# Write the Taylor expansion of order $2$ at $x=0$ of $h(x)=g^{-1}(x+\sin(x))$, for $g(x)=x\ln(2+x^2)$

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:

$$g(x)\,=\,x\ln(2+x^2)$$

• 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 $$0$$: $$h(x) \simeq \left(0+\frac{1}{\ln(2)}(x-0)-0+\mathcal{o}(x-0)^2\right)\left(x+x+\mathcal{o}(x)\right)= \frac{2x}{\ln(2)}+\mathcal{o}(x)^2$$

## 1 Answer

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 $$h(x)=\frac{2x}{\ln(2)}+o(x^2).$$

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