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**nevinsmith****Member**- Registered: 2010-05-19
- Posts: 39

Hi guys, say that I want to expand ln(1+x) using maclarin series,

How do i find what limits the expansion will be valid for?

what if I use taylor series and center it around 1,2,3... etc

I know how to do the expansions and stuff but not really how to find what values of x the expansion will converge for.

please help me!

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi;

The Mclaurin series expansion of ln(1+x) is

One way of knowing the radius of convergence is by your general knowledge of series. It is easy to see the when x>1 that series will not converge. Why? Because the terms will not be getting smaller. Same thing for x<-1.

Now let's look at x = 1.

That is the well known alternating harmonic series and is known to sum to ln(2).

How about at x = -1?

The series on the RHS is the well known harmonic series and is known to diverge.

So we have for an interval of convergence

-1 < x ≤ 1

This is one way of doing it, there are others. I consider this the minimum amount of knowledge that you should know for radii of convergence problems.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**nevinsmith****Member**- Registered: 2010-05-19
- Posts: 39

bobbym, thanks for helping me

I still don't entirely understand this,

'One way of knowing the radius of convergence is by your general knowledge of series. It is easy to see the when x>1 that series will not converge. Why, because the terms will not be getting smaller. Same thing for x<-1.'

How did you see that the series will not converge if x>1? is it like a guess and check thing?

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**nevinsmith****Member**- Registered: 2010-05-19
- Posts: 39

Is it that the absolute value of the f(x) exponentialed term must be always be less than 1?

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi;

No, the numerator is increasing by a power and the denominator is increasing by 1. Let's say you had chosen x = 2.

The absolute value of the terms would look like this.

The numerator is growing faster than the denominator which you could verify by taking the limit of (2^n)/n.

What I am trying to do is to first give you an intuitive sense ( one based on general knowledge ) of when these series converge or not. There are tests that should have been covered before they rushed you into Taylor Series.

Have you seen these tests?

http://en.wikipedia.org/wiki/Convergent_series

If you are familiar with these then we can go on to the ratio test which will do this problem in a more methodical way.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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