Can you simplify this geometric series for computing the z-transform?

[yoran]

Hi,

I have a problem with computing this geometric series.

Homework Statement

I have to compute
\sum_{i=0}^\infty{(\frac{1}{2z})^{2k}} + \sum_{i=0}^\infty{(\frac{1}{3z})^{2k+1}}.
It's for computing the z-transform of
f[k]=0 for k<0
f[k]=(\frac{1}{2})^k for k=0,2,4,6,...
f[k]=(\frac{1}{3})^k for k=1,3,5,...

Homework Equations

The Attempt at a Solution

It's the 2k and 2k+1 that annoys me in the sum.
I tried
\sum_{i=0}^\infty{(\frac{1}{2z})^{2k}}=\sum_{i=0}^\infty{(\frac{1}{4z^2})^{k}}
but I don't know if that helps?

Thanks,

Yoran

 

[HallsofIvy]

Hi,

I have a problem with computing this geometric series.

Homework Statement

I have to compute
\sum_{i=0}^\infty{(\frac{1}{2z})^{2k}} + \sum_{i=0}^\infty{(\frac{1}{3z})^{2k+1}}.
It's for computing the z-transform of
f[k]=0 for k<0
f[k]=(\frac{1}{2})^k for k=0,2,4,6,...
f[k]=(\frac{1}{3})^k for k=1,3,5,...

Homework Equations

The Attempt at a Solution

It's the 2k and 2k+1 that annoys me in the sum.
I tried
\sum_{i=0}^\infty{(\frac{1}{2z})^{2k}}=\sum_{i=0}^\infty{(\frac{1}{4z^2})^{k}}
but I don't know if that helps?

Yes, that helps a great deal! The sum of a geometric series is given by
\sum_{i=0}^\infty ar^i= \frac{a}{1- r}
In this case a= 1 and r= 4z².

Similarly, for the second sum
\sum_{i=0}^\infty{(\frac{1}{3z})^{2k+1}}
You can factor out 1 1/3z and then take the "2" from "2k" 'inside' to get
\sum_{i=0}^\infty{\frac{1}{3z}(\frac{1}{9z^2})^k}
Now you have a= 1/3z and r= 1/9z^2.

Thanks,

Yoran