We have the following equation

\begin{equation} x(k+1)=\arccos\bigg( -\frac{1}{2(Dr^{\frac {|\sin(2x(k)+\theta)|}{M\sin x(k)\sqrt{A+2B\cos(2x(k)+\theta)}}}+1)} \bigg) \nonumber \end{equation}

$A,B,D,r,M,\theta$ are constants.

x[k + 1] ==
        ArcCos[-1/(Abs[Sin[2 x[k] + θ]]/(
        2 (d r^(m Sin[x[k]] Sqrt[a + 2 b Cos[2 x[k] + θ]])) + 1))]

The equation above can be written as $x_{n+1}=f(x_{n})$. How do we show that it converges using Mathematica?

How do we find the first derivative of the equation given above using Mathematica?

Also, is that possible to show that the first derivative is bounded?

  • 1
    $\begingroup$ Please at least write down the Mathematica code for your sequence. And show what have you tried, if possible. $\endgroup$ – Dr. belisarius Oct 18 '13 at 11:58
  • $\begingroup$ As a rule of thumb do not use upper case letters for names of parameters, variables, functions, etc. In your case D is a built-in predicate. $\endgroup$ – Sektor Oct 18 '13 at 12:19
  • $\begingroup$ What would you consider normal values for {a, d, r, [Theta], m, b}? You need to constrain these, since otherwise the answer from the right hand side is complex-valued. $\endgroup$ – bill s Oct 18 '13 at 13:58
  • $\begingroup$ @bills $\theta <2\pi/3$ and $A,B,D>0$, $0\leq r \leq 2$ and $\pi/2 < x_{k}<2\pi/3$ $\endgroup$ – Harry Oct 18 '13 at 20:26
  • $\begingroup$ @bills math.stackexchange.com/questions/514426/… $\endgroup$ – Harry Oct 18 '13 at 20:40

The derivative is straightforward: write the equation explicitly as a function of x:

f[x_] := ArcCos[-1/(Abs[Sin[2 x + θ]]/(2 (d r^(m Sin[x] Sqrt[ a + 2 b Cos[2 x + θ]]))+1))]

and take the derivative:

FullSimplify[D[f[x], x]]

Evan after simplification it's not very simple.

Using a set of parameters that seems to lie in the designated set {a, d, r, θ, m, b} = {1, 1, 1, Pi/3, 1, 1} I get f[5 Pi/8] as 3.14159 - 1.79944 I so either something is wrong with the f[] as written or the bounds on the parameters are incorrect.

  • $\begingroup$ is that possible to show that $D[f[x]]$ bounded? $\endgroup$ – Harry Oct 19 '13 at 0:48
  • $\begingroup$ Please check the equation. It is returning complex values -- this means that there is something seriously wrong with the setup. $\endgroup$ – bill s Oct 19 '13 at 1:10

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