I would like to solve the following equation: $$x\cdot e^{(x-C)^2}=1.$$ I tried to use Lambert W function but unsuccessfully. Is there any way to get an exact (or approximate for some non-zero $C$) solution to the equation above?
My attempt was to reduce it to the form of $\tau \cdot e^\tau = B$ which can be solved as $\tau = W(B)$. Yet the $-C$ term in the exponent does not allow me to do this and the trick works only for $C=0$.
Best Answer
As noted by OP, this can be solved for $C=0$ by squaring both sides and multiplying by $2$, yielding $$ 2x^2 e^{2x^2}=2, $$ or $2x^2=W(2)$, or $x=\sqrt{W(2)/2}\approx 0.652919$. For $C=1$, of course, the solution is $x=1$. More generally, note that the solution is a fixed point of $$ f(x) = \exp\left(-(x-C)^2\right); $$ since this function is a contraction of $\mathbb{R}$ onto $(0,1]$, it has a unique fixed point, and iterating $x\rightarrow f(x)$ (starting at $0$, say) will rapidly converge to the solution for any $C$. In particular, the initial iterate of $e^{-C^2}$ is an excellent approximation for even moderately large $|C|$... it differs from the exact solution by less than $10\%$ for $C=\pm 2$, by about $1\%$ for $C=\pm 2.5$, and by less than $0.1\%$ for $|C|\ge 3$.
Expanding in powers of $x$, we have $$ x e^{(x-C)^2}=xe^{C^2}e^{-2Cx + x^2}=1, $$ or $$ xe^{C^2}=e^{2Cx-x^2}=1+(2Cx-x^2)+\frac{1}{2!}(2Cx-x^2)^2+\frac{1}{3!}(2Cx-x^2)^3+\ldots. $$ This has a self-consistent asymptotic solution of $x \sim e^{-C^2}+\sum_{k=2}^{\infty}b_k(C)e^{-kC^2}$, where the coefficients are polynomials in $C$ and can be found by substitution and equating like terms.