Geometric series weighted by a cosine function

Question

[Edited, corrected notation!]

I'm dealing with following series $$ f_d(\omega) := \frac{1-d}{1+d}\sum_{j=-\infty}^{\infty} d^j cos(2\pi j \omega) = \frac{1-d}{1+d}\Big[1 + 2 \sum_{j=1}^{\infty} d^j cos(2\pi j \omega)\big], $$ where $0<d<1$ and $\omega>0$. In particular, I would like to prove that, for every $d_1, d_2 \in (0,1)$, $$ f_{d_1}(\omega) \leq f_{d_2}(\omega) \text{ if } d_1 \geq d_2. $$ Moreover, do you think is possible to compute the value of the sum for a given $\omega$? One can observe that, for $\omega = n \in \mathbb{N}$, $f_d(\omega) = 1$ and that, for $\omega = n + \frac{1}{2} \in \mathbb{N}$, $f_d(\omega) = \Big(\frac{1-d}{1+d}\Big)^2$. Is there a general expression for any value of $\omega$? Can I at least say that $ \lim_{d \to 0} f_d(\omega) = 1?$

Here is a link to a plot of $f_d$ for many values of $d$.

Answer 1

Here's an outline how to get started to find the sum $\sum_{j=1}^nd^j\cos(2\pi j \omega).$You will need to apply the summation by parts formula twice. We'll use the forward difference notation: $\Delta f(j)=f(j+1)-f(j)$. Then the summation by parts formula is $$\sum_{j=a}^b f(j)\Delta (g(j))=f(b+1)g(b+1)-f(a)g(a)-\sum_{j=a}^b\Delta (f(j))g(j+1) $$To simplify notation,let $c=2\pi \omega$. Put $f(j)=\cos(cj)$. Note that $$d^j=\Delta (\frac{d^j}{d-1})$$ Thus $$\sum_{j=1}^nd^j\cos(2\pi j \omega)=\sum_{j=1}^n\cos(cj)\Delta (\frac {d^j}{d-1}) $$. Note that$$\Delta \cos(cj)=\cos(c(j+1))-\cos(cj)$$ $$=\cos(c(j+1/2)+c/2)-\cos(c(j+1/2)-c/2)$$ $$=-2\sin(c(j+1/2))\sin(c/2)$$Thus when you applythe summation by parts formula, you get an expression similar to the one you started with, but "cos" has been replaced by "sin" and j has been shifted to j+1/2. Now apply the summation by parts formula again. Now you will get an expression in which "sin" has been replaced by "cos" and j+1/2 has been shifted to j+1. Use $$\sum_{j=1}^nd^{j+1}\cos(2\pi (j+1) \omega)=\sum_{j=1}^nd^j\cos(2\pi j \omega)-d\cos(2\pi \omega)+d^{n+1}\cos(2\pi \omega (n+1))$$ Thus, after some algebra, you will have an equation of the form $$\sum_{j=1}^nd^j\cos(2\pi j \omega)=A+B\sum_{j=1}^nd^j\cos(2\pi j \omega) $$ which you can solve for $\sum_{j=1}^nd^j\cos(2\pi j \omega).$ Do the same for $\sum_{j=n'}^1d^j\cos(2\pi j \omega).$ and take limits as $n \rightarrow \infty, n' \rightarrow - \infty$

Answer 2

Thanks a lot! It took several hours, but I finally did it. I'll share here the results, should anyone be interested. After following the suggested steps taking the limit for $n\to\infty$, one gets \begin{align*} \sum_{j=1}^{\infty} d^{j} \cos(c) &= \Bigg[-\frac{d}{d -1} \cos(c) - \frac{2d^2}{(d -1)^2}\sin\Big(\frac{c}{2}\Big)\sin\Big(\frac{3}{2}c\Big) + \frac{4d^2}{(d -1)^2}\sin^2\Big(\frac{c}{2}\Big)\cos(c) \Bigg] \cdot \bigg[ 1 + \frac{4d}{(d -1)^2} \sin^2\Big( \frac{c}{2}\Big)\bigg]^{-1} \\&= \frac{d\cos(c)-d^2}{d^2-2d\cos(c)+1}\end{align*}

Answer 3

Another way is to write (Re means real part)

$\cos(aj) = Re(e^{iaj})$ so the sum becomes

$\begin{array}\\ s(d, a) &=\sum d^j\cos(ja)\\ &=Re(\sum d^je^{iaj})\\ &=Re(\sum (de^{ia})^j)\\ &=Re(\dfrac1{1-de^{ia}})\\ &=Re(\dfrac1{1-d(\cos(a)+i\sin(a))})\\ &=Re(\dfrac1{1-d\cos(a)-id\sin(a)})\\ &=Re(\dfrac1{1-d\cos(a)-id\sin(a)}\dfrac{1-d\cos(a)+id\sin(a)}{1-d\cos(a)+id\sin(a)})\\ &=Re(\dfrac{1-d\cos(a)+id\sin(a)}{(1-d\cos(a))^2+d^2\sin^2(a)})\\ &=Re(\dfrac{1-d\cos(a)+id\sin(a)}{1-2d\cos(a)+d^2\cos^2(a)+d^2\sin^2(a)})\\ &=Re(\dfrac{1-d\cos(a)+id\sin(a)}{1+d^2-2d\cos(a)})\\ &=\dfrac{1-d\cos(a)}{1+d^2-2d\cos(a)}\\ \end{array} $

Answer 4

$$ f_d(\omega) := \frac{1-d}{1+d}\sum_{j=-\infty}^{\infty} d^j \cos(2\pi j \omega) \ne \frac{1-d}{1+d}\Big[1 + 2 \sum_{j=1}^{\infty} d^j \cos(2\pi j \omega)\big], $$

$$ f_d(\omega) := \frac{1-d}{1+d}\sum_{j=-\infty}^{\infty} d^j \cos(2\pi j \omega) $$

$$= \frac{1-d}{1+d}\Big[1 + \sum_{j=1}^{\infty} \left(d^j + d^{-j} \right)\cos(2\pi j \omega)\big], $$

Since $0<d<1$, we can write $x=\ln d$, so that $d=e^x$:

$$ f_d(\omega) = \frac{1-e^x}{1+e^x}\Big[1 + 2\sum_{j=1}^{\infty} \cosh jx \cos(2\pi j \omega)\Big] =\frac{e^{-x/2}-e^{x/2}}{e^{-x/2}+e^{x/2}}\Big[1 + 2\sum_{j=1}^{\infty} \cosh jx \cos(2\pi j \omega)\Big] $$

$$ f_d(\omega) = -\tan \left(\frac{x}{2}\right) \Big[1 + 2\sum_{j=1}^{\infty} \cosh (jx )\cos(2\pi j \omega)\Big] $$

$$= -\tan \left(\ln d^{1/2}\right) \Big[1 + 2\sum_{j=1}^{\infty} \cosh (j\ln d )\cos(2\pi j \omega)\Big]$$

Not a proof, but experimental evidence: Mathematica says that for $\omega \in \mathbb{N}$, the series for $f_d(\omega)$ does not converge, otherwise $f_d(\omega)=0.$