<s>[INST] trouble expanding taylor series about a point other than zero using geometric series I'm trying to understand how to use a Taylor series expansion to correctly expand a population growth function about a point other than zero using the geometric series.
For expansion about $t=0$, I get:
$$
\begin{align}
p(t) &= 2 ln (t+2) \\
&= 2  \int \frac{1}{t+2} dt \\
&= 2  \int \frac{1}{ 2(\frac{t}{2}+1)} dt \\
&=   \int \frac{1}{ \frac{t}{2}+1} dt \\
&=   \int \frac{1}{ 1 - (-\frac{t}{2} )} dt \\
&=   \int  (1 + (\frac{-t}{2})+(\frac{-t}{2})^2+(\frac{-t}{2})^3+\cdots) dt \\
&=   \int  (1 - \frac{1}{2}t+\frac{1}{4}t^2-\frac{1}{8}t^3+\cdots) dt \\
&=   t - \frac{1}{4}t^2+\frac{1}{12}t^3-\frac{1}{16}t^4+\cdots + C \\
p(0) &= 2\ln 2 = C\\
p(t) &=   2\ln 2+t - \frac{1}{4}t^2+\frac{1}{12}t^3-\frac{1}{16}t^4+\cdots \\
\end{align}
$$
I'm not sure how to go about expanding $p(t) = 2ln(t+2)$ about $t=2$ using the above method. In Wolfram Alpha, the answer it spits out is:
$$\ln 16 +\frac{x-2}{2}-\frac{1}{16}(x-2)^2+\frac{1}{96}(x-2)^3 + O((x-2)^4)$$
 [/INST] You can expand $2\log(t+4)$ around $t=0$ and then plug in $t = x-2$.
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