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I am relatively new to infinite series and have a question about the Lagrange Inversion Theorem.

Can you use it to dervive Taylor series of inverse functions? For example, knowing the Taylor series for $e^x$ can I obtain the Taylor series for $\ln x$? For $\sin x$, can I obtain $\arcsin x$? For $\cos x$, can I obtain $\arccos x$?

I am trying to learn about the Lagrange Inversion Theorem on my own and there are not much sources online about it. Any help will be greatly appreciated!

Henrik supports the community
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  • Yes, you can derive Taylor expansions for inverse functions, but it isn't obtained from the Taylor expansion of the original function (I think). So knowing the Taylor series for $e^x$ doesn't particularly help much for finding the expansion of $\ln(x)$. – Simply Beautiful Art Jul 15 '17 at 22:28
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    This is what the Lagrange inversion theorem says : for $|z|$ small enough, if $f(z)$ is analytic and $f(0)=0, f'(0) \ne 0$ then $f^{-1}(z)$ is analytic, and it gives a recursion formula for the coefficients. – reuns Jul 15 '17 at 22:29
  • So then what I am saying is not possible? –  Jul 15 '17 at 22:30
  • When $f'(0) =0$ or when $f$ is not analytic. – reuns Jul 15 '17 at 22:32
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    Is there a method to determine the inverse of a Taylor series without restrictions besides calculating the derivatives at the centre of the series? –  Jul 15 '17 at 22:38
  • The restrictions that you have been told are equivalent to the existence of the inverse (locally). When they don't hold there is no inverse to find. – user463383 Jul 15 '17 at 22:49
  • @ualberta It is unclear what you mean. For $f(z) = e^z = f'(z)$ we can find $g(z) = f^{-1}(z)$ from $g'(z) = \frac{1}{f'(f^{-1}(z))} = \frac{1}{f(f^{-1}(z))} = \frac{1}{z}$ – reuns Jul 15 '17 at 22:49
  • https://en.wikipedia.org/wiki/Lagrange_inversion_theorem ? – Emanuel Landeholm Mar 01 '23 at 08:14

2 Answers2

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I have the feeling that, for truncated series, you are looking for series reversion which is well described in the given link.

Using a simple example $$y=\sin(x)=x-\frac{x^3}{6}+\frac{x^5}{120}-\frac{x^7}{5040}+\frac{x^9}{362880}-\frac{x^{11}}{3 9916800}+O\left(x^{13}\right)$$ Aplying the method you will get $$x=\sin^{-1}(y)=y+\frac{y^3}{6}+\frac{3 y^5}{40}+\frac{5 y^7}{112}+\frac{35 y^9}{1152}+\frac{63 y^{11}}{2816}+O\left(y^{13}\right)$$ You can check that the result is the Taylor series of the inverse function.

A more complex example (but the same method applies) $$y=1+e^x \log (x+1) \sin (x)=1+x^2+\frac{x^3}{2}+\frac{x^4}{6}-\frac{x^5}{12}+\frac{x^6}{36}-\frac{2 x^7}{45}+\frac{x^8}{28}+O\left(x^{9}\right)$$ would lead to $$x=\sqrt{y-1}-\frac{y-1}{4}+\frac{7(y-1)^{3/2}}{96} +\frac{(y-1)^2}{24} -\frac{233 (y-1)^{5/2}}{2048}+\frac{449 (y-1)^3}{2880}-\frac{3442697 (y-1)^{7/2}}{20643840}+O\left((y-1)^4\right)$$ The problem is that you need to solve quite large sets of equations.

Claude Leibovici
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We have $e^0=1$. From the Taylor series for $e^x$ centered at $x=0$, we may obtain the Taylor series for $\ln y$ centered at $y=1$.

Another quirk here. We get the terms of the inverse only one at a time, recursively. So, for example, the Taylor series for $\arctan x$ at $x=0$ has a nice formula for the terms, $$ \arctan x = x -\frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \dots $$ but the Taylor series for $\tan y$ at $y=0$ doesn't... those terms involve Bernoulli numbers.

GEdgar
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