# Inverse of $f(x)=x^2\log(1+x^2)$

How can one find the inverse of the funtion $$f:[0, \infty) \rightarrow[0, \infty)$$ defined as $$f(x)=x^2\log(1+x^2)$$ using $$log_2$$?

The general process of finding the inverse for functions that involve the e.g. $$log(1+x^2)$$ is clear to me. I struggle with with the $$x^2\log(\cdots)$$ term. I tried reformulating using logarithm rules. However, it didn't bring me further.

Any hint is highly appreciated.

• There is no reason to think this inverse function can be expressed in terms of known functions. Jan 4, 2020 at 21:43

Using $$\log_2$$ or using $$\log$$ is the same thing, and in any case you cannot invert $$x^2\log(1+x^2)$$ over $$\mathbb{R}^+$$ using elementary functions. You can by using something similar to the Lambert $$W$$ function, i.e. the inverse function of $$xe^x$$, which has a nice power series centered at the origin, by the Lagrange inversion theorem. In order to invert $$x^2\log(1+x^2)$$ is it enough to invert $$x\log(1+x)$$, or $$\sqrt{x\log(1+x)}$$, which meets the criteria for the application of the previous theorem. Anyway the coefficients of the power series giving a local inverse are not nice, since they depend on the derivatives at the origin of $$\frac{1}{\log(1+x)^k}$$, i.e. on generalized Gregory coefficients.
• Why would you say the logarithm of $x$ is invariant under change of base? Jan 5, 2020 at 7:01
• @Allawonder: of course not, but the ratio between $\log(x)$ and $\log_2(x)$ is constant, so if we are able to invert a function through $\log$ we are also able to do the same through $\log_2$ and vice-versa. Jan 5, 2020 at 14:46
From a numerical point of view, letting $$x^2=t$$, consider that you look for the zero of function $$f(t)=t \log(1+t)-k$$ For the case where $$k$$ is large (and so $$t$$), using Taylor we have $$f(t)=-k+t \log \left({t}\right)+1-\frac{1}{2 t}+O\left(\frac{1}{t^2}\right)$$ and a first approximation is given by $$t_0=\frac{k-1}{W(k-1)}$$ where appears Lambert function, as already mentioned in Jack D'Aurizio's answer.
Now, using Newton method, the first iterate will be given by $$t_1=t_0+\frac{(1+t_0) (k-t_0 \log (1+t_0))}{t_0+(1+t_0) \log (1+t_0)}$$ which, as usual, could be repeated until we reach the desired accuracy.
For illustration purposes, let $$k=2^p$$ and compare $$\left( \begin{array}{cccc} p & t_0 & t_1 & \text{exact} \\ 1 & 1.763222834 & 1.888858495 & 1.886631698 \\ 2 & 2.857390784 & 2.925571241 & 2.925214625 \\ 3 & 4.592135677 & 4.629608267 & 4.629550385 \\ 4 & 7.462896125 & 7.483295792 & 7.483286698 \\ 5 & 12.33757459 & 12.34851487 & 12.34851350 \\ 6 & 20.76851364 & 20.77429598 & 20.77429578 \\ 7 & 35.56161152 & 35.56463055 & 35.56463052 \\ 8 & 61.82786270 & 61.82942399 & 61.82942399 \\ 9 & 108.9372842 & 108.9380859 & 108.9380859 \\ 10 & 194.1651995 & 194.1656089 & 194.1656089 \end{array} \right)$$