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Math Help - Find equation w/ 3 Fixed points

  1. #1
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    Find equation w/ 3 Fixed points

    Hello

    I am returning to school after a long break. One of my assignments is a mix of ODE and discrete math and is clogging my mind.

    I have been struggling on my 12 question homework for 4 hours now and have made it 1/4 of the way through. Luckily i have found this forum and hope someone(s) can help.

    The question is this:
    Find a continuous function:
    f : [0, 1] --> [0, 1] with exactly three fixed points and draw its graph.

    The fixed points is throwing me for a loop.

    Any suggestions??
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  2. #2
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by npthardcorebmore View Post
    Hello

    I am returning to school after a long break. One of my assignments is a mix of ODE and discrete math and is clogging my mind.

    I have been struggling on my 12 question homework for 4 hours now and have made it 1/4 of the way through. Luckily i have found this forum and hope someone(s) can help.

    The question is this:
    Find a continuous function:
    f : [0, 1] --> [0, 1] with exactly three fixed points and draw its graph.

    The fixed points is throwing me for a loop.

    Any suggestions??
    Start with definition: x is a fixed point of a function f if and only if f(x) = x.

    All you have to do is draw a continuous function in the region x \in [0,1], y \in [0,1] that intersects with the line y=x three times. If you start at the point (0,0) and end at the point (1,1) then you can just have it cross the line y=x once for x \in (0,1).

    You can make it piecewise...
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  3. #3
    Senior Member roninpro's Avatar
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    How about f(x)=x^3? We have f(-1)=-1, f(0)=0, f(1)=1. These are all of the fixed points since x^3=x has only three solutions, by the Fundamental Theorem of Algebra.
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  4. #4
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    Quote Originally Posted by roninpro View Post
    How about f(x)=x^3? We have f(-1)=-1, f(0)=0, f(1)=1. These are all of the fixed points since x^3=x has only three solutions, by the Fundamental Theorem of Algebra.
    But when restricted to the domain [0,1], f(x)=x^3 only has two fixed points.
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by undefined View Post
    But when restricted to the domain [0,1], f(x)=x^3 only has two fixed points.
    Scale, f\left(\frac{x+1}{2}\right)
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  6. #6
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by Drexel28 View Post
    Scale, f\left(\frac{x+1}{2}\right)
    Maybe I'm misinterpreting.. f(x) = x^3 so f\left(\frac{x+1}{2}\right)=\left(\frac{x+1}{2}\ri  ght)^3 which has only two fixed points for x \in [0,1] given by x \in \{-2+\sqrt{5},1\}.

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  7. #7
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by undefined View Post
    Maybe I'm misinterpreting.. f(x) = x^3 so f\left(\frac{x+1}{2}\right)=\left(\frac{x+1}{2}\ri  ght)^3 which has only two fixed points for x \in [0,1] given by x \in \{-2+\sqrt{5},1\}.

    Oh! fixed points! ....why don't you just construct a polygonal line that intersects the line y=x three times?
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  8. #8
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    Quote Originally Posted by Drexel28 View Post
    Oh! fixed points! ....why don't you just construct a polygonal line that intersects the line y=x three times?
    that would be a piecewise function correct?

    i was just hoping for something a little more "pretty" than that
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  9. #9
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by npthardcorebmore View Post
    that would be a piecewise function correct?

    i was just hoping for something a little more "pretty" than that
    My original thought, before the piecewise idea, was to transform the sine curve so that it's rotated 45 degrees and can be stretched or compressed (along the line y=x) to produce as many fixed points as desired (greater than or equal to 2)... but since I figured you might ask how this is done, and I'd have to think about that, I just went the easier route.
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  10. #10
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    would a piecewise function preserve continuity though?
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  11. #11
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    Yes. It would preserve continuity but not differentiability in the "transition" points (you can see this by comparing the one-sided limits..)
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