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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?

intersections

gt6989b
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1 Answers1

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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 $x\approx0.450763652$

Rhys Hughes
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  • but how can we get to $x≈0.45$ by this function?: $x_{n+1}=\sqrt {ln(x_n)+1}$ –  Jun 25 '18 at 20:25
  • You can't get that result from that iteration. Instead you have to use the other iteration. Usually each iteration rearrangement of an equation only yields one solution, if any, See my own question here to understand it better. – Rhys Hughes Jun 25 '18 at 20:49
  • @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