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I have to find the intersection(s) of two functions $\frac{1}{\sqrt{\ln(x)}}$ and $5-x^2+\sqrt[3]{x}$ with the fixed point iteration method. I suppose that I have to find the roots of the equation f1-f2 = 0, but when I try to iterate this equation is repels and goes to inf or -inf. I also tried to rearrange both equations to x, but I can't see how can I rearrange the equation with the logarithmic function. Could you please point me in the right direction, how these equations should be solved and how should I write the fixed point function? Thank you!

Sorry if this is a very newbie question but I just had to get back to mathematics and I didn't use it for a long time.

B3ne
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3 Answers3

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Write the equation as $$x^2=\sqrt[3]{x}-\frac{1}{\sqrt{\log (x)}}+5$$ that is to say $$x_{n+1}=\sqrt{\sqrt[3]{x_n}-\frac{1}{\sqrt{\log (x_n)}}+5 } $$ For sure, because of the square root and the logarithm, you must start with $x_0>1$

Claude Leibovici
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i dont know about fixed point iteration. Didnt hear about it. But maybe you can just try to rewrite in terms of "x". i.e. $$\begin{cases} y=\frac{1}{\sqrt{lnx}}\\y=5-x^2+x^{1/3}\end {cases}$$ $\implies\begin{cases} x=e^{\frac{1}{y^2}}\\y=5-x^2+x^{1/3}\end {cases} \\ $ then try to plug x into second equation.

Mukhammad
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Start by sketching the two functions. It should be clear that the first (red) only gives real values when $x >0$, and that there are two intersections

enter image description here

You can certainly find the first of these by fixed point iteration: $f_1(x)= \frac{1}{\sqrt{\ln(x)}}$ has an inverse $g_1(y)=\exp\left(\frac{1}{y^2}\right)$ so if you try $x_{n+1} = g_1(f_2(x_n))$ iteratively then you will find you get convergence to about $1.042037$ from almost any starting point: for example starting with $x_0=2$ you get about $1.216284, 1.048651, 1.042242, 1.042044, 1.042037, \ldots$

That does not give you the second intersection as it is not easy to invert $f_2(x)=5-x^2+x^{1/3}$. So we may need a different approach such as Newton's method with $x_{n+1}=x_n - \dfrac{f_1(x_n)-f_2(x_n)}{f_1'(x_n)-f_2'(x_n)}$. It turns out this does converge on the second intersection of around $2.283985$ if you start at $1.337$ or greater: for example starting at $x_0=2$ you get about $2.315419, 2.284270, 2.283985, \ldots$

Henry
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  • Thanks, super helpful! Why is it converges to the second intersection when I use the equation Claude Leibovici provided below? Couldn't I somehow rearrange the first equation or I must stick to this inverse method? – B3ne Dec 18 '20 at 13:04
  • Edit.: I could solve for the first intersection using @Mukhammad answer. I rewrote the first equation like $x = e^{\tfrac{1}{(5-x^2+\sqrt[3]{x})^2}}$ – B3ne Dec 18 '20 at 13:21
  • @B3ne your second comment is precisely my $x_{n+1} = g_1(f_2(x_n))$ – Henry Dec 18 '20 at 13:56
  • Right, I see it now! Thanks for helping me out! I was confused a bit about these inverse functions, but now I've some memories about them! – B3ne Dec 18 '20 at 16:09