Taylor Series of f(x) = 1/(1-6x) at c=6

Soccerdude
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Homework Statement



Find the Taylor Series for f(x) = 1/(1-6) centered at c=6

Homework Equations




Ʃ Fn(a)(x-a)/n!
n=0

The Attempt at a Solution



I believe that the nth derivative of 1/(1-6x) is

(-6)n-1n!/(1-6x)n+1

So i figured that the taylor series at c=6 would be

(-6)n-1(x-6)n/(1-6x)n+1

What am I doing wrong here?
 
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Soccerdude said:

Homework Statement



Find the Taylor Series for f(x) = 1/(1-6) centered at c=6
Did you mean: $$f(x) = \frac{1}{1-6x}$$ ... from below, it appears so.

Homework Equations




Ʃ Fn(a)(x-a)/n!
n=0

The Attempt at a Solution



I believe that the nth derivative of 1/(1-6x) is

(-6)n-1n!/(1-6x)n+1
What leads you to believe that?
So i figured that the taylor series at c=6 would be

(-6)n-1(x-6)n/(1-6x)n+1

What am I doing wrong here?
Start from the definition of the Taylor series.
Try writing out the 1st 3-4 terms and see if you spot a pattern.
 
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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