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What is the inverse function of $f(x) = x^5 + 2x^3 + x - 1?$ I have no idea how to find the inverse of a polynomial, so I would greatly appreciate it if someone could show me the steps to solving this problem. Thank you in advance!

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  • $\begingroup$ please enclose your math in \$ \$ to have it typeset nicely $\endgroup$ – Valentin Oct 10 '12 at 21:41
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    $\begingroup$ Even Mathematica can't find inverse function, but you can be confident - inverse function does exist $\endgroup$ – Norbert Oct 10 '12 at 21:42
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    $\begingroup$ Your polynomial is increasing, and its range is all reals, so there is an inverse. Finding a pleasant expression for the inverse is another matter. But one can find information about the derivative of the inverse without knowing a formula. $\endgroup$ – André Nicolas Oct 10 '12 at 22:05
  • $\begingroup$ I think, it's very hard. Is it homework? Or where is this problem from? $\endgroup$ – Berci Oct 10 '12 at 22:21
  • $\begingroup$ This question has been asked before, albeit in greater detail. Most likely we're having trouble answering it because we're missing this information: math.stackexchange.com/questions/85226/… $\endgroup$ – Gyu Eun Lee Dec 15 '12 at 3:42
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This is an experimental way of working out the inverse.

We can treat the polynomial like an expansion \begin{equation} f(x) = -1 + x + 0x^2 + 2x^3 + 0x^4 + x^5 + 0x^6 + 0x^7 + \cdots \end{equation} then we can perform a Series Reversion on this to give the inverse series (as an infinite expansion) \begin{equation} f^{-1}(x) = (1+x) -2(1+x)^3 +11(1+x)^5-80(1+x)^7+665(1+x)^9-\cdots \end{equation} at a glance this doesn't seem to helpful, but if we search the coefficients in the OEIS we seem to get a hit! It would appear (conjecture) that the general coefficient is then \begin{equation} a(n) = \binom{5n+1}{n}\frac{(-1)^n}{2n+1} \end{equation} and we could write that \begin{equation} f^{-1}(x) = \sum_{n=0}^\infty \binom{5n+1}{n}\frac{(-1)^n}{2n+1}(x+1)^{2n+1} \end{equation} if we evaluate that in Mathematica, it gives \begin{equation} f^{-1}(x)=(1+x)\;_4F_3\left(\frac{2}{5},\frac{3}{5},\frac{4}{5},\frac{6}{5}\bigg|\frac{3}{4},\frac{5}{4},\frac{6}{4}\bigg|-\frac{5^5}{4^4}(1+x)^2 \right) \end{equation} a generalised hypergeometric function. If we plot the composition of these two i.e $f(f^{-1}(x))$ or $f^{-1}(f(x))$ the plot seems to indicate \begin{equation} f(f^{-1}(x))=f^{-1}(f(x))=x \end{equation} of course this is not a proof, and evaluating the inverse function may become numerically unstable if $x$ becomes too large.

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Generally, you say $y=$ your polynomial and solve for $x$. Fifth degree polynomials are generally not solvable. The general approach for a quadratic would be essentially the quadratic formula. Given $y=ax^2+bx+c$, you find $x=\frac {-b \pm \sqrt{b^2-4a(c-y)}}{2a}$. You need to pick one sign to get a function.

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  • $\begingroup$ Thank you! I will try it again. $\endgroup$ – Jaden M. Oct 10 '12 at 23:47
  • $\begingroup$ The solution that is expected to be x = f(y) does not contain y. Substitute c by c-y ? $\endgroup$ – h22 Nov 23 '18 at 8:02
  • $\begingroup$ @h22: yes, that is correct. Thanks $\endgroup$ – Ross Millikan Nov 23 '18 at 15:07
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As others indicated, there is no algebraic formula for the inverse function $f^{-1}$. The inverse functions exists (since $f$ is increasing), but there are serious algebraic obstructions to solving $y=x^5 + 2x^3 + x - 1$ for $x$.

But we can find particular values of $f^{-1}$ and of its derivative. For example, to find $f^{-1}(3)$ we would just have to note that $f(1)=3$. Therefore, $f^{-1}(3)=1$. (The number $3$ is lucky here; if asked about $f^{-1}(4)$, one would need a numerical method, e.g., a calculator.)

Also, the inverse function theorem can be used to find the derivative of $f^{-1}$ at $3$: $$ (f^{-1})'(3) = \frac{1}{f'(1)} = \frac{1}{5+6+1} = \frac{1}{12} $$


Related question.

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