I'm doing an exercise in which I need to show that the DCT of $\tilde{x} = (x_{N-1}, x_{N-2}, ..., x_1, x_0) $, with $\tilde x_m = x_{N-m-1}$, is equal to $ \tilde{X}_k = (-1)^{k}X_{k}$, but I have difficulty with the indices in the cosine manipulation and $\tilde x$:

\begin{eqnarray*} \tilde X_k &=& \sqrt{\frac{2}{N}}\sum_{m=0}^{N-1}\tilde x_{m}\cos\bigg(\frac{\pi k(N-m-1+\frac{1}{2})}{N}\bigg) \\ &=& \sqrt{\frac{2}{N}}\sum_{m=0}^{N-1}\tilde x_{m} \cos\bigg(\frac{\pi k(N-m-\frac{1}{2})}{N}\bigg) \\ &=& \sqrt{\frac{2}{N}}\sum_{m=0}^{N-1}\tilde x_m \cos\bigg(\frac{\pi k(-m-\frac{1}{2})}{N} + \pi k \bigg) \\ &=&\sqrt{\frac{2}{N}}\sum_{m=0}^{N-1}\tilde x_m \cos\bigg(\pi k - \frac{\pi k(m + \frac{1}{2})}{N} \bigg) \\ &=&\sqrt{\frac{2}{N}}\sum_{m=0}^{N-1} x_m (-1)^{k}\cos\bigg(\frac{\pi k(m + \frac{1}{2})}{N} \bigg) \\ &=&(-1)^{k}\sqrt{\frac{2}{N}}\sum_{m=0}^{N-1} x_m \cos\bigg(\frac{\pi k(m + \frac{1}{2})}{N} \bigg) \\ &=& (-1)^k X_k \end{eqnarray*}

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    $\begingroup$ Isn't a tilde missing in $x_k = x_{N-k-1}$? $\endgroup$ – Laurent Duval May 17 '20 at 20:15
  • $\begingroup$ I have tried to simplified some notations, hoping this is correct $\endgroup$ – Laurent Duval May 17 '20 at 20:39

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