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I was looking for the root of the following function on the domain $x\geq 0$:

$$F(x)=(x+a)e^{x^2}(1−erf(x))−\frac{b}{\sqrt π}$$

where

$$erf(x)=\frac{2}{\sqrt \pi}\int_0^x e^{-t^2}dt$$

is the familiar error function. Also $a>0$, $0<b<1$.

I tried several numerical solutions for different values of $a$ and $b$. It seems that there is at most one root on $[0,\infty)$. However I am not able to prove it since $F(x)$ is not monotone in $x$. Is there any hints about the proof?

What I know about this function is

$$F(0)=a-\frac{b}{\sqrt\pi},F(\infty)=\frac{1-b}{\sqrt\pi}>0.$$

Thank you!

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Since the standard asymptotic expansion as $x \to \infty$ is $1-erf(x) =\dfrac{e^{-x^2}}{x\sqrt{\pi}}(1-\dfrac1{2x^2}+O(\dfrac1{x^4})) $ (see https://en.wikipedia.org/wiki/Error_function#Asymptotic_expansion ),

$\begin{array}\\ F(x) &=(x+a)e^{x^2}(1−erf(x))−\frac{b}{\sqrt π}\\ &=(x+a)e^{x^2}(\dfrac{e^{-x^2}}{x\sqrt{\pi}}(1-\dfrac1{2x^2}+O(\dfrac1{x^4})))−\frac{b}{\sqrt π}\\ &=\dfrac{(x+a)}{x\sqrt{\pi}}((1-\dfrac1{2x^2}+O(\dfrac1{x^4})))−\frac{b}{\sqrt π}\\ &=\dfrac{1+a/x}{\sqrt{\pi}}((1-\dfrac1{2x^2}+O(\dfrac1{x^4})))−\frac{b}{\sqrt π}\\ &=\dfrac{1-b}{\sqrt{\pi}}+\dfrac{a}{x\sqrt{\pi}}-\dfrac{1}{2\sqrt{\pi} x^2}+O(\dfrac1{x^3})\\ \end{array} $

For a root, approximately $0 =(1-b)+a/x $ or $x = -a/(1-b) $.

This seems to show that there is no root for large $x$.

If another term is taken, $0 =\dfrac{1-b}{\sqrt{\pi}}+\dfrac{a}{x\sqrt{\pi}}-\dfrac{1}{2\sqrt{\pi} x^2} $ or $0 =x^2(1-b)+ax-\frac12 $.

The root of this is $x =\dfrac{-a+\sqrt{a^2+2(1-b)}}{2(1-b)} $.

You might try the power series expansions for small $x$ of $erf(x) = \dfrac{2}{\sqrt{\pi}}(x-\dfrac{x^3}{3}+O(x^5)) $ and $e^{-x^2} =1-x^2+O(x^4) $.

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  • $\begingroup$ I think you are missing a $a/x$ for the last two term in the last equality but still THANK YOU!! This is very helpful insight for large and small x. This will reduce to intermediate x then. $\endgroup$ – Jackie Lu Nov 8 '17 at 21:36
  • $\begingroup$ If you like it, why not accept it? $\endgroup$ – marty cohen Nov 8 '17 at 21:37
  • $\begingroup$ Oh actually my last comment was wrong. You were right about the last equality. $\endgroup$ – Jackie Lu Nov 8 '17 at 22:59
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    $\begingroup$ Here is another answer of the same problem: mathoverflow.net/questions/285515/… $\endgroup$ – Jackie Lu Nov 10 '17 at 20:22
  • $\begingroup$ That's good also. $\endgroup$ – marty cohen Nov 10 '17 at 20:59

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