Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I will be to grateful if help me find a tight lower bound $g(x)$ over the following convex function: $$f(x) = \sqrt{1+4x^2} -1 + \log(\sqrt{1+4x^2}-1) - \log(2x^2) \geq g(x),$$ where $g(x)$ is preferably a polynomial with degree of at least $2$. The taylor expansion around the point ($x = 0$) of this function is given by: $$f(x) = x^2 - \frac{x^4}{2} + \frac{2x^6}{3} - \frac{5x^8}{4} + \frac{14x^{10}}{5} - \ldots$$

share|improve this question
    
@Farzad: First, note that you need $x > 0$. Also, what is wrong with having $g(x)$ be the Taylor expansion you mention? –  JavaMan May 31 '11 at 6:18
    
Using that $(\sqrt{1+4x^2}-1)(\sqrt{1+4x^2}+1)=4x^2$, we can simplify this as: $$\sqrt{1+4x^2} -1 + \log 2 - \log(\sqrt{1+4x^2}+1)$$ –  Thomas Andrews May 31 '11 at 6:41
1  
Do you really want the inequality to be true for all $x$, or just small $x$? Because the function is bounded above by $2x$, so it cannot be bounded below by any polynomial of degree greater than 1. –  Thomas Andrews May 31 '11 at 6:47
    
@ Thomas, thanks a lot for your great comment. –  Farzad May 31 '11 at 6:59
    
@ THomas, Can you help me for the case $\lvert x \rvert \leq 1$? –  Farzad May 31 '11 at 7:07
show 1 more comment

1 Answer 1

up vote 1 down vote accepted

For the case $|x|\leq 1$ I propose the following: Write $g(x):= a x^2- b x^4$ and choose $a$ and $b$ such that $f(1)=g(1)$, $f'(1)=g'(1)$. I did this using Mathematica and obtained $$a=2\Bigl({3\over 1+\sqrt{5}} +\log {2\over 1+\sqrt{5}}\Bigr)\ \qquad b= {2\over 1+\sqrt{5}} +\log{2\over 1+\sqrt{5}}\ .$$ Plotting the resulting graph of $f-g$ one gets the impression that $$0\leq f(x)-g(x)\leq 0.0117 \qquad(0\leq x\leq 1)\ .$$

share|improve this answer
add comment

Your Answer

 
discard

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.