The function is $\dfrac {z-z^3}{\sin {\pi z}} $. How to find the radius of convergence in $ z=0 $?
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Yiorgos S. Smyrlis
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Elensil
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What did you try? – Harald Hanche-Olsen Dec 24 '13 at 10:07
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I tried expanding this into Taylor series. – Elensil Dec 24 '13 at 10:09
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Ah, but that is a pretty hopeless task. You need to search for the largest disk where the function is analytic, with removable singularities allowed. This function has three ef them. – Harald Hanche-Olsen Dec 24 '13 at 10:14
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@MhenniBenghorbal Looks like three to me. But let's give the OP a chance to work it out for himself. – Harald Hanche-Olsen Dec 24 '13 at 10:32
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Well, the first singularity is at z=0, but it is removable, since the limit is 0. The singularities at z=+-1 are removable too. So the answer is 2? – Elensil Dec 24 '13 at 10:44
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@HaraldHanche-Olsen: I should be $2$. – Mhenni Benghorbal Dec 24 '13 at 10:44
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@Elensil: The limit at $z=0$ is $1/\pi$, surely? But yes, since there is a limit, the singularity is removable. – Harald Hanche-Olsen Dec 24 '13 at 10:51
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@MhenniBenghorbal $z=0$, $z=\pm1$. I count three still. What am I missing? – Harald Hanche-Olsen Dec 24 '13 at 10:51
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Yes, you are right. The limit is $\frac {1}{\pi} $ of course. My stupidity – Elensil Dec 24 '13 at 10:56
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@HaraldHanche-Olsen: I am talking about the radius of convergence. – Mhenni Benghorbal Dec 24 '13 at 11:18
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@MhenniBenghorbal Oooooh, I see. Sorry, I misunderstood. Your comment came right after mine, so I thought it was a response to that one. – Harald Hanche-Olsen Dec 24 '13 at 12:44
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1@Harald: I expected that. – Mhenni Benghorbal Dec 24 '13 at 13:11
2 Answers
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It suffices to find which is the larger disk centered at the origin, i.e., $D(0,r)$, with $r$ maximum, in which this function is analytic. And as both numerator and denominator are entire functions, the fraction is analytic in those points where the denominator does not vanish. Unless numerator and denominator vanish simultaneously at some point, and the order of the zero of the denominator does not exceed the one of the numerator.
Yiorgos S. Smyrlis
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2
The radius of convergence is $2$ which is the nearest singularity to $0$. The point $z=2$ is a pole of order $1$.
Mhenni Benghorbal
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