I cannot understand how the summation range has changed via substitution of m.


Full problem with solution (for context):

Original Problem

  • \$\begingroup\$ The Photon, Why am I substituting k for m when m=n-k and not m=k? And how does n become the new upper limit? Explain a bit more. \$\endgroup\$
    – howland12
    Jan 9 '17 at 19:26
  • \$\begingroup\$ You're making the substitution m=n-k, and summing over some values of m. But you want to sum the same terms you had before. So, when k=0, what does m equal? That's one of your limits when summing over m. And when k = infinity, what does m equal? That's your other limit in the new summation. \$\endgroup\$
    – The Photon
    Jan 9 '17 at 20:19
  • \$\begingroup\$ Okay. So the prime purpose of the substitution was to manipulate the summand by substituting with an "arbitrary" m in order to match one sampling equation involving delta which I have (which is in the same form of the bottom equation utilizing the m substitution), and the aftermath was the sum limits being changed, correct? The goal is changing the summand, not the limits (limit change is just the effect of changing the summand, which I should take into account for proper evaluation of the sum). \$\endgroup\$
    – howland12
    Jan 9 '17 at 23:13
  • \$\begingroup\$ Probably, but you haven't included enough further steps in the calculation for me to tell. \$\endgroup\$
    – The Photon
    Jan 9 '17 at 23:18
  • \$\begingroup\$ Question is edited to include original context of my summation question. \$\endgroup\$
    – howland12
    Jan 9 '17 at 23:27

When you substituted m = n-k, that also swaps the limits.

Because when \$k=\infty\$ (the old upper limit), \$n-k=-\infty\$. Since that is lower than any other index in the summation, it becomes the new lower limit.


I think the problem you were given is itself a bit wrongheaded since you can just pull out the first term of the first summation and get

$$\begin{align} \sum_{m=0}^n (n-m) - \sum_{m=1}^n(n-m) &= (n-0) + \sum_{\color{red}{m=1}}^n (n-m) - \sum_{m=1}^n(n-m)\\ &= n \end{align}$$

No mucking about with delta functions is needed to get the desired result.


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