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Can you justify the existence of a $x_{*}$ solving $$\DeclareMathOperator{\li}{li}\DeclareMathOperator{\erf}{erf}\li(x_{*})=\erf(x_{*})?$$

Here $\li(x)$ is a special function, the so called logarithmic integral, I believe that the formula that you need for $\li(x)$ then is $$\li(x)=\int_\mu^x\frac{dt}{\log t}$$ where $\mu$ as you've read is Soldner's constant, and $$\erf(x)=\frac{2}{\sqrt{\pi}}\int_0^xe^{-t^2}dt$$ is the error function.

I would like to see a numerical method computing a good approximation of such $x_*$, since my purpose is learn and refresh mathematicas, I would like to know/rememer what method works to get an approximation of such problem. I don't need a good approximation since I've used an online tool to compute this as $1.9653631831359533312687921...$, but I would like to know an explanation of such method to compute a good approximation for solve an integral equation (I don't take the derivative since I don't know how handle the derivative of $\li(x)$, wich is defined as a P.V.).

Question. Can you justify the existence of a $x_{*}$ solving $$\li(x_{*})=\erf(x_{*}),$$ and give a explanation of a numerical method to compute this?

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    $\begingroup$ $\li(\mu)=0<\erf(\mu)$ and $\li\to \infty$ while $\erf\to 1$ so the solution must exist. Derivative is $\li'(x)=\frac 1{\log x}$. $\endgroup$ – A.S. Feb 20 '16 at 10:50
  • $\begingroup$ Very thanks much @A.S. also for your notes, the graph show it, but the graph that I did in my paper was confuse to me. $\endgroup$ – user243301 Feb 20 '16 at 10:52
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    $\begingroup$ A Cauchy principal value often just differentiates in the same way as a normal integral; you can go back to the original definition of it to see this. Actually, since in your definition, the lower limit is the Soldner-Ramanujan constant, it is not even a PV integral anymore. $\endgroup$ – J. M. ain't a mathematician Feb 20 '16 at 10:53
  • $\begingroup$ Thank you for clarifying it, also to A.S., I'll wait an answer to my question, but I much appreciate your early remarks @J.M. I would like learn it. $\endgroup$ – user243301 Feb 20 '16 at 10:57
  • $\begingroup$ You know that you can just use Newton-Raphson for this, right? $\endgroup$ – J. M. ain't a mathematician Feb 20 '16 at 11:02
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If you consider the function $$f(x)=\text{li}(x)-\text{erf}(x)$$ one easy way to find its zero is Newton method. Starting from a reasonable guess $x_0$, it will be updated accoreding to $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ $$x_{n+1}=x_n-\frac{\text{li}(x_n)-\text{erf}(x_n)}{\frac{1}{\log (x_n)}-\frac{2 e^{-x_n^2}}{\sqrt{\pi }}}$$ Starting using $x_0=2$, Newton method will then generate the following iterates $$x_1=1.9649503996039002033$$ $$x_2=1.9653631233745275939$$ $$x_3=1.9653631831359520789$$ $$x_4=1.9653631831359533313$$ which is the solution for twenty significant figures.

You could have a faster convergence using Halley method (it requires $f''(x)$). Starting again from $x_0=2$, the iterates will be $$x_1=1.9653642096145336397$$ $$x_2=1.9653631831359533313$$ which is the solution for twenty significant figures.

You could even have a faster convergence using higher order methods (Newton is quadratic, Halley is cubic). Let me give you below the first iterate $x_1^{(n)}$ for this methods $$\left( \begin{array}{cc} n & x_1^{(n)} \\ 2 & 1.9649503996039002033 \\ 3 & 1.9653642096145336397 \\ 4 & 1.9653631809125822565 \\ 5 & 1.9653631829431668086 \\ 6 & 1.9653631831480595920 \\ 7 & 1.9653631831361504849 \\ 8 & 1.9653631831359467245 \\ 9 & 1.9653631831359532274 \end{array} \right)$$

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  • $\begingroup$ Very thanks much, now I am starting to read it. $\endgroup$ – user243301 Feb 21 '16 at 8:38
  • $\begingroup$ Very thanks much, I believe that your answer and the remarks of *A.S., and J.M. give me the best answer, and I can remember in details how work this numerical methods. too thans to kamil09875 for the edit. Very thanks much. $\endgroup$ – user243301 Feb 21 '16 at 8:44
  • $\begingroup$ @user243301. You are very welcome ! $\endgroup$ – Claude Leibovici Feb 21 '16 at 9:44

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