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How do we go from

\begin{equation*} \frac{1}{1+(\frac{x}{2})+(\frac{x^2}{3})+(\frac{x^3}{4})+\dots} \end{equation*}

to

\begin{equation*} 1-\left(\frac{x}{2}\right)-\left(\frac{x^2}{12}\right)-\dots? \end{equation*}

If I could consolidate the series in denominator in some form, then I could use binomial expansion in numerator by raising it to power $-1$, but I don't see the one in denominator in any known form other than converting it to $\log.$

gymbvghjkgkjkhgfkl
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  • In the denominator of what? In the numerator of what? – Mariano Suárez-Álvarez Sep 25 '10 at 21:50
  • I mean the first series is in denominator of 1. –  Sep 25 '10 at 21:54
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    If it's helpful for anyone else, I understand the question to be asking for an explanation of $$\frac{1}{\sum_{k=0}^{\infty}\frac{x^k}{k+1}}=1-\frac{x}{2}-\frac{x^2}{12}-\frac{x^3}{24}-\frac{19x^4}{720}-\frac{3x^5}{160}-\frac{863x^6}{60480}-\cdots$$ – Isaac Sep 25 '10 at 22:15
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    @Isaac: I interpreted it the same way too and added an answer with that interpretation. – Aryabhata Sep 25 '10 at 22:17
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    Well, I was not very subliminally asking for clarification in the question itself... – Mariano Suárez-Álvarez Sep 25 '10 at 22:44
  • @Mariano: I was expecting the OP will do it himself... Anyway, I went ahead and made the change. – Aryabhata Sep 25 '10 at 22:48
  • @Moron: Thanks for editing. Pardon me for being new! –  Sep 25 '10 at 22:56
  • @Harpreet: No worries! btw, Welcome to this site. I hope you will find it useful. – Aryabhata Sep 25 '10 at 23:03
  • I'm late to this particular party, but I'd have asked the same question(s) as Mariano myself. @Harpreet, FYI, what you wanted to do is sometimes termed reciprocation of a power series (i.e. trying to find the reciprocal of a given power series). – J. M. ain't a mathematician Sep 25 '10 at 23:21
  • @Moron, @Mariano: I was both trying to clarify as the question now reads and asserting some more of the terms of the result (as generated using Mathematica) because the first three terms didn't give me the slightest realistic clue what the pattern might be. – Isaac Sep 26 '10 at 00:12

2 Answers2

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Using $$-\log(1-x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \dots$$

We have that

$$ - \frac{x}{\log(1-x)} = \frac{1}{1 + \frac{x}{2} + \frac{x^2}{3} + \dots}$$

Now the series for $$\frac{x}{\log(1-x)}$$ is well known.

See the question asked on this very site here: Formula for the harmonic series $H_n = \sum_{k=1}^n 1/k$ due to Gregorio Fontana

And the page here: http://en.wikipedia.org/wiki/Euler-Mascheroni_constant. (search the page for Gregory).

The series expansion is given by

$$\frac{x}{\log(1-x)} = \sum_{k=0}^{\infty} C_{k} x^{k} = -1 + \frac{x}{2} + \frac{x^2}{12} + \frac{x^3}{24} + \dots$$

The $C_{k}$ are called as Gregory coefficients. The wiki page I linked above tells you how they can be calculated using a recursive formula.

So to answer your question, we get

$$\frac{1}{1 + \frac{x}{2} + \frac{x^2}{3} + \dots} = - \sum_{k=0}^{\infty} C_{k} x^k = 1 - \frac{x}{2} - \frac{x^2}{12} - \frac{x^3}{24} - \dots$$

Community
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Aryabhata
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1

Let $f(x) = \sum f_n x^n$ be a formal power series. To compute the inverse of $1 - x f(x)$, write

$$\frac{1}{1 - x f(x)} = \sum_{n \ge 0} x^n f^n(x).$$

To compute the coefficient of $x^k$ it suffices to compute the contributions from the first $k$ terms on the RHS, so this results in a finite algorithm which will let you compute any particular coefficient of the inverse. If $f$ has special properties (for example it is a polynomial or more generally meromorphic) then often one can even find a nice closed form.

Qiaochu Yuan
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