Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am trying to minimize the function in the form of $f(x) = (1-a^x)^x$ where $0 < a < 1$ with respect to $x$ (for $x > 0$) and I am stuck!

Unfortunately the derivative is not nice enough to use the traditional method of setting it equal to zero.

Some quick plots show that the minimizer should be something around $-1 / \log(a)$ but not exactly that. (Indeed $x^* = - 1 / \log(a)$ is the minimizer of $1 - x a^x$ which approximates $f(x)$ if $a \ll 1$).

I appreciate if you someone can give me hints or ideas about how to minimize such a function?


share|cite|improve this question
The solution seems to be $x=\log_a \frac12$, though I couldn't tell you how to get there. – Rahul Jan 15 '13 at 13:59
@RahulNarain: Thank you very much for the hint. I guess your answer can be right (at least according to plots) and to get there one may define $u = (1-a^x)$ and observe that $f(u) = u^{\log_a(1-u)}$ is symmetric around $u = 1/2$. (More precisely $f(u) = f(1-u)$) Then if we show this function is convex in $0 < u <1$ we can conclude that $u = 1/2$ is the minimizer. – MBP Jan 15 '13 at 14:13
Oh, neat! You should post that as the answer. P.S. I notice that the symmetry can be made explicit by writing it as $f(u) = a^{\log_a(u)\log_a(1-u)}$. – Rahul Jan 15 '13 at 14:17
up vote 2 down vote accepted

The derivative is not nice, but having a clue to the solution helps tame it.

$$ \frac{d}{dx} \log{f(x)} = \log{(1-a^x)} - \frac{x a^x}{1-a^x} \log{a} $$

Set the above to zero (because $\log$ is monotonic), and let $x = \log_a{y}$. A little manipulation shows that the equation needed to find the critical point is

$$y \log{y} = (1-y) \log{(1-y)} $$

The equation holds when $y = 1-y$, or $y=1/2$, as pointed out above. You can compute the second derivative of the above to show that this is indeed a minimum at this point.

share|cite|improve this answer

Differentiating, you get $$f'=\left(1-a^x\right)^x \left(\frac{a^x x \log(a)}{a^x-1}+\log(1-a^x)\right)=0$$ Since $a\ne1$, you also have $a^x\ne1$ and you can forget about the prefactor. You're left with $$\frac{a^x \log(a^x)}{a^x-1}+\log(1-a^x)=0\ ,$$ which is better written as $$a^x \log(a^x)-(1-a^x)\log(1-a^x)=0\ .$$ Now define $y\equiv a^x$ and write that as $$y \log(y)=(1-y)\log(1-y)\ .$$ Defining $g(y)=y\log(y)$, you get that you actually need to find a $y$ that satisfies $g(y)=g(1-y)$ and this obviously happens for $y=\frac{1}{2}$, or $$x=\log_a\left(y\right)=\log_a\left(\frac{1}{2}\right)$$. You can convince yourself that the only other solutions to $g(y)=g(1-y)$ are $y=0,1$.

share|cite|improve this answer

One other possible solution (following the suggestions of Rahul) is the following:

Take $u = 1 - a^x$, then $f(u) = u^{\log_a (1-u)} = a^{\log_a(u) \log_a(1-u)}$.

Now take a look at $\phi(u) = \log_a(f(u)) = \log_a(u) \log_a(1-u)$. Since $\log_a(\cdot)$ is convex when $a < 1$, $\phi(u)$ is also convex for $u \in [0,1]$. Furthermore $\phi(u)$ is symmetric around $u = 1/2$, i.e., $\phi(u) = \phi(1-u)$. Then its minimum must be at $u = 1/2$.

This shows that the minimizer of $f(x)$ is indeed $x^* = \log_a(1/2)$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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