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In my attempt to complete this answer, I hit a snag in showing that

$$\lim_{t\to 0} \dfrac{\mathrm d^{k-1}}{\mathrm dt^{k-1}}\left(\frac{t\sqrt{1+t}}{\log\sqrt{1+t}}\right)^k=2(k+2)^{k-1}$$

This shows up when trying to apply Lagrangian inversion to the function $\dfrac{\log\sqrt{1+x}}{\sqrt{1+x}}$. My sticking point here is that I am unable to find a convenient expression for the derivatives. Is there an easy proof for this?

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@Asaf, sorry; I needed something to reflect my general mood, and this suited me nicely. –  J. M. May 31 '13 at 1:34
    
I can relate... –  Asaf Karagila May 31 '13 at 1:40
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1 Answer

up vote 7 down vote accepted

Complex analysis comes to the rescue. Using Cauchy differentiation formula: $$ f^{n-1}(0) = \frac{(n-1)!}{2 \pi i} \oint \frac{f(z)}{z^n} \mathrm{d} z $$ Now $$ \begin{eqnarray} \lim_{t\to 0} \dfrac{\mathrm d^{k-1}}{\mathrm dt^{k-1}}\left(\frac{t\sqrt{1+t}}{\log\sqrt{1+t}}\right)^k &=& \frac{(k-1)!}{2 \pi i} \oint\left(\frac{t\sqrt{1+t}}{\log\sqrt{1+t}}\right)^k \frac{\mathrm{d} t}{t^k} \\ &=& \frac{(k-1)!}{2 \pi i} \oint \left(\frac{\sqrt{1+t}}{\log\sqrt{1+t}}\right)^k \mathrm{d} t \end{eqnarray} $$ Now performing the change of variable $t = \mathrm{e}^u-1$: $$ \begin{eqnarray} \lim_{t\to 0} \dfrac{\mathrm d^{k-1}}{\mathrm dt^{k-1}}\left(\frac{t\sqrt{1+t}}{\log\sqrt{1+t}}\right)^k &=& \frac{(k-1)!}{2 \pi i} \oint \left( \frac{\exp(u/2)}{u/2} \right)^k \mathrm{e}^u \mathrm{d} u \\ &=& 2^k \left[ \frac{(k-1)!}{2 \pi i} \oint \frac{\exp( u(k/2+1)}{u^k} \mathrm{d} u \right] \\ &=& 2^k \lim_{u \to 0} \dfrac{\mathrm d^{k-1}}{\mathrm du^{k-1}} \mathrm{e}^{ u \left(\frac{k}{2}+1\right) } = 2^k \left(\frac{k}{2}+1\right)^{k-1} = 2 (k+2)^{k-1} \end{eqnarray} $$

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Thanks! Cauchy slipped my mind. This helps to justify the series expansion of $W(-x)^2$ in the answer I linked to. –  J. M. Sep 21 '11 at 15:02
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