Evaluating a taylor series around a given point

Question

So I'm having some trouble with the problem: Given that $\ln(x+1)=\sum_{n=1}^{\infty } \frac{(-1)^{n+1}}{n}x^{n}, -1<x\leq 1$, find the Taylor series of ln(x) around 3. For which x is this series valid?

What I've figured out is that $\ln(x)=\sum_{n=1}^{\infty } \frac{(-1)^{n+1}}{n}(x-1)^{n}, 0<x\leq 2$

Since the sum is restricted to $\ 0<x\leq 2$ I'm having problems to find the taylor series for ln(x) at x=3, it seems $\ x> 2$ diverges which makes it impossible to approximate?

Can it be that what's asked for simply is the sum $\ln(3)=\sum_{n=1}^{\infty } \frac{(-1)^{n+1}}{n}(2)^{n} $ , even though it diverges?

Answer 1

You can set $x=3+4y$ then $$\ln(x+1)=\ln(4+4y)=ln4+ln(1+y)$$ Now you only need to expand $ln(1+y)$ at point $y=0$: $$\ln(y+1)=\sum_{n=1}^{\infty } \frac{(-1)^{n+1}}{n}y^{n}, -1<y\leq 1$$

Finally you may substitute $y$ by $(x-3)/4$.

Hope it helps. -mike

Answer 2

No thats not a good idea...

First of all note that you cannot get a power series around 3 immediately from your power series around 1 but you have to annoyingly first get one around 2 and then the one around 3 (or one around 1.71 and then one around 2.34 and then one around 2.62 and then the one around 3).

Now you get the k-th term of your power series around 2 by differentiating the sum k-times(!) and then evaluating it at $x=2$ and weight it with the k-th factorial $\frac{1}{k!}$ - but do it termwise so for $\frac{(-1)^{n+1}}{n}(x-1)^n$ but for all of these ones and then sum all these together.

For example for $k=5$ and $n=6$ you get $(\frac{(-1)^{6+1}}{6}(x-1)^6)^{(5)}=\frac{-1}{6}6\cdot 5\cdot 4\cdot 3\cdot 2(x-1)^1$ and so at $x=2$ this is $\frac{-1}{6}6\cdot 5\cdot 4\cdot 3\cdot 2(2-1)^1=(-1)5\cdot 4\cdot 3\cdot 2$ and so the weighted becomes $\frac{1}{5!}(-1)5\cdot 4\cdot 3\cdot 2=-1$, that is: $$a_{k=5}=\{\ldots +(-1)+\ldots\}\text{ where the left out are the remaining terms for } n=0,\ldots,5,7,\ldots$$

I know that is super annoying but it gets simpler if you rather compute it not explicitely but in general not substituting specific numbers for $k$ and $n$.