# fixed point theory for $\ln x=x^2-1$

if we choose $$X_{n+1} = \sqrt {1+\ln(x_n)} \quad \text{or} \quad X_{n+1} = \sqrt {1-\ln(x_n)}$$ for the fixed point theory, you draw the graphs of $\ln(x)$ and $x^2-1$ you'll see two intersection in range $[0,1]$ and if you choose $x_0=0.5$ as the start point of fixed point theory, you will get to the $x=1$ for the answer which is correct.

With any different $x_0$ you will get to the $x=1$ again but there's another answer for the equation and i don't know how i'm supposed to get to that. does anybody have an idea of what should i do?

You're missing the other way to iterate it. You have: $$\ln(x)=x^2-1\to x=\sqrt{\ln(x)+1}$$ which you transformed into the iterative $$x_{n+1}=\sqrt{\ln(x_n)+1}, x_0=\frac12$$ which we can show converges to $x=1$, and check this to confirm it is a solution.
However, we can also isolate that $x$ on the LHS, as such: $$\ln(x)=x^2-1\to x=e^{x^2-1}$$ which is transformed to the iterative $$x_{n+1}=e^{x_n^2-1}, x_0=\frac12$$ and converges to the other solution $x\approx0.450763652$
• but how can we get to $x≈0.45$ by this function?: $x_{n+1}=\sqrt {ln(x_n)+1}$ – zahra Niosha Afsharikia Jun 25 '18 at 20:25
• @zahraNioshaAfsharikia to see what I mean, try iterating $$x_{n+1}=\sqrt{\ln(x)+1}, x_0=0.450763652$$ You'll see that $x_{n+1}<x_n$ until a few iterations later when $x_n<\frac1e$ and you get a math error. – Rhys Hughes Jun 26 '18 at 23:04