Conference rusure::math

59.0. "Trigonometric Sum" by HARE::STAN () Tue Apr 24 1984 01:35

I've recently been playing around with the sum

		  n
		-----
		\
	S(n) =   >	F   sin( kx )
		/        k
		-----
		 k=1

where F  is the kth Fibonacci number.
       k

I've finally evaluated this in closed form (with a lot of help from
the computer).  The answer is:


S(n) =

-2 sin x - F    sin (n-1)x - F    sin nx + F    sin (n+1)x + F  sin (n+2)x
	    n+1               n+2           n-1               n
--------------------------------------------------------------------------  .
			2 cos 2x  -  3

This is probably not in simplest form, but it was the best I could do.

You probably couldn't care less about this result; however, if there is
anyone out there who is interested in my method of attack, you can give
me a call.  You might even play around with the simpler sum

		  n
		-----
		\
	T(n) =   >	sin( kx )
		/
		-----
		 k=1

which has appeared before in the literature, but still isn't very easy to
calculate. (Hint: I think it can be made to telescope.)
If anyone knows of any other results similar to these, please let me know.