Prove that a function does not have a fixed point

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Discussion Overview

The discussion revolves around proving that the function f(x) = 2 + x - tan-1(x) does not have a fixed point. Participants explore the implications of the derivative condition |f'(x)| < 1 and the potential existence of fixed points, particularly focusing on the equation y = 2 + y - tan-1(y).

Discussion Character

  • Debate/contested

Main Points Raised

  • One participant claims to have found a fixed point at y = tan(2), suggesting the original question may be incorrect.
  • Another participant hints that tan-1(tan(2)) does not equal 2, implying a need for further clarification on the fixed point concept.
  • A later reply explains that due to the periodic nature of the tangent function, there are infinitely many x values for each y, but the arctangent function returns principal values within a specific interval, leading to tan-1(tan(2)) = 2 - π.

Areas of Agreement / Disagreement

Participants express disagreement regarding the existence of a fixed point, with some asserting that the function does have one while others challenge this claim based on the properties of the arctangent function.

Contextual Notes

Participants note the importance of understanding the principal values of the arctangent function and the implications of periodicity in the tangent function, which may affect the determination of fixed points.

Who May Find This Useful

This discussion may be useful for those studying fixed point theorems, properties of trigonometric functions, and their inverses, particularly in the context of calculus and mathematical analysis.

Amer
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it is a question in my book said

Prove that the function f(x) = 2 + x - \tan ^{-1} x has the property \mid f&#039;(x)\mid &lt; 1
Prove that f dose not have a fixed point

but i found that this function has a fixed point

y = 2 + y - \tan ^{-1} y

y = \tan 2

is it right that the question is wrong
 
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Re: Prove that a function dose not have a fixed point

Hint: $\tan^{-1}(\tan 2)\ne 2$.
 
Re: Prove that a function dose not have a fixed point

Evgeny.Makarov said:
Hint: $\tan^{-1}(\tan 2)\ne 2$.

can you explain more please
 
Re: Prove that a function dose not have a fixed point

arctan.png


Since tan(x) is periodic, for each y (such as y = tan(2)) there exists an infinite number of x such that tan(x) = y. Therefore, a convention is needed to select a single value for $\tan^{-1}(y)$. By definition, arctangent returns values in the interval $(-\pi/2,\pi/2)$, called principal values. Therefore, $\tan^{-1}(\tan(2))=2-\pi$.
 

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