How do I find the fourier series approximation to the even extension of g(t), where g(t) is given as shown below using the fact that :
$\displaystyle sin(2pir/3)=(-1)^r+1sin(pir/3)$
I was thinking that perhaps since the period is 6 that we need to change the intervals so they are equal to something like 0<t<1/2
Alternatively, it may have something to do with the range -3<t<3
In which case the even extensions will be those that have positive y and the odd will have negative y.
Its really difficult - hope someone on this forum can help out with this one.
I think you are right about the range being -3<t<3, as in part a of the question the function f(t)=f(t + 2pi)
2pi was the period of the function, so if you look at the formulas for even and odd extension they use the letter $\displaystyle L $ and $\displaystyle 2L$ is defined as the period.
If $\displaystyle L$ is equal to 3 as the fundamental interval is $\displaystyle [-L,L]$ then 2x3 equals 6 giving us the desired period...... i think.
That makes sense - though when i substitute this into formula
$\displaystyle F(t)=A_0 + \sum_{n=0}^{\infty}A_rcos(2r\pi/T)$
and then try and work our $\displaystyle A_0 and A_n$
I still can see how we should use
$\displaystyle sin(2r\pi/3)=-1^r+1sin(r\pi/3)$
Yeah
I think for part i) it is only asking you to write the extensions in function form.
So I have something like
G(odd) (t) =
-1 (-3<= t <-2)
0 (-2<= t < -1)
1 (-1<= t < 0)
then
G(even) (t) =
1 (-3<= t <-2)
0 (-2<= t < -1)
-1 (-1<= t < 0)
Then draw each graph...
Does that make some sense? (sorry i'm still not up to speed on Latex)