Arctan convergence rate

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  • Thread starter jostpuur
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  • #1
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What do you think about the claim that

[tex]
\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
[/tex]

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.
 
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Answers and Replies

  • #2
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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
 
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  • #3
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I see, thanks. If [itex]f[/itex] reached negative values, then [itex]f'[/itex] should have a third zero somewhere.

You made a typo with the quantity [itex]\big(\frac{\pi}{2}x\big)^2[/itex] in an intermediate step.
 

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