Approximating $\sin x$ with given polynomials . I am required to approximate function $  sin (\pi x) \text{ on} -1 \leq x \leq 1 $ with the function $$ f_{N}(x) = \displaystyle\sum_{k=0}^{N‎} a_{k}x^{2k+1}$$ where coefficients $\ a_{k},\ k=0,1,..N$ are chosen to minimize integral
$$F(a_{0},a_{1},...a_{N}) = \displaystyle \int_{-1}^{1} [f_{N}(x)  - sin(\pi x)]^2  dx $$ 
I need to show the coefficients $ a_{k},\ k= 0,1,..N,\ $ are given by the solution of linear  equations
$$ \displaystyle\sum_{k=0}^{N} \frac{a_k}{2(k+j)+3 } = I_j ,  \text{where } I_j = 
 \displaystyle\int_{0}^{1} x^{2j+1} sin(\pi x)\ dx , $$ 
 for   $j=0,1,..N$,  
I am out of ideas how can summation with 'a' coefficient in LHS I can produce 'sin' on RHS
 A: This reduces to a standard least squares fit problem.
Let $b_k(x) = x^{2k+1}$, $k=0,..,N$, and $f(x) =\sin(\pi x)$. The problem is to solve
$\min_{a \in \mathbb{R}^{n+1}} \| \sum_k a_k b_k -f \|^2 = \min_{a \in \mathbb{R}^{n+1}} \| Ba -f \|^2$, where
$B: \mathbb{R}^{n+1} \to L^2[-1,1]$ is given by
$(B a)(x) = \sum_k a_k b_k(x)$.
Since
\begin{eqnarray}
\| Ba -f \|^2 &=& \langle Ba -f , Ba -f \rangle \\
&=&  \langle Ba, B a \rangle -2\langle Ba ,f \rangle  + \|f\|^2 \\
&=& \langle a, B^*B a \rangle -2\langle Ba ,f \rangle  + \|f\|^2 \\
&=& \langle a, B^*B a \rangle -2\langle a ,B^*f \rangle  + \|f\|^2 \\
\end{eqnarray}
we see that this is a convex quadratic problem in $\mathbb{R}^{n+1}$, and the first order
conditions give
$B^*Ba = B^* f$.
To compute $B^*$, use the definition
\begin{eqnarray}
\langle B^*g,a \rangle &=& \langle g,Ba \rangle \\
 &=& \int g(x) (Ba)(x)dx \\
&=& \sum_k a_k \int g(x) x^{2k+1} dx \\
&=& \sum_k a_k c_k \\
&=& \langle c,a \rangle
\end{eqnarray}
so $[B^* g]_k = c_k = \int g(x) x^{2k+1} dx$.
Note that $[B^*Ba]_j = \int (Ba)(x) x^{2j+1}dx = \sum_k a_k \int x^{2(k+j)+2} dx$
and
$[B^*f]_j = \int \sin(\pi x) x^{2j+1} dx$.
Now evaluate the $[B^*Ba]_j$ and use oddness to get the $I_j$ formula.
Note:
The whole dual thing can be avoided by noting that
\begin{eqnarray}
\|B a -f \|^2 &=& \int (\sum_k a_k x^{2k+1} - f(x))^2 dx \\
&=& \sum_j \sum_k a_ja_k \int x^{2(k+j)+2}dx - 2 \sum_j a_j \int x^{2j+1} f(x)dx + \int f(x)^2 dx \\
&=& \langle a, Ga \rangle - 2 \langle a , c \rangle + \|f\|^2
\end{eqnarray}
where $[G]_{jk} = \int x^{2(k+j)+2}dx$ and $c$ is as above. At a solution
the first order conditions will give
$Ga=c$ which is equivalent to the desired result.
A: The non-negative function $F(a_0,a_1,\ldots,a_N)=\int_{-1}^1(\sum_{k=0}^Na_kx^{2k+1}-\sin\pi x)^2\,dx$ is quadratic in each of its variables, and therefore is minimized when each of its derivatives is $0$. The derivative with respect to $a_j$ is
$$\begin{align}
\int_{-1}^12x^{2j+1}\left(\sum_{k=0}^Na_kx^{2k+1}-\sin\pi x\right)\,dx
&=2\sum_{k=0}^Na_k\int_{-1}^1x^{2(k+j)+2}\,dx-2\int_{-1}^1x^{2j+1}\sin\pi x\,dx\\
&=4\sum_{k=0}^N{a_k\over2(k+j)+3}-2\int_{-1}^1x^{2j+1}\sin\pi x\,dx
\end{align}$$
Setting this equal to $0$ implies
$$\sum_{k=0}^N{a_k\over2(k+j)+3}={1\over2}\int_{-1}^1x^{2j+1}\sin\pi x\,dx=\int_0^1x^{2j+1}\sin\pi x\,dx$$
where the final step uses the fact that $x^{2j+1}\sin\pi x$ is an even function.
