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Convergence test

  • Thread starter arpon
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Homework Statement


##P(z) = 1 - \frac{z}{2} + \frac{z^2}{4} - \frac{z^3}{8} + ... ##
Determine if the series is convergent or divergent if ## |z| = 2 ##, where, ## z## is a complex number.

Homework Equations


##1+r+r^2+r^3+...+r^{N-1}=\frac{1-r^N}{1-r}##

The Attempt at a Solution


Let, ##z = 2 exp (i \theta)##[/B]
For the first ##N## terms, the summation is,
##P_N (z) = \frac{1-(-1)^N exp(iN\theta)}{1+exp(i \theta)}=\frac{1-(-1)^N \cos (N\theta) - i (-1)^N \sin (N \theta)}{1 + exp (i \theta)}##
As ## N \rightarrow \infty##, ##\cos (N \theta)## and ## \sin (N \theta)## do not converge to a particular value.
So, I conclude that the series is not convergent for ## |z|=2##
But the answer says that the series converges to ##\frac{1}{1+\frac{z}{2}}##
 
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Answers and Replies

  • #2
Samy_A
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Homework Statement


##P(z) = 1 - \frac{z}{2} + \frac{z^2}{4} - \frac{z^3}{8} + ... ##
Determine if the series is convergent or divergent if ## |z| = 2 ##, where, ## z## is a complex number.

Homework Equations


##1+r+r^2+r^3+...+r^{N-1}=\frac{1-r^N}{1-r}##

The Attempt at a Solution


Let, ##z = 2 exp (i \theta)##[/B]
For the first ##N## terms, the summation is,
##P_N (z) = \frac{1-(-1)^N exp(iN\theta)}{1+exp(i \theta)}=\frac{1-(-1)^N \cos (N\theta) - i (-1)^N \sin (N \theta)}{1 + exp (i \theta)}##
As ## N \rightarrow \infty##, ##\cos (N \theta)## and ## \sin (N \theta)## do not converge to a particular value.
So, I conclude that the series is not convergent for ## |z|=2##
But the answer says that the series converges to ##\frac{1}{1+\frac{z}{2}}##
The general term of the series is ##a_n=\frac{(-1)^nz^n}{2^n}##.
Then, for ##|z|=2##, ##|a_n|=|\frac{z^n}{2^n}|=\frac{|z|^n}{2^n}=1##. How can this series converge if the sequence ##a_n## doesn't converge to 0?

I think you are correct.
 
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  • #3
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The general term of the series is ##a_n=\frac{(-1)^nz^n}{2^n}##.
Then, for ##|z|=2##, ##|a_n|=|\frac{z^n}{2^n}|=\frac{|z|^n}{2^n}=1##. How can this series converge if the sequence ##a_n## doesn't converge to 0?

I think you are correct.
Thanks for your comments. This was an example from 'Mathematical Methods for Physics and Engineering' by Riley, Hobson and Bence (page 133). This is a very reliable book. I think I lack some important concept here.
 
  • #4
Samy_A
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Thanks for your comments. This was an example from 'Mathematical Methods for Physics and Engineering' by Riley, Hobson and Bence (page 133). This is a very reliable book. I think I lack some important concept here.
I see. They write that it converges for all ##z## with ##|z|=2## except ##z=-2##.
Let's take ##z=2## then. The series becomes ##1-1+1-1 ...##, so the partial sums alternate between ##0## and ##1##.
A few lines before this they give an example with the same oscillating series and conclude (correctly in my book) that it doesn't converge:
limit.jpg


Call me baffled.
 
  • #5
PeroK
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##\frac{1}{1+ z/2} = (1 + \frac{z}{2})^{-1} = 1 - \frac{z}{2} + \frac{z^2}{4} - \frac{z^3}{8} ... \ (|z| < 2)##
 
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  • #6
Samy_A
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##\frac{1}{1+ z/2} = (1 + \frac{z}{2})^{-1} = 1 - \frac{z}{2} + \frac{z^2}{4} - \frac{z^3}{8} ... \ (|z| < 2)##
But the book claims that this series also converges for all ##z## with ##|z|=2## except ##z=-2##. This is clearly a highly regarded book, but still, I can't see how the series converges when ##|z|=2##.
 
  • #7
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But the book claims that this series also converges for all ##z## with ##|z|=2## except ##z=-2##. This is clearly a highly regarded book, but still, I can't see how the series converges when ##|z|=2##.
It doesn't. I thought you'd proved that.
 
  • #8
Samy_A
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It doesn't. I thought you'd proved that.
I thought so too, but given the reputation of the book I started to fear I was missing something. Thanks for the second opinion.

Thanks for your comments. This was an example from 'Mathematical Methods for Physics and Engineering' by Riley, Hobson and Bence (page 133). This is a very reliable book. I think I lack some important concept here.
It's not you lacking some important concept here. It surely is a reliable book, but on this point you were right and they made a mistake. Happens to the best.
 
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