Derriving the taylor series for $e^{\frac{\ln{(x+1)}}{x}}$ So I'm trying to find the Taylor polynomial of $e^{\frac{\ln{(x+1)}}{x}}$ around $x=0$ up to order $1$, however I encounter quite the problem.
We know that:
$$\ln{(1+x)}=x-\frac{x^2}{2}+O(x^3)$$
Hence, $$\frac{\ln{(1+x)}}{x}=1-\frac{x}{2}+O(x^2)$$
Now, the Taylor expansion for $\exp$ is:
$$e^ u=1+ u + \frac{u^2}{2}+...$$
Thus,
$$e^{\frac{\ln{(x+1)}}{x}} =1+ {\frac{\ln{(x+1)}}{x}} + \frac{1}{2} \left({\frac{\ln{(x+1)}}{x}}\right)^2+...$$
Now,
$$\left({\frac{\ln{(x+1)}}{x}}\right)^2=\left(1-\frac{x}{2}+O(x^2)\right)\left(1-\frac{x}{2}+O(x^2)\right)=1-x+O(x^2)$$
Continuing this way, we observe that all $\left({\frac{\ln{(x+1)}}{x}}\right)^n$ will contain a $x^1$ term, how can we then actually find the coefficient of the taylor expansion of $x^1$, and the constant term, since we will always have a $1$ in the beginning.
 A: It is not correct to expand $f(x)=\exp\left(\frac{\ln(1+x)}x\right)$ this way because $g(x)=\frac{\ln(1+x)}x$ doesn't have limit 0 at $x=0$.
Here is the correct way to get the desired Taylor expansion :
\begin{eqnarray*}
f(x) & = & \exp\left(1-\frac x2+O(x^2)\right)\\
& = & e \cdot \exp\left(-\frac x2+O(x^2\right)\\
& = & e\cdot\left(1-\frac x2+O(x^2)\right)\\
& = & e-\frac{ex}2+O(x^2)
\end{eqnarray*}
A: You can calculate the derivative of $e^{\frac{\ln{(x+1)}}{x}}$ and evaluate them at $0$ to get the expansion:
$$\frac{d}{dx}e^{\frac{\ln{(x+1)}}{x}}=(\frac{d}{dx}\frac{\ln{(x+1)}}{x})e^{\frac{\ln{(x+1)}}{x}}$$
You can see that the derivative at $0$ of $e^{\frac{\ln{(x+1)}}{x}}$ is equal to the derivatives of $\frac{\ln(x+1)}{x}$ at $0$ multiplied by $e$ since $\lim_{x\to 0}e^{\frac{\ln{(x+1)}}{x}}=e$
Hence the taylor expansion of $e^{\frac{\ln{(x+1)}}{x}}$ up to order one is $e-\frac{ex}{2}+O(x^2)$.
A: It is easier to compose the series from inside to outside
$$y=e^{\frac{\ln{(x+1)}}{x}} \implies \log(y)=\frac{\ln{(x+1)}}{x}$$ Starting as you did
$$\log(y)=1-\frac{x}{2}+\frac{x^2}{3}+O\left(x^3\right)$$Now, continuing with Taylor
$$y=e^{\log(y)}=e-\frac{e x}{2}+\frac{11 e x^2}{24}+O\left(x^3\right)$$
