# Homework Help: Simple series but my answer looks wrong

1. Oct 30, 2005

### Benny

I have doing a question which I cannot get the answer to. I would like some help with it if it can be done.
a) Derive a Taylor series about the origin for the function f(x) = log(1+x).
b) Find the interval of convergence for the Taylor series, justifying your answer.
c) Use the result of 'a' to deduce the Taylor series about the origin of the function $$\arctan h\left( x \right) = \frac{1}{2}\log \frac{{\left( {1 + x} \right)}}{{\left( {1 - x} \right)}}$$
d) On what interval does your series in 'c' converge?

I obtained:
a) $$\log \left( {1 + x} \right) = \sum\limits_{n = 1}^\infty {\left( { - 1} \right)^{n + 1} \frac{{x^n }}{n}}$$

b) I = (-1,1]

c) $$\arctan h\left( x \right) = \frac{1}{2}\log \frac{{\left( {1 + x} \right)}}{{\left( {1 - x} \right)}} = \frac{1}{2}\left( {\log \left( {1 + x} \right) - \log \left( {1 - x} \right)} \right)$$

$$\log \left( {1 + x} \right) = \sum\limits_{n = 1}^\infty {\left( { - 1} \right)^{n + 1} \frac{{x^n }}{n}} \Rightarrow \log \left( {1 - x} \right) = - \sum\limits_{n = 1}^\infty {\frac{{x^n }}{n}}$$

$$\arctan h\left( x \right) = \frac{1}{2}\sum\limits_{n = 1}^\infty {\left[ {\left( { - 1} \right)^{n + 1} \frac{{x^n }}{n} + \frac{{x^n }}{n}} \right]}$$

$$= \sum\limits_{n = 1}^\infty {\left[ {\left( {\left( { - 1} \right)^{n + 1} + 1} \right)\frac{{x^n }}{{2n}}} \right]}$$

That doesn't look right.

d) I know(or rather, remember seeing somewhere) that if you have a product of series then the radius of convergence of the product is the smaller of the radii of convergence of the individual series. I don't think that applies here through since I've got a sum of series and not a product. I'm stuck on part 'c' mainly. Can someone have a look through my answer and tell me where I'm going wrong?

2. Oct 30, 2005

### benorin

Good work, almost there

Parts a and b are correct, without question.

As for part c: The last sum is good except that $$\frac{{x^n }}{{2n}}}$$ should be $$\frac{{x^n }}{{n}}}$$. From there, note that

$$\left( {\left( { - 1} \right)^{n + 1} + 1} \right) = \left( 1 - (-1)^{n} \right)$$, which is 0 for even n and 2 for odd n, hence your sum reduces to

$$\mbox{arctanh}(x) = \sum\limits_{n = 1}^{\infty} {\left[ {\left( {\left( { - 1} \right)^{n + 1} + 1} \right)\frac{{x^n }}{{n}}} \right]} = \frac{1}{2}\sum_{n=1,3,5,...}^{\infty} (2)\frac{x^{n}}{n} = \sum_{n=1,3,5,...}^{\infty} \frac{x^{n}}{n} = \sum_{n=1}^{\infty} \frac{x^{2n-1}}{2n-1}$$

as it must.

As for part d, the interval of convergence of the above series is $$-1 < x < 1$$, that is $$x\in (-1,1)$$. I obtained this by using the ratio test and testing both end points (neither of them converge).

Last edited: Oct 30, 2005
3. Oct 30, 2005

### Benny

Thanks for the help benorin.

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