How to solve $x^2 + \ln(x) = 0$ I was just investigating $y = f(x) = e^{-x^2}$ and then went ahead to plot $x=f(y), y=-f(x), and x=-f(y)$, and what I got was interest rounded square shape, and I think we can calculate this area using integration.
However to get the bounds I must solve the equation $x^2 + \ln(x) = 0$. Is there a way to solve this equation without using the graph?
Shape
Not really. Just as you can't solve $2^x=3$ using algebra (you need to create a new kind of function; logarithms), you can't solve tihs one without creating a new kind of special function. The one we have for this is called the Lambert W function, the inverse function of $xe^x$. Then we can solve
$$ x^2+\ln x=0 $$
$$ x^2+\frac{1}{2}\ln x^2=0 $$
$$ 2x^2+\ln x^2=0 $$
$$ 2x^2+\ln(2x^2)=\ln(2) $$
$$ 2x^2e^{2x^2}=2 $$
$$ 2x^2=W(2) $$
$$ x=\sqrt{\frac{W(2)}{2}}. $$
Let $x=e^{-u/2}$, which turns the equation to solve into $u=2e^{-u}$. Iterating the function
$$f(u)={u+2e^{-u}\over2}$$
starting with, say $u=1$, leads to $u\approx0.8526055020$, from which $x\approx0.6529186404$ follows.
To be explicit about the iteration (and to indicate how quickly/slowly things converge),
$$\begin{align}
f(1)&amp;\approx0.8678794412\\
f(0.8678794412)&amp;\approx0.8537806239\\
f(0.8537806239)&amp;\approx0.8526923995\\
f(0.8526923995)&amp;\approx0.8526119077\\
f(0.8526119077)&amp;\approx0.8526059741\\
f(0.8526059741)&amp;\approx0.8526055368\\
f(0.8526055368)&amp;\approx0.8526055046\\
f(0.8526055046)&amp;\approx0.8526055022\\
f(0.8526055022)&amp;\approx0.8526055020\\
f(0.8526055020)&amp;\approx0.8526055020
\end{align}$$
This approach seems to pick up about one decimal point per iteration. It'd be nice to see a swifter approach.