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I need to find the convergence properties of the Taylor Expansion of

$$f(z)=\frac{z}{z-1}$$


I found the Taylor Series:

$$\sum_{j=1}^\infty \frac{(-1)^{j+1}(z-i)^{j-1}}{(i-1)^j}$$

Then I used the "ratio test" to find out when it converges:

$$\lim_{j\to\infty}\left| \frac{\frac{(-1)^{j+2}(z-i)^{j}}{(i-1)^{j+1}}}{\frac{(-1)^{j+1}(z-i)^{j-1}}{(i-1)^j}} \right| = \left| \frac{(-1)(z-i)}{(i-1)} \right| < 1$$

So the series converges when $|z|<1$.

However wolfalpha disagrees.

And I cannot spot my error?!

EDIT:

Forgot to mention: Expanding at $z=i$ and fixed typo.

EDIT: Wolframalpha says the expansion converges when $\sqrt{2}|z-i|<2$.

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  • $\begingroup$ Should converge for $|z|<1$ not $z>1$. Is that the confusion, or just a typo? $\endgroup$ Jan 5 '16 at 16:19
  • $\begingroup$ Wait, are you expanding at $z=0$ or around $z=i$? $\endgroup$ Jan 5 '16 at 16:20
  • $\begingroup$ Cuz that's not the expansion for $z=0$. $\endgroup$ Jan 5 '16 at 16:21
  • $\begingroup$ @GregoryGrant yes at z=i Sorry forgot to mention. $\endgroup$
    – user302915
    Jan 5 '16 at 16:22
  • $\begingroup$ Ah I see. It must converge then for a disk around $z=i$, so not $|z|<1$ which is a disk around $0$. Should be more like $|z-i|<R$ for some $R$. Probably $R=i$ but I didn't work it out. $\endgroup$ Jan 5 '16 at 16:26
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You have done this right. You need to find the $z$ for which $\left|\frac{z-i}{i-1}\right|<1$. Now $|i-1|=\sqrt{2}$ so this is the same as

$|z-i|<\sqrt{2}$.

Since $\sqrt{2}=2/\sqrt{2}$ this is the same as woflram's answer.

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  • $\begingroup$ OK that clears things up! Thanks $\endgroup$
    – user302915
    Jan 5 '16 at 16:50

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