# Trigonometric system and orthogonality 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 $$\frac{0}{0}$$?

## 3 Answers

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.

• huh ok, weird.. – 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 $$j=k$$.

• 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? – MinYoung Kim Jul 12 at 9:59
• What I am saying is for $j=k$ prove (4.9) without referring to previous line. Just see what LHS is. – Kavi Rama Murthy Jul 12 at 10:05
• 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? – MinYoung Kim Jul 12 at 10:08
• 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. – Kavi Rama Murthy Jul 12 at 10:11
• 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. – MinYoung Kim Jul 12 at 10:12