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Given the sequence $\{ f(i) \}_{0 \leq i \leq n}$ and the array $\{ g_{n}(i,j) \}_{0 \leq i, j \leq n}$ such that $f(i) = \sum_{j = 0}^{n} h_{j} \, g_{n}(i,j)$ for some integers $h_{0}, \dots, h_{n}$, is there a way to write $h_{j}$ as a sum over $f(i)$ and $g_{n}(i,j)$ using the calculus of finite differences?

Thanks.

Update: The first post below answers the question of writing $h_{j}$ as a function of $f$ and $g$, but I'd like to avoid matrix inversion if at all possible, hence calculus of finite differences.

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  • $\begingroup$ Keeping all the parameters $g_n(i,j)$ fixed (and the matrix $g$ defined in my post invertible), any formula for $h_0$ as a function of the parameters $f(i)$ valid for every $(f(i))_i$ must be the one I gave in my answer. So, you could explain a little more how calculus of finite differences, or any other sweet technology, could enter the picture at all. $\endgroup$ – Did Mar 7 '11 at 6:39
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Using the vector notations $f=(f(i))_i$ and $h=(h_i)_i$ and the matrix notation $g=(g_n(i,j))_{i,j}$, you impose that $f=g\cdot h$ and you ask for an expression of $h$ as a function of $f$ and $g$. Let $D=\det g$. If $g$ is not invertible, that is, if $D=0$, either this is impossible or the solution is not unique. If $g$ is invertible, that is, if $D\ne 0$, the unique solution is $h=g^{-1}\cdot f$ and you know that $D\cdot g^{-1}$ is the comatrix $c$ of $g$.

Thus, in the invertible case $h_i=\displaystyle\frac1D\sum_{j=0}^nc(i,j)f(j)$ for every $0\le i\le n$.

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  • $\begingroup$ Excellent post! $\endgroup$ – user02138 Mar 6 '11 at 21:23
  • $\begingroup$ @user02138 Then you might want to accept it. $\endgroup$ – Did Mar 11 '11 at 22:29
  • $\begingroup$ Sure. ;) $\endgroup$ – user02138 Mar 11 '11 at 23:39

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