Determinant of a series of Hadamard matrix

Given: $$H=\ \left[ \begin{array}{cc|r} 1 & 1 \\ 1 & -1 \end{array} \right]$$

a Hadamard $H_2$ matrix.

and the series:

$$S=\sum_{k=0}^{N}{\frac{H^k}{k!}}$$

Is it possible to calculate the quantity $D=\det(S)$ as a function of N?

Thanks

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Have you computed $H^2$? $H^3$? etc just to see what $H^k$ looks like? –  Dilip Sarwate Feb 14 '12 at 15:41
This is a good example of the power of diagonalization. $$H = A \begin{pmatrix} 2 & 0 \\ 0 & -2 \end{pmatrix} A^{-1}$$ where it doesn't matter what $A$ is. Then $$S = A \begin{pmatrix} \sum_{k=0}^N 2^k/k! & 0 \\ 0 & \sum_{k=0}^N (-2)^k/k! \end{pmatrix} A^{-1}$$ so $$\det S = \left( \sum_{k=0}^N 2^k/k! \right) \left( \sum_{k=0}^N (-2)^k/k! \right).$$
For $N$ large, the first sum approaches $e^2$ and the second sum approaches $e^{-2}$, so the determinant approaches $1$. For a more accurate bound, we have $$\sum_{k=0}^N 2^k/k! \approx e^2 - 2^{N+1}/(N+1)!$$ and similarly for the other sum, so $$\left( \sum_{k=0}^N 2^k/k! \right) \left( \sum_{k=0}^N (-2)^k/k! \right) \approx (e^2 - 2^{N+1}/(N+1)!)(e^{-2} - (-2)^{N+1}/(N+1)!)$$ $$\approx 1 - \frac{2^{N+1}}{(N+1)!}(e^{-2} + (-1)^{N+1} e^2) + \mbox{lower order terms}.$$