Prove F(x) Maps [0,1] into Itself and Not Contraction

In summary, to prove that the function F(x) = 4x(1-x) maps [0,1] into itself, we need to find the range of F on the interval and show that it includes all values in [0,1]. A simpler way to prove that F(x) is not a contraction is by finding a number in [0,1] such that the first derivative exceeds 1. This can be done by finding the critical points and using the IVT to show that the function is continuous and maps [0,1] onto [0,1]. Another way to prove it is by setting c as a value in [0,1] and solving for x using the quadratic formula, which will also
  • #1
Amer
259
0
Prove that the function
[tex]F(x) = 4x(1-x) [/tex] maps [0,1] into itself and it not contraction

to prove it is not contraction it is enough to prove that there exist a number in [0,1] such that the first derivative exceed 1

[tex]F'(x) = 4(1-x) - 4x = 4 - 8x [/tex]

[tex]4-8x > 1 \Rightarrow \frac{3}{8} > x [/tex]

choose x = 2/8. is this right

how to prove that F(x) maps [0,1] into itself ?
 
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  • #2
Your proof of non-contraction seems fine. As for proving that F maps [0,1] into itself, you need to find the range of F on the interval. Find the max and min of the function using good ol' Calc I techniques. What do you get?
 
  • #3
A simpler way to prove that F(x) is not a contraction is to sketch the graph of F(x) and note that F(1/2) - F(0) = 1 > 1/2 - 0.
 
  • #4
ok the derivative is [tex]4-8x [/tex]
the critical points we have one 1/2, the bound of the interval increasing in [0,1/2] and decreasing in (1/2,1)
we have absolute maximum at 1/2 f(1/2) = 1, and we have local minimum at 1,0 which is zero
and since the function is continuous then it is maps [0,1] onto [0,1]
 
  • #5
Looks good. Might want to invoke the IVT explicitly to explain why a continuous function has to hit all the values in the interval.
 
  • #6
let [tex] c \in [0,1] [/tex]

[tex]4x - 4x^2 = c \Rightarrow 4x^2 - 4x + c = 0 [/tex]

[tex]x = \frac{4 \mp \sqrt{16 - 16c}}{8} [/tex]
 
  • #7
Amer said:
let [tex] c \in [0,1] [/tex]

[tex]4x - 4x^2 = c \Rightarrow 4x^2 - 4x + c = 0 [/tex]

[tex]x = \frac{4 \mp \sqrt{16 - 16c}}{8} [/tex]

That works, too.
 

Related to Prove F(x) Maps [0,1] into Itself and Not Contraction

1. How do you prove that F(x) maps [0,1] into itself?

To prove that F(x) maps [0,1] into itself, we need to show that for any value of x in the interval [0,1], the output of F(x) is also within the interval [0,1]. This can be done by either showing that F(x) is always greater than or equal to 0 and less than or equal to 1, or by using the intermediate value theorem.

2. What is the importance of proving that F(x) maps [0,1] into itself?

Proving that F(x) maps [0,1] into itself is important because it ensures that the function is well-defined and does not produce any values outside of the specified interval. This is crucial in many mathematical proofs and applications, as it helps to establish the validity and reliability of the function.

3. How is the concept of contraction related to F(x) mapping [0,1] into itself?

A function is said to be a contraction if it reduces the distance between any two points in the domain. In the case of F(x) mapping [0,1] into itself, if the function is a contraction, it means that it will always produce outputs that are closer to each other than the corresponding inputs. This property is important in proving the existence of fixed points in a function.

4. What are some common methods for proving that F(x) is not a contraction?

There are several methods for proving that F(x) is not a contraction. One approach is to show that the function does not satisfy the definition of a contraction, either by producing outputs that are farther apart than the corresponding inputs, or by not satisfying the Lipschitz condition. Another method is to find counterexamples, where the function fails to be a contraction for certain inputs.

5. Can F(x) map [0,1] into itself and still be a contraction?

Yes, it is possible for F(x) to be a contraction and still map [0,1] into itself. This can occur if the function satisfies the definition of a contraction, but also has a fixed point within the interval [0,1]. In this case, the function will still reduce the distance between any two points in the domain, but it will also have a point where the output is equal to the input, making it a self-mapping function.

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