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

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Discussion Overview

The discussion focuses on proving that the function F(x) = 4x(1-x) maps the interval [0,1] into itself and is not a contraction. Participants explore methods to demonstrate these properties, including calculus techniques and graphical analysis.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant suggests that to prove F is not a contraction, it suffices to show that there exists a number in [0,1] where the first derivative exceeds 1, calculating F'(x) = 4 - 8x.
  • Another participant agrees with the non-contraction proof and advises finding the range of F on [0,1] using calculus techniques to establish that it maps into itself.
  • A different approach is proposed, suggesting that sketching the graph of F(x) can illustrate that F is not a contraction by comparing F(1/2) - F(0) to 1/2 - 0.
  • One participant identifies critical points, noting that F is increasing on [0,1/2] and decreasing on (1/2,1), with an absolute maximum at 1/2 and local minima at 0 and 1, concluding that F maps [0,1] onto [0,1].
  • Another participant suggests invoking the Intermediate Value Theorem (IVT) to explain why a continuous function must attain all values in the interval.
  • Participants discuss solving the equation 4x - 4x^2 = c for c in [0,1], leading to a quadratic formula solution for x.

Areas of Agreement / Disagreement

Participants generally agree on the approach to proving that F(x) maps [0,1] into itself, but there are multiple methods proposed for demonstrating that F is not a contraction, indicating some disagreement on the best approach.

Contextual Notes

Some participants reference calculus techniques and the Intermediate Value Theorem without fully resolving the implications of these methods or their assumptions.

Amer
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Prove that the function
F(x) = 4x(1-x) 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

F'(x) = 4(1-x) - 4x = 4 - 8x

4-8x > 1 \Rightarrow \frac{3}{8} > x

choose x = 2/8. is this right

how to prove that F(x) maps [0,1] into itself ?
 
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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?
 
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.
 
ok the derivative is 4-8x
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]
 
Looks good. Might want to invoke the IVT explicitly to explain why a continuous function has to hit all the values in the interval.
 
let c \in [0,1]

4x - 4x^2 = c \Rightarrow 4x^2 - 4x + c = 0

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

4x - 4x^2 = c \Rightarrow 4x^2 - 4x + c = 0

x = \frac{4 \mp \sqrt{16 - 16c}}{8}

That works, too.
 

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