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unless $k-r=mN$ the sum is, as you say: $$ (1/N)\frac{1-e^{2\pi j(k-r)}}{1-e^{2\pi j(k-r)/N}} = 0 $$ if $k-r=mN$ then this formula is indeterminate, but since each term in the sum is 1, the sum of $N$ terms, divided by $N$ is $1$


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Your calculation, if I read it correctly, is right. However, $$e^{(j)(2\pi)(k-r)}=1,$$ since $k-r$ is an integer. It follows that the expression $1-e^{(j)(2\pi)(k-r)}$ that we get on top when summing the geometric series is equal to $0$. If $k-r$ is not of the shape $mN$, the result is $0$, since the denominator is non-zero. If $k-r=mN$, the formula for ...


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HINT: From the geometrical sequence, you know that $$ \sum_{n=0}^Nq^n =\frac{1-q^{N+1}}{1-q}, \quad q\neq 1. \tag1 $$ By differentiating with respect to $q$ $$ \sum_{n=0}^Nnq^{n-1} =\partial_q \left(\frac{1-q^{N+1}}{1-q}\right), \quad q\neq 1. \tag2 $$ Then apply it with $q=e^{2i\pi x}$, observing that $\displaystyle ...


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For the second equality, you can use the formula for a geometric sum: $$ \sum_{n=0}^N q^n = \frac{1-q^{N+1}}{1-q}. $$ In this case, $q=e^{2\pi i x}$, so you get (notice that the zeroth term $e^{2\pi i 0 x}=1$ is missing from the sum) $$ \sum_{n=1}^N e^{2\pi i n x} = \frac{1- e^{2\pi i (N+1) x}}{1-e^{2\pi i x}} - 1 = \frac{e^{2\pi i x} - e^{2\pi i (N+1) ...


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The character sum involves the polynomial $$ f(a,b,c)=\frac{1}{12}(a^3-c^3)+(\frac{b}{2}+\frac{1}{8})(a^2-c^2)+(\frac{b}{2}+b^2)(a-c)+kc-ja. $$ Staring at this for a while suggested to me to look at $$ f(A+C,B-A/2-1/4,A-C)=\frac{C^3}6+C\left(-\frac18+2B^2-j-k\right)+A(-j+k) $$ instead (ain't Mathematica's Simplify wonderful!). In the interest of avoiding ...


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For any $n$, $$1+1+1+1+1+1+1+1+1+1=10.$$ Alternatively, denoting $z=e^{i\theta}$, $$S=\frac{z^{\frac12}(z^{10}-1)}{z^{\frac12}-z^{-\frac12}}=\frac{z(z^{10}-1)}{z-1}.$$ By L'Hospital, $$\lim_{z\to1}\frac{z^{10}-1}{z-1}=\lim_{z\to1}10z^{9}.$$ Or, if you want to do it the hard way, ...


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If $\theta=2n\pi$, $e^{ki\theta}=e^{2\pi(nk)i}=1$ if $k\in\Bbb Z$. Thus we have $$S=10.$$



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