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My question is a little bit general: what we can do if the inverse of our function can not be founded explicitly?

For example, let consider the function $$f(x)=x \cos (x)\quad\quad \quad\mbox{ for }\;x\in [0, \frac12]$$ this function is clearly invertible on $[0, \frac12]$, but we can not have an explicit formula of the inverse (you can try !).

Let $y \in f([0, \frac12])$; Is there any approximation or something to do to have an expression of $y$ such that $$x=f^{-1}(y)$$

Integrand
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user745750
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    thanks, yes the inverse of the values can be determined, but I speak about the function in the hull interval – user745750 Jul 08 '20 at 15:10
  • How accurate do you want the approximation to be? – Andrei Jul 08 '20 at 15:19
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    In certain cases, you can determine the series expansion of the inverse function. An archetypal example is the inverse of $f(x)=xe^x$ around the origin, in which the Lagrange inversion theorem gives a very neat formula for the Taylor series of the inverse of $f$ around the origin. – Sangchul Lee Jul 08 '20 at 15:22
  • @Andrei the best possible approximation you can obtain but not only for this function in a general case. as the answer of Sangchul Lee but this is just in a neighborhood of a point – user745750 Jul 08 '20 at 15:27
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    @user745750 That's a very general problem. For this particular case I would write the Taylor series around $0$: $f(x)\approx x-x^3/2=y$, then use the formula for cubic equation to find $f^{-1}(y)=x(y)$. Alternatively, you can find a numeric solution for $f(x)-y=0$ for a given $y$. Newton's method would probably work fine. – Andrei Jul 08 '20 at 15:58
  • There is no general answer, because this depends on why exactly you need the inverse function. –  Jul 09 '20 at 11:14
  • Very similar question except for the general title of this one – Тyma Gaidash Dec 20 '22 at 03:20

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$$y=x \cos (x)\quad\quad \quad\mbox{ for }\;x\in [0, \frac12]$$

The function is so linear that Taylor series built around $x=0$ is very tempting. now, using series reversion $$x=y+\frac{y^3}{2}+\frac{17 y^5}{24}+\frac{961 y^7}{720}+\frac{116129 y^9}{40320}+\frac{3488503 y^{11}}{518400}+\frac{7935695921 y^{13}}{479001600}+O\left (y^{15} \right)$$ is almost at the level of machine $\epsilon$.

If we use the expansion of $x$ up to $O\left (y^{2n+1} \right)$ in $y-x \cos(x)$, the result is $O\left (y^{2n+3} \right)$.

Similarly, if $T_n$ is the Taylor expansion of $x \cos(x)$ to $O\left (x^{2n+1} \right)$, computing the norm $$\Phi_n=\int_0^{\frac 12} \big[x \cos(x)-T_n\big]^2\,dx$$ we have the following results $$\left( \begin{array}{cc} n & \log(\Phi_n) \\ 0 & -8.2167 \\ 1 & -16.393 \\ 2 & -26.272 \\ 3 & -37.328 \\ 4 & -49.290 \\ 5 & -61.988 \\ 6 & -75.306 \\ 7 & -89.161 \\ 8 & -103.49 \\ 9 & -118.24 \\ 10 & -133.37 \end{array} \right)$$

Claude Leibovici
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