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Let's have function $$ \psi (x) = -\frac{1}{ax} - \frac{b}{a^2}\ln(x) + \text{const} + O(x). $$ I have read that the inverse function is written in a form $$ \psi^{-1}(t) = -\frac{1}{at} - \frac{b}{a^3}\frac{\ln(t)}{t^2} + \frac{c}{t^2} + O\left(\frac{\ln^2(t)}{t^3} \right). $$ How to prove that?

Michael Hardy
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John Taylor
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1 Answers1

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Have you tried the Lagrange Inversion formula:http://en.wikipedia.org/wiki/Lagrange_inversion_theorem.

You can also solve by hand by writing $\psi^{-1}(t)=a_0+a_1t+a_2t^2/2!+\cdots$, where $a_i$ are derivatives of $\psi^{-1}(t)$ evaluated at $t=0$. But if $y=\psi^{-1}(t)$, then $t=\psi(y)$ or 1=$\psi'(y)y'(t)$. So $y'(t)=1/\psi'(t)$. and you can read off $\psi'(0)$ from your series expansion of $\psi(x)$. Now rinse and repeat for higher derivatives. By the way, in your case you're expanding around $1/t$ instead.

Alex R.
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