So after the half range fourier series of sinx i have got the left hand side.
But my teachers answer has got this on the right how do i get it to equal this?
Thanks
hmmm...
$\displaystyle \frac{1}{ \pi } [ \frac{1}{n+1} (-1)^n - \frac{1}{n-1} (-1)^n + \frac{1}{n+1} - \frac{1}{n-1} ] $
Lets work with the inside for now,
$\displaystyle \frac{1}{n+1} (-1)^n - \frac{1}{n-1} (-1)^n + \frac{1}{n+1} - \frac{1}{n-1} $
Group terms
$\displaystyle \frac{1}{n+1} ( 1 +(-1)^n) - \frac{1}{n-1} ( 1 + (-1)^n) $
Note that if $\displaystyle n = odd $ then the above must equal 0. So n must be even, which means
$\displaystyle \frac{1}{n+1} ( 1 +(-1)^n) - \frac{1}{n-1} ( 1 + (-1)^n) \to \frac{2}{n+1} - \frac{2}{n-1} $
This gives us
$\displaystyle \frac{ 2(n-1) -2(n+1) }{(n+1)(n-1)} $
$\displaystyle - \frac{ 4} {(n+1)(n-1)} $
So my answer will be,
$\displaystyle - \frac{4}{ \pi} \frac{ 1} {(n+1)(n-1)} $
Is this the same as theres? Lets change it around
If you sub in $\displaystyle n = 2 $ into their equation and mine you arrive at $\displaystyle - \frac{4}{3 \pi } $
So I think they are equal. let's equate them
$\displaystyle - \frac{4}{ \pi} \frac{ 1} {(n+1)(n-1)} = -\frac{2}{ \pi } \frac{ (-1)^n +1 } {n^2 - 1} $
$\displaystyle 2 \frac{ 1} {n^2 - 1} = \frac{ (-1)^n + 1 }{n^2 - 1} $
Note how these are the same if n=even!
However, if n=odd the right side is equal to 0, which is the same result as I achieved, however i had the stipulation before making my equation that n cannot be odd.
Thus, left side = right side!
$\displaystyle \frac{1}{n+1}-\frac{1}{n-1} = \frac{n-1 - (n+1)}{n^2-1} = \frac{-2}{n^2-1}$
So when n is even you have two of these inside the brackets, making overall $\displaystyle \frac{-4}{\pi(n^2-1)}$ and when n is odd you have a positive and negative of these canceling out to 0.