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How do I find Taylor expansion of(around 1):

$$f(x)=x^x-1$$

The answer should be:

$$(x-1)+(x-1)^2+\frac 12(x-1)^3+\cdots$$

How the answer was obtained?

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Do you know what $x^x$ is? –  Git Gud Feb 23 '13 at 11:44
    
The title and the question itself are misleading: what you show there is the expansion around $\,x=1\,$ of the function, not the "binomial expansion" of it. –  DonAntonio Feb 23 '13 at 11:46
    
it's just a function without any name, i think –  Bek Abdik Feb 23 '13 at 11:47
    
@DonAntonio Yes, there is a misleading. Because, I translated it from a Russian book. –  Bek Abdik Feb 23 '13 at 11:48
    
@BekAbdik Do you know where $x^x$ can be defined? –  Git Gud Feb 23 '13 at 11:48
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4 Answers

up vote 2 down vote accepted

Use the fact that the derivative of $x^x$ is $(\ln x + 1)x^x$, and expand around $x=1$: $$y(x) = x^x,\ y'(x) =(\ln x + 1)x^x,\ y''(x)=\left((\ln x + 1)^2+\frac{1}{x}\right)x^x$$ $$y(1) = 1,\ y'(1) = 1,\ y''(x)=2$$ $$y(x) \sim 0 + (x-1)+(x-1)^2+...$$

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Using the power series for $\log(1+x)$, we get $$ \begin{align} (1+x)\log(1+x) &=(1+x)\left(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\dots\right)\\ &=x+\frac{x^2}{2}-\frac{x^3}{6}+\frac{x^4}{12}-\frac{x^5}{20}+\frac{x^6}{30}-\dots\tag{1} \end{align} $$ Plugging $(1)$ into the power series for $e^x$ yields $$ (1+x)^{1+x}=1+x+x^2+\frac{x^3}{2}+\frac{x^4}{3}+\frac{x^5}{12}+\frac{3x^6}{40}-\frac{x^7}{120}+\dots\tag{2} $$


Alternatively, use the binomial theorem to get $$ (1+x)^{x+1}=1+\tfrac{x+1}{1}x+\tfrac{(x+1)x}{2}x^2+\tfrac{(x+1)x(x-1)}{6}x^3+\dots\tag{3} $$ which also produces $(2)$.


Plug $x-1$ into $(2)$ and subract $1$ to get the Taylor Series for $x^x-1$ at $x=1$: $$ x^x-1=(x-1)+(x-1)^2+\frac{(x-1)^3}{2}+\frac{(x-1)^4}{3}+\frac{(x-1)^5}{12}+\dots\tag{4} $$

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With $z:=x-1$, thus $x=z+1$, use Taylor expansions for $e^z$ and $\ln(z+1)$: $$f(x)+1=x^x=(z+1)^{z+1}=(z+1)\cdot e^{z\ln(z+1)}=\\ = (z+1)\cdot\sum_{n\ge 0}\frac{(z\cdot\ln(z+1))^n}{n!} = \\ =(z+1)\cdot\sum_{n\ge 0}\frac1{n!}z^n\left(z-\frac12z^2+\frac13z^3\mp\dots\right)^n$$ Well, it seems a bit ugly, but perhaps you can calculate the coefficient of each $z^n$ from that, using binomial and factorial tricks...

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Another approach

$$\begin{align*}f(x):=x^x-1&,\;\;\;f(1)=0\\f'(x)=x^x(\log x+1)&,\;\;f'(1)=1\\f''(x)=x^x\left((\log x+1)^2+\frac{1}{x}\right)&,\;\;f''(1)=2\ldots\,\,etc.\end{align*}$$

Thus, the Taylor expansion around $\,x=1\,$ is

$$f(x)=(x-1)+\frac{2(x-1)^2}{2!}+\frac{3(x-1)^3}{3!}\ldots=(x-1)+(x-1)^2+\frac{(x-1)^3}{2}+\ldots$$

Note: Try to find, perhaps inductively, an expression for the $\,n-$th derivative of the function...

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Good catch on that typo, @robjohn. Thanks. –  DonAntonio Feb 23 '13 at 13:26
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