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An interesting function June 16, 2010

Posted by choonyee in Mathematics.
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This post discusses the function $f(x)=x^x$ and another similar function of this form (I’m unsure if there is a name associated with this family of functions). The first encounter with this function is probably from Calculus on the study of indeterminate form $0^0$ .

First of all, we can show that $f$ is well-defined for all $x>0$ by taking natural logarithm,

$\log f(x) = \log (x^x) = x\log x$ . (1)

Then, we can evaluate the indeterminate form $0^0$ by taking limit

$\displaystyle\lim_{x\to 0^+} x^x = \displaystyle\lim_{x\to 0^+} \exp(x \log x)$ .

Now we encounter another indeterminate form of $0\times \infty$ which can be dealt with by L’Hopital’s Rule, giving

$\displaystyle\lim_{x\to 0^+} \frac{\log x}{1/x} = \displaystyle\lim_{x\to 0^+} \frac{1/x}{-1/x^2} = 0$ .

Hence, we have the following result

$\displaystyle\lim_{x\to 0^+} x^x = e^0 = 1$ .

So far all the discussions above can be found from standard Calculus textbook. Now let us explore this function further by considering a few problems. A natural problem to look at is the derivative of $f$ . Is $f$ differentiable for $x>0$ ? Or in the first place, is $f$ continuous for $x>0$ ? The answers to both questions are the same — YES. We can of course invoke the rigorous $\varepsilon-\delta$ definition but in this case it is much easier to take advantage of equation (1). Note that for $x>0$ , $x^x = \exp(x\log x)$ is just a composition of continuous and differentiable functions.

It is important to emphasize that conventional differentiation rules for powers/exponents do not apply here since both the base and exponent are variable now. For instance,

$\frac{d}{dx} (x^x) \neq x(x^{x-1})$ .

Nonetheless, the derivative $f'$ can still be computed by implicitly differentiating equation (1),

$\frac{1}{f(x)}f'(x) = x(\frac{1}{x})+\log x$ ,

$f'(x) = (1+\log x)x^x$ .

After obtaining the derivative, we can find the stationary point of $f$ by letting $f'(x)=0$ . Since $f$ is never 0, we have $1+\log x = 0$ which gives $x=e^{-1}$ . Thus, $f$ has only one stationary point $(e^{-1},e^{-e^{-1}})$ . To determine whether this is a min or max point, we need to compute the second derivative

$f''(x) = [\frac{1}{x}+(1+\log x)^2]x^x.$

At $x=e^{-1}, f''(x)>0$ implies that it is a minimum point. Now we can combine all the information obtained to sketch the graph of $f$ for $x>0$ . It will look like a skewed U-shaped curve with a min point and goes to infinity as $x$ increases.

Next, let’s turn our attention to negative values of $x$ . In particular, we would like to verify whether the following claim (posted in a math forum) is true

For $x<0, x^x$ is real-valued if and only if $x$ is negative integer.

One of the direction is easy to verify, i.e. if $x$ is negative integer, then $x^x$ is real-valued. However, the other direction is rather tricky. For example,

$(-\frac{1}{2})^{-1/2} = -\sqrt{2} \, i$ ,

shows that $x^x$ can be complex-valued. Take another example says $x=-\frac{1}{3}$ , then we’ll encounter the n-th root of unity, namely $(-1)^{1/3}$ , which yields one real and two complex values. This suggests that $x^x$ becomes a multi-valued function and further discussion will wander far off Calculus and drift into the realm of Complex Analysis.

After experimenting with a few more examples show that $x^x$ is complex-valued for most of negative $x$ except at integer points. However, providing examples does not secure a mathematical proof and thus a solid proof is still sought after. My guess is we do need to use complex analysis to prove that $x^x$ is real-valued for negative $x$ only at integer points.

From the above discussion, we see that $x^x$ is rather wild at the negative side. Why don’t we deal away with the negative values by considering the following function

$g(x) = (x^2)^{x^2}$ ,

which is well-defined for all values of $x\in\mathbb{R}-\{0\}$ . At the origin, we encounter the indeterminate form $0^0$ again. But this time we can evaluate both one-sided limits which, after similar computations as above, give

$\displaystyle\lim_{x\to 0^-} g(x) = \displaystyle\lim_{x\to 0^+} g(x) = 1$ .

Hence, in parallel to conventions, it is wise to define $0^0 = 1$ so that the function $g(x)$ is now continuous and differentiable everywhere. The graph of $g$ will look like a W-shaped (with smooth corners) curve, with three turning points and symmetrical about y-axis since $g$ is an even function.

i-maginary World

An interesting function June 16, 2010

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