Equation with tangent and powers 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!
 A: 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 $x \approx 18986,12832$
A: 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 $\sigma$.
