0
$\begingroup$

Given Legendre polynomial generating function \begin{equation} \sum_{n=0}^\infty P_n (x) t^n = \frac{1}{\sqrt{(1-2xt+t^2)}} \end{equation} Show that $$ P_n (1)=1 $$ and $$ P_n (-1)=(-1)^n $$

Not sure where to start with either

$\endgroup$
1
$\begingroup$

Substituting $ x=1 $ in the given equation yields

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

which gives $P_n(1)=1$. You can do the same with the other one.

$\endgroup$
1
$\begingroup$

substitute $x=1$: $$ \sum_{n=0}^\infty P_n (1) t^n = \frac{1}{(1-2t+t^2)^{\frac12}} = \frac{1}{((1-t)^2)^{\frac12}} = \frac{1}{1-t} $$ and can you expand that in powers of $t$?

Similar method for $x=-1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.