I need to solve this equation for x:

$$2000 \sigma = 1 - \frac{20x}{\pi^2x^2 + 100} - \frac{2 \arctan(\frac{\pi x}{10})}{\pi} $$

$\sigma$ is a known value. I need to solve this for $ \sigma = 10^{-5}$, $\sigma = 10^{-10}$ and $\sigma = 10^{-15}$

As $x$ is inside the $\arctan$ and squared I cannot get a solution.

Can anybody help me?

Thank you!

  • $\begingroup$ what is $\sigma$? $\endgroup$ – Dr. Sonnhard Graubner Oct 16 '14 at 15:10
  • 1
    $\begingroup$ You cannot get any analytical solution for this mixture of functions. Only numerical methods would do it. If you want, I can elaborate; just post (and give me a value for $\sigma$). $\endgroup$ – Claude Leibovici Oct 16 '14 at 15:11
  • $\begingroup$ $\sigma $ is a known value. I need to solve this for $ \sigma = 10^{-5}$, $\sigma = 10^{-10}$ and $\sigma = 10^{-15}$ $\endgroup$ – Maria Oct 16 '14 at 15:13
  • $\begingroup$ Yes, please, I would really appreciate if you can elaborate a little bit. Thanks! $\endgroup$ – Maria Oct 16 '14 at 15:15

i have found with Maple
$\sigma=10^{-5}$ and $x\approx 8,34475638$
$\sigma=10^{-10}$ and $x \approx 409,0338318$
$\sigma=10^{-15}$ and $x \approx 18986,12832$


Let us consider the equation for which we look for the root $$f(x)= 1 - \frac{20x}{\pi^2x^2 + 100} - \frac{2 \arctan(\frac{\pi x}{10})}{\pi}-2000 \sigma$$ A simple method is Newton which, starting from a "reasonable" guess $x_0$ will update it according to $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ Let me be very lazy and start, for $\sigma=10^{-5}$, at $x_0=1$. The successive iterates will be : $2.82332$, $4.43589$, $6.05230$, $7.43493$, $8.18384$, $8.33939$, $8.34473$, $8.34474$ which is the solution for six significant figures.

This was to show you how this works.

Now, let us look at what happens when $x$ is large. An expansion of the function is $$f(x)=-2000 \sigma +\frac{4000}{3 \pi ^4 x^3}+O\left(\left(\frac{1}{x}\right)^4\right)$$ which means that an estimate $$x_0=\sqrt[3]{\frac{2}{3\pi^4 \sigma}}$$ is probably good. For the previous case with $\sigma=10^{-5}$, this would lead to $x_0=8.81258$ which is very good. Using the same estimation formula, we should have $x_0=409.044$ for $\sigma=10^{-10}$ and $x_0=18986.1$ for $\sigma=10^{-15}$. Please notice that these estimates are almost the solutions given by Dr. Sonnhard Graubner in his/her answer.

Now, since you have estimates and a simple way to generate them, apply the method.

Added later

A deeper look at this problem reveals that it would be better to solve $$g(x)=\log\Big( 1 - \frac{20x}{\pi^2x^2 + 100} - \frac{2 \arctan(\frac{\pi x}{10})}{\pi}\Big)-\log(2000 \sigma)$$ For $\sigma=10^{-5}$, starting at $x_0=1$ as before, the successive iterates of Newton method are $7.48118$, $8.30867$, $8.34467$, $8.34474$.

Just for curiosity, plot functions $f(x)$ and $g(x)$ for any value of $\sigma$.


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