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I wanted to solve the following equation. $$x \times e^{x^2}=1$$

I heard about the $W$ Lambert function but what I can see on Wikipedia can only solve the equations of type $x \times e^{x}=\lambda$. How can I do?

Online solvers told me the solution was $x=e^{-\frac{W(2)}{2}}$ but I don't understand why...

xam4lor
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2 Answers2

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When attempting to solve equations using the Lambert $W$ function, one usually has to exploit the fact that different manipulations affect the "coefficient" in front of $e$ and the exponent above $e$ differently.

In this case, you have $x^2$ in the exponent, so you want that in the coefficient as well. We can try squaring our equation and see where that gets us: $$ x^2e^{2x^2} = 1 $$ OK. This squared the coefficient but doubled the exponent. So now we have $x^2$ both places. However, the exponent is $2x^2$ instead of $x^2$. So, we multiply our equaiton by $2$ to get $$ 2x^2e^{2x^2} = 2 $$ Now we can apply the Lambert $W$ function to get $$ 2x^2 = W(2)\\ x = \sqrt{W(2)/2} $$

Arthur
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    Or more precisely $x = \pm \sqrt{W(2)/2}$. But of course, if you're using the principal branch of $W$ the positive square root will give you a solution of $x \exp(x^2) = 1$ while the negative square root solve $-x \exp(x^2) = 1$. – Robert Israel Mar 11 '19 at 15:02
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Firstly, you want to use the fact that $x = e^{ln x }$, therefore the expression becomes $$x e^{(e^{lnx})^2} = 1, $$ combining it over a single exponential $$\Rightarrow e^{lnx + e^{2 ln x}} = 1.$$ Taking the natural logarithm, we then get $$ln x + e^{2 ln x} = 0 \Rightarrow e^{2 ln x} = -ln x .$$ Now we can rearrange the expression to get it into a form usable by the Lambert W functions by dividing through by $exp(2 lnx)$ and multiplying by $2$ $$\Rightarrow 2 = -2 ln x e^{-2 ln x}$$ $$ \Rightarrow W(2) = -2 ln x$$ $$\Rightarrow x = e^{-W(2)/2}$$

user1066
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