# Arctan convergence rate

• I
What do you think about the claim that

$$\frac{x}{\frac{1}{a} + \frac{x}{b}} \;<\; \frac{2b}{\pi}\arctan\Big(\frac{\pi a}{2b}x\Big),\quad\quad\forall\; a,b,x>0$$

First I thought that if this is incorrect, then it would be a simple thing to find a numerical point that proves it, and also that if this is correct, then it would be a simple thing to prove this via some derivatives or integrals. For some reason I didn't succeed in either objective, and now I'm not sure why I feel confused about this.

I like Serena

I like Serena
Homework Helper
Hi jostpuur,

Since we can scale the graphs horizontally and vertically, without loss of generality, we can assume that a=1 and b=1.
Let:
$$f(x)=\frac 2\pi \arctan\left(\frac \pi 2 x\right) - \frac x{1+x}$$
Then:
$$f'(x)=\frac{1}{1+(\frac\pi 2 x)^2} - \frac 1{(1+x)^2} = \frac{x(8-(\pi^2-4)x)}{(4+\pi^2 x^2)(1+x)^2}$$
The derivative is zero at ##x=0## and ##x=\frac{8}{\pi^2-4}##.
And we have a horizontal asymptote at ##y=0## when ##x\to\infty##.
It follows that the graph ##y=f(x)## has a minimum at ##x=0##, a maximum at ##x=\frac{8}{\pi^2-4}##, and then descends to its asymptote ##y=0##.
QED

Last edited:
I see, thanks. If $f$ reached negative values, then $f'$ should have a third zero somewhere.
You made a typo with the quantity $\big(\frac{\pi}{2}x\big)^2$ in an intermediate step.