Intermediate value theorem problem

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1. How can you prove if F is continuous, then there exists a fixed point of F in [0,1]?



I know F:[0,1] ---> [0,1] an bijective, but what is f(c)=c mean?
 
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Have you tried drawing a picture? It might be enlightening.
 


I think it means more or less exactly what it says. There is a c in [0,1] such that F(c)=c. Use the intermediate value theorem. If F is bijective there is an a such that F(a)=0 and a b such that F(b)=1. What happens in between? Apply the IVT to F(x)-x on the interval [a,b].
 


Dick said:
I think it means more or less exactly what it says. There is a c in [0,1] such that F(c)=c. Use the intermediate value theorem. If F is bijective there is an a such that F(a)=0 and a b such that F(b)=1. What happens in between? Apply the IVT to F(x)-x on the interval [a,b].

Does that mean I need to pick a point like .5, which is between [0,1].

Also, would there be any point where F isn't continuous and a fixed point may not exist.

Thanks guys
 


I don't think you really understood what I wrote. As morphism said, draw a picture. Look up the IVT.
 


No, you can't "pick" a point. And you are told that the function is continuous. Why are you asking if it isn't?

If F is from [0, 1] what can you say about F(0)? What can you say about F(1)?

What can you say about 0- F(0) and 1- F(1)?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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