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So I'm trying to follow the derivation in a book on Analytic combinatorics. We have the function $$P(z,u)=(1-z)^{-u}$$ and I need to take the partial derivative with respect to $u$, then evaluate at $u=1$. So, my attempt: $$\ln (P(z,u))=-u\ln(1-z)$$ $$\frac {P_u(z,u)}{P(z,u)}=-\ln (1-z)$$ $$P_u(z,u)=-(1-z)^{-u}\ln (1-z)$$ then evaluating at $u=1$, ending up with $$-(1-z)^{-1}\ln (1-z)$$

The only problem is, the author has no negative sign in his expression....did I make a mistake in the chain rule or something?

Alan
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  • You're correct. Is it possible that the author wrote $(z-1)^{-1}\log(1-z)$ and you missed it? – Git Gud Mar 17 '15 at 22:21
  • Hmm, the book is at http://algo.inria.fr/flajolet/Publications/book.pdf and it's equation 11 on page 160, for the expected value of the stirling cycle distribution – Alan Mar 17 '15 at 22:24
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    I don't know about the maths involved, but $-(1-z)^{-1}\log(1-z)=\dfrac{1}{1-z}\log\left(\dfrac{1}{1-z}\right)$ for all $z<1$, and this is similar to what's in the book. Edit: If you see this comment in time, please specify that you meant page 160 of the book, which is page 176 of the PDF. – Git Gud Mar 17 '15 at 22:28
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    Note that $-\ln\left(1-z\right)=\ln\frac{1}{1-z}$, because the $-1$ goes to the exponent of the inside of the logarithm. – Brian Mar 17 '15 at 22:49
  • Doh, that's what I missed. Thanks everyone! – Alan Mar 17 '15 at 23:25

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No, you didn't make a mistake. Your answer is correct. That's clearly a typo in the book.

J.R.
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