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Math Help - Identity of a sum

  1. #1
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    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 $$\underset{i=1}{\overset{N}{\sum}}\frac{x_{i}}{2i +1}.$$
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  2. #2
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    Re: Identity of a sum

    I know that I have to use the inverse of the Hilbert matrix, but I can understand how.
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  3. #3
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    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}
    Last edited by Idea; June 20th 2014 at 01:14 AM.
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  4. #4
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    Re: Identity of a sum

    Quote Originally Posted by Idea View Post
    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 $H^{-1}?$
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  5. #5
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    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.
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  6. #6
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    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?
    Last edited by Peppo; July 8th 2014 at 02:59 AM.
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