Polynomial solution,Generating function of Legendre differential equation

Polynomial solution :

For polynomial solution, we proceed from solution for k=n, we have

$$ a_{2j}= \frac{(-1)^j n(n-1)(n-2)\dotsm (n-2j+1)a_0}{2\cdot 4\dotsm 2j(2n-1)(2n-3)\dotsm(2n-2j+1)}$$

The solution is

$$y=\sum_{j=0}^\infty a_{2j}x^{n-2j}$$

$$y=\sum_{j=0}^\infty\frac{(-1)^j n(n-1)(n-2)\dotsm (n-2j+1)a_0 x^{n-2j}}{2\cdot 4\dotsm 2j(2n-1)(2n-3)\dotsm(2n-2j+1)}$$

$$=\sum_{j=0}^\infty\frac{(-1)^j n(n-1)(n-2)\dotsm (n-1j+1)x^{n-2j}}{2^j J! (2n-1)(2n-3)\dotsm (2n-2j+1)}\times \frac{1\cdot 3\cdot 5\dotsm(2n-1)}{n!}$$

$$=\sum_{j=0}^\infty \frac{(-1)^j(2n-2j-1)\dotsm 5\cdot 3\cdot 1}{2j J!(n-2j) \dotsm 3\cdot 2\cdot 1}x^{n-2j}$$

$$=\sum_{j=0}^\infty \frac{(-1)^j(2n-2j-1)\dotsm 5\cdot 3\cdot 1}{2j J!(n-2j) !}x^{n-2j}\times \frac{2n-2j\dotsm 6\cdot 4\cdot 2}{(2n-2j)\dotsm 6\cdot 4\cdot 2}$$

$$=\sum_{j=0}^\infty \frac{(-1)^j (2n-2j)! x^{n-2j}}{2^j J! (n-2j)! 2^{n-j} (n-j)!}$$

$$=\sum_{j=0}^\infty \frac{(-1)^j (2n_2j)! x^{n-2j}}{2^nJ! (n-2j)!(n-j)!}$$

$$or\;\; P_n(x)=\sum_{j=0}^{\frac n2} \frac{(-1)^j\cdot (2n-2j)! x^{n-2j}}{2^n J! (n-2j)! (n-j)!}$$

The polynomial solution is

$$P_n(x) = \sum_{j=0}^N \frac{(-1)^j (2n-2j)! x^{n-2j}}{2^n J! (n-2j)! (n-j)!}$$

Where N=n/2 for even and n-1/2 for odd

Put n=0

$$P_0(x)=\sum_{j=0}^0 \frac{(-1)^J (2n-2j)! x^{n-2j}}{2^n J! (n-j)! (n-2j)!}$$

$$=\frac{1\times 0! \times x^0}{1\times 0! \times 0! 0!}=1$$

Put n=1

\(N=\frac{1-1}{0}=0\)

$$P_1(x)= \sum_{j=0}^0 \frac{(-1)^j (2n-2j)! x^{n-2j}}{2^nJ!(n-j)! (n-2j)!}$$

Similarly, \( P_2(x)= \frac13 (3x^2-1)\)

and \(P_3(x)= \frac12 (5s^3-3x)\)

Prove: \(P_n(-x)= (-1)^n P_n(x)\)

We know the polynomial solution for Legendre differential equation

$$P_n(x) = \sum_{j=0}^N \frac{(-1)^j (2n-2j)! x^{n-2j}}{2^n J! (n-2j)! (n-j)!}$$

Where N=n/2 for even and n-1/2 for odd

substituting x\(\to\) -x

$$P_n(x = \sum_{j=0}^N \frac{(-1)^j (2n-2j)! (-x)^{n-2j}}{2^n J! (n-j)! (n-2j)!}$$

$$=\sum_{j=0}^N\frac{(-1)^j (2n-2j)! (-1)^{n-2j} x^{n-2j}}{2^n J! (n-J)! (n-2J)!}$$

$$=(-1)^n \sum_{j=0}^n \frac{(-1)^j (2n-2j)! x^{n-2j}}{2^n J! (n-J)! (n-2J)!}$$

Generating function of Legendre differential equation:

Prove that the generating function fo Legendre differential equation is

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

L.H.S

$$(1-2xt+t^2)^{-\frac12}$$

$$=[1-(2xt-t^2)]^{-\frac12}$$

$$=1+\frac{1}{1!}+\frac{1}{2} (2xt-t^2) +\frac{1}{2!} \frac12\frac32(2xt-t^2)^2+\frac13\frac12\frac32\frac52(2xt-t^2)^3+\dotsm+\frac{1}{5} \frac{1\cdot\3\cdot5\dotsm(2s-1)}{2s}$$

$$=\sum_{s=0}^\infty \frac{1}{s!}\frac{1.3.5... 2s-1}{2s} (2xt-t^2)^s \times(2xt -t^2)^s\times \frac{2.4.6...2s}{2.4.6...2s}$$

$$=\sum_{s=0}^\infty \frac{1}{s!} \frac{2s}{2^s 2^s s!}(2st-t^2) ^s$$

Let us have,

$$(2xt-t^2)^s=c(s,0)(2xt)^s(t^2)^0+c(s,1)(2xt)^{s-1} (-t)^1 + c(s,2)(2xt)^{s-2} (-t^2)^2+\dotsm + c(s,t)(2xt)^{s-1} (-t)^1+\dotsm c(s,s) (2xt)^0 (-t^2)^s$$

$$(2xt-t^2)^s=\sum_{j=0}^s c(s,j) (2xt)^{s-j} (-t^2)^j$$

$$=\sum_{j=0}^s \frac{s!}{(s-j)!j!} (2xt)^{s-j} (-t^2)^j$$

$$=\sum_{j=0}^s \frac{(-1)^j s!}{(s-j)!j! } 2^{s-j} x^{s-j} t^{s+j}$$

Now,

L.H.S

$$\sum_{s=0}^\infty \frac{1}{s!} \frac{(2s)!}{2^{2s}S!}\sum_{j=0}^s \frac{(-1^j s! 2^{s-j} x^{s-j} t^{s+j}}{(s-j)! J!}$$

$$=\sum_{s=0}^\infty \sum_{j=0}^s \frac{(-1))^j (2s!) x^{s-j}+ t^{s+j}}{s!(s-j)! j! 2^{s+j}}$$

Put, s+j=n , s=n-j

$$=\sum_{n=0}^\infty \biggl[ \sum_{j=0}^{\frac n2} \frac{(-1)^j (2n-2j)! x^{n-2j} t^n}{J! (n-j)! (n-2j)! 2^n}\biggr]$$

$$\sum_{n=0}^\infty t^n p_n(x)=R.H.S$$

Reference:

1. Dass, H. K.Mathematical Physics. New Delhi: S. Chand, 2005.
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3. Butkov, Eugene.Mathematical Physics. Reading, MA: Addison-Wesley Pub., 1968.
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.