Generalization of minimisation problem First I would like introduce my problem ! There is an easy way to solve this one :

Find the value of $$
\inf_{(a,b)\in \mathbb{R}^2} \int_0^1 (t^2-at-b)^2 dt 
$$ and precise for which values $a$ and $b$ it is reaching.

Well by considering this dot product for two polynoms in $\mathbb{R}[X]^2$, $P$ and $Q$
$$
(P|Q)=\int_0^1 P(t)Q(t)dt
$$
By Gram-Schimdt process we find that 
$$
\inf_{(a,b)\in \mathbb{R}^2} \int_0^1 (t^2-at-b)^2 dt= \int_0^1 \left(t^2-t+\frac{1}{6}\right)^2 dt =\frac{1}{180}
$$
Well I'm capable to find this in less 20 minutes with a pen and a papper, now imagine we want to know the value of :
$$
\inf_{(a_k)\in \mathbb{R}^n} \int_0^1 \left(t^n-\sum_{k=0}^{n-1} a_k \cdot t^k\right)^2 dt
$$
Have you got any idea to solve this one for $n=100$ by example without computer and also find if it exist is limit when $n$ comes to infinity ? 
 A: You can write the polynomial in the integral in the orthogonal basis of (shifted) 
Legendre polynomials:
$$
t^n-\sum_{i=0}^{n-1} a_i t^i = \sum_{i=0}^n c_i \cdot \bar{P}_i(t).
$$
We will need that the coefficient of $\overline{P}_k$ is $\frac{2^k\cdot(2k-1)!!}{k!}$ and $\int_0^1 \overline{P}_k=\frac1{2k+1}$.
The leading coefficient enforces $c_n=\frac{n!}{2^n\cdot (2n-1)!!}$; the remaining coefficients $c_0,\ldots,c_{n-1}$ are free.
From the orthogonality we have
$$
\int_0^n \left(\sum_{i=0}^n c_i\bar{P}_i(t)\right)^2 dt =
\sum_{i=0}^n |c_i|^2\int_0^n \big(\bar{P}_i(t)\big)^2 dt \\ \ge
|c_n|^2\int_0^n \big(\bar{P}_n(t)\big)^2 dt 
= \left(\frac{n!}{2^n\cdot (2n-1)!!}\right)^2 \cdot \frac1{2n+1}
= 
\frac{(n!)^2}{4^n\cdot (2n-1)!!\cdot (2n+1)!!}.
$$
The minimum is attained when $c_0=c_1=\ldots=c_{n-1}=0$.
Therefore, the minimum is 
$$
\frac{(n!)^2}{4^n\cdot (2n-1)!!\cdot (2n+1)!!} .
$$
The minimum is attained when 
$$
t^n-\sum_{i=0}^{n-1} a_i t^i = c_n \overline{P}_n(t) = 
\frac{n!}{(2n)!} \cdot \big(x^n(1-x)^n\big)^{(n)},
$$
so
$$
a_i = 
\frac{n!}{(2n)!} \cdot 
(-1)^{n-i+1}\binom{n}{i} (i+1)(i+2)\ldots(i+n).
$$
A: You can represent the polynomial $t^n-\sum_{k=0}^{n-1}a_kt^k$ as $$t^n-a^Tb(t)$$ where $a=[a_0\ a_2\cdots\ a_{n-1}]^T$ and $b(t)=[1\ t\ \cdots\ t^{n-1}]^T$. Then, the function that you want to minimize becomes $$f(a)=\int_{0}^1(t^n-a^Tb(t))(t^n-a^Tb)^Tdt\\=\int_{0}^1 (a^Tb(t)b(t)^Ta-2t^na^Tb(t)+t^{2n})dt\\=a^TQa-2a^Tp+1/(2n+1)$$ where $Q=\int_{0}^1b(t)b(t)^Tdt\implies Q=[q_{ij}],\ q_{ij}=\int_{0}^1 t^{i-1}t^{j-1}dt=\frac{1}{i+j-1}$ i.e. $Q$ is a Hilbert matrix. $p=\int_{0}^1t^n b(t) dt=c$ where $c=\left[\frac{1}{n+1}\ \frac{1}{n+2}\ \cdots\ \frac{1}{2n}\right]^T$. Then, minimizing $f(a)$ w.r.t $a$ will give the minimizer $a=Q^{-1}p$ 
A: This is only a partial answer. It describes how to compute the coefficients $(a_k)$
For a null gradient, we need
$$
\begin{align}
0
&=\frac{\partial}{\partial a_j}\int_0^1\left(t^n-\sum_{k=0}^{n-1} a_k \cdot t^k\right)^2\,\mathrm{d}t\\
&=2\int_0^1\left(t^{n+j}-\sum_{k=0}^{n-1}a_k\cdot t^{k+j}\right)\,\mathrm{d}t\\
&=2\left(\frac1{n+j+1}-\sum_{k=0}^{n-1}\frac{a_k}{k+j+1}\right)\tag{1}
\end{align}
$$
for $0\le j\le n-1$. Written as a matrix, $(1)$ is
$$
\begin{bmatrix}
\frac1{n+1}\\
\frac1{n+2}\\
\frac1{n+3}\\
\vdots\\
\frac1{2n}
\end{bmatrix}
=
\begin{bmatrix}
1&\frac12&\frac13&\cdots&\frac1n\\
\frac12&\frac13&\frac14&\cdots&\frac1{n+1}\\
\frac13&\frac14&\frac15&\cdots&\frac1{n+2}\\
\vdots&\vdots&\vdots&\ddots&\vdots\\
\frac1n&\frac1{n+1}&\frac1{n+2}&\cdots&\frac1{2n-1}
\end{bmatrix}
\begin{bmatrix}
a_0\\
a_1\\
a_2\\
\vdots\\
a_{n-1}
\end{bmatrix}\tag{2}
$$
The square matrix in $(2)$ is the Hilbert Matrix. The determinant is not $0$ and there is an explicit formula for the inverse; the $(i,j)$ element is
$$
(-1)^{i+j}(i+j+1)\binom{n+i}{n-j-1}\binom{n+j}{n-i-1}\binom{i+j}{i}^2\tag{3}
$$
where $i,j$ are $0$-based. $(2)$ and $(3)$ give explicit values for $a_j$.
