What have you done so far? Have you tried small examples, like N = 2 or N = 3?
RGV
#3
nrgyyy
3
0
Well its easy to see that it works with examples like N=2 or N=3. For example for N=2 the value of both sides is f(1)+f(-1)+2f(0). Same for N=3. I am thinking that maybe I should do a variables change in the first double sums, to end up to a more common summation formula, but I am kinda stuck.
EDIT: We can consider that s and t are integers, or that the arbitrary f() function represents a discrete time signal.
Looks to me like there is a change of index at work there. Since the left side of the equation involves f(t- s) and the other side f(τ), you should immediately think of τ= t- s.
#5
nrgyyy
3
0
could you please help on how the limits of the left side sums would be in that case (τ=t-s)?
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Hello,
This is the attachment, the steps to solution are pretty clear. I guess there is a mistake on the highlighted part that prompts this thread.
Ought to be ##3^{n+1} (n+2)-6## and not ##3^n(n+2)-6##. Unless i missed something, on another note, i find the first method (induction) better than second one (method of differences).