How to algebraically solve this equation?

Question

The original question is: How many solutions does a given equation have.

I understand that we can quickly draw graphs of both equations and see how many times they cross, but how would you solve this equation algebraically.

$\ln(x)=\frac{x^2}{2}-1$

Answer 1

Lambert W solution $$ \log(x) = \frac{x^2}{2}-1 \\ x = e^{(x^2/2)-1} \\ x^2 = e^{x^2-2} \\ x^2 e^{-x^2}=e^{-2} \\ -x^2 e^{-x^2} = -e^{-2} \\ -x^2 = W\left(-e^{-2}\right) \\ x = \left(-W(-e^{-2})\right)^{1/2} $$ Two real solutions from branches of $W$: $$ (-W_0(-e^{-2})\big)^{1/2} \approx 0.3982390482,\qquad (-W_{-1}(-e^{-2})\big)^{1/2} \approx 1.773751172 $$

Answer 2

Solving $\ln x=x^2/2-1$ is same as solving $\ln x^2-x^2+2=0$ which is same as solving $\ln u-u+2=0$ and using $x^2=u$ to solve for $x$. Using Newton's method we get the following iteration formula which if you apply a good number of times starting with $u_0=4$ once and $u_0=1.5$ the next time will land you fairly close to the square of the actual solutions.

$$u_{n+1}=u_n-\dfrac{\ln u_n-u_n+2}{1/u_n-1} \land u_0=\{1.5, 4\}$$

Can you carry out the calculations?

Answer 3

Consider that you look for the zero(s) of function $$f(x)=\log(x)-\frac{x^2}{2}+1$$ Then $$f'(x)=\frac{1}{x}-x \qquad \text{and} \qquad f''(x)=-\frac{1}{x^2}-1 <0 \,\, \forall x$$ The first derivative cancels at $x=\pm 1$ but $x >0$ because of the logarithm.

Close to $0$, $f(x) \to -\infty$ and the function increases. For $x=1$, $f(1)=\frac 12$ and it is a maximum. Since $x^2$ grows faster than $\log(x)$, at a point, $f(x)$ will become negative.

So, to summarize, there is one root in $(0,1)$ and another root $>1$.

To get a rough idea about the roots, build a Taylor expansion around $x=1$; this would give $$f(x)=\frac{1}{2}-(x-1)^2+O\left((x-1)^3\right)$$ Ignoring the higher order terms, this gives, as estimates, $x_1=\frac{2-\sqrt{2}}{2} $ and $x_2=\frac{2+\sqrt{2}}{2}$.

Now, you can start Newton method and get the following iterates $$\left( \begin{array}{cc} n & x_n \\ 0 & 0.292893 \\ 1 & 0.379664 \\ 2 & 0.397648 \\ 3 & 0.398238 \\ 4 & 0.398239 \end{array} \right)$$ $$\left( \begin{array}{cc} n & x_n \\ 0 & 1.70711 \\ 1 & 1.77639 \\ 2 & 1.77375 \end{array} \right)$$ wich are the solutions for six significant figures.

Answer 4

You can not solve this equation algebraically. But you can solve it with numerical methods, like the Newton method or maybe the Lambert W-function.