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I am looking at the following function on the domain $x\geq 0$:

$$F(x)=(x+a)e^{x^2}(1-\mathrm{erf}(x))-\frac{b}{\sqrt\pi},$$

where $a>0$, $0<b<1$ are parameters. From plotting this function for different values of $a$ and $b$ it seems that there is at most one root on $x\in[0,\infty)$. But how to prove it?

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    $\begingroup$ Looks a normal question for me, why downvote and close? $\endgroup$ – Fedor Petrov Nov 8 '17 at 9:11
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    $\begingroup$ So 5 people know how to solve it in under 5 minutes and nobody gave the slightest hint to the OP before voting to close. Isn't that just rude? $\endgroup$ – fedja Nov 8 '17 at 14:14
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Indeed, there is at most one root on $x\in[0,\infty)$ -- actually, for each real $a$. Consider the equivalent equation \begin{equation} r(x):=\frac{f(x)}{g(x)}=\frac{\sqrt\pi}b \tag{1} \end{equation} for real $x>z_a:=0\vee(-a)$, where $0<b<1$ and \begin{equation} f(x):=\frac{e^{-x^2}}{x+a},\quad g(x):=1-\text{erf}(x). \end{equation} Consider the "derivative ratio" \begin{equation} \rho(x):=\frac{f'(x)}{g'(x)}=\frac{\sqrt{\pi } \left(1+2 a x+2 x^2\right)}{2 (x+a)^2}. \end{equation} Then \begin{equation} \rho'(x)=\frac{\sqrt{\pi } \left(a^2+a x-1\right)}{(x+a)^3}, \end{equation} so that, for real $x>z_a$, we have $$\rho'(x)>0 \iff a>1\text{ or } \big(0 < a \le 1\ \&\ x > x_a:=(1 - a^2)/a\big).$$

Now we are ready to use l'Hospital-type rules for monotonicity.

Case 1: $a\le0$, so that $z_a=-a$. Then $\rho$ is (strictly) decreasing (on the entire interval $(z_a,\infty)$), whence, by Proposition 4.1 in the mentioned paper, $r$ is decreasing, and so, equation (1) has at most one root (in $(z_a,\infty)$). In fact, in this case there is exactly one root, since $r(z_a+)=\infty$ and $r(\infty-)=\sqrt\pi<\frac{\sqrt\pi}b$.

Case 2: $0<a\le1$, so that $z_a=0$. Then $\rho$ is decreasing on $(z_a,x_a)$ and increasing on $(x_a,\infty)$, whence, by Proposition 4.3 in the same paper, $r$ is decreasing on $(z_a,c_a)$ and increasing on $(c_a,\infty)$, for some real $c_a\ge z_a$. So, $r<r(\infty-)=\sqrt\pi<\frac{\sqrt\pi}b$ on $[c_a,\infty)$. Hence, equation (1) has at most one root in $(z_a,\infty)$, and that possible root must be in $(z_a,c_a)$. In fact, if $0<a\le b/\sqrt\pi$, then there is exactly one root in $[0,\infty)$, since $r(0)=\frac1a\ge\frac{\sqrt\pi}b$ and $r(c_a)<r(\infty-)=\sqrt\pi<\frac{\sqrt\pi}b$. Accordingly, if $b/\sqrt\pi<a\le1$, then there is no root.

Case 3: $a>1$. Then $\rho$ is increasing (on the entire interval $(z_a,\infty)$), whence, by the mentioned Proposition 4.1, $r$ is increasing. So, $r<r(\infty-)=\sqrt\pi<\frac{\sqrt\pi}b$. Thus, equation (1) has no roots (in $(z_a,\infty)$) in this case.

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  • $\begingroup$ Wow Thank you!!! I didn't finish reading yet but a first glance at your paper already shows how powerful and relevant it is! Thank you for pointing out such a relevant reference! $\endgroup$ – Jackie Lu Nov 10 '17 at 1:45
  • $\begingroup$ I am glad this was of help. Thank you for the suggested edit concerning the mistake in the last sentence in the consideration of Case 2 in the previous version of this answer. $\endgroup$ – Iosif Pinelis Nov 10 '17 at 2:42
  • $\begingroup$ I have now introduced $z_a:=0\vee(-a)$, to simplify the writing a bit. $\endgroup$ – Iosif Pinelis Nov 10 '17 at 3:40

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