Can anyone tell me whether I carried out properly this exercise and where are mistakes? Thank you.

Let be $g: \mathbb{R} \to \mathbb{R}$ the function defined by:


  • Show that $\exists \,\, \delta\gt0$ s.t. the function $g: \,(-\delta,\delta)\to g(-\delta,\delta)$ is invertible

  • Write the taylor polynomial of degree $2$ centered at $x=0$ of the following function: $$h(x)\,=\,g^{-1}(x+\sin(x))$$

About the first point, basically we have to show that $g$ is injective in a neighborhood $(-\delta,\delta)$ of $0$. So $\,\,g(0)=0$, $\,\,\,\,\,\lim_{x\to +\infty}g(x)=+\infty\,\,$ and $\,\,\,\,\lim_{x\to -\infty}=-\infty$.

Moreover $g'(x)=\frac{2x^2}{2+x^2}+\ln(2+x^2)\,\,\gt0$,$\,\,\,\,\forall x \in \mathbb{R}\,\,\,\,$ so $g$ is monotonic and strictly increasing and $g\in C^{\infty}(\mathbb{R})\,\,\,$which prove that $g$ is a diffeomorphism.

From the theory, the taylor polynomial of degree $2$ centered at $x=0$ of $g$ is $$g^{-1}(y)=g^{-1}(0)+(g^{-1})'(0)(y-0)+\frac{(g^{-1})''(0)}{2!}(y-0)^2+\mathcal{o}(y-0)^2$$

Since $g(0)=0$ we have $g^{-1}(0)=0$

$(g^{-1})'(0)=\frac{1}{g'(0)}=\frac{1}{\ln(2)}\,\,\,\,$ and

$\,\,\,(g^{-1})''(0)=-\frac{g''(0)}{(g'(0))^3}=0\,\,\,$ since$\,\,g''(x)=\frac{4x(2+x^2)-2x^2(2x)}{(2+x^2)^2}+\frac{2x}{2+x^2}$

hence, in a neighborhood of $0$: $$h(x) \simeq \left(0+\frac{1}{\ln(2)}(x-0)-0+\mathcal{o}(x-0)^2\right)\left(x+x+\mathcal{o}(x)\right)= \frac{2x}{\ln(2)}+\mathcal{o}(x)^2$$


You are almost correct, in your result there is just an extra factor $x$. At the last step you have to consider the composition, not the multiplication $$h(x)=\left(0+\frac{1}{\ln(2)}(2x+o(x^2))-0+o(2x+o(x^2))^2\right)\\= \frac{2x}{\ln(2)}+o(x^2).$$

This is a shorter way based on the uniqueness of the Taylor expansion: $$g(x)=x\ln(2+x^2)=x\ln(2)+x\ln(1+x^2/2)=x\ln(2)+o(x^2).$$ Let $h(x)=a+bx+cx^2+o(x^2)$, then $$g(h(x))=a+b(x\ln(2)+o(x^2))+c(x\ln(2)+o(x^2))^2+o(x^2)\\ =a+b\ln(2)x+cx^2\ln(2)+o(x^2)$$ On the other hand, $h(x)=g^{-1}(x+\sin(x))$ implies that $$g(h(x))=x+\sin(x)=2x+o(x^2).$$ Therefore, by comparing the coefficients, we get $a=0$, $b=2/\ln(2)$, and $c=0$. Hence $$h(x)=\frac{2x}{\ln(2)}+o(x^2).$$

  • $\begingroup$ Yes, thank you! The multiplication was an oversight due to the rush $\endgroup$ – F.inc Feb 8 at 10:53
  • 1
    $\begingroup$ @F.inc Well done!! $\endgroup$ – Robert Z Feb 8 at 10:54

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