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Math Help - Series (2 related exercises)

  1. #1
    MHF Contributor arbolis's Avatar
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    Series (2 related exercises)

    Given the series \sum _{n=-\infty}^{+ \infty} \frac{z^n}{10^{|n|}}, calculate its region of convergence and the function it represents.
    Attempt: \sum _{n=-\infty}^{\infty} \frac{z^n}{10^{|n|}}=2 \sum _{n=0}^{+ \infty} \frac{z^n}{10^n}-1=2\left (\frac{1}{1-\frac{z}{10}} \right ) -1 for |z|<10. Which is equal to \frac{10+z}{10-z}=f(z). The radius of convergence is 10, so the region is a disk (without the border) centered at 0 with radius 10. Is that good?
    2)Is \sum _{n=-\infty}^{\infty} z^n a Laurent series of a function? Why?
    Attempt: No, it isn't. Because \sum _{n=-\infty}^{\infty} z^n=\sum _{n=0}^{+\infty} z^n+ \left ( \sum _{n=-\infty}^{0} z^n \right ) -1 which would be equal to \frac{1}{1-z}+\frac{z}{z-1}-1. The problem is that \sum _{n=0}^{+\infty} z^n=\frac{1}{1-z} for |z|<1 and \sum _{n=-\infty}^{0} z^n = \frac{z}{z-1} for |z|>1. Since the modulus cannot be greater and lesser than 1 at a time, the series doesn't converge to any function. (Yeah I know, I didn't prove it well. How would you solve the exercise?)
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  2. #2
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    You have a mistake in your first series since when you reduced it from a doubly infinite to an infinite series, you forgot that the sign flips on the z^n.

    The calculation is actually:

    [LaTeX ERROR: Convert failed]

    [LaTeX ERROR: Convert failed]

    Your analysis on the radius of convergence is correct for the calculation you made. Now for this one that I've shown, what is the radius of convergence? I like your analysis in the attempt for the second part and as far as I can tell it has the right idea. Here is what I think:

    [LaTeX ERROR: Convert failed]

    Now, the left sum only converges for |z|<1 and the right sum only converges if |z|>1, so the overal expression diverges for all z in the complex plane. The series doesn't converge at all, let alone to some function.

    P.S. I notice you are a bit iffy with the indices of summation, so be careful.
    Last edited by Vlasev; August 1st 2010 at 10:13 PM.
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  3. #3
    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by Vlasev View Post
    You have a mistake in your first series since when you reduced it from a doubly infinite to an infinite series, you forgot that the sign flips on the z^n.

    The calculation is actually:

    [LaTeX ERROR: Convert failed]

    [LaTeX ERROR: Convert failed]
    Ok thanks, I see.

    Your analysis on the radius of convergence is correct for the calculation you made. Now for this one that I've shown, what is the radius of convergence?
    Hmm which series? For \sum _{n=0}^{\infty} \frac{1}{(10z)^{n}}, I get that it's worth 1/10. While for \sum _{n=0}^{\infty} \frac{z^n}{10^{n}}, I get 10... I'm obviously wrong. Hmm it's impossible, what do you get?
    I like your analysis in the attempt for the second part and as far as I can tell it has the right idea. Here is what I think:

    [LaTeX ERROR: Convert failed]

    Now, the left sum only converges for |z|<1 and the right sum only converges if |z|>1, so the overal expression diverges for all z in the complex plane. The series doesn't converge at all, let alone to some function.

    P.S. I notice you are a bit iffy with the indices of summation, so be careful.
    Ok.
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    Quote Originally Posted by arbolis View Post
    Ok thanks, I see.

    Hmm which series? For \sum _{n=0}^{\infty} \frac{1}{(10z)^{n}}, I get that it's worth 1/10. While for \sum _{n=0}^{\infty} \frac{z^n}{10^{n}}, I get 10... I'm obviously wrong. Hmm it's impossible, what do you get?

    Ok.
    You are partially right. To get the answer though, consider the region of convergence for each series. The second series converges for |z|<10 as you said, because if  0<|z|<10, then 0<|z/10|<1 and then the geometric sum converges. Now for the other one you have to be careful. Here's how I would do it:

    You want  0<|1/(10z)|<1. Then take the exponent -1 of it to get:

    \infty >|10 z| > 1

    Then divide out by 10 to finally get

    \infty > |z |> 1/10

    So in this case you want |z| > 1/10.

    Finally, you just have the region of convergence being 1/10 < |z| < 10, since this is where both series converge.
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  5. #5
    MHF Contributor arbolis's Avatar
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    Wow, thanks a lot, that was very helpful and useful to know.
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