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Math Help - Finding a monotone function crossing three given points

  1. #1
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    Finding a monotone function crossing three given points

    Hi,
    I have been trying from quite a while to solve the following problem.
    Find a real differentiable monotone function f(x) in the interval [0, c], such that:
    f(0) = 0
    f(b) = 1/2
    f(c) = 1,
    where 0 < b < c.
    This is a particular case of more general problem:
    Find a real differentiable monotone function f(x) in the interval [a, c], such that:
    f(a) = A
    f(b) = B
    f(c) = C,
    where a < b < c and A < B < C.
    I tried several different ways but ended up with some bad looking equations. My method was to define f as a function of some type with three parameters and try find these parameters from the above equations.
    Example, f(x) = pe^{qx} + r, where p, q and r are real parameters.
    If applied to the more particular first case, I end up with the equation
    2e^{bq} - e^{cq} - 1 = 0.
    I would be very grateful if you know such a function and share it.
    Last edited by Sogartar; July 1st 2008 at 05:33 AM. Reason: Typo
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Sogartar View Post
    Hi,
    I have been trying from quite a while to solve the following problem.
    Find a real differential monotone function f(x) in the interval [0, c], such that:
    f(0) = 0
    f(b) = 1/2
    f(c) = 1,
    where 0 < b < c.
    This is a particular case of more general problem:
    Find a real monotone function f(x) in the interval [a, c], such that:
    f(a) = A
    f(b) = B
    f(c) = C,
    where a < b < c and A < B < C.
    I tried several different ways but ended up with some bad looking equations. My method was to define f as a function of some type with three parameters and try find these parameters from the above equations.
    Example, f(x) = p*Exp(qx) + r, where p, q and r are real parameters.
    If applied to the more particualr first case, I end up with the equation
    2*Exp(bq) - Exp(cq) - 1 = 0.
    I would be very grateful if know such a function and share it.
    Maybe I am misunderstanding you but

    Let f(x)=x

    We see that f(x) is monotonically increasing \forall{x}\in\mathbb{R}

    We also see that if we let b=\frac{1}{2} and c=1

    We have that

    [0,1]

    Where 0<\frac{1}{2}<1

    and we also have that

    f(0)=0

    and

    f\left(\frac{1}{2}\right)=\frac{1}{2}

    and

    f(1)=1
    Last edited by Mathstud28; June 30th 2008 at 05:25 PM. Reason: typo
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  3. #3
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    Quote Originally Posted by Mathstud28 View Post
    We also see that if we let b=\frac{1}{2} and c=1
    No, b and c are given. You can’t choose your own values for them.
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  4. #4
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    Solution

    Thanks for your consideration on my problem.
    Today while thinking on it, I decided to apply the old Roman tactic, divide and conquer. So my plan was to find two functions. The first in the interval [0, b], crossing (0,0) and (b,1/2). The second in the interval (b, c], crossing (b,1/2) and (c, 1). Also I needed those functions to have equal derivatives in point b.
    So, my first guess was two parabolas F(x)\in[0, b] and G(x)\in(b, c], where F'(b) = G'(b) = 0. This is a solution of the problem at hand, but the characteristics of this function were not what I was looking for. I needed changes in x around b to yell greater changes in f(x), than around the ends of the interval [0, c]. So I defined F(x) and G(x) such that, F^{-1}(x) and G^{-1}(x) to be parabolas with the above requirements.
    Another solution was to use ellipses. Taking the correct quarter of each ellipse where, the first one is with center (0, 1/2) and passes through (0,0) and (b,1/2). The second is with center (c,1/2) and passing through (b, 1/2) and (c, 1). This defines a nice monotone differentiable function in (0, c).
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  5. #5
    Lord of certain Rings
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    Quote Originally Posted by Sogartar View Post
    Hi,
    I have been trying from quite a while to solve the following problem.
    Find a real differentiable monotone function f(x) in the interval [0, c], such that:
    f(0) = 0
    f(b) = 1/2
    f(c) = 1,
    where 0 < b < c.
    EDIT: The function I found is not necessarily monotone, so sorry.

    Try defining such a function:

    Let f(x) = A(x-b)(x-c) + Bx(x-b)+Cx(x-c), where A, B, C are constants we will work out.

    f(0) = Abc = 0\Rightarrow A = 0
    f(b) = Cb(b-c) = \frac12 \Rightarrow C = \frac{1}{2b(b-c)}
    f(c) = Bc(c-b) = 1 \Rightarrow C = -\frac{1}{b(b-c)}

    I think this should be differentiable since its a polynomial function. All that remains is to restrict the function to the interval [0,c].
    Last edited by Isomorphism; July 1st 2008 at 09:28 AM. Reason: Thanks algebraictopology
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  6. #6
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    Quote Originally Posted by Sogartar View Post
    Hi,
    I have been trying from quite a while to solve the following problem.
    Find a real differentiable monotone function f(x) in the interval [0, c], such that:
    f(0) = 0
    f(b) = 1/2
    f(c) = 1,
    where 0 < b < c.
    This is a particular case of more general problem:
    Find a real differentiable monotone function f(x) in the interval [a, c], such that:
    f(a) = A
    f(b) = B
    f(c) = C,
    where a < b < c and A < B < C.
    I tried several different ways but ended up with some bad looking equations. My method was to define f as a function of some type with three parameters and try find these parameters from the above equations.
    Example, f(x) = pe^{qx} + r, where p, q and r are real parameters.
    If applied to the more particular first case, I end up with the equation
    2e^{bq} - e^{cq} - 1 = 0.
    I would be very grateful if you know such a function and share it.
    Try finding constants: c_1>0, c_2>0, c_3, such that:

    f(x)=c_1e^{c_2 x}+c_3

    that satisfy the conditions.

    RonL

    ronL
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  7. #7
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    Quote Originally Posted by Isomorphism View Post
    Try defining such a function:

    Let f(x) = A(x-b)(x-c) + Bx(x-b)+Cx(x-c), where A, B, C are constants we will work out.

    f(0) = Abc = 0\Rightarrow A = 0
    f(b) = Cb(b-c) = \frac12 \Rightarrow C = \frac{1}{2b(b-c)}
    f(c) = Bc(c-b) = 1 \Rightarrow C = -\frac{1}{b(b-c)}

    I think this should be differentiable since its a polynomial function. All that remains is to restrict the function to the interval [0,c].
    It may not be monotone though.

    I did think of CaptainBlack’s method, but I got some rather horrible-looking expressions for the coefficients and gave up.
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  8. #8
    Lord of certain Rings
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    Quote Originally Posted by algebraic topology View Post
    It may not be monotone though.
    Whoops! My bad... I did not notice that
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  9. #9
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    Well, here’s a modification of CaptainBlack’s suggestion.

    Join the first two points ( (a,A) and (b,B)) by a straight line. This will have gradient \frac{B-A}{b-a}. Then join (b,B) to (c,C) by the function y=f(x)=C_1\mathrm{e}^{C_2x}+C_3, where f(b)=B, f(c)=C, and f'(b)=\frac{B-A}{b-a}.

    Hopefully that might makes things a bit simpler.

    By the way, CaptainBlack, the condition C_1,C_2>0 will make f(x) convex, and this will only work if (b,B) lies below the straight line joining (a,A) and (c,C). If (b,B) lies above that line, then we need a concave function, which means we need both C_1 and C_2 to be negative ( C_2<0 to make the function concave and C_1<0 to make it increasing).
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  10. #10
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    Quote Originally Posted by algebraic topology View Post
    Well, here’s a modification of CaptainBlack’s suggestion.

    Join the first two points ( (a,A) and (b,B)) by a straight line. This will have gradient \frac{B-A}{b-a}. Then join (b,B) to (c,C) by the function y=f(x)=C_1\mathrm{e}^{C_2x}+C_3, where f(b)=B, f(c)=C, and f'(b)=\frac{B-A}{b-a}.

    Hopefully that might makes things a bit simpler.

    By the way, CaptainBlack, the condition C_1,C_2>0 will make f(x) convex, and this will only work if (b,B) lies below the straight line joining (a,A) and (c,C). If (b,B) lies above that line, then we need a concave function, which means we need both C_1 and C_2 to be negative ( C_2<0 to make the function concave and C_1<0 to make it increasing).
    I expect my mental image of this was convex, so it works for what I was trying to do, its just that what I was trying to do was only half the problem!

    In fact we need no restrictions on the constants, with the given relations between the values at the given points a function of the given form which passes through the data will be monotone increasing.

    RonL
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