Let $a>0$ be a fixed parameter. I would like to find the (I think there are only two) $x\in \mathbb{R}$ such that $$(x-a)e^{-\frac{1}{2}(x-a)^2} = (x+a)e^{-\frac{1}{2}(x+a)^2}.$$
I know this might not be possible, or maybe it is possible in terms of exotic functions. If not, is there a general technique to study the locations of the zeros? I mean, if $x_1(a)$ and $x_2(a)$ are the two zeros and without loss of generality $x_1(a) \leq x_2(a)$ for all $a>0$ fixed. Then something like $$\underline{B}(a) \leq x_1(a) \leq x_2(a) \leq \overline{B}(a),$$ hopefully being $\underline{B}$ and $\overline{B}$ nicely close for this case :D
I do not even have an intuition of how the curves $x_1(a)$ and $x_2(a)$ look like and I am interested in the behaviour when $a\to 0$, so asymptotics would be nice too.
Any idea? :) thanks a lot!
You can discuss the curve $y=z^{\frac{1}{z}}$ with $z\ge 0$. Choose $z_1:=(1+\frac{1}{t})^t:=e^{(x\mp a)^2}$ and $z_2:=(1+\frac{1}{t})^{t+1}:=e^{(x\pm a)^2}$ with $a>0$. We get $1<e^{(x\mp a)^2}\le e\le e^{(x\pm a)^2}$ because the maximum of $z^{\frac{1}{z}}$ is $e^{\frac{1}{e}}$. If you reduce $z_1^{\frac{1}{z_1}}=z_2^{\frac{1}{z_2}}$ by logarithm and positive square root, you get your equation. Now you can discuss the values for $x$.