}\biggr]^{-\frac12}$$

$$= [1(1-te^{-i\theta})-te^{-i\theta}(1-te^{i\theta}]^{-\frac12}$$

$$=[(1-te^{-i\theta})(1-te^{i\theta})]^{-\frac12}$$

$$=\biggl[1+\frac{1}{1!}\frac12te^{-i\theta}+\frac{1}{2!}\frac12\frac32t^2e^{-2i\theta}+\frac{1}{3!}\frac12\frac32\frac52t^3e^{-3i\theta}+\dotsm]\times[1+\frac{1}{1!}\frac12 te^{i\theta}+\frac{1}{2!}+ \frac{1}{2!}\frac12\frac32 t^2e^{2i\theta}+\frac{1}{3!}\frac12\frac32\frac52t^3e^{3i\theta}+\dotsm\biggr]$$

Comparing the coefficient of \(t^0\) on both sides, we get

or, \(P_0(cos\theta)=1\)

or, \(P_1(cos\theta)=\frac12 e^{i\theta}+\frac12 e^{-i\theta}\)

or, \(p_1(coos\theta)=\frac12 (e^{i\theta}+e^{i\theta})=cos\theta\)

or, \(p_2(cos\theta)=\frac{1}{2!}\frac12\frac32 e^{2i\theta}+\frac{1}{2!}\frac12\frac32e^{-2i\theta}+\frac14(e^{-i\theta}e^{i\theta}\)

or, \(p_2(cos\theta)= \frac34 \biggl(\frac{e^{2i\theta}+e^{-2i\theta}}{2}\biggr)+\frac14\)

or \(p_2(cos\theta)= \frac34 cos2\theta+\frac14\)

Rodrigue's formula:

\(P_n(x)=\frac{1}{2^nn!} \frac{d^n}{dx^n}(x^2-1)^n\)

Suppose \(\phi=(x^2-1)^n\)

Differentating both sides with respect to x, we get

$$\frac{d\phi}{dx}=n(x^2-1)^{n-1} \times 2x$$

$$=2nx(x^2-1)^{n-1}$$

Multiplying both sides by \((x^2-1)\), we get,

$$or,\;\; (x^2-1) \frac{d\phi}{dx}=2nx(x^2-1)n$$

$$or,\;\; (x^2-1)\frac{d\phi}{dx}=2nx\phi$$

$$or,\;\; -(1-x^2)\frac{d\phi}{dx}=2nx\pi$$

$$or(1-x^2)\frac{d\phi}{dx}+2nx\phi=0$$

Now, diff. both sides with respect to x, we get,

$$or,\;\; (1-x^2)\frac{d^2\phi}{dx^2}-2x\frac{d\phi}{dx}+2nx\frac{d\phi}{dx}+2n\phi=0$$

$$or,\;\; (1-x^2)\frac{d^2\phi}{dx^2}-2x(1-n)\frac{d\phi}{dx}+2n\phi=0$$

Diff booth sides with respect to x for n times using leibnitz theorem.

$$or,\;\; (1-x^2)\frac{d^{2+n}\phi}{dx^{2_n}}+n(-2x) \frac{d^{1+n}\phi}{dx^{1+n}}+\frac{n(n-1)}{2}\cdot (-2)\frac{d^n\phi}{dx^n}-2x(1-n)\frac{d^{1+n}\phi}{dx^{1+n}} -2(1-n)\cdot n\frac{d^n\phi}{dx^n}+2n\cdot \frac{d^n\phi}{dx^n}=0$$

Put\(\frac{d^n}{dx^n}=\psi\)

or, \( P_n(x)=\frac{d^n\phi}{dx^n}\)

or, \(cP_n(x)=\frac{d^n}{dx^n}(x^2-1)^n\dotsm(1)\)

or \(cP_n(x)=\frac{d^n}{dx^n}[(x-1)^n(x+1)^n]\)

or \(CP_n(x)= (x-1)^n \frac{d^n}{dx^n}(x+1)^n+(x+1)^n\frac{d^n}{dx^n}(x-1)^n\)

or \(CP_n(x)= (x-1)^n n! +(x+1)^n n!\)

or \(CP_n(x)= (x-1)^n n!+(x+1)^n n!\)

put x=1

or \(CP_n(1)= 2^n n!\)

or\( C\cdot 1= 2^n.n!\)

or \( C= 2^n\cdot n!\)

Now from equation (a)

\(CP_n(x)=\frac{d^n}{dx^n}(x^2-1)n\)

\(P_n(x)=\frac{1}{2^n\cdot n!}\frac{d^n}{dx^n}(x^2-1)^n\)

Reference:

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4. Carroll, Robert W.Mathematical Physics. Amsterdam: North-Holland, 1988.
5. Adhikari, Pitri Bhakta, and Dya Nidhi Chhatkuli.A Textbook of Physics. Third Revised Edition ed. Vol. III. Kathmandu: SUKUNDA PUSTAK BHAWAN, 2072.