That makes more sense, but I'm not yet seeing how that relates to my question. I wanted to make sure the following is true...
\sum_{n=1}^{\infty}{\frac{(-1)^n}{n^2}}=\sum_{n=1}^{\infty}\left({\frac{1}{2n^2}-\frac{1}{(2n-1)^2}}\right)
The summation on the right side is what allows me to...
You said add them but then you subtracted instead. Did you mean to write...
\sum_{n=1}^{\infty}{\frac{1}{n^2}}-\sum_{n=1}^{\infty}{\frac{1^n}{n^2}}=\sum_{n=1}^{\infty}{\frac{1-(-1)^n}{n^2}}
With the subtracted summation
(1)...
I know the first summation is equal to pi^2/6 so I just replaced the summation with it.
\frac{\pi^2}{6}-\sum^{\infty}_{n=1} \frac{(-1)^n}{(n)^2}
Are you saying these two summations are equivalent to...
\sum^{\infty}_{n=1} \frac{1}{(2n-1)^2}
[SIZE="1"]\sum^{\infty}_{n=1} \frac{1}{(n)^2}-\sum^{\infty}_{n=1} \frac{(-1)^n}{(n)^2}
Is this what you are saying? For some reason it's showing this on 3 lines. It should be the summation minus the second summation.
Find the Fourier Series for f(x)=x^2 evaluate f(0) and show that the summation
\sum^{\infty}_{n=1}\frac{1}{(2n-1)^2}=\frac{\pi^2}{8}
The first part of this problem asked that I find a_{n} and b_{n}
Since x^2 is an even function b_n=0 and for a_n I got \frac{4(-1)^2}{n^2}
for a_0 I got...
I realize my mistake. I completely missed that part about the blocks sticking together after the collision. That would make this a "completely Inelastic Collision" meaning that I should have used this formula below (not sure if it will show) as the initial velocity. In equation 7, I changed...
Homework Statement
Homework Equations
The Attempt at a Solution
Here is the work that I've done. This is an even problem in my book so the answer is not given. The professor did gave the answer to this problem as .33m. The most recent answer I got is .4m which means block b...