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Computing a geometric series

  • Thread starter yoran
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Hi,

I have a problem with computing this geometric series.

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

2. Homework Equations

3. The Attempt at a Solution
It's the [tex]2k[/tex] and [tex]2k+1[/tex] that annoys me in the sum.
I tried
[tex]\sum_{i=0}^\infty{(\frac{1}{2z})^{2k}}=\sum_{i=0}^\infty{(\frac{1}{4z^2})^{k}}[/tex]
but I don't know if that helps?

Thanks,

Yoran
1. Homework Statement



2. Homework Equations



3. The Attempt at a Solution
 

Answers and Replies

HallsofIvy
Science Advisor
Homework Helper
41,732
893
Hi,

I have a problem with computing this geometric series.

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

2. Homework Equations

3. The Attempt at a Solution
It's the [tex]2k[/tex] and [tex]2k+1[/tex] that annoys me in the sum.
I tried
[tex]\sum_{i=0}^\infty{(\frac{1}{2z})^{2k}}=\sum_{i=0}^\infty{(\frac{1}{4z^2})^{k}}[/tex]
but I don't know if that helps?
Yes, that helps a great deal! The sum of a geometric series is given by
[tex]\sum_{i=0}^\infty ar^i= \frac{a}{1- r}[/tex]
In this case a= 1 and r= 4z2.

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

Yoran
1. Homework Statement



2. Homework Equations



3. The Attempt at a Solution
 

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