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I'm having trouble understanding the 2nd line. So, just looking at the 3rd line (4.9), I can see that this is true, ie, each term in the series will $= 1$ if $j=k$ by orthogonality of the trig system, and summing that N times gives N if $j = k$, but all terms 0 if $j \neq k$

But then, using the partial sum equation in line 2, if $j = k$, the partial sum is $\frac{0}{0}$?


In the given proof, there were two minor omissions, which may have been the source of your doubts about the proof.

First omission . . .

The line in the proof which started with $$ \text{For}\;\,0\le j\le N-1\;\text{. . .} \qquad\qquad\qquad\; $$ should have been $$ \text{For}\;\,0\le j,k\le N-1\;\text{. . .} \qquad\qquad\;\;\;\;\; $$

Second omission . . .

Let $N$ be a positive integer, and let $r\in\mathbb{C}$.

The stated formula should have been $$ \sum_{n=0}^{N-1}r^n = \begin{cases} {\Large{\frac{1-r^N}{1-r}}}&&\text{if}\;r\ne 1\\[4pt] N&&\text{if}\;r=1\\ \end{cases} \qquad\;\;\; $$ With those corrections, let $$ r=e^{2i\pi(j-k)/N} \qquad\qquad\qquad\qquad\qquad $$ Then for the case $j=k$, we have $r=1$, so the identity $$ \sum_{n=0}^{N-1}\left[e^{2i\pi(j-k)/N}\right]^n=N \qquad\qquad\qquad\;\;\; $$ is immediate.

For the case $j\ne k$, since $0\le j,k\le N-1$, it follows that $2\pi(j-k)/N$ is not an integer multiple of $2\pi$.

Hence $r\ne 1$, so \begin{align*} &\sum_{n=0}^{N-1}\left[e^{2i\pi(j-k)/N}\right]^n\\[4pt] =&\sum_{n=0}^{N-1}r^n\\[4pt] =&\frac{1-r^N}{1-r}\\[4pt] =&\;0\;\;\;\text{[since $r^N=e^{2i\pi(j-k)}=1$, and $1-r\ne 0$]}\\[4pt] \end{align*}


Well large part of the issue seems that the first line is completly false. The identity does not hold when r=1 and also does not seem to hold when r is negative. Since this is a proof such information about r should be contained in hypothesis, and in that case the case of j=k has to be treated separately.

  • $\begingroup$ huh ok, weird.. $\endgroup$ – MinYoung Kim Jul 12 at 10:20

You have to handle the case $j=k$ separately. It is obvious that the sum is $N$ when $j=k$ because each term is just $1$.

The line just above (4.9) is not valid when $j=k$.

  • $\begingroup$ For the partial sum equation in line 2, if $j = k$, we get $\frac{1-e^0}{1-e^0} = \frac{0}{0}$. But clearly that is not correct? $\endgroup$ – MinYoung Kim Jul 12 at 9:59
  • $\begingroup$ What I am saying is for $j=k$ prove (4.9) without referring to previous line. Just see what LHS is. $\endgroup$ – Kavi Rama Murthy Jul 12 at 10:05
  • $\begingroup$ I get that just looking at (4.9) with $j=k$, we get sum = N. But if we use equation in line 2, we don't get it. My question is why is there a discrepancy? $\endgroup$ – MinYoung Kim Jul 12 at 10:08
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    $\begingroup$ Equation in line 2 is not valid when $r=k$. Division by $0$ is not allowed in Mathematics. So there is a mistake in that line. $\endgroup$ – Kavi Rama Murthy Jul 12 at 10:11
  • $\begingroup$ oh ok. so equation 2 is just a "general" form, but when we are considering j =k, it's not applicable because we divide by zero. $\endgroup$ – MinYoung Kim Jul 12 at 10:12

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