What is the Taylor polynomial for x^x around the point a=1?

danik_ejik
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hello,
please help to calculate the taylor polynomial for
[URL]http://latex.codecogs.com/gif.latex?f(x)=x^{x}-1[/URL] around the point a=1

i thought to write it as g(x)=x^x
and then f(x)=g(x)-1
and then find the polynomial for g(x) as lng(x)=xln(x)
but it seems incorrect.
 
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In order to do this you have to calculate the derivative if y=x^{x}$, take logs of this equation and differentiate that using implicit differentiation and that will help you, or you could write:
<br /> x^{x}=e^{x\log x}<br />
And use the chain rule
 
thanks,
successfully managed by directly calculating the taylor polynomial when
[URL]http://latex.codecogs.com/gif.latex?f(x)=e^{xlnx}[/URL]
 
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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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