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Homework Help: Mathematical Modeling Question using a 2-cycle

  1. Apr 6, 2016 #1
    1. The problem statement, all variables and given/known data
    a. Use algebra and calculus to find the positive fixed point p(c) (in terms of c) and identify its exact interval of stability.
    b. Use algebra and calculus to find the exact interval of stability of this fixed point
    c. Use algebra to find the point p1(c), p2(c) of the 2-cycle (in terms of c), and identify its exact interval of stability.
    d. Use algebra and calculus to find the exact interval of stability of this 2-cycle.

    2. Relevant equations
    fc(x) = 4/x + x/2 - c for 0<c<2.3

    3. The attempt at a solution
    I did part a and solved that 4/x + x/2 - c = x and got -c + sqrt(c^2+8).
    For part b I was supposed to take the derivative of 4/x + x/2 - c which I got to be -4/x^2 + 1/2. I was then supposed to replace x with -c+sqrt(c^2+8) and make it into an inequality -1<-4/(-c+sqrt(c^2+8))<1 and solve and I broke it up and found c<1 (not sure if this is correct).
    For part c I was supposed to solve 4/(4/x + x/2 -c) + .5(4/x + x/2 -c) = x and I got -12c^2x^2 - 4cx^3 + 64cx + 3x^4 - 16x^2 - 64. I was then supposed to divide x-(-c+sqrt(c^2+8) into that problem via long division and I keep trying but I am never to find the exact solution.
  2. jcsd
  3. Apr 8, 2016 #2


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    Homework Helper

    You can rearrange [tex]\left| -\frac{4}{x^2} + \frac12\right| < 1[/tex] as [tex]|x^2 - 8| < 2x^2[/tex] or [tex]8 -2x^2 < x^2 < 8 + 2x^2.[/tex] Now [itex]x^2 < 8 + 2x^2[/itex] holds for all real [itex]x[/itex], but the other inequality yields [itex]x^2 > \frac83[/itex]. Now setting [itex]x = \sqrt{c^2 + 8} - c[/itex] will yield [tex]
    \frac83 + c^2 > c\sqrt{c^2 + 8}[/tex] and as [itex]c[/itex] is positive it is legitimate to conclude that [tex]
    \left(\frac83 + c^2\right)^2 > c^2(8 + c^2).[/tex] Now the [itex]c^4[/itex] terms cancel and you have an inequality for [itex]c^2[/itex].

    I don't get a term in [itex]x^3[/itex], but the rest of the expression I agree with. This might explain why you can't divide by what should be a known factor.

    However multiplying polynomials is much easier than dividing them. I would suggest writing [tex]
    3x^4 - (12c^2 +16)x^2 + 64cx - 64 = (3x^2 + Ax + B)(x^2 + 2cx - 8)[/tex] as you know it must factorize in that way. Now multiply out the right hand side and compare coefficients of powers of [itex]x[/itex] to determine [itex]A[/itex] and [itex]B[/itex] in terms of [itex]c[/itex]. Then solve [tex]3x^2 + Ax + B = 0.[/tex]
    Last edited: Apr 8, 2016
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