What is the Legendre expansion of $\sqrt{1-x^2}$ ? I need a closed form. Thx.
I have tried to reform it
$$ \sqrt{1-x^2}=\left. \sqrt{1-2tx+x^2} \right|_{t=x} $$
while the generating function of Legendre reads
$$ G\left( t,x \right) =\frac{1}{\sqrt{1-2tx+x^2}}=\sum_{\ell =0}^{\infty}{\mathrm{P}_{\ell}\left( x \right) t^{\ell}} $$
what should I do next?
I wonder if I can take $t\equiv x$ at last. First, I have tried like this $$ \sqrt{1-x^2}=\frac{1-x^2}{\sqrt{1-x^2}}=\left. \frac{1-x^2}{\sqrt{1-2tx+x^2}} \right|_{t=x} $$ then I had the coefficient $c_\ell$ reads
$$ c_{\ell}=\frac{2\ell +1}{2}\sum_{n=0}^{\infty}{\left[ t^n\int_{-1}^1{\left( 1-x^2 \right) \mathrm{P}_n\left( x \right) \mathrm{P}_{\ell}\left( x \right) \mathrm{d}x} \right] _{t=x}}=x^{\ell}-\frac{2\ell +1}{2}\sum_{n=0}^{\infty}{\left[ t^n\int_{-1}^1{x^2\mathrm{P}_n\left( x \right) \mathrm{P}_{\ell}\left( x \right) \mathrm{d}x} \right] _{t=x}}=\frac{2\ell +1}{2}\sum_{n=0}^{\infty}{\left[ t^n\int_{-1}^1{\left[ x\mathrm{P}_n\left( x \right) \right]}\cdot \left[ x\mathrm{P}_{\ell}\left( x \right) \right] \mathrm{d}x \right] _{t=x}} $$
and using the recurrence relations, I had
$$ c_{\ell}=x^{\ell}-\frac{1}{2}\sum_{n=0}^{\infty}{\left\{ t^n\int_{-1}^1{\left[ \frac{n+1}{2n+1}\mathrm{P}_n\left( x \right) +\frac{n}{2n+1}\mathrm{P}_{n-1}\left( x \right) \right] \cdot \left[ \left( \ell +1 \right) \mathrm{P}_{\ell +1}\left( x \right) +\ell \mathrm{P}_{\ell -1}\left( x \right) \right] \mathrm{d}x} \right\} _{t=x}}=x^{\ell}-\frac{1}{2}\sum_{n=0}^{\infty}{\left\{ t^n\left[ \frac{\left( \ell +1 \right) \left( n+1 \right)}{2n+1}\frac{2}{2\left( n+1 \right) +1}\delta _{n+1,\ell +1}+\frac{\left( n+1 \right) \ell}{2n+1}\frac{2}{2\left( n+1 \right) +1}\delta _{n+1,\ell -1}+\frac{n\left( \ell +1 \right)}{2n+1}\frac{2}{2\left( n-1 \right) +1}\delta _{n-1,\ell +1}+\frac{n\ell}{2n+1}\frac{2}{2\left( n-1 \right) +1}\delta _{n-1,\ell -1} \right] \right\} _{t=x}}=x^{\ell}-\left[ \frac{\left( \ell +1 \right) ^2}{\left( 2\ell +1 \right) \left( 2\ell +3 \right)}t^{\ell}+\frac{\ell \left( \ell -1 \right)}{\left( 2\ell -1 \right) \left( 2\ell -3 \right)}t^{\ell -2}+\frac{\left( \ell +1 \right) \left( \ell +2 \right)}{\left( 2\ell +3 \right) ^2}t^{\ell +2}+\frac{\ell ^2}{\left( 2\ell -1 \right) \left( 2\ell +1 \right)}t^{\ell} \right] _{t=x} $$
then I had the result(?)
$$ c_{\ell}=x^{\ell}\left\{ \left[ 1-\frac{\left( \ell +3 \right) ^2}{\left( 2\ell +5 \right) \left( 2\ell +7 \right)}-\frac{\left( \ell +2 \right) ^2}{\left( 2\ell +3 \right) \left( 2\ell +5 \right)} \right] x^2-\frac{\left( \ell +2 \right) \left( \ell +1 \right)}{\left( 2\ell +3 \right) \left( 2\ell +1 \right)}-\frac{\left( \ell +3 \right) \left( \ell +4 \right)}{\left( 2\ell +7 \right) ^2}x^4 \right\} $$
where $\ell \rightarrow \left(\ell+2\right) $ and $\ell = 0,2,4,6,\cdots $. I just didn't know if I was wrong.
Try another way to integral by substitution: First step, put $\sqrt{1-x^2}$ into $\mathrm{d}x$, then we have the form reads
$$ c_n=\frac{2n+1}{2}\left\{ \left[ \frac{1}{4}\sin \left( 2\mathrm{arc}\sin x \right) +\frac{1}{2}\mathrm{arc}\sin x \right] P_n\left( x \right) \right\} _{-1}^{1}-\frac{2n+1}{2}\int_{-1}^1{\frac{\mathrm{d}P_n\left( x \right)}{\mathrm{d}x}\sqrt{1-x^2}\mathrm{d}x}, $$
and notice that the first item of $c_n$ reduces to $\pi$, so it can be reformed as
$$ c_n=\frac{2n+1}{2}\left[ 1\pi +\left( -1 \right) ^1\int_{-1}^1{\sqrt{1-x^2}\left( \frac{\mathrm{d}}{\mathrm{d}x} \right) ^1P_n\left( x \right) \mathrm{d}x} \right] $$
Second step is to repeat the substitution $k$ times, using mathematical induction to prove it make sense. Then we have
$$ c_n=\frac{2n+1}{2}\left[ k\pi +\left( -1 \right) ^k\int_{-1}^1{\sqrt{1-x^2}\left( \frac{\mathrm{d}}{\mathrm{d}x} \right) ^kP_n\left( x \right) \mathrm{d}x} \right] , $$
if the item $ \left( \frac{\mathrm{d}}{\mathrm{d}x} \right) ^kP_n\left( x \right) $ still contains the factory of the power of $x$, then we should repeat the step 2, until $k=n$, then we have
$$ c_n=\frac{2n+1}{2}\left[ n\pi +\left( -1 \right) ^n\int_{-1}^1{\sqrt{1-x^2}\left( \frac{\mathrm{d}}{\mathrm{d}x} \right) ^{2n}P_n\left( x \right) \mathrm{d}x} \right] =\frac{2n+1}{2}\left[ n\pi +\frac{\left( -1 \right) ^n\left( 2n \right) !}{2^nn!}\int_{-1}^1{\sqrt{1-x^2}\mathrm{d}x} \right] =\frac{2n+1}{2}\left[ n\pi +\frac{\left( -1 \right) ^n\left( 2n \right) !}{2^nn!}\frac{\pi}{2} \right] , $$
At the same time, noticed that the function $f\left(x\right)$ is an even function, using the properties of Legendre integration, the parameter $n$ above should be even, then let $n=2\ell$, we finally have the closed form of the coefficient of Legendre expansion series reads
$$ c_{\ell}=\frac{4\ell +1}{2}\left[ 2\ell \pi +\frac{\pi \left( 4\ell \right) !}{2^{2\ell +1}\left( 2\ell \right) !} \right] =\left[ \left( 4\ell +1 \right) \ell +\frac{\left( 4\ell +1 \right) !}{4^{\ell +1}\left( 2\ell \right) !} \right] \pi . $$
So, the complete series reads
$$ \sqrt{1-x^2}\cong \pi \sum_{\ell =0}^{\infty}{\left[ \left( 4\ell +1 \right) \ell +\frac{\left( 4\ell +1 \right) !}{4^{\ell +1}\left( 2\ell \right) !} \right] P_{2\ell}\left( x \right)}. $$