How can I expand $x^x$ in Taylor series if it isn't defined in $x=0$?

$x^x=e^{x \ln x}$, so when I apply Taylor's formula at $x_0=0$ the first term $f(x_0)$ is not defined, but this is a known formula where the first term is $1$.

How is that possible?

Thank you.

  • $\begingroup$ You can't expand it around $x = 0$. $\endgroup$ – embedded_dev Jun 14 '17 at 17:01
  • 3
    $\begingroup$ A Taylor series always has an almost symmetric interval of convergence about its center point (other than at the endpoints), whereas $x^x$ is not defined for $x<0$. $\endgroup$ – Daniel Schepler Jun 14 '17 at 17:16
  • $\begingroup$ After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark ✓ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?. $\endgroup$ – Clement C. Jun 16 '17 at 18:51

The right-hand limit of $x^x$ is $\lim_{x \searrow 0} x^x = 1$, so one could either ask about the function $f$ defined to have value $1$ at $0$ and $x^x$ elsewhere where defined, or just use the convention that $0^0 = 1$. In this way, there's no problem with computing the zeroth-order term of the expansion.

On the other hand, for $x > 0$ we have $\frac{d}{dx} (x^x) = (1 + \log x) x^x$, so $\lim_{x \searrow 0} \frac{d}{dx} (x^x) = -\infty$, and hence there is no first-order Taylor approximation to the function.

Now, we can write $x^x = \exp (x \log x)$ and so expand $x^x$ in a series of a slightly different form that converges to $x^x$ for $x \geq 0$, namely, $$x^x \sim \sum_{k = 0}^{\infty} \frac{1}{k!} (x \log x)^k = 1 + x \log x + \frac{1}{2} x^2 \log^2 x + O(x^3 \log^3 x) .$$

| cite | improve this answer | |
  • $\begingroup$ What formula do we use for the lastone expansion? $\endgroup$ – pter26 Jun 14 '17 at 17:17
  • $\begingroup$ This is simply writing out the usual Taylor expansion for $\exp u$ about $u = 0$ and substituting $u = x \log x$ (which makes sense, since $\lim_{x \searrow 0} x \log x = 0$). $\endgroup$ – Travis Willse Jun 14 '17 at 18:27

Since $x\ln x\xrightarrow[x\to 0]{} 0$, you can use the expansion of $e^u$ (when $u\to0$) to write $$ e^{x\ln x} = 1+x\ln x+\frac{x^2\ln^2 x}{2} + o(x^2\ln^2 x) $$ (and have extra terms if you want; I stopped at order $2$). But that does not follow immediately from Taylor's theorem. (And indeed, Taylor's theorem would only give you a polynomial approximation: here, you get terms involving logarithms.)

| cite | improve this answer | |
  • $\begingroup$ For completeness: what was used here is that $$e^u = 1+u + \frac{u^2}{2}+\frac{u^3}{6} + \dots+\frac{u^k}{k!}+o(u^k)$$ when $u\to 0$, for any fixed integer $k\geq 0$. $\endgroup$ – Clement C. Jun 14 '17 at 17:26
  • $\begingroup$ The firt term of e^u is defined because of sositution? $\endgroup$ – pter26 Jun 14 '17 at 17:30
  • $\begingroup$ The above is the Taylor expansion of $u\mapsto e^u$ around $0$, and is well-defined. Since $x\ln x \to 0$ when $x\to 0$, one can "plug in $u=x\ln x$" in this expansion (technically, compose it) to get the result (intuitively, "if $f(u) \approx g(u)$ whenever $u$ is small, then $f(x\ln x) \approx g(x\ln x)$ whenever $x$ is small enough, because then $x\ln x$ is small too"). But the composition of the two, while valid, is not a Taylor series expansion -- since otherwise it would only have powers of $x$, not also logarithms. $\endgroup$ – Clement C. Jun 14 '17 at 17:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.