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Could you please explain how to expand $(1 - \frac1x)^{-n}$ into a sum of powers of $x$?

Thank you in advance.

RHS
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    Use the binomial theorem. – Chris Eagle Jul 25 '13 at 14:28
  • If you mean positive powers, then its impossible – Yurii Savchuk Jul 25 '13 at 14:29
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    You need to be clear about what you want. If you want a formal (infinite) expression, you need to choose between $\mathbf Z[[x]]$ (unbounded positive powers, see the answer by Mark Bennet, and the comment by @Yuri Savchuk is wrong) or $\mathbf Z[[x^{-1}]]$ (unbounded negative powers, answer by Ron Gordon); you cannot have it both ways as there is no such ring as $\mathbf Z[[x,x^{-1}]]$. If $x$ is a concrete value, you need to choose between an expression for $|x|<1$ and one for $|x|>1$ (basically the same choice as before), again you cannot have it both ways. – Marc van Leeuwen Jul 26 '13 at 12:54

3 Answers3

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It depends a bit what you want to achieve. A little rewriting gets a form which may be useful.

Note that $$(1-\frac1x)^{-n}=\left(\frac {x}{(x-1)}\right)^n=(-x)^n(1-x)^{-n}$$ You can expand this using the binomial theorem, with the usual restrictions on the value of $x$ for convergence. This does give you an expansion in positive powers of $x$.

Mark Bennet
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Hint:

$$(1-t)^{\alpha} = \sum_{k=0}^{\infty} { \alpha \choose k }(-1)^k t^k. $$

Mhenni Benghorbal
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Consider the Taylor expansion of $(1-1/x)^{-n}$ about $1/x=0$:

$$\left ( 1-\frac{1}{x}\right)^{-n} = 1+ (-n) \left ( -\frac{1}{x}\right)+\frac{1}{2!}(-n)(-n-1)\left ( -\frac{1}{x}\right)^2 + \frac{1}{3!} (-n)(-n-1)(-n-2)\left ( -\frac{1}{x}\right)^3+\cdots$$

Note also that you may define the negative binomial coefficients $\binom{-n}{k}$ to match the above coefficients to resemble a binomial expansion.

Ron Gordon
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  • How about the - sign is taking out from each coeff e.g. the second coeff is -nC1 etc. Is it possible? – RHS Jul 25 '13 at 14:41
  • @RHS: I am afraid I do not follow you. A quantity such as $\binom{-n}{1}$ is a mnemonic defined to match the above expression. In the meantime, the most general way to think about this is to consider what the Taylor expansion of the binomial expression is about $x=0$. That is where my expression comes from. – Ron Gordon Jul 25 '13 at 14:47
  • @RonGordon Just figured it out it is 1 + nC1(1/x) + (n+1)C2(1/x^2) + (n+2)C3(1/x^3) + ... – RHS Jul 25 '13 at 14:54
  • @RHS: yes, I believe it works out that way. – Ron Gordon Jul 25 '13 at 14:55
  • @RonGordon I did not mean it to do that way . If you doubt I can remove my answer . – Harish Kayarohanam Jul 25 '13 at 14:57
  • @RonGordon Thanks. – RHS Jul 25 '13 at 14:57
  • btw the confusing form appeared at nearly the end of page 4 of this paper http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.aoms/1177731715 – RHS Jul 25 '13 at 15:02
  • I just jumped in to help him with that pdf which answers exactly what he asks. Ok no problem . I removed my comment even . Your comment "hijack" hurts . No probs. Cheers . – Harish Kayarohanam Jul 25 '13 at 15:03
  • @HarishKayarohanam Thanks anyway. But all I want to look for is the form I provide above(in my comment) which appeared in that paper. – RHS Jul 25 '13 at 15:06
  • @HarishKayarohanam: you can communicate with the OP within the comments after the question or your own answer. You can also use the comments here to add your input to the discussion about this answer, no worries. But your comment attempted to direct discussion away from here and toward your own answer. No doubt there is an element of competition here, but such blatant behavior (which I assume you did not really intend, it just came out that way) does not help the OP and infuriates the person whose answer is being hijacked. And, yes, the word "hijack" applies, sorry. – Ron Gordon Jul 25 '13 at 15:08
  • and RonGordon's explicit form helps me get to that conclusion so I accepted his answer. – RHS Jul 25 '13 at 15:12