# Using the generating function for the Legendre polynomials

• May 17th 2011, 08:37 AM
garunas
Using the generating function for the Legendre polynomials
Using the identity:

$\displaystyle \sum_{n=0}^\infty P_{n}(x)r^n = \frac{1}{\sqrt {1-2xr+r^2} }$ with:

$\displaystyle x \in [-1,1], r \in (-1,1)$

Derive an expression for $\displaystyle P_{2}(x)$ the legendre polynomial of degree 2.

Any help here will be amazing as I absolutely haven't got a clue!

Also from this it also asks to evaluate $\displaystyle P_{3}'(0)$ but i take that if you can find $\displaystyle P_{2}(x)$ then you can also find $\displaystyle P_{2}'(0)$ and use the same method for $\displaystyle P_{3}(x)$

Thank you
• May 17th 2011, 09:02 AM
TheEmptySet
Quote:

Originally Posted by garunas
Using the identity:

$\displaystyle \sum_{n=0}^\infty P_{n}(x)r^n = \frac{1}{\sqrt {1-2xr+r^2} }$ with:

$\displaystyle x \in [-1,1], r \in (-1,1)$

Derive an expression for $\displaystyle P_{2}(x)$ the legendre polynomial of degree 2.

Any help here will be amazing as I absolutely haven't got a clue!

Also from this it also asks to evaluate $\displaystyle P_{3}'(0)$ but i take that if you can find $\displaystyle P_{2}(x)$ then you can also find $\displaystyle P_{2}'(0)$ and use the same method for $\displaystyle P_{3}(x)$

Thank you

you need to expand the right hand side in a Taylor series around zero

so let

$\displaystyle f(r)=(1-2xr+r^2)^{-\frac{1}{2}}$

Now you get

$\displaystyle 1+f'(0)r+\frac{f''(0)}{2!}r^2+\frac{f'''(0)}{3!}r^ 3+...$

Now just equate coefficients this gives

$\displaystyle P_2(x)r^2=\frac{f''(0)}{2!}r^2 \implies P_2(x)=\frac{f''(0)}{2!}$

Can you finish from here?
• May 17th 2011, 09:41 AM
garunas
...i can sure finish from here. Thanks a lot! :)
• May 17th 2011, 10:28 AM
Opalg
Quote:

Originally Posted by garunas
Using the identity:

$\displaystyle \sum_{n=0}^\infty P_{n}(x)r^n = \frac{1}{\sqrt {1-2xr+r^2} }$ with:

$\displaystyle x \in [-1,1], r \in (-1,1)$

Derive an expression for $\displaystyle P_{2}(x)$ the legendre polynomial of degree 2.

Any help here will be amazing as I absolutely haven't got a clue!

Also from this it also asks to evaluate $\displaystyle P_{3}'(0)$ but i take that if you can find $\displaystyle P_{2}(x)$ then you can also find $\displaystyle P_{2}'(0)$ and use the same method for $\displaystyle P_{3}(x)$

$\displaystyle \frac{1}{\sqrt {1-2xr+r^2} } = \bigl(1-(2xr-r^2)\bigr)^{-1/2}$. Use the first few terms of the binomial expansion of $\displaystyle (1-t)^{-1/2}$ (with $\displaystyle t=2xr-r^2$) to find the coefficient of $\displaystyle r^2$.

You could get $\displaystyle P_3'(0)$ by the same method. Alternatively, it might be slightly easier to start by differentiating $\displaystyle \frac{1}{\sqrt {1-2xr+r^2} }$ with respect to x, then put x=0, and finally use the binomial series to find the coefficient of $\displaystyle r^3$.