Advanced Calculus: Proving x=cosx for x in (0,π/2)

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Homework Help Overview

The problem involves proving that there exists an x in the interval (0, π/2) such that x equals cos(x). This falls under the subject area of advanced calculus, specifically dealing with functions and their intersections.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss defining a function f(x) = x - cos(x) and the implications of its limits. There are questions about the validity of using the behavior of function values at the endpoints of the interval to draw conclusions about root existence.

Discussion Status

The discussion is ongoing, with participants exploring the application of the Intermediate Value Theorem and questioning the reasoning behind function behavior. Some guidance has been provided regarding the theorem's relevance, but no consensus has been reached on the approach to the problem.

Contextual Notes

Participants are considering the implications of function values at specific points and how these relate to the existence of roots, while also questioning the general applicability of their reasoning to other functions.

kimkibun
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Homework Statement



Show that x=cosx, for some xε(0,∏/2).

Homework Equations





The Attempt at a Solution



Define f(x)=x-cosx, i want to show that for some aε(0,∏/2), limx→af(x)=0. is this correct?
 
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kimkibun said:

Homework Statement



Show that x=cosx, for some xε(0,∏/2).

Homework Equations


The Attempt at a Solution



Define f(x)=x-cosx, i want to show that for some aε(0,∏/2), limx→af(x)=0. is this correct?

No, that won't help you.
If the y value at x=0 is negative, and the y value at x = pi/2 is positive (these values can be shown because it's easy to compute them), then what can we conclude from this?

Does this logic extend to every function? Think about y=1/x, at x=-1 we have y=-1, and at x=1 we have y=1, but the function doesn't cross the x-axis at all.
 
Mentallic said:
No, that won't help you.
If the y value at x=0 is negative, and the y value at x = pi/2 is positive (these values can be shown because it's easy to compute them), then what can we conclude from this?

Does this logic extend to every function? Think about y=1/x, at x=-1 we have y=-1, and at x=1 we have y=1, but the function doesn't cross the x-axis at all.


do you have a better explanation sir?
 
kimkibun said:
do you have a better explanation sir?

I suppose.

Take the function y=2x. How do we show it crosses the x-axis between x=-1 and x=1?

Well, what is the y value at x=-1? y=2(-1)=-2. So at x=-1, the function is below the x-axis.
What about at x=1? y=2(1)=2, which is above the x-axis. So since the function went from below the x-axis at x=-1 to above the x-axis at x=1, does this mean we can conclude that it must've crossed the x-axis somewhere in between? Yes!

Why? Well again, think about the function y=1/x and try using the same procedure I just showed you. Everything seems to be the same, except that this function doesn't cross the x-axis. What's different?
 
It is a direct application of a known theorem - I guess it was discussed during lecture or is mentioned in your book.

Mentallic tries to guide you to the intuitive understanding behind this theorem.
 
Borek said:
It is a direct application of a known theorem - I guess it was discussed during lecture or is mentioned in your book.

Mentallic tries to guide you to the intuitive understanding behind this theorem.

Right, it was silly of me not to mention the theorem involved in solving this problem.

kimkibun, the Intermediate Value Theorem is what you're looking for.
 

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