# Is the Arctan Convergence Rate Claim Valid for Positive Values of a, b, and x?

• I
• jostpuur
In summary, the conversation discusses the claim that for all positive values of a, b, and x, the expression \frac{x}{\frac{1}{a} + \frac{x}{b}} is less than \frac{2b}{\pi}\arctan\Big(\frac{\pi a}{2b}x\Big). The discussion includes attempts to prove or disprove this claim using derivatives and integrals, and a summary of the possible behavior of the function f(x) = \frac 2\pi \arctan\left(\frac \pi 2 x\right) - \frac x{1+x}.
jostpuur
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
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.

## 1. What is the definition of convergence rate?

The convergence rate refers to the speed at which a sequence or series approaches a limit or converges. It is a measure of how quickly the terms in the sequence or series approach the final value.

## 2. How is convergence rate related to Arctan?

The Arctan function, also known as the inverse tangent function, is used in many mathematical and scientific applications to find the angle or arc length of a right triangle. In terms of convergence rate, the Arctan function is important because it can be used to approximate the convergence rate of certain series or sequences.

## 3. What is the formula for calculating the convergence rate using Arctan?

The formula for calculating the convergence rate using Arctan is given by:

Convergence rate = 1 / (pi * Arctan(x))

Where x is the value at which the series or sequence converges.

## 4. How does the value of x affect the convergence rate using Arctan?

The value of x has a significant impact on the convergence rate using Arctan. As the value of x increases, the convergence rate decreases, meaning that the series or sequence converges more quickly. On the other hand, as the value of x decreases, the convergence rate increases, indicating a slower convergence.

## 5. What are some real-world applications of Arctan convergence rate?

Arctan convergence rate has many real-world applications in fields such as physics, engineering, and finance. It is used to analyze the rate of convergence of numerical methods, such as the Newton-Raphson method, in solving equations. It is also used in modeling and predicting stock market trends and in optimizing control systems in engineering.

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