Distance in polynomial spaces

57 Views Asked by At

I'm trying to solve some exercises proposed by ENS of Paris. In particular the last one (which can be seen at http://www.ens.fr/IMG/file/SI2015/Sujets%20SCiences/Math2-version%20anglaise.pdf).

Since there are no solutions, I have a doubt!

I'm referring to second exercise.

1) In order to find $a_0$ I transformed the numerator as follows:

$(x-1)\cdot \dots \cdot (x-n) = (x+2-3)\cdot \dots \cdot (x-n) = (x+2)(x-2)\cdot \dots \cdot (x-n) + \dots$, where $\dots$ means the rest of the multiplication by $-3$. Then:

$=(x+2)(x+3-5)\cdot \dots \cdot (x-n) + \dots = (x+2)(x+3)\cdot \dots \cdot (x-n) + \dots$ and so on.

Then $a_0$ must be equal to $1$. Right?

For the second point I computed the integral of $T\cdot x^j$ for $1\leq j \leq n$ because if $T$ is orthogonal to the all generators, then it will be orthogonal to the whole space $F$. Then:

$\left< T, x^j \right> = \int_0^1\left( T(x) x^j \right)dx = \dots = \frac{a_n}{n+j+1} + \dots + \frac{a_0}{j+1}$ and this is zero because for $x = j$ then $S(X)= 0$. Right?

May I have a hint for the (3)?

Thank you very much!