What is the Closed Form of a Summation of Sinusoidal Functions?

Frillth
Messages
77
Reaction score
0

Homework Statement



I am looking for a closed form of the summation:
sin(x) + sin(3x) + sin(5x) + ... + sin((2n-1*)x)

Homework Equations



None.

The Attempt at a Solution



Through a complete stroke of luck, I believe I have arrived at the correct solution: sin^2(nx)/sin(x)
I have tested this for many different cases, and I believe it is correct. However, I am having a hard time proving that it is. Can anybody point me in the right direction?
 
Physics news on Phys.org
How did you prove that?


[EDIT: The last term should be sin(2n+1)x not 2n-1
 
Seems to me that mathematical induction would be an obvious thing to try...
 
I'm a little rusty on my induction skills. Is this what I need to do?

1. Show that:
sin^2(nx)/sin(x) + sin((2(n+1)-1)x) = sin^2((n+1)x)/sin(x)

2. Show that my formula works for any specific case.
 
You need to show that

\sum_{n=0} ^N sin(2n-1)x= \frac{sin^2(Nx)}{sinx}then add the (N+1)th term to each side and show that is can be written as

\frac{sin^2((N+1)x)}{sinx}
 
Last edited:
I've been trying to get these two sides equal, but I'm not coming up with anything. Which identities should I be using to tackle this problem?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top