Least square fit using Legendre polynomials I would like to apply Legendre polynomials to least square approximation. Therefore I would like the function: 
$$L_n (x)=\sum_{k=0}^n a_k P_k (x)$$
to fit $f(x)$ defined over $[-1,1]$ in a least square sense.
We should minimize:
$$I(a_0, ..., a_n)= \int_{-1}^1 [f(x) - L_n (x)]^2 \; dx\tag1$$ 
and so we must set
$$\frac{\partial I}{\partial a_r} = 0,\qquad r=0,1, \ldots,n\tag2$$
Using equations $(1)$ and $(2)$
$$\int_{-1}^1 P_r(x) \left[f(x) - \sum_{k=0}^n a_k P_k (x)\right]dx = 0,\qquad r=0,1, \ldots,n$$ 
should be an equivalent term.
My question now is: why is that true? 
I would be glad if someone could illustrate the last step with more details.
Thanks, Rainier.
 A: You have
$$
\frac{\partial}{\partial a_r}\int_{-1}^1 [f(x) - L_n (x)]^2dx=0\\
\int_{-1}^1 \frac{\partial}{\partial a_r}[f(x) - L_n (x)]^2dx=0\\
\int_{-1}^1 2[f(x) - L_n (x)]\frac{\partial}{\partial a_r}[f(x) - L_n (x)]dx=0\\
\int_{-1}^1 2[f(x) - L_n (x)][0 - \frac{\partial}{\partial a_r}L_n (x)]dx=0\\
-\int_{-1}^1 2[f(x) - L_n (x)]\frac{\partial}{\partial a_r}\sum a_kP_k(x)dx=0\\
-\int_{-1}^1 2[f(x) - L_n (x)]P_r(x)dx=0
$$
A: It is even easier than that.  Because the Legendre polynomials are orthogonal, you can get the coefficients just from $a_n=\frac {2n+1}2\int_{-1}^1f(x)P_n(x)dx$
A: This is coming late but for the benefit of others, let me post my opinion:
$$\frac{\partial\sum_{k=0}^n a_k P_k (x)}{\partial a_r} = \frac{\partial}{\partial a_r} [a_0 P_0 + a_1 P_1 + ... + a_r P_r + ... + a_k P_k] = P_r (x)$$
This is because all other coefficients of $$P_k (x) $$ are treated as constants except for when $$k=r$$
The Orthogonality of the Legendre Polynomial was what was responsible for the final step
$$\int_{-1}^1 P_r(x)P_k (x)dx = 0, \qquad r=0,1, \ldots,n$$
The above equation is true as long as r is not equal to k
