# Lower bound over a convex function

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: $$f(x) = x^2 - \frac{x^4}{2} + \frac{2x^6}{3} - \frac{5x^8}{4} + \frac{14x^{10}}{5} - \ldots$$

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@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
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

## 1 Answer

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 $$0\leq f(x)-g(x)\leq 0.0117 \qquad(0\leq x\leq 1)\ .$$

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