recurrence relation in approximating theory I am stuck in a little part of a problem:
I wish to give an approximation of $\sin(\pi x)$ on $-1 \leq x \leq 1$, when using the polynomial $$F_N(x)=\sum_{k=0}^{N}a_kx^{2k+1}$$
with the coefficients $a_k$ chosen to minimize the integral:
$$F(a_0, a_1, \cdots a_N)=\int_{-1}^{1}(F_N(x)   - \sin \pi x)^2dx$$
I showed that:
$$\sum_{k=0}^{N}\frac{a_k}{2(k+j)+3}= I_j = \int_{0}^{1}x^{2j+1} \sin \pi x dx$$ for $j=0,1, \cdots N$.
Now I wish to show that:
$$I_n = \frac{1}{\pi}- \frac{2n(2n+1)}{\pi^2}I_{n-1}$$ 
where $I_0= \frac {1}{\pi} $ and to deduce that 
$$I_n = \sum_{k=0}^{n} \frac{(-1)^k}{\pi^{2k+1}} \frac{(2n+1)!}{(2n-2k+1)!}$$
To be honest I don't know where to start.
Any help appreciated
 A: You use integration by parts twice with $u$ the power of $x$ and $dv$ the trig function
$$I_n = \int_{0}^{1}x^{2n+1} \sin \pi x dx=-\frac 1\pi x^{2n+1}\cos \pi x|_0^1+\frac 1\pi \int_0^1(2n+1)x^{2n}\cos\pi x dx\\
=\frac 1\pi +\frac 1{\pi^2}(2n+1)x^{2n}\sin \pi x|_0^1-\frac 1{\pi^2}(2n+1)(2n)\int_0^1x^{2n-1}\sin \pi x dx\\
=\frac 1\pi-\frac 1{\pi^2}(2n+1)(2n)I_{n-1}$$
Then we need $I_0=\int_0^1x \sin \pi x dx=\frac 1{\pi}$  We prove the final sum by induction.  You can verify your final sum expression works for $I_0$.  Assume it is good up to $n-1$, then $$I_n = \frac 1\pi-\frac 1{\pi^2}(2n+1)(2n)I_{n-1}\\=\frac 1\pi-\frac 1{\pi^2}(2n+1)(2n)\sum_{k=0}^{n-1} \frac{(-1)^k}{\pi^{2k+1}} \frac{(2n-1)!}{(2n-2k-1)!} \\=\sum_{k=0}^{n} \frac{(-1)^k}{\pi^{2k+1}} \frac{(2n+1)!}{(2n-2k+1)!}$$
because the leading $\frac 1{\pi^2}(2n+1)(2n)$ promotes each term in the sum from $k$ to $k+1$ and the $\frac 1\pi$ fills in the $k=0$ term from the bottom.  
Added:  consider the $k=m$ term of the sum in the next to last line.  When we multiply it by the factors in front of the sum we get
$$-\frac 1{\pi^2}(2n+1)(2n)\frac {(-1)^m}{\pi^{2m+1}}\frac {(2n-1)
!}{(2n-2m-1)!}=\frac{(-1)^{m+1}}{\pi^{2(m+1)+1}}\frac{(2n+1)!}{(2 n-2(m+1)+1)!}$$
which is the $k=m+1$ term of the bottom sum.  This accounts for all the terms from $k=1$ through $k=n$ in the bottom sum and the $k=0$ term is just the $\frac 1\pi$
