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Dear MOs,

Here is a calculus problem which bored me for sometime. Let $a>0$ and $b<0$ be fixed.Define the following function (EDIT: Following the comment by Barry Cipra, you may only consider the case where $a=1$)

$$ W_{a,b} (x)= e^{-2 b x}\left( \Phi^2\left(\frac{a b -x}{\sqrt{a}}\right)-\Phi\left(\frac{2 a b -x}{\sqrt{a}}\right)\right), $$

where

$$ \Phi(x): = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^x e^{-y^2/2} d y = \frac{1}{2} \left(Erf\left(\frac{x}{\sqrt{2}}\right)+1\right). $$

The question is whether the following equation has only three zeors: $x=0$ and $x=\pm\infty$:

$$ W_{a,b}(x) = W_{a,b}(-x). $$

Plotting the function $W_{a,b}(x)$ suggests that the answer is right. But to find a proof seems quite hard.

Here are some graphs of the functions: $W_{1,-1}(x)$ (the blue one), $W_{1,-1}(-x)$ (the red one), and $W_{1,-1}(x)-W_{1,-1}(-x)$ (the one crossing the origin).

alt text


Here is the original problem. Define

$$ E_{a,b}(x) = e^{-b x}\Phi\left(\frac{ab-x}{\sqrt{a}}\right)+e^{b x}\Phi\left(\frac{ab+x}{\sqrt{a}}\right). $$

We wish to prove that for $a>0$, $b<0$,

$$ E_{a,b}^2(x)\ge E_{a,2b}(x),\quad\text{for all $x\in R$.} $$

If one define

$$ F_{a,b}(x) =E_{a,b}^2(x)- E_{a,2b}(x), $$

then

$$ \frac{d F_{a,b}(x)}{d x} = -b \left( W_{a,b}(x) - W_{a,b}(-x)\right). $$

Hence, this problem reduces to the above question.

Thank you very much for any suggestions!

Anand

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1  
It looks like the derivative of $W_{a, b}(x)-W_{a, b}(-x)$ isn't too nasty, can show that it only changes sign twice? –  Noah S Mar 26 '13 at 15:08
    
Dear Noah S. The derivatives of $W_{a,b}(x)-W_{a,b}(-x)$ is a bit nasty. To prove it changes sign twice is not that easy, there are some recursions. –  Anand Mar 26 '13 at 15:18
1  
This may be of no help in solving the problem, but writing $b=c/\sqrt{a}$ and $x=\sqrt{a}u$ certainly simplifies the look of the expression. –  Barry Cipra Mar 26 '13 at 16:22
    
Dear Barry Cipra, thanks for your comments. Please simply choose $a=1$ and a negative $b$. –  Anand Mar 26 '13 at 16:24
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