The standard type III discrete cosine transformation (DCT) is defined as [follows](https://en.wikipedia.org/wiki/Discrete_cosine_transform):


$${X\_k} = \frac{1}{2}{x\_0} + \sum\limits\_{n = 1}^{N - 1} {{x\_n}} \cos \left[ {\frac{\pi }{N}n\left( {k + \frac{1}{2}} \right)} \right]\quad \quad k = 0, \ldots ,N - 1.$$


It transforms a list of $\{x\_n\}$ into list of $\{X\_k\}$


Now I encountered a summation in a paper like this:


$${X\_k} = \frac{1}{2}{x\_0} + \sum\limits\_{n = 1}^{N - 1} {{x\_n}} \cos \left[ {\frac{\pi }{{\tilde N}}n\left( {k + \frac{1}{2}} \right)} \right]\quad \quad k = 0, \ldots ,\tilde N - 1.$$


The only difference is that now $\tilde N = 2N$.


The paper says this summation could exploit the power of fast DCT. However, it tooks me days, still can't figure out how to relate it to the standard DCT-III. How to do it?



How to relate this summation to standard discrete cosine transformation?
in the second transformation the integer $k$ runs from $0$ to $\tilde{N}-1$, not to $N-1$; so you have twice as many data points $X\_k$ than you have moments $x\_n$, which is perfectly OK (oversampling); to use the conventional formulas for the DCT, just extend the list of moments $x\_n$ by padding it with zeros ($x\_n=0$ for $N\leq n\leq \tilde{N}-1$, so that you have $\tilde{N}$ moments $x\_n$; these then give you $\tilde{N}$ values $X\_n$.


