Taylor Series for ln(1-3x) about x = 0 | Homework Question

ganondorf29
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Homework Statement


Determine the Taylor Series for f(x) = ln(1-3x) about x = 0

Homework Equations



ln(1+x) = \sum\fract(-1)^n^+^1 x^n /{n}

The Attempt at a Solution



ln(1-3x) = ln(1+(-3x))

ln(1+(-3x)) = \sum\fract(-1)^n^+^2 x^3^n /{n}

Is that right?
 
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The -1 is in the right place, but I'm not sure why the 3 migrated to the exponent.
 
So is it:
<br /> \sum\fract(-1)^n^+^2 3x^n /{n}<br />
 
You check it yourself by computing the first couple of terms in the Taylor series.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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