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Hi I'd like to find the inverse of: $$ y=(1/3)(x^{4} + 4x^{3}) $$

I have learned to do inverses using the following example: $$y=2x-1$$ $$x=2y-1$$ $$x+1=2y$$ $$(x+1)/2=y$$ $$f^{-1}(x)=(x+1)/2$$

Process:

  • Change the x's into y's and the y's into x's
  • Rearrange the equation to get a single y (formerly x) by itself on one side
  • Replace y with $f^{-1}(x)$

But the equation I am working with seems too complicated. I can't get x by itself on one side because the terms are to the power of 2 and 4.

Can anyone suggest a way forward? Thanks.

UPDATE:

As has been pointed out by some people the inverse of the said function is actually quite complicated and it turns out I was over complicating things myself. The main thing I am working on is to show that the roots of the equation $$x^{3}+4x^{2}-3=0$$ can be found (approximately) using iteration formulae which are rearrangements of the equation. After some trial and error I cam up with the following rearrangements which between them cover the three roots. $$x_{n+1}=(3-4x^{2})^{1/2}$$ $$x_{n+1}=(1/4)(3/x - x^{2})$$ $$x_{n+1}=((1/4)(3-x^{3}))^{1/2}$$ That's all I needed to do. Thanks everyone.

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It ain't pretty: Wolfram Alpha – Bill Cook Dec 22 '11 at 20:37
Thanks @BillCook but the solutions produced by wolframalpha are ridiculously complicated. Surely there must be a more simple answer? – NSDigital Dec 22 '11 at 20:40
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But that's how it is. At least there is a closed-form solution; when you move to the fifth power there usually isn't one at all. What do you need the inverse for? Perhaps there's a better way to achieve your eventual goal. – Henning Makholm Dec 22 '11 at 20:47
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So you have transformed a somewhat difficult problem (your concrete third degree equation) into an extremely difficult problem (inverting a fourth degree polynomial). That does not look like progress to me. – Henning Makholm Dec 22 '11 at 21:41
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Note that inverting a polynomial is a strictly harder problem than finding a root, since you can get a root easily by plugging $0$ into the inverse. – Paolo Capriotti Dec 22 '11 at 21:44
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1 Answer

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As has been pointed out by some people the inverse of the said function is actually quite complicated and it turns out I was over complicating things myself. The main thing I am working on is to show that the roots of the equation $$x^{3}+4x^{2}-3=0$$ can be found (approximately) using iteration formulae which are rearrangements of the equation. After some trial and error I came up with the following rearrangements which between them cover the three roots. $$x_{n+1}=(3-4x^{2})^{1/2}$$ $$x_{n+1}=(1/4)(3/x - x^{2})$$ $$x_{n+1}=((1/4)(3-x^{3}))^{1/2}$$ That's all I needed to do. No need for tricky inverses in the end. :0) Thanks everyone.

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