Which function have unique fixed point on the stated intervals?
balabalabala...
iii. h(x)=x^3 1 x in [1,2]
Why this isn't chose?
The answer says 'The function has unique fixed point when it maps [a,b] into [a,b] and f'(x)<=1, the function in iii doesn't satisfy the condition.'
Why it needs to satisfy this condition f'(x)<=1 ?
Is fixed points just let the f(x)=x? And in this equation, h(1)=1 and the other points in [1,2] doesn't satisfy h(x)=x.
So I think its right.........
Anyone ca help me with it?
Fixed point
Re: Fixed point
A fixed point of a function f is a value c for which f(c) = c.
Consider h(x) = x³  1, where x is in the interval [1,2].
If c is a fixed point, we must have h(c) = c, so c³  1 = c. Rearranging this, have c³  c  1 = 0. So let's say g(x) = x³  x  1; if we can guarantee that g has exactly one zero in [1,2], then h(x) has exactly one fixed point on [1,2].
First, notice that g(1) = 1 and g(2) = 5. Since g is continuous on [1,2], it must have at least one zero on that interval by the Intermediate Value Theorem, and so h must have at least one fixed point on that interval.
But is that fixed point unique? Let's look at g'(x), which is 3x²  1. Notice that g'(1) = 2, and g' is increasing for x > 0, so g'(x) is positive for all x in the interval [1,2]. This means that g is strictly increasing on the interval [1,2]. That's enough to guarantee that the zero of g, and therefore the fixed point of h, is unique.
( As a quick sanity check, WolframAlpha shows that the only real fixed point of h is about 1.3247. http://www.wolframalpha.com/input/?i=x%5E3++1+%3D+x )
Consider h(x) = x³  1, where x is in the interval [1,2].
If c is a fixed point, we must have h(c) = c, so c³  1 = c. Rearranging this, have c³  c  1 = 0. So let's say g(x) = x³  x  1; if we can guarantee that g has exactly one zero in [1,2], then h(x) has exactly one fixed point on [1,2].
First, notice that g(1) = 1 and g(2) = 5. Since g is continuous on [1,2], it must have at least one zero on that interval by the Intermediate Value Theorem, and so h must have at least one fixed point on that interval.
But is that fixed point unique? Let's look at g'(x), which is 3x²  1. Notice that g'(1) = 2, and g' is increasing for x > 0, so g'(x) is positive for all x in the interval [1,2]. This means that g is strictly increasing on the interval [1,2]. That's enough to guarantee that the zero of g, and therefore the fixed point of h, is unique.
( As a quick sanity check, WolframAlpha shows that the only real fixed point of h is about 1.3247. http://www.wolframalpha.com/input/?i=x%5E3++1+%3D+x )
Re: Fixed point
Uhh, you're right. Can you post the full problem?
f(x) = x^3  x  1 has a zero in [1, 2] by IVT. f(1) = 1, f(2) = 5. If it had 2 zeroes (meaning >= 2, I know a cubic can't have exactly 2 zeroes), call them a and b, then f'(x) = 0 somewhere in (a, b) by Rolle's Theorem. f'(x) = 3x^2  1 = 0, so x^2 = 1/3, so x = + 1/sqrt(3). But these aren't in [1, 2].
The f'(x) < 1 is not necessary for a fixed point, but for what's known as an attracting fixed point. (I think the proof of this uses MVT.) This means there's an interval around the fixed point so if you pick a point from that interval and start iterating the hell out of the function (meaning look at f(x), f(f(x)), f(f(f(x))), etc.), you stay in the interval and in fact approach the fixed point.
Here's some more information: http://www.ms.unimelb.edu.au/~andrewr/6 ... fp_sum.pdf
And here's a picture so you get the idea of the iteration tending to the fixed point: https://upload.wikimedia.org/wikipedia/ ... _point.svg
As for uniqueness, look at the proof here: https://www.math.ucdavis.edu/~thomases/ ... ce_sol.pdf (problem 1). But your function isn't on [1, 2] > [1, 2]. 1 > 0, and 2 > 7. And the derivative, 3x^2, is strictly positive on [1, 2], so the range is [0, 7].
f(x) = x^3  x  1 has a zero in [1, 2] by IVT. f(1) = 1, f(2) = 5. If it had 2 zeroes (meaning >= 2, I know a cubic can't have exactly 2 zeroes), call them a and b, then f'(x) = 0 somewhere in (a, b) by Rolle's Theorem. f'(x) = 3x^2  1 = 0, so x^2 = 1/3, so x = + 1/sqrt(3). But these aren't in [1, 2].
The f'(x) < 1 is not necessary for a fixed point, but for what's known as an attracting fixed point. (I think the proof of this uses MVT.) This means there's an interval around the fixed point so if you pick a point from that interval and start iterating the hell out of the function (meaning look at f(x), f(f(x)), f(f(f(x))), etc.), you stay in the interval and in fact approach the fixed point.
Here's some more information: http://www.ms.unimelb.edu.au/~andrewr/6 ... fp_sum.pdf
And here's a picture so you get the idea of the iteration tending to the fixed point: https://upload.wikimedia.org/wikipedia/ ... _point.svg
As for uniqueness, look at the proof here: https://www.math.ucdavis.edu/~thomases/ ... ce_sol.pdf (problem 1). But your function isn't on [1, 2] > [1, 2]. 1 > 0, and 2 > 7. And the derivative, 3x^2, is strictly positive on [1, 2], so the range is [0, 7].
Re: Fixed point
I would proceed much the same way as DMAshura and padic:
(1) By definition, the function h(x)=x^31 has a fixed point when x^31=x. This is equivalent to finding the zeroes of the polynomial g(x)=x^3x1.
(2) Since g(1)=1 and g(2)=5, by the IVT g(x) has at least one zero (fixed point of h(x)) in the interval [1,2].
(3) By Descartes' rule of signs, g(x) has only one sign change therefore it has only one positive root. This guarantees uniqueness of the fixed point.
(1) By definition, the function h(x)=x^31 has a fixed point when x^31=x. This is equivalent to finding the zeroes of the polynomial g(x)=x^3x1.
(2) Since g(1)=1 and g(2)=5, by the IVT g(x) has at least one zero (fixed point of h(x)) in the interval [1,2].
(3) By Descartes' rule of signs, g(x) has only one sign change therefore it has only one positive root. This guarantees uniqueness of the fixed point.

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 Joined: Wed Sep 09, 2015 9:42 pm
Re: Fixed point
Thanks for all your replies. I agree that the answer is wrong.
Thanks every one.
Thanks every one.