I am attempting to simplify the following summation and would appreciate being pointed in the correct direction on how to do this, or if even possible. When I say simplify, I specifically mean removing the summation, in a manner similar to Lagrange Trig Identities.
$$f(x)=\sum_{k=1}^{n/2-1} \frac{\sin((2x+1)\frac{\pi} {n} k))} {\sin(\frac{\pi} {n} k)}$$ Where:
- $n$ is an integer power of $2$ such that $n \geq 4$ (so the summation's upper limit makes sense): ${4,8,16,32,...}$
- $x$ is a non-negative integer: ${0,1,2, 3,...}$
What I have tried:
$f(x)$ can be rewritten in terms of Chebyshev Polynomials of the 2nd kind.
$$f(x) = \sum_{k=1}^{n/2-1} U_{2x}(\cos(\frac{\pi}{n}k)) = \sum_{k=1}^{n/2-1} U_{2x}(\sin( \frac{\pi}{n}k) )$$
P.S. The $\cos(\frac{\pi}{n}k)$ can be changed to $\sin(\frac{\pi}{n}k)$ above if you consider expanding the summation in the opposite order (from $n/2-1$ to $1$). Happy to explain more on this if unclear.
I then substituted the summation form of Chebyshev Polynomials of the second kind (Eq 16) in the hope that reordering the summations might yield something useful but that seemed to not simplify further.
To summarise, the question is: How can I simplify the following so that the summation sign can be removed (in a Lagrange Trig Identity fashion)?
$$f(x)=\sum_{k=1}^{n/2-1} \frac{\sin((2x+1)\frac{\pi}{n}k))} {\sin(\frac{\pi} {n} k)}= \sum_{k=1}^{n/2-1} U_{2x}(\cos( \frac{\pi}{n}k) ) = \sum_{k=1}^{n/2-1} U_{2x}(\sin( \frac{\pi}{n}k) )$$
Other notes:
- Perhaps the fact that $(2x+1)$ is always odd and $n$ is always a power of 2 is helpful.