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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.

Thanks,

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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

1 Answer 1

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 $0<q<1$.

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@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

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