Question

Find the Maclaurin series with the general term for f(x)=ln(e+x).

Answer 1

\[\ln(e+x)=\ln\left(e\left(1+\frac{x}{e}\right)\right)=\ln e+\ln\left(1+\frac{x}{e}\right)=1+\ln\left(1+\frac{x}{e}\right)\]
If you happen to know the Maclaurin series for \(\ln(1+x)\), you can replace \(x\) with \(\dfrac{x}{e}\).

However, if you are supposed to derive the series for the given function, find your derivatives:
\[\begin{align*}
f(x)&=\ln(e+x)&f(0)&=1\\\\
f'(x)&=\frac{1}{e+x}&f'(0)&=\frac{1}{e}\\\\
f''(x)&=-\frac{1}{(e+x)^2}&f''(0)&=-\frac{1}{e^2}\\\\
&~~\vdots&&~~\vdots\\\\
f^{(n)}(x)&=\frac{(-1)^{n+1}}{(e+x)^n}&f^{(n)}(0)&=\frac{(-1)^{n+1}}{e^n}
\end{align*}\]

Answer 2

Slight typo:
\[\begin{align*}
f(x)&=\ln(e+x)&f(0)&=1\\\\
f'(x)&=\frac{1}{e+x}&f'(0)&=\frac{1}{e}\\\\
f''(x)&=-\frac{1}{(e+x)^2}&f''(0)&=-\frac{1}{e^2}\\\\
&~~\vdots&&~~\vdots\\\\
f^{(n)}(x)&=\frac{(-1)^{n+1}}{(e+x)^n}&f^{(n)}(0)&=\frac{(-1)^{n+1}\color{red}{(n-1)!}}{e^n}
\end{align*}\]

Answer 3

How does the general term account for the first two terms?

Answer 4

Actually, I got it. Thanks!