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 came up with this problem, which I cannot solve myself.

Consider the function:
$\displaystyle f(x) = x^{\ln(|\pi \cos x ^ 2| + |\pi \tan x ^ 2|)}$, which has singularities at $\sqrt{\pi}\sqrt{n + \dfrac{1}{2}}$, with $n \in \mathbb{Z}$. Looking at its graph: Graph of f(x)

we can see it is globally increasing:

f(x) again

I was wondering if there exists a function $g(x)$, such that $f(x) - g(x) \ge 0, \forall x \in \mathbb{R^{+}}$ and that best fits the "lowest points" of $f(x)$.
Sorry for the inaccurate terminology but I really don't know how to express this concept mathematically. Here is, for example, $g(x) = x ^ {1.14}$ (in red): f(x) and g(x)

Actually $g(x)$ is not correct because for small values of $x$ it is greater than $f(x)$. g(x) at small values

Is it possible to find such a $g(x)$, given that the "nearest" is $g(x)$ to $f(x)$'s "lowest points" the better it is? Again, sorry for my terminology, I hope you could point me in the right direction.


share|cite|improve this question
Note that $f$ can also be written as $$f(x)=(\pi|\cos(x^2)| +\pi|\tan(x^2)|)^{\ln x}\ .$$ Now try to find a good "lower envelope" of the oscillating expression in parentheses. – Christian Blatter Jul 8 '12 at 20:12
I think the appropriate technical term is "greatest convex minorant", also known as "convex envelope". It can be defined as $g(x)=\sup_{m,b} (mx+b)$ where the supremum is taken over all pairs $(m,b)$ such that $f(x)\ge mx+b$ for all $x$. Of course, you'll want something more explicit for this particular problem. – user31373 Jul 8 '12 at 21:00
Following up on @ChristianBlatter suggestion: since the minimum of $|\cos t|+|\tan t|$ is equal to $1$ (attained when $\cos t=\pm 1$), it follows that $f(x)\ge \pi^{\ln x}=x^{\ln \pi}$. Note that $\ln \pi = 1.14472989\dots$ – user31373 Jul 8 '12 at 21:08
@ChristianBlatter: That's interesting, I didn't notice that. What do you mean by "lower envelope"? I noticed that $\pi|\cos(x^2)| +\pi|\tan(x^2)|$ should never go below $\pi$. – rubik Jul 9 '12 at 8:53
@LeonidKovalev: Following my previous comment, I should have $\pi ^{\ln x} = x ^{\ln \pi}$, but that function is greater than $f(x)$ over the interval $]0;1[$. – rubik Jul 9 '12 at 8:55
up vote 2 down vote accepted

As $a^{\ln b}=\exp(\ln a\cdot\ln b)=b^{\ln a}$ the function $f$ can be written in the following way: $$f(x)=\bigl(\pi|\cos(x^2)|+\pi|\tan(x^2)|\bigr)^{\ln x}\ .$$ Now the auxiliary function $$\phi:\quad{\mathbb R}\to[0,\infty],\qquad t\mapsto \pi(|\cos(t)|+|\tan(t)|)$$ is periodic with period $\pi$ and assumes its minimum $\pi$ at the points $t_n=n\pi$. The function $$\psi(x):=\phi(x^2)=\pi|\cos(x^2)|+\pi|\tan(x^2)|\bigr)$$ assumes the same values as $\phi$; in particular it is $\geq\pi$ for all $x\geq0$ and $=\pi$ at the points $x_n:=\sqrt{n\pi}$ $\ (n\geq0)$. Therefore $$f(x)=\bigl(\psi(x)\bigr)^{\ln x}\geq \pi^{\ln x}=x^{\ln\pi}\qquad(x\geq1)$$ and $=x^{\ln\pi}$ at the $x_n>1$. For $0<x<1$ the inequality is the other way around because $y\mapsto q^y$ is decreasing when $0<q<1$.

share|cite|improve this answer
@Rubik ... which means that for x<1 we should look for the maximum of cos+tan, attained at 1. So our minorant will be piecewise defined, with the first small piece being $x^q$ with $q=\ln \pi+\ln(\cos 1+\tan 1)$, about 1.8855. – user31373 Jul 9 '12 at 12:00
Thank you to both. I have only one question: why do we check for maximum for $x<1$? Is it because $q^y$ is decreasing there? – rubik Jul 9 '12 at 13:16
@LeonidKovalev: Is that right? – rubik Jul 10 '12 at 15:47
@rubik It's seen from the formula $f(x)=(\pi|\cos(x^2)| +\pi|\tan(x^2)|)^{\ln x}$, for which we seek a lower bound. When the exponent $\ln x$ is positive, we should estimate the base $\pi|\cos(x^2)| +\pi|\tan(x^2)|$ from below. But when the exponent $\ln x$ is negative, we estimate the base from above, since a larger number raised to a negative power gives a smaller result. – user31373 Jul 10 '12 at 15:53
@LeonidKovalev: Oh right, now it's all clear! Thank you so much! – rubik Jul 10 '12 at 16:56

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.