Starting a Taylor series problem, .

ArcainineFalls531
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Compute the Taylor series for f(x)= sq root (x) about x=1. Determine where the series sconverges absolutely, converges conditionally, and diverges. Hint: 2(k!)=2*4*6...(2k-2)*2k. Also 1<2, 3<4, 5<6,..., 2k-1<2k should help you out with a comparision.
 
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Use Taylor's theorem. f(x)=\sum_{n=0}^{\infty}\frac{f^{n}(a)*(x-a)^n}{n!}.
 

Thread 'Finding the nth roots of a complex number'

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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