I am stuck in the following 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: $$F(a_0, a_1, \cdots a_N)=\int_{-1}^{1}(F_N(x) - \sin \pi x)^2dx$$ Is it possible to show 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 $j=0,1, \cdots N$.
I started by $$F(a_0, a_1, \cdots a_N)=\int_{-1}^{1}\left(\sum_{k=0}^{N}a_kx^{2k+1}- \sin \pi x\right)^2dx$$
But to be honest, I haven't got a clue where to begin and where to continue. I hope somebody can help me
I think your start is correct. Consider $$ F(a_0, \dots, a_N) = \int \limits_{-1}^1 \left( \sum \limits_{k=0}^N a_k x^{2k+1} - \sin(\pi x) \right)^2 dx $$ Using the $L^2$ dot product $$ \langle f, g \rangle := \int \limits_{-1}^1 f(x) g(x) dx $$ and the corresponding $L^2$-norm $\| f \|_2 := \sqrt{\langle f, f \rangle}$, you can rewrite your starting expression as (using the abbreviation $\sum \dots$ for the sum and assuming that all coefficients are real, i.e. $a_j \in \mathbb{R}$ for all $j$) $$ F(a_0, \dots, a_N) = \int \limits_{-1}^1 \left( \sum \limits_{k=0}^N a_k x^{2k+1} - \sin(\pi x) \right)^2 dx = \left\| \sum \dots - \sin(\pi x) \right\|_2^2 = \\ = \langle \sum \dots, \sum \dots \rangle - 2 \langle \sum \dots, \sin(\pi x) \rangle + \langle \sin(\pi x), \sin(\pi x) \rangle $$ Due to the linearity of the dot product, you can write this as $$ \sum \limits_{k,j=0}^N a_k a_j \int \limits_{-1}^1 x^{2(k+j)+2} dx - 2 \sum \limits_{j=0}^N a_j \int \limits_{-1}^1 \sin(\pi x) x^{2j+1} dx + \int \limits_{-1}^1 \sin^2(\pi x) dx $$ which evaluates to $$ \sum \limits_{k,j=0}^N a_k a_j \int \limits_{-1}^1 x^{2(k+j)+2} dx - 2 \sum \limits_{j=0}^N a_j \int \limits_{-1}^1 \sin(\pi x) x^{2j+1} dx + 1 $$ The first integral simply has the value $\frac{2}{2(k+j)+3}$ by elementary integration. Now factoring out the sum with index $j$ finally yields: $$ 2 \sum \limits_{j = 0}^N a_j \left( \sum \limits_{k=0}^N \frac{a_k}{2(k+j)+3} - \int \limits_{-1}^1 \sin(\pi x) x^{2j+1} dx \right) + 1 $$ The expression in brackets yields exactly the term $I_j$ you're looking for.