Why do we need f(-1)=f(15) = -1 ?
We have f(-1)=f(15) by periodic extension. And f(-1)= -f(1) = -0.8315.
We're talking about odd meaning antisymmetric about the f axis (n=0), correct?
I understand... except the part where you say the function is not exactly odd. Based on what you're saying, if we extend the signal beyond n=0..N-1 to make it periodic, then isn't it exactly odd (regardless of any discontiniuity)?
Thanks all for the responses!
I'm confused about the DFT of the data, fn = cos(3\pin/N) for n=0,1,...,N. It is definitely an even function, and I read that the Fourier coefficients of an even function is real. But when I take the FFT of this in Matlab I get complex numbers, not real numbers. What am I missing?
Thanks ...
Sorry I realize that was a little incomplete. Let \mathcal{X} and \mathcal{Y} be complex Euclidean spaces with dim(\mathcal{X})=dim(\mathcal{Y})=n.
Define
M(\rho) = \max\left\{<u u^{\ast},\rho>\,:\,u\in\mathcal{X}\otimes\mathcal{Y}\;\text{is maximally entangled}
\right\}...
Consider the maximum inner product of a density operator with a maximally entangled state. (So, given a density operator, we're maximizing over all maximally entangled states.)
I'm pretty sure the minimum value (a lower bound on the maximum) for this is 1/n2, using the density operator 1/n2...
I'm trying to prove that the Holevo quantity does not increase when a channel is applied to the ensemble of states.
So, if
\Phi(ε) = { (p(a), \Phi(ρa)) : a\in\Gamma},
then I want to prove that
\chi(\Phi(ε)) ≤ \chi(ε)
where \chi refers to the Holevo quantity. I'm trying an...