I have the following expression:
$$ \phi_x(e^{j\omega_n})=\sum_{\tau=0}^{N-1}\sum_{k=0}^{N-1} x(k)x(k-\tau)e^{-j\omega_n k}e^{j\omega_n(k-\tau)} $$
I need help with justifying the following steps (where I need clarification, I'll put write a bold question):
Step 1) Perform the change of variables $l=k-\tau$
$$ \phi_x(e^{j\omega_n})=\sum_{\tau=0}^{N-1}\sum_{k=0}^{N-1} x(k)x(k-\tau)e^{-j\omega_n k}e^{j\omega_n(k-\tau)}= \sum_{k=0}^{N-1}\sum_{\tau=0}^{N-1} x(k)x(k-\tau)e^{-j\omega_n k}e^{j\omega_n(k-\tau)} $$
$$ \phi_x(e^{j\omega_n})=\sum_{k=0}^{N-1}\sum_{l=k}^{k-(N-1)} x(k)x(l)e^{-j\omega_n k}e^{j\omega_nl} $$
Q1: how come we can only "preferentially" replace variables here? For example the sum $\sum_{k=0}^{N-1}$ is left untouched - yet one would think that the variable change concerns it as well, in other words I would have transformed it into $\sum_{l=-\tau}^{(N-1)-\tau}$, yet my professor does not do this. Why?
Step 2) Separate the sums
$$ \phi_x(e^{j\omega_n})=\sum_{k=0}^{N-1}\biggl(x(k)e^{-j\omega_n k}\biggl(\sum_{l=k}^{k-(N-1)}x(l)e^{j\omega_nl} \biggr)\biggr) $$
We actually have the property of periodicity, that $x(k)=x(k+N)$, so we can write the above as:
$$ \phi_x(e^{j\omega_n})=\sum_{k=0}^{N-1}\biggl(x(k)e^{-j\omega_n k}\biggl(\sum_{l=0}^{N-1}x(l)e^{j\omega_nl} \biggr)\biggr) $$
Now my professor does something really funky, splitting these sums:
$$ \phi_x(e^{j\omega_n})=\biggl(\sum_{k=0}^{N-1}x(k)e^{-j\omega_n k}\biggr)\biggl(\sum_{l=0}^{N-1}x(l)e^{j\omega_nl} \biggr) $$
Q2: how come we can just "split" these two sums? I understand how this can be done when the elements of the inner sum are independent of the elements of the outer sum (then we just use distributivity), however here we have $l=k-\tau$ so $l$ is actually implicitly a function of $k$, so I do not understand how these sums can be split. Could you please explain?
Thanks a lot for your help!
Q1: In this case, he is just replacing $\tau$ wherever it occurs with $k - l$ (which is what you get when you solve $l = k - \tau$), so $$\begin{align}\phi_x(e^{j\omega_n})&= \sum_{k=0}^{N-1}\sum_{\tau=0}^{N-1} x(k)x(k-\tau)e^{-j\omega_n k}e^{j\omega_n(k-\tau)} \\&=\sum_{k=0}^{N-1}\quad\sum_{(k - l)=0}^{(k - l) =N-1} x(k)x(k-(k - l))e^{-j\omega_n k}e^{j\omega_n(k-(k - l))} \\&=\sum_{k=0}^{N-1}\quad\sum_{l=k}^{k-(N-1)} x(k)x(l)e^{-j\omega_n k}e^{j\omega_nl} \end{align}$$ There is no $\tau$ to replace in the first sum. Further, your version would still leave you $\tau$ in the equation, so you would now have three indices instead of two - not very helpful.
Q2: user586907 has already pointed this out, but let me expand on it a little:
$\sum_{l=0}^{N-1}x(l)e^{j\omega_nl}$ does not depend on $k$ at all. So from the standpoint of the sum over $k$, we can just set $A = \sum_{l=0}^{N-1}x(l)e^{j\omega_nl}$, and rewrite the expression as $$\begin{align}\phi_x(e^{j\omega_n})&=\sum_{k=0}^{N-1}x(k)e^{-j\omega_n k}A\\ &=A\sum_{k=0}^{N-1}x(k)e^{-j\omega_n k}\end{align}$$ by a simple application of the distributive law.