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

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SUMMARY

The Taylor polynomial for the function \( f(x) = x^x - 1 \) around the point \( a = 1 \) can be calculated by first expressing it as \( g(x) = x^x \) and then finding the polynomial for \( g(x) \). The correct approach involves using the logarithmic transformation \( \ln(g(x)) = x \ln(x) \) and applying implicit differentiation. The final result is obtained by directly calculating the Taylor polynomial for \( f(x) = e^{x \ln x} \).

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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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