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Here, i have a function for an iterative series.

Next value = x + sin(x). converging on a value

I want to make it so that i can find the current value, when i know the convergence value, The only way i can do that is to reverse the function.

Here is my shot at it: x = arcsin(Next value - x)

I'm not so sure how to do this, could someone help?

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  • $\begingroup$ You have $x_{n+1} = x_n + \sin(x_n)$ and you want $x_n = f(x_{n+1})$ where is $f(\cdot)$ is some function. To me this seems quite hard in this case. $\endgroup$ – Ritz Nov 13 '14 at 14:12
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It actually can't be done unless you originally start at the convergence value.

For if the iterator function is invertible you have, for a fixed point\convergence value $x_f$:

$$f(x_f)=x_f\Rightarrow f^{-1}(x_f)=x_f,$$

so iterating you just have

$$f^{-n}(x_f)=x_f.$$

If the iterator function is invertible on $[a,b]$ and you start at $x_0\in[a,b]$ not fixed, the iterates can converge to $x_f$ but don't equal it eventually.

Proof: Consider the orbit of $x_0$ under $f$:

$$\text{orb}(x_0)=\{x_0,f(x_0),f^2(x_0),\dots\}.$$

Suppose that $f^N(x_0)=x_f$. Now by applying $f^{-N}$ we have $$x_0=f^{-N}(x_f)=x_f$$ by the above calculation. But this is a contradiction as $x_0$ is assumed be not fixed.

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  • $\begingroup$ I see what you mean, the convergent value of pi is for any value 0 < x < 2 pi so getting x would be impossible, unless it can be defined by the number of steps it takes. $\endgroup$ – It'sRainingMen Nov 13 '14 at 14:40
  • $\begingroup$ What? I am saying that if you reverse from a convergent value such as $x=\pi$ you will just stay at $\pi$. $\endgroup$ – JP McCarthy Nov 13 '14 at 14:47
  • $\begingroup$ i know, if any value kπ is used you will have sin(π) or just 0. all values will converge on the value (2n+1) pi. $\endgroup$ – It'sRainingMen Nov 13 '14 at 15:07
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I am not sure I understood you question, but I will try to answer anyway. Take any $x_0\in\mathbb{R}$ and define $$ x_{n+1}=x_n+\sin x_n,\quad n\ge0. $$ If $\lim_{n\to\infty}x_n=\hat x$, then, since $\sin$ is a continuous function, we get $$ \sin\hat x=0\implies \hat x=k\,\pi,\quad k\in\mathbb{Z}. $$ The convergence values are the fixed points of the function $f(x)=x+\sin x$. It is not difficult to prove that there are two possibilities:

  1. $x_0=2\,k\,\pi$, $k\in\mathbb{Z}$. Then $\hat x=x_0$.
  2. $x_0\ne2\,k\,\pi$, $k\in\mathbb{Z}$. Then $\hat x=(2\,n+1)\,\pi$ for some $n\in\mathbb{Z}$.

There is no way to know the current value from the convergence value, except when the convergence value is an even multiple of $\pi$.

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  • $\begingroup$ Could you have not just written $k\pi$ for $k\in\mathbb{Z}$? $\endgroup$ – JP McCarthy Nov 13 '14 at 14:23
  • $\begingroup$ I wanted to distinguish the stable fixed points (odd multiples of $\pi$) from the unstable (even multiples of $\pi$.) If the initial value is, say, in $(0,2\pi)$, then the convergence value is $\pi$. $\endgroup$ – Julián Aguirre Nov 13 '14 at 14:27
  • $\begingroup$ Ah yes I misunderstood what you were trying to show. $\endgroup$ – JP McCarthy Nov 13 '14 at 14:30

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