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## Main Question or Discussion Point

Hello fellow nerds,

I've come across a math problem, where I'd like to find the solution vector of a NxN square matrix. It is possible to find a solution for a given N, albeit numbers in the matrix become very large for any N>>1, and numbers in the solution vector become very small. So it's not easy to compute solutions. Anyhow I realized that the components of the solution vector seem to converge and I'm asking myself if it's possible to write down the solution for N->∞.

Here's the matrix:

\begin{equation}

\begin{pmatrix}

1&-2&[-2]^2&\dots&[-2]^{N-1}\\

1&-6&[-6]^2&\dots&[-6]^{N-1}\\

1&-12&[-12]^2&\dots&[-12]^{N-1}\\

\vdots&\vdots&\vdots&\ddots&\vdots\\

1&-N(N+1)&[-N(N+1)]^2&\dots&[-N(N+1)]^{N-1}

\end{pmatrix}

\begin{pmatrix}

a_1\\

a_3\\

a_5\\

\vdots\\

a_{2N-1}

\end{pmatrix}=

\begin{pmatrix}

1/3\\

1/5\\

1/7\\

\vdots\\

\frac 1 {2N+1}

\end{pmatrix}

\end{equation}

or:

\begin{eqnarray}

\sum_{k=1}^{N}a_{2k-1} \left[-l(l+1)\right]^{k-1}=\frac 1{2l+1}\;,

\end{eqnarray}

for all l≤N.

I'm looking for the vector

But I cannot think of a way to find these values without resorting to numerics.

Cheers, Max

I've come across a math problem, where I'd like to find the solution vector of a NxN square matrix. It is possible to find a solution for a given N, albeit numbers in the matrix become very large for any N>>1, and numbers in the solution vector become very small. So it's not easy to compute solutions. Anyhow I realized that the components of the solution vector seem to converge and I'm asking myself if it's possible to write down the solution for N->∞.

Here's the matrix:

\begin{equation}

\begin{pmatrix}

1&-2&[-2]^2&\dots&[-2]^{N-1}\\

1&-6&[-6]^2&\dots&[-6]^{N-1}\\

1&-12&[-12]^2&\dots&[-12]^{N-1}\\

\vdots&\vdots&\vdots&\ddots&\vdots\\

1&-N(N+1)&[-N(N+1)]^2&\dots&[-N(N+1)]^{N-1}

\end{pmatrix}

\begin{pmatrix}

a_1\\

a_3\\

a_5\\

\vdots\\

a_{2N-1}

\end{pmatrix}=

\begin{pmatrix}

1/3\\

1/5\\

1/7\\

\vdots\\

\frac 1 {2N+1}

\end{pmatrix}

\end{equation}

or:

\begin{eqnarray}

\sum_{k=1}^{N}a_{2k-1} \left[-l(l+1)\right]^{k-1}=\frac 1{2l+1}\;,

\end{eqnarray}

for all l≤N.

I'm looking for the vector

__a__(the numbering of the index is indeed arbitrary). For instance for N=1, one finds a_{1}=1/3. For N=2 one finds a_{1}=2/5 and a_{3}=1/30. And so on. I already found that a_{1}→1/2 for N→∞. When I look at the numerical solutions, the other a_{i}also seem to converge slowly, see the following plot:But I cannot think of a way to find these values without resorting to numerics.

Cheers, Max

Last edited: