Fixed point

Forum for the GRE subject test in mathematics.
STATISTICS
Posts: 12
Joined: Wed Sep 09, 2015 9:42 pm

Fixed point

Postby STATISTICS » Wed Sep 09, 2015 9:52 pm

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?

User avatar
DMAshura
Posts: 36
Joined: Sat Jul 05, 2014 7:53 am

Re: Fixed point

Postby DMAshura » Wed Sep 09, 2015 10:33 pm

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 )

p-adic
Posts: 96
Joined: Fri Mar 27, 2015 6:42 pm

Re: Fixed point

Postby p-adic » Wed Sep 09, 2015 10:35 pm

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].

Ivanjam
Posts: 60
Joined: Tue Mar 17, 2015 2:29 am

Re: Fixed point

Postby Ivanjam » Thu Sep 10, 2015 6:18 am

I would proceed much the same way as DMAshura and p-adic:
(1) By definition, the function h(x)=x^3-1 has a fixed point when x^3-1=x. This is equivalent to finding the zeroes of the polynomial g(x)=x^3-x-1.
(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.

STATISTICS
Posts: 12
Joined: Wed Sep 09, 2015 9:42 pm

Re: Fixed point

Postby STATISTICS » Tue Sep 15, 2015 8:42 pm

Thanks for all your replies. I agree that the answer is wrong.
Thanks every one.




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