# Evaluating a taylor series around a given point

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?

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Since the coefficients of a Taylor series are given by the derivative evaluated at chosen point you have to differentiate the infinite sum. But this works for power series termwise. See also the answer to 'Moving Centre of Power Series': math.stackexchange.com/a/780768/79762 – Alexander Frei Jun 1 '14 at 14:17
By the way $\ln(3)\neq\sum_{n=1}^{\infty }\frac{(-1)^{n+1}}{n}(2)^{n}$, that is the function cannot be expressed by this power series for values outside the radius of convergence. (There is really nothing that prevents the series from diverging like cancellation by phases or signs of subsequent terms.) – Alexander Frei Jun 1 '14 at 14:24
Thank you Freeze_S I appreciate your answer! So if I know that $\ln(x)=\sum_{n=1}^{\infty } \frac{(-1)^{n+1}}{n}(x-1)^{n}, 0<x\leq 2$, is the center for this one 1? So should I simply move the center to 3, hence (x-3)^n and 2<x<=4? – user154566 Jun 1 '14 at 14:32
...wait let me put it as answer (not enough characters available by a comment) – Alexander Frei Jun 1 '14 at 15:41

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

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Thats cheating! ;) The question was: "Given that [...] $=\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}x^n$, [...], find the Taylor series [...] around 3." So no preknowledge of $\ln(\cdot)$ assumed/allowed. – Alexander Frei Jun 1 '14 at 14:10
@Freeze_S OK. Now I set $x=3+3y$ and $x^n=3^n(1+y)^n$. Can I use binomial formula to expand $(1+y)^n$? ;) – mike Jun 1 '14 at 14:29

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 $n$.

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