Summation question

[lilshai]

Hello...
I have this summation as follows (PI = 3.14...):

1/PI times the summation going from n=2 to infinity with the following inside the summation:

[[(-1)ⁿ⁺¹-1]*cos(nt)]/(n²-1)

Somehow, all of that is equivalent to:

-2/PI times the summation going from n=1 to infinity with the following inside the summation:

cos(2nt)/[4n²-1]

I know they changed the index to have the summation start from n=1 instead but I don't see how everything else changed...

 

[pka]

There is no mystery to this.
If n is even (−1)<SUP>n+1</SUP>−1=−2, if n is odd it is 0.
As you said, just change the index.

 

[lilshai]

Can you please elaborate? I get what you said about the even and odd part, but I still don't see how the following is equivalent:

1/PI times the summation going from n=2 to infinity with the following inside the summation:

[[(-1)ⁿ⁺¹-1]*cos(nt)]/(n²-1)

IS EQUIVALENT TO:

-2/PI times the summation going from n=1 to infinity with the following inside the summation:

cos(2nt)/[4n²-1]

 

[soroban]

Hello, lilshai!

If you write out the first few terms, it becomes clear . . .

I have this summation: . \(\displaystyle \L\frac{1}{\pi} \sum^{\infty}_{n=2}\,[(-1)^{n+1} - 1]\cdot\frac{\cos(nt)}{n^2 - 1}\)

Somehow, all of that is equivalent to: . \(\displaystyle \L-\frac{2}{\pi}\sum^{\infty}_{n=1}\frac{\cos(2nt)}{4n^2 - 1}\)

The first sum is:

. . . \(\displaystyle \L\frac{1}{\pi}\left\[(-2)\frac{\cos(2t)}{2^2-1}\,+\,(-2)\frac{\cos(4t)}{4^2-1}\,+\,(-2)\frac{\cos(6t)}{6^2-1}\,+\,(-2)\frac{\cos(8t)}{8^2-1}\,+\,\cdots \right]\)


Factor out -2 and rewrite the denominators:

. . . \(\displaystyle \L -\frac{2}{\pi}\left[\frac{\cos(2t)}{4\cdot1^2 - 1}\,+\,\frac{\cos(4t)}{4\cdot2^2 - 1}\,+\,\frac{\cos(6t)}{4\cdot3^2 - 1}\,+\,\frac{\cos(8t)}{4\cdot4^2 - 1}\,+\,\cdots\right]\)


And this becomes:

. . . \(\displaystyle \L-\frac{2}{\pi}\sum^{\infty}_{n=1} \frac{\cos(2nt)}{4n^2 - 1}\)

 

[lilshai]

Hi soroban!

Thanks for clarifying that and showing me clearly how they arrived at that answer. I always try to do the problem first and attempt it on my own before I post the question here. It doesn't help that our Professor doesn't have office hours and doesn't really stick around to answer questions (this is a graduate-level course btw) so this forum is a God-send.