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

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SUMMARY

The discussion centers on proving that the equation x = cos(x) has a solution for x in the interval (0, π/2). Participants define the function f(x) = x - cos(x) and explore the behavior of this function at the endpoints of the interval. They conclude that since f(0) is negative and f(π/2) is positive, the Intermediate Value Theorem guarantees at least one solution exists in the interval. The theorem is crucial for understanding the continuity and behavior of functions in calculus.

PREREQUISITES
  • Understanding of the Intermediate Value Theorem
  • Basic knowledge of limits and continuity
  • Familiarity with trigonometric functions, specifically cosine
  • Ability to analyze function behavior over an interval
NEXT STEPS
  • Study the Intermediate Value Theorem in detail
  • Learn about the properties of continuous functions
  • Explore the application of limits in calculus
  • Investigate other methods for finding roots of equations, such as the Bisection Method
USEFUL FOR

Students studying calculus, particularly those focusing on function analysis and theorems related to continuity and roots of equations.

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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