# How can I solve for $x\,$? $:\;\;x-x_r=(x-x_1)e^{\large -(x-x_1)^2}$

I want to find $x$ for given values of $x_r$ and $x_1\,$ (domain $\mathbb{R}$):

$$x-x_r=(x-x_1)e^{\large -(x-x_1)^2}$$

Thanks

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I think you will have to content yourself with a numerical solution. –  Eckhard Jan 13 '13 at 16:06

You can rewrite the equation as ${{x-x_r}\over{x-x_1}}=e^{-(x-x_1)^2}$ and graph each side. The left is a hyperbola with a vertical asymptote at $x=x_1$ and a horizontal one at $y=1$. The right side is a bell shaped curve with max at $(x_1,1)$. Now decide based on $x_r>x_1$ or $x_r<x_1$ to see where the unique root is going to be. Next you my try to get some asymptotics of the answer. It all depends.
You will have an easier time if you change your variables. Let $x=x_r+y$ and $a=x_r-x_1$ then your equation becomes $y=(y+a)e^{-(y+a)^2}$. Now you have just one parameter to worry about namely $a$. –  Maesumi Jan 14 '13 at 5:33
$y$ becomes an odd function of $a$ meaning if you change sign of $a$ then sign of $y$ will also change. So you can concentrate on positive $a$'s. Many software or online pages/apps solve this type of equation for you. In Maple you can use: for i from 0 by .1 to 4 do z := fsolve(x-(x+i)*exp(-(x+i)^2) = 0, x); print(i, z) end do; to get 0.1, 0.3821323792; 0.2, 0.4225239009; 0.3, 0.4284936468; 0.4, 0.4188046129; 0.5, 0.4002917456; 0.6, 0.3763617819; 0.7, 0.3490298472; 0.8, 0.3196340536; and so on. –  Maesumi Jan 14 '13 at 6:13
Yes that expansion does provide useful answer. For the first attempt I get $y\approx a^{1/3}$ which seems to agree with numerical answers for very small values of $a$. For example for $a=0.004$, $y=.1554124270$, while $a^{1/3}=.158$. –  Maesumi Jan 14 '13 at 13:29
For large values of $a$ you can get $y\approx a e^{-a^2}$. –  Maesumi Jan 14 '13 at 13:38