Taylor approximation of the function $\ln(x+(x^2+1)^{1/2})$

Question

My task is to approximate this function in a way $f(x)=T2(x)+R2(x)$ and so far I came to:

$$(x^2+1)^{1/2}=1+\frac{1}{2}x^{2}+o(x^2),$$ which gives us $$\ln\left(1+ x +\frac{1}{2}x^2+o(x^2)\right)= x+\frac{1}{2}x^2+o(x^2),$$ but wolfram says that it is only $x$ (without this $\frac{1}{2}x^2$), where am I mistaken?

Answer 1

Your expansion of the parenthesis is OK. Use $1+y\approx 1+x+x^2/2$. For logarithm expansion, go to the second order $\ln(1+y)\approx y-y^2/2$. If you plug in $y$ from the first equation, you get $$y-\frac{y^2}{2}=x+\frac{x^2}{2}-\frac{1}{2}\left(x+\frac{x^2}{2}\right)^2=x+\frac{x^2}{2}-\frac{x^2}{2}+\mathcal{O}(x^3)=x+\mathcal{O}(x^3)$$

Answer 2

Since: $$ f(x)=\ln\left(x+\sqrt{x^2+1}\right)=\text{arcsinh}\left(x\right) $$ $$ \frac{\partial}{\partial x}f(x)=\frac{1}{\sqrt{1+x^2}} $$ $$ \frac{\partial^2}{\partial x^2}f(x)=-\frac{x}{\sqrt{(1+x^2)^3}} $$ $$ \frac{\partial^3}{\partial x^3}f(x)=\frac{2 x^2-1}{\left(x^2+1\right)^{5/2}} $$ We have: $$ f(x)=x-\frac{x^3}{6}+\mathcal{O}\left(x^4\right) $$

I think the error is due to the chain rule: $$ f(x)=f(x_{0})+\left(x-x_{0}\right)f'(x_{0})+\frac{1}{2}\left(x-x_{0}\right)^{2}f''(x_{0})+\mathcal{O}\left(\left(x-x_{0}\right)^{3}\right) $$

$$ g(x)=g(x_{0})+\left(x-x_{0}\right)g'(x_{0})+\frac{1}{2}\left(x-x_{0}\right)^{2}g''(x_{0})+\mathcal{O}\left(\left(x-x_{0}\right)^{3}\right) $$

So we have:

$$ f(g(x))=f(g(x_{0}))+\left(g(x)-g(x_{0})\right)f'(g(x_{0}))+\frac{1}{2}\left(g(x)-g(x_{0})\right)^{2}f''(x_{0})+\mathcal{O}\left(\left(g(x)-g(x_{0})\right)^{3}\right)= $$

$$ =f(g(x_{0}))+\left(\left(x-x_{0}\right)g'(x_{0})+\frac{1}{2}\left(x-x_{0}\right)^{2}g''(x_{0})\right)f'(g(x_{0}))+ $$

$$ +\frac{1}{2}\left(\left(\left(x-x_{0}\right)g'(x_{0})+\frac{1}{2}\left(x-x_{0}\right)^{2}g''(x_{0})\right)\right)^{2}f''(g(x_{0}))+\mathcal{O}\left(\left(x-x_{0}\right)^{3}+\left(g(x)-g(x_{0})\right)^{3}\right) $$

Answer 3

Your calculation of $\ln(1+x+\frac12x^2+o(x^2))$ is incorrect. Remember the Taylor expansion for $\ln(1+y)$: $\ln(1+y) = y-\frac12y^2+o(y^2)$. Plugging in $x+\frac12x^2+o(x^2)$ for $y$, the result you gave corresponds to the $y$ term in the expansion, but the $-\frac12y^2$ term gives a term of size $\theta(x^2)$ that you failed to account for.