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