# Thread: Identity of a sum

1. ## Identity of a sum

Let $x_{1},\dots,x_{N}$ real numbers such that $$\underset{i=1}{\overset{N}{\sum}}\frac{x_{i}}{i+ j}=\frac{1}{2j+1}$$ for each natural number $j\in\left[1,N\right]$. Find (in function of $N$) the value of $$\underset{i=1}{\overset{N}{\sum}}\frac{x_{i}}{2i +1}.$$

2. ## Re: Identity of a sum

I know that I have to use the inverse of the Hilbert matrix, but I can understand how.

3. ## Re: Identity of a sum

Well, yes but...
First

$X = H^{-1} B$

where H is (like) the Hilbert matrix and

B is the matrix with entries $\frac{1}{2j+1}$

Then evaluate
$B^t H^{-1} B=\frac{n(n+1)}{(2n+1)^2}$

4. ## Re: Identity of a sum

Originally Posted by Idea
Well, yes but...
First

$X = H^{-1} B$

where H is (like) the Hilbert matrix and

B is the matrix with entries $\frac{1}{2j+1}$

Then evaluate
$B^t H^{-1} B=\frac{n(n+1)}{(2n+1)^2}$
Yes I see what you say... but how I can prove that $B^{t}H^{-1}B=\frac{N\left(N+1\right)}{\left(2N+1\right)^{2} }?$ Have I to use the explicit expression of $H^{-1}?$

5. ## Re: Identity of a sum

To start, let me say that I don't know how to solve this problem.
I arrived at the answer by checking the results for some small values of n and then making a wild guess.

If you have an explicit expression for the i-j entry of the inverse of H then that would be one way.
There is such a formula for the inverse of the Hilbert matrix in terms of binomial coefficients and it is quite complicated.

Another way would be to use induction on the size of H.

6. ## Re: Identity of a sum

I'm not able to use the explicit inverse of the Cauchy (Hlibert) matrix. This is a new "hint" for this problem: if you consider the Shifted Legendre polynomials $$P_{n}\left(x\right)=\frac{1}{n!}\frac{d^{n}}{dx^ {n}}\left(x-x^{2}\right)^{n}$$ you have $$\int_{0}^{1} P_{n}\left(x\right)P_{m}\left(x\right)dx=\frac{ \delta_{mn} }{2n+1}$$ where $\delta_{mn}$ is the Kronecker delta. How can use it? Can someone help me?