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Say you want to find the inverse of a function, and it's not expressible in terms of known functions, so you want to find the taylor series for that inverse. But, you at least know the taylor series for your original function f(x), defined as $$f(x)=\sum_{n=0}^{\infty} \frac{g(x,n)h(n)}{n!}$$ and it converges for a sufficient radius. Since Taylor series stem from derivatives, and there is a theorem called the inverse function theorem that relatives derivatives to a functions' inverse $$(f^{-1})'(f(x))=\frac{1}{f'(x)},$$

is there any consequential theorem where you can derive a new Taylor series for the inverse of a function by using the original Taylor series for that original function? In other words, $$f^{-1}(x)=\sum_{n=0}^{\infty} \frac{???}{n!}$$

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Yes it is possible to find the inverse of a Taylor Series. You can use series reversion. It can be used to find the inverse of Taylor Series given the forward function. Here is a URL: http://mathworld.wolfram.com/SeriesReversion.html

  • Okay, thank you for bringing that to my attention. It would be helpful if there was an example because for some reason that summation is unexpectedly of several variables at once, it doesn't look like the end result is a nice Taylor series like I'm looking for. Just to be clear, you're not inverting or undoing the summation operator itself, you're actually defining the inverse of the original function explicitly as a taylor series by using a series reversion? –  Jul 30 '17 at 04:48
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    Sorry for the late reply. Yes I believe the inverse function of a Taylor Series can be strictly defined on series reversion (although tedious, it is the most natural definition in my opinion). –  Jul 30 '17 at 05:16
  • @DaneJoe. Have a look at https://math.stackexchange.com/questions/2360037/inverse-function-for-a-taylor-series/2360186#2360186 for an example. – Claude Leibovici Jul 30 '17 at 05:16
  • @ClaudeLeibovici The problem is that example doesn't actually give any of the intermediate steps, so it doesn't really help anything. When I look for any examples of how to apply this series reversion theorem, there's literally no results whatsoever on the entire internet, there is only a statement of the start of the theorem with no shown application, nothing more. So I have absolutely no way of verifying if I'm interpreting the process correctly even if I arrive at something correct, because it could just be coincidence. –  Jul 30 '17 at 05:45
  • @DaneJoe. Apply the method by yourself on something like $y=\sin(x)$ to get the Taylor series for $\sin^{-1}(y)$. – Claude Leibovici Jul 30 '17 at 05:58
  • @ClaudeLeibovici That's the thing, I don't even know if I'm applying the first step correctly. I don't even know what the first step is suppose to be, no one really says what it's suppose to be. There's no intermediate steps that explains how it works, no paper on how it was derived. I can't see any reason why an explicit inverse would suddenly explode with a shotgun of random variables that were never present in the original function. If I start with x, y and n, then I expect to end with x, y and n. –  Jul 30 '17 at 06:05
  • @DaneJoe. I shall write the example. – Claude Leibovici Jul 30 '17 at 06:19
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Considering $y=\sin(x)$, let us develop it as a truncated Taylor series $$y=x-\frac{x^3}{6}+\frac{x^5}{120}+\cdots\tag 1$$ Now, following the steps given here, let $$x=a_1y+a_2y^2+a_3y^3+a_4y^4+a_5y^5+\cdots\tag 2$$ and replace.

Expand all terms and group the terms for a given power of $y$. You will get $$y=a_1 y+a_2 y^2+\left(a_3-\frac{a_1^3}{6}\right) y^3+\left(a_4-\frac{1}{2} a_1^2 a_2\right) y^4+\frac{1}{120} \left(a_1^5-60 a_3 a_1^2-60 a_2^2 a_1+120 a_5\right) y^5+\cdots\tag 3$$ Now, identify the powers to get $$a_1=1$$ $$a_2=0$$ $$a_3-\frac{a_1^3}{6}=0\implies a_3=\frac 16 $$ $$a_4-\frac{1}{2} a_1^2 a_2=0\implies a_4=0$$ $$a_1^5-60 a_3 a_1^2-60 a_2^2 a_1+120 a_5=0\implies a_5=\frac 3 {40}$$

Claude Leibovici
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  • I guess it's not too terribly different than the normal process for deriving a Taylor series. I will look at it more in depth tomorrow when I have time, thank you for explaining it I'll get back to you on this. –  Jul 30 '17 at 06:36
  • @DaneJoe. It is different since the coefficients are no more related to the derivatives. But, as you see, you just need to solve a few equations (one at the time as I did here). – Claude Leibovici Jul 30 '17 at 06:50
  • Alright thanks it works out. It is confusing when WA says "plug in" because that makes it seem literally like every single instance of "x" is replaced with the entire infinite series defined by y. –  Jul 31 '17 at 03:33