Expand the following function in Legendre polynomials on the interval [-1,1] :
$$f(x) = |x|$$
The Legendre polynomials $p_n (x)$ are defined by the formula :
$$p_n (x) = \frac {1}{2^n n!} \frac{d^n}{dx^n}(x^2-1)^2$$
for $n=0,1,2,3,…$
My attempt :
we have using the fact that $|x|$ is an even function.
$$a_0 = \frac {2}{\pi}$$
$$ a_n= \frac {2}{π} \int_{-1}^{1}x\cos(nx)\,dx$$
Then what is the next step ?
Best Answer
We know that the Fourier-Legendre series is like $$ f(x)=\sum_{n=0}^\infty C_n P_n(x) $$
where $$ C_n=\frac{2n+1}{2} \int_{-1}^{1}f(x)P_n(x)\,dx $$
So now we are going to calculate the result of $$ \frac{2n+1}{2} \int_{-1}^{1}|x|P_n(x)\,dx $$ As $|x|$ is an even function, and the parity of $P_n(x)$ depends on the parity of $n$, We can write
$$ \int_{-1}^{1}|x|P_n(x)\,dx $$ as $$ \int_{-1}^{1}|x|P_{2k}(x)\,dx\ \ k=0,1,2... $$ and $$ \int_{-1}^{1}|x|P_{2k}(x)\,dx \\ =\int_{-1}^{0}-xP_{2k}(x)\,dx + \int_{0}^{1}xP_{2k}(x)\,dx\\ =2\int_{0}^{1}xP_{2k}(x)\,dx $$ As $$ (n+1)P_{n+1}(x)-x(2n+1)P_{n}(x)+nP_{n-1}(x)=0 $$ we get $$ 2\int_{0}^{1}xP_{2k}(x)\,dx\\ =2(\frac{2k+1}{4k+1}\int_{0}^{1}P_{2k+1}(x)\,dx+\frac{2k}{4k+1}\int_{0}^{1}P_{2k-1}(x)\,dx) $$ As $$ \int_{0}^{1}P_{n}(x)\,dx=\begin{cases} 0& n=2k\\ \frac{(-1)^k (2k-1)!!}{(2k+2)!!}& n=2k+1 \end{cases} $$ We get $$ C_{2k}=(2k+1)\frac{(-1)^k (2k-1)!!}{(2k+2)!!}+n\frac{(-1)^{k-1} (2k-3)!!}{(2k)!!}\\ =\begin{cases} \frac{1}{2}& k=0\\ \frac{(-1)^{k+1} (4k+1)}{2^{2k}(k-1)!}\frac{(2k-2)!}{(k+1)!}& k>0 \end{cases} $$ So $$ |x|=\frac{1}{2}+\sum_{k=1}^\infty \frac{(-1)^{k+1} (4k+1)}{2^{2k}(k-1)!}\frac{(2k-2)!}{(k+1)!} P_{2k}(x) $$