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I'm reading through a proof in Analytic Combinatorics by Flajolet/Sedgewick and I have come across this:

We have the asymptotic expansion:

$(1+\frac{t}{n})^{-n-1}=e^{-(n+1)\log(1+\frac{t}{n})}=e^{-t}[1+\frac{t^2-2t}{2n}+\frac{3t^4-20t^3+24t^2}{24n^2}+...]$

How do I go from second expression to the third?

glS
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MathStudent
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    $$\log (1+x) = x - \frac{1}{2}x^2 + \frac13 x^3 - \frac14 x^4 + \dotsb$$ and $$e^y = 1 + y + \frac12 y^2 + \frac16 y^3 + \frac{1}{24}y^4 + \dotsb$$ then compute as far as you need. – Daniel Fischer Sep 25 '13 at 13:37
  • Okay, I've been looking at this since you posted it... I still don't see how to expand these and get $e^{-t}$ – MathStudent Sep 25 '13 at 14:07

1 Answers1

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Using the Taylor expansion of the logarithm, we obtain

$$\begin{align} (n+1)\log \left(1 + \frac{t}{n}\right) &= (n+1)\left(\frac{t}{n} - \frac{t^2}{2n^2} + \frac{t^3}{3n^3} + O\left(\frac{1}{n^4}\right)\right)\\ &= t + \frac{t}{n} - \frac{t^2}{2n} - \frac{t^2}{2n^2} + \frac{t^3}{3n^2} + \frac{t^3}{3n^3} - \frac{t^4}{4n^3} + O\left(\frac{1}{n^4}\right)\\ &= t - \frac{t^2-2t}{2n} - \frac{3t^2-2t^3}{6n^2} + O\left(\frac{1}{n^3}\right), \end{align}$$

and hence, using $e^{x+y} = e^x\cdot e^y$ and the Taylor expansion of the exponential function:

$$\begin{align} e^{-(n+1)\log \left(1+\frac{t}{n}\right)} &= \exp \left(-t + \frac{t^2-2t}{2n} + \frac{3t^2-2t^3}{6n^2} + O(n^{-3})\right)\\ &= e^{-t}\exp\left(\frac{t^2-2t}{2n} + \frac{3t^2-2t^3}{6n^2} + O(n^{-3})\right)\\ &= e^{-t}\left(1 + \frac{t^2-2t}{2n} + \frac{3t^2-2t^3}{6n^2} + \frac12\left(\frac{t^2-2t}{2n}\right)^2 + O(n^{-3})\right)\\ &= e^{-t}\left(1 + \frac{t^2-2t}{2n} + \frac{12t^2 - 8t^3 + 3(t^4 - 4t^3 + 4t^2)}{24n^2} + O(n^{-3})\right)\\ &= e^{-t}\left(1 + \frac{t^2-2t}{2n} + \frac{3t^4 - 20t^3 + 24t^2)}{24n^2} + O(n^{-3})\right). \end{align}$$

Daniel Fischer
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