The power series above is the Taylor series....

nfcfox
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Homework Statement


http://imgur.com/1aOFPI7

PART 2

Homework Equations


Taylor series form

The Attempt at a Solution


My thought process is that the answer is 3 because using the geometric series equation (1st term)/(1-R) then you can get the sum. In this case R would be x+2 where x is -2 so 0. 1st term is 3/1 so the answer would be 3. I feel like this isn't what I'm supposed to do as it's saying the power series is the taylor series of some function, f but I have no idea how I could find a sum for that.
 
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I will try to help, but the PF rules don't allow me to give the complete answer. You are already very close to the correct answer using S=A/(1-R). Your R=x+2 is also correct. One hint is that the (x-a) shows up repeatedly in the Taylor expansion about x=a. You need to determine what "a" is in their Taylor expansion. Then try expanding your S(x) about x=a. Does this duplicate their function?
 
Charles Link said:
I will try to help, but the PF rules don't allow me to give the complete answer. You are already very close to the correct answer using S=a/(1-R). Your R=x+2 is also correct. One hint is that the (x-a) shows up repeatedly in the Taylor expansion about x=a. You need to determine what "a" is in their Taylor expansion. Then try expanding your S(x) about x=a. Does this duplicate their function?
So it's 3? A is -2...
 
nfcfox said:
So it's 3? A is -2...
I just edited my post so that "A" and "a" are not confused. Yes, the expansion is about a=-2. For x=-2 the answer is 3. (This question wasn't clear in the link-it wants the sum of the series for...and the next character or two I couldn't see.) They also want to know the interval of convergence. That should be evident from your geometric series expression.
 
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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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