So, I'm trying to prove the following sum formula, and I can't really get it to work out correctly :
where ab = $2\pi$ and G = Fourier Transform of g. How would I approach a proof of this formula? Usually, I see this type of formula presented very differently in a lot of different resources,, and it's throwing me off.
Here is what my textbook is doing. I don't really understand the substitution for the cosine partial sum or the N->infinity part. Any clue?
On
Express the sum as a geometric series.
$$\sum_{n=1}^N \cos{n z} = \operatorname{Re}{\left [\sum_{n=1}^N e^{i n z} \right ]} $$
$$\begin{align}\sum_{n=1}^N e^{i n z} &= e^{i z}\frac{e^{i N z}-1}{e^{i z}-1} \\ &= e^{i z/2} e^{i N z/2} \frac{e^{i N z/2}-e^{-i N z/2}}{e^{i z/2}-e^{-i z/2}}\\ &= e^{i (N+1) z/2} \frac{\sin{(N z/2)}}{\sin{(z/2)}}\end{align} $$
Taking real parts, we have
$$\begin{align}\sum_{n=1}^N \cos{n z} &= \frac{\sin{(N z/2)} \cos{((N+1) z/2)}}{\sin{(z/2)}}\end{align} $$
Use the fact that
$$2 \sin{a} \cos{b} = \sin{(a+b)}+\sin{(a-b)}$$
to see that
$$\sin{(N z/2)} \cos{((N+1) z/2)} = \frac12 \left [\sin{((2 N+1) z/2)} - \sin{(z/2)} \right ]$$
Therefore
$$\begin{align}\sum_{n=1}^N \cos{n z} &= \frac{\sin{((2 N+1) z/2)}}{2 \sin{(z/2)}} - \frac12\end{align}$$
Let $z=\lambda a$ and the result follows.
One Normalization of the Fourier Transform
There are a several ways to normalize the Fourier Transform. The way I find nicest is to define $$ \hat{f}(\xi)=\int_{-\infty}^\infty f(x)\,e^{-2\pi ix\cdot\xi}\,\mathrm{d}x\tag{1} $$ Define $$ g(x)=\sum_{k=-\infty}^\infty f(x+k)\tag{2} $$ Then $g$ has period $1$. Plugging $x=0$ into $(2)$ gives $$ g(0)=\sum_{k=-\infty}^\infty f(k)\tag{3} $$ For $n\in\mathbb{Z}$, the coefficients of the Fourier Series are $$ \begin{align} \hat{g}(n) &=\int_0^1g(x)\,e^{-2\pi inx}\,\mathrm{d}x\\ &=\sum_{k=-\infty}^\infty\int_0^1f(x+k)\,e^{-2\pi inx}\,\mathrm{d}x\\ &=\int_{-\infty}^\infty f(x)\,e^{-2\pi inx}\,\mathrm{d}x\\ &=\hat{f}(n)\tag{4} \end{align} $$ Inverting the Fourier Series, we get $$ g(x)=\sum_{n=-\infty}^\infty\hat{g}(n)e^{2\pi inx}\tag{5} $$ Plugging $x=0$ into $(5)$ and applying $(4)$ yields $$ \begin{align} g(0) &=\sum_{n=-\infty}^\infty\hat{g}(n)\\ &=\sum_{n=-\infty}^\infty\hat{f}(n)\tag{6} \end{align} $$ $(3)$ and $(6)$ give the Poisson Summation Formula: $$ \sum_{n=-\infty}^\infty f(n)=\sum_{n=-\infty}^\infty\hat{f}(n)\tag{7} $$
Let $f(x)=g(ax)$. Then we have $$ \begin{align} \hat{f}(n) &=\int_{-\infty}^\infty f(x) e^{-2\pi inx}\,\mathrm{d}x\\ &=\int_{-\infty}^\infty g(ax) e^{-2\pi inx}\,\mathrm{d}x\\ &=\frac1a\int_{-\infty}^\infty g(x) e^{-2\pi inx/a}\,\mathrm{d}x\\ &=\frac1a\hat{g}(n/a)\tag{8} \end{align} $$ Applying $(8)$ gives the scaled version of $(7)$: $$ \sqrt{a}\sum_{n=-\infty}^\infty g(an)=\frac1{\sqrt{a}}\sum_{n=-\infty}^\infty\hat{g}(n/a)\tag{9} $$ Equation $(9)$ is the equivalent of the equation in the question for the Fourier Transform defined in $(1)$. To get the equation in the question, we need to use a different normalization of the Fourier Transform.
Another Normalization of the Fourier Transform
If we define the Fourier Transform as $$ \hat{f}(\xi)=\int_{-\infty}^\infty f(x)\,e^{-ix\cdot\xi}\,\mathrm{d}x\tag{1a} $$ and define $$ g(x)=\sum_{k=-\infty}^\infty f(x+2\pi k)\tag{2a} $$ then $g$ has period $2\pi$ and plugging $x=0$ into $\mathrm{(2a)}$ gives $$ g(0)=\sum_{k=-\infty}^\infty f(2\pi k)\tag{3a} $$ For $n\in\mathbb{Z}$, the coefficients of the Fourier Series are $$ \begin{align} \hat{g}(n) &=\int_0^{2\pi}g(x)\,e^{-inx}\,\mathrm{d}x\\ &=\sum_{k=-\infty}^\infty\int_0^{2\pi}f(x+2\pi k)\,e^{-inx}\,\mathrm{d}x\\ &=\int_{-\infty}^\infty f(x)\,e^{-inx}\,\mathrm{d}x\\ &=\hat{f}(n)\tag{4a} \end{align} $$ Inverting the Fourier Series, we get $$ g(x)=\frac1{2\pi}\sum_{n=-\infty}^\infty\hat{g}(n)e^{inx}\tag{5a} $$ Plugging $x=0$ into $\mathrm{(5a)}$ and applying $\mathrm{(4a)}$ yields $$ \begin{align} g(0) &=\frac1{2\pi}\sum_{n=-\infty}^\infty\hat{g}(n)\\ &=\frac1{2\pi}\sum_{n=-\infty}^\infty\hat{f}(n)\tag{6a} \end{align} $$ $\mathrm{(3a)}$ and $\mathrm{(6a)}$ give the Poisson Summation Formula: $$ \sum_{n=-\infty}^\infty f(2\pi n)=\frac1{2\pi}\sum_{n=-\infty}^\infty\hat{f}(n)\tag{7a} $$
Let $f(x)=g(ax)$. Then we have $$ \begin{align} \hat{f}(n) &=\int_{-\infty}^\infty f(x) e^{-inx}\,\mathrm{d}x\\ &=\int_{-\infty}^\infty g(ax) e^{-inx}\,\mathrm{d}x\\ &=\frac1a\int_{-\infty}^\infty g(x) e^{-inx/a}\,\mathrm{d}x\\ &=\frac1a\hat{g}(n/a)\tag{8a} \end{align} $$ Applying $\mathrm{(8a)}$ gives the scaled version of $\mathrm{(7a)}$: $$ \sqrt{2\pi a}\sum_{n=-\infty}^\infty g(2\pi an)=\frac1{\sqrt{a}}\sum_{n=-\infty}^\infty\hat{g}(n/a)\tag{9a} $$ $\mathrm{(9a)}$ is the equation in the question.