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Is DCT differentiable? I.e. is $\frac{\mathrm{d}\mathrm{DCT}(x)}{\mathrm{d} x}$ defined?

I've seen a few implementations in automatic differentiation software packages, so I suppose the answer is positive, but I'm still missing a mathematical proof.

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The Discrete Cosine Transform of a discrete sequence $x$ can be defined as:

$$ {\tt DCT(}x{\tt )} = X_k = \frac{1}{2} (x_0 + (-1)^k x_{N-1}) + \sum_{n=1}^{N-2} x_n \cos \left[\frac{\pi}{N-1} n k \right] \quad \quad k = 0, \dots, N-1. $$

If, by differentiable, you mean: $$ \frac{\partial {\tt DCT(}x{\tt )}}{\partial x} $$ then isn't it just $$ \frac{\partial {\tt DCT(}x{\tt )}}{\partial x} = \frac{\partial X_k}{\partial x} =\left [ \begin{array}{ccccc} \frac{1}{2} & \cdots & \cos \left[\frac{\pi}{N-1} n k \right] & \cdots & (-1)^k\frac{1}{2} \end{array} \right ] $$ where $x = [ x_0, x_1, \ldots, x_{N-2}, x_{N-1} ]^T$ ?

What do you mean by a "mathematical proof"?

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