An equation is given to me which has to be solved by direct iteration method: $$\sin(x) = {x+1 \over x-1}$$ or $$f(x)=\sin(x)-{x+1 \over x-1} = 0$$

I follow the following procedure with reasons given along:

Rearrange the equation as: $$x=\sin^{-1}{{x+1\over x-1}}=g(x)$$

Then, we get $$g'(x) = {1 \over (1-x)\sqrt{-x}}$$

$$\implies \forall x < -1, \left|g'(x)\right| < 1$$


$\forall x < -1, 0 < {x+1 \over x-1} < 1$. But, $\sin{x}$ oscillates between 1 and -1 infinitely over the whole domain $\mathbb{R}$ and hence also its subset, the interval $(-\infty, -1)$. Thus, $\sin{x}$ intersects infinitely with ${x+1 \over x-1}$, at least once each time in its interval of size $\pi$. These intersection points are exactly the roots of $f(x)$ over the interval $\left(-\infty, -1\right)$.

Thus, if we choose arbitrary points $a,b < -1 : a - b \ge \pi$, then $\sin{x}$ goes from -1 to 1 at least once, and since ${x+1 \over x-1}$ is always between 0 and 1, the period contains a root of $f(x)$.

Thus, to summarize, we have got a function $g(x)$ which is defined over the negative number line, and is bounded and continuous there. On the negative number line, we also have an interval $(a, b)$ where we have a root of given equation. Also, derivative of $g(x)$ is numerically less than one in the given interval.

So, ideally, I should be able to take any value of $x = x_0$ in $(a, b)$ and should be able to get increasing better iterates.

But the fact that my function $g(x)$ is $\sin^{-1}{{x + 1 \over x - 1}}$ means that any negative value of $x < -1 $yields me a positive value of iterate in the next step, which in turn leads to domain error.

Where am I wrong? Or more precisely, are there any more conditions that are required for the fixed-point iterative method that are not known to me. One step which seems dubious to me is the very first step where I take $\sin^{-1}$ of original equation on the both sides, to get the $g(x)$, but I am not able to pinpoint how it can derail my method.

Update: Renamed the method from direct iterative method to fixed-point iteration method. Also, as is noted in the comments below, problem seems to be in my choice of $g(x)$ only. I am interested in what it really is. How can I avoid the same in future too?

  • $\begingroup$ What is the 'direct iteration method'? $\endgroup$ – copper.hat Sep 6 '13 at 15:24
  • $\begingroup$ @copper.hat I meant fixed-point iteration method. Sorry for confusion; I have updated the post accordingly. $\endgroup$ – Anurag Kalia Sep 6 '13 at 16:10
  • $\begingroup$ @Amzoti Thanks for the reference and the tip; I have blindly used the method till now. But the fact remains this is a homework queston where the function and the method are pre-specified. Still, plotting another function g(x) = (sin(x) + 1)/(sin(x) - 1) on the graph is yielding me infinite roots as promised by the original equation. I am still working on its solution, but the fact remains I still don't know why my original method failed. $\endgroup$ – Anurag Kalia Sep 6 '13 at 17:22
  • $\begingroup$ @Amzoti Thanks! $\endgroup$ – Anurag Kalia Sep 6 '13 at 17:30
  • $\begingroup$ @Amzoti I have answered my own question because it seemed satisfactory enough to me. Please check it if there are any discrepancies in the answer; I can't seem to find one. Also: what is the protocol regarding answering your own question? $\endgroup$ – Anurag Kalia Sep 6 '13 at 18:47

The problem is in the choice of $g(x)$ as initially doubted in the question itself. The Banach fixed-point theorem, on which this method is based, says:

Let $\left(X, d\right)$ be a non-empty complete metric space with a contraction mapping $T : X \to X$. Then $T$ admits a unique fixed-point $x^*$ in $X$ (i.e. $T(x^*) = x^*$). Furthermore, $x^*$ can be found as follows: start with an arbitrary element $x_0$ in $X$ and define a sequence $\{x_n\}$ by $x_n = T(x_{n−1})$, then $x_n \to x^*$.

Here, the equation chosen is $$x = g(x) = sin^{-1}{{x+1 \over x-1}}$$ thus, $$g : (-\infty, 0) \to (0, 1)$$

Therefore, there is no sub-interval $X$ in $g$'s domain where $g : X \to X$ is true. Since this condition is required for the definition of contraction mapping, this $g(x)$ can't be a contraction mapping over real number line, and since Banach fixed-point theorem talks about contraction mappings, we can't use this $g(x)$ for fixed-point iteration method.

  • $\begingroup$ The reason it fails is because it may start as a real iteration but it silently breaks into a complex one. In complex iteration the basin of attraction for each iteration is switched as you say, and a branch cut separates the two basins ($\sqrt{-x}$), so the limit will be a two-cycle modulo conjugates, and the claim of the fixed point theorem is valid for absolute values of the complex numbers involved in the (complex) iteration. $\endgroup$ – Yiannis Galidakis Aug 22 '15 at 22:14

I am not convinced that in your case the following is true:

$$\forall x < -1, \left|g'(x)\right| < 1$$

I tried to run the numbers quickly and here is what I got, it is clear that there is no conversion.

enter image description here


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