I'm calculating the 3rd degree Taylor polynomial for $y =$ $x^{x^x}$ in $x_0=1$. I calculated the first derivative, which is $e^{lnx*x^x}*(\frac{x^x}{x}+lnx*e^{lnx*x}*(1+lnx))$. Finding the first degree derivative in $x_0=1$ is very easy, but calculating the second and third derivative is not.

Is there any way I can simplify it or do I really have to find all the lengthy second and third degree derivatives of $y$?


  • $\begingroup$ $x_0=0$???? That isn't even in the domain of the function. Do you perhaps mean $x_0=1$? I would not do this problem by calculating derivatives. Do you have any other techniques to consider? $\endgroup$ Nov 17 '19 at 17:33
  • $\begingroup$ @TedShifrin Oh shoot, I made a typo. $x_0=1$.. Sorry, fixed it. And I'm not sure I know about any other ways. Could you help? $\endgroup$
    – user714814
    Nov 17 '19 at 17:39
  • $\begingroup$ Here's an easier one. Do you know how to give me the T.P. of degree $4$ centered at $x_0=0$ of $e^{x^2}$ without computing derivatives? (I assume you know the T.P. of $e^x$.) $\endgroup$ Nov 17 '19 at 17:41
  • $\begingroup$ @TedShifrin Not sure how without calculating derivatives. $\endgroup$
    – user714814
    Nov 17 '19 at 17:47
  • $\begingroup$ You substitute $x^2$ for $x$ in the T.P. of degree $2$ of $e^x$. Can you figure out (prove?) why this works? $\endgroup$ Nov 17 '19 at 17:49

To make life a bit easier, what I would do is $$y=x^{x^x}\implies z=\log(y)=x\log(x^x)$$for which $$\frac {dz}{dx}=x^x \left(\frac{1}{x}+\log ^2(x)+\log (x)\right)\to 1$$ $$\frac {d^2z}{dx^2}= x^{x-2} (2 x+x \log (x) (x+x \log (x) (\log (x)+2)+3)-1)\to 1$$ For sure, this is tedious but doable.

We should arrive to $$z=(x-1)+\frac{1}{2} (x-1)^2+\frac{5}{6} (x-1)^3+\frac{1}{12} (x-1)^4+O\left((x-1)^5\right)$$ Now

$$y=e^{z}=1+(x-1)+(x-1)^2+\frac{3}{2} (x-1)^3+\frac{4}{3} (x-1)^4+O\left((x-1)^5\right)$$


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