Taylor Series centered at 2 Question

Question

I need a little bit of assistance with the following problem:

Calculate general Taylor Series and the first 3 terms of the Taylor series centered at 2 of the function

f(x) = ln(x^2 + 2x + 1)

I know the first step would be calculating multiple derivatives as below:

f'(x)=2/(x+1)

f"(x)= -2/(x+1)^2

f"'(x)=4/(x+1)^3

f""(x)= -12/(x+1)^4

f""'(x)=48/(x+1)^5

What would be the next step after doing this? I guess my issue is turning this into a taylor series using the formula.

José Carlos Santos · Accepted Answer · 2019-04-28 14:43:17Z

2

Hint: Since $f(x)=\log\bigl((x+1)^2\bigr)=2\log(x+1)$, then$$f'(x)=\frac2{x+1}=\frac2{3+(x-2)}=\frac23\cdot\frac1{1+\frac{x-2}3}.$$

answered Apr 28, 2019 at 14:43

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$\begingroup$ where did you get the 3 from? and how would you turn this into a taylor series? $\endgroup$
– Emily Richards
Apr 28, 2019 at 14:46
$\begingroup$ The $3$ comes from the difference between $x+1$ and $x-2$. And$$\frac1{1+\frac{x-2}3}=1-\frac{x-2}3+\frac{(x-2)^2}{3^2}-\frac{(x-2)^3}{3^3}+\cdots$$ $\endgroup$
– José Carlos Santos
Apr 28, 2019 at 14:48
$\begingroup$ and how does this turn into a taylor series? $\endgroup$
– Emily Richards
Apr 28, 2019 at 14:57
$\begingroup$ Since $f'(x)$ is the sum of that power series and since $f(2)=\log9$, then$$f(x)=\log9+2(x-2)-\frac{2(x-2)^2}{2\times3}+\frac{2(x-2)^3}{3\times3^2}-\frac{2(x-2)^4}{4\times3^3}+\cdots$$ $\endgroup$
– José Carlos Santos
Apr 28, 2019 at 14:59
1

$\begingroup$ The summation index doesn't matter. The $(x-2)^n$ terms are given instead by $(x-2)^k$ terms in the expansion I gave. $\endgroup$
– Peter Foreman
Apr 28, 2019 at 15:19

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