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Write down the power series expansion of the function $f(z) =\frac{1}{z^2}$ in a neighbourhood of $z = −1$.

My attempt : I know that $1/z^2$ is a pole of order $2$ , first I take $d(1/z)/dz = -1/z^2$ then I try to make $1/(a-a+(z^2))$ and I tried to make Taylor expansion series but it seem useless. I don"t know from where I have to start

If anybody help me I would be very thankful… and please tell me the solution

José Carlos Santos
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jasmine
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3 Answers3

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What happens at $0$ is irrelevant. What matters here is what happens at $-1$. Note that\begin{align}f(z)=z^{-2}&\implies f(-1)=1\\f'(z)=-2z^{-3}&\implies f'(-1)=2\\f''(z)=6z^{-4}&\implies f''(-1)=6\\&\cdots\end{align}It is easy to prove by induction that$$(\forall n\in\mathbb{Z}^+):f^{(n)}(z)=(-1)^n(n+1)!z^{-n-2}.$$So, $f^{(n)}(-1)=(n+1)!$ and therefore $\frac{f^{(n)}(-1)}{n!}=(n+1)$. It follows that$$\frac1{z^2}=1+2(z+1)+3(z+1)^2+4(z+1)^3+\cdots$$

José Carlos Santos
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  • im not getting @ jose Carlos Santos ..why u take (z - 1),,? – jasmine Sep 12 '17 at 11:23
  • @lomberlego My mistake. I've edited my answer. I wrote $z-1$, but I should have written $z+1(=z-(-1))$. – José Carlos Santos Sep 12 '17 at 11:25
  • thanks a lots and May god bless u @ JOse carlos santos....im now voting up ur solution – jasmine Sep 12 '17 at 11:28
  • @lomberlego Thank you. – José Carlos Santos Sep 12 '17 at 11:29
  • Please correct me if I am wrong. Answer should be $$\frac1{z^2}=1+2(z+1)-3(z+1)^2+ 4(z+1)^3+\cdots$$ – Sitanshu Sep 12 '17 at 11:54
  • @Sitanshu You are right. I've edited my answer. – José Carlos Santos Sep 12 '17 at 11:56
  • The crucial step "So, $f^{(n)}(-1)=(-1)^n(n+1)!$" has no proof. Note that the approach suggested is rather cumbersome and is most probably not the one expected from the OP. (Why the avalanche of upvotes, one wonders.) – Did Sep 12 '17 at 11:59
  • You have made it all positive. Never the less, your approach to question was right. Cheers. – Sitanshu Sep 12 '17 at 12:00
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    @Did Since the OP wrote “thanks a lots and May god bless u”, I would like to know why do you assert that my answer “is most probably not the one expected from the OP”. Did you downvote my answer? – José Carlos Santos Sep 12 '17 at 12:10
  • I see you modified your answer, which seems to indicate you are well aware the previous version was not addressing the question. Your answer “(was) most probably not the one expected from the OP" because it was not addressing the question. Re votes on the site, your confidence in their meaning is touching, to say the least... – Did Sep 12 '17 at 15:50
  • @Did Anybody can see the editions that an answer had. Anyway, what I asserted was that the OP had thanked me, which seemed to indicate that my answer addresses her question. So, again, why do you assert that my answer “is most probably not the one expected from the OP”? Besides, I mentioned nothing concerning my confidence, or lack thereof, in the meaning of votes in this site. I just asked you if you downvoted my answer. It is a simple and straightforward question. And, again, you did not answer. – José Carlos Santos Sep 12 '17 at 17:05
  • FYI, the votes of other users are none of your business (I tried to be polite about this the first time, but since you insist...). // Re the pertinence of your answer, I find your "the OP had thanked me, which seemed to indicate that my answer addresses her question" surprising, to say the least, coming from a mathematician (are you going to force me to expand on this as well? I hope not). Here is a suggestion: why not, from now on, stop the diversions and stay on the mathematical correctness/relevance of the posts you write on the site? Deal? – Did Sep 12 '17 at 18:58
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Let $w=z+1$ then for $|w|<1$, $$\frac{1}{z^2}=\frac{1}{(1-w)^2}=\frac{d}{dw}\left(\frac{1}{1-w}\right) =\frac{d}{dw}\left(\sum_{n=0}^{\infty}w^n\right).$$ Can you take it from here?

Robert Z
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$$ \begin{align} \frac1{z^2} &=\frac1{(1-(z+1))^2}\tag{1}\\ &=\sum_{k=0}^\infty(-1)^k\binom{-2}{k}(z+1)^k\tag{2}\\ &=\sum_{k=0}^\infty\binom{k+1}{k}(z+1)^k\tag{3}\\ &=\sum_{k=0}^\infty(k+1)(z+1)^k\tag{4} \end{align} $$ Explanation:
$(1)$: center at $z=-1$
$(2)$: apply the Binomial Theorem with exponent $-2$
$(3)$: convert from a Negative Binomial Coefficient
$(4)$: evaluate the binomial coefficient

robjohn
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  • Perhaps someone thinks that the binomial theorem is too advanced for someone quoting complex analytic ideas (although incorrectly). I can only assume when there is a downvote without a comment. – robjohn Sep 12 '17 at 19:56