# Now this baffles me.

I have a simple question. Does there exists a finite constant $${\alpha}_{0}>0$$, Such that the sequence $$\lbrace {a}_{n} \rbrace_{n=0}$$, defined by the iteration :

$${a}_{n} = {a}_{n-1}+\alpha\sin({a}_{n-1})$$

Not converges to any single finite value for $${a}_{0}=1 , \alpha > {\alpha}_{0}$$

What inspires me to ask this is the analysis I have done on my laptop :

What I did is that for different values of $$\alpha$$ , I calculated $$n$$ such that $${a}_{n}$$ gets within $$0.01 %$$ the value of $$\pi$$. I got this :

$$\alpha = 1.99 , n=622$$

$$\alpha = 1.991 , n=686$$

$$\alpha = 1.992 , n=766$$

$$\alpha = 1.993 , n=867$$

$$\alpha = 1.994 , n=1000$$

$$\alpha = 1.995 , n=1183$$

$$\alpha = 1.996 , n=1453$$

$$\alpha = 1.997 , n=1893$$

$$\alpha = 1.998 , n=2743$$

$$\alpha = 1.999 , n=5150$$

And the last one is :

$$\alpha = 2 , n=15198162$$

Also when I tried for $$\alpha = 2.001$$ , it just won't show any value for half an hour.

It is very weird as rate of converge gets immediately slow and I believe convergence must stop at some value of $$\alpha$$.

I don't know can someone do a computer analysis of this, or tell me the maths of the underlying situation.

Note by Ronak Agarwal
3 years, 4 months ago

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Thanks, the analysis was an interesting read.

Staff - 3 years, 4 months ago

Aah! @Ronak Agarwal I found this too! Before I posted my question, I put this in c++ and the thing failed to converge as soon as $$\alpha$$ became $$>2$$. And I think I know what's happening.

From what I saw in my program, the value of $$a_n$$ when $$\alpha>2$$ as $$n\to \infty$$ seems to be shuttling between two values. What may be happening is that the value overshoots $$\pi$$ then becomes less than $$\pi$$ and so on. It might be possible that for some value $$a>\pi,b<\pi$$, $$b=a+\alpha sin(a)$$ and $$a=b+\alpha sin(b)$$. Thus giving us an infinite non convergent loop. And yes, wolfram alpha agrees with me:

I put in $$x=x+2.001sin(x)+2.001sin(x+2.001sin(x))$$

- 3 years, 4 months ago

@Raghav Vaidyanathan , guess what I entered the same thing into wolfram, thought exactly the same way and as I am typing this, I have tab of wolfram-alpha currently opened in which I entered EXACTLY the same thing you have entered, I figured out what is happening it is oscillating between two values, the two values I found by putting the same expression in wolfram-alpha, I love mathematical programming.

I was intending to put the same snapshot before I read your comment.

- 3 years, 4 months ago

Great! Notice that the value might have converged to pi if the initial value of $$a_0$$ was more than the infinite loop root value. (or so I think)

- 3 years, 4 months ago

Guess what, it again sticks to it's two-value cycle , check this Wolfram Alpha Query

- 3 years, 4 months ago

Interestingly when $$\alpha$$ increases, our iteration begins to oscillate between cycles of increasing length.

For example for $$\alpha=3$$, it begin to oscillate between $$3$$ values, and so on.

- 3 years, 4 months ago

I was just studying point of inflection for board exam... God! this is interesting. Okay, so I took our equation and tried to find the conditions for which we have real roots (other than $$\pi$$) for $$x=x+\alpha sin(x)+\alpha sin(x+\alpha sin(x))$$. After simplification:

$$f(x)=-(\pi -x)+(\alpha /2)sin(x)=0$$)

Now we get to the interesting part. This curve is symmetric about the point $$(\pi,0)$$similar to the way $$x^3$$ is symmetric about $$(0,0)$$.

$$f^{'}(x)=1+(\alpha /2)cos(x)$$

Observe the above equation. When $$\alpha /2=1$$, $$(\pi,0)$$ is a point of inflection.

Now if we increase $$\alpha /2$$ to more than $$1$$.

$$\Rightarrow f^{'}(\pi)<0$$ and also$$f(\pi)=0$$

But there are roots for $$f^{'}(x)$$ on either side of $$\pi$$ and equidistant from it. All these things pretty much guarantee us a root for $$f(x)$$ on either side of $$\pi$$ when $$\alpha>2$$

Check it out:

It's like a cubic.

- 3 years, 4 months ago

I did this analysis too , well it's very interesting.

- 3 years, 4 months ago

- 3 years, 4 months ago

Here are my thoughts,

consider newtons approximation of root, let us assume that the iteration is actually of the form

$${ a }_{ n-1 }-\frac { y(a_{n-1} }{ y'(a_{n-1}) }$$=$$a_{n})$$

then comparing we require

$$-\alpha \frac { y' }{ y } =cosec(x)\\$$

$$y=(cot(x)+cosec(x))^{ \frac { 1 }{ \alpha } }$$

now after sufficient iterations, newtons method takes us to the closest root,

let us see where our expression tends to go as $$x \rightarrow \pi$$

$$\frac { 1+cos(x) }{ sin(x) } \quad =\quad \frac { 2cos(\frac { x }{ 2 } ) }{ 2sin(\frac { x }{ 2 } ) } \simeq 0\quad for\quad x\quad close\quad to\quad \pi$$

as long as $$\alpha \geq 0$$

hence i think it should converge for all values of $$\alpha$$ given $$a_{0} = 1$$ to $$\pi$$

also the rate at which we reach the root may be affected however by alpha,

as , alpha tends to $$\infty$$ , the whole equation tends to 1, and hence it approaches 0 perhaps more slowly and hence you see the result (not very sure of this)

you should also see raghav's method in the post where a similar question is asked with $$\alpha =1$$

- 3 years, 4 months ago

I think we should check the conditions where newton's raphson's methos fails.

- 3 years, 4 months ago

But this is just amazing at $$\alpha=2$$ , the convergence rate goes awfully slow.

- 3 years, 4 months ago

$$\left(\frac{1}{\tan \left(x\right)}+\frac{1}{\sin \left(x\right)}\right)^{\frac{1}{t}}$$

type this in desmos and adjust the parameter 't', you will see for yourself why it is so

https://www.desmos.com/calculator

- 3 years, 4 months ago

I have done this in fooplot , but I can't see anything that interests me or helps me in understanding this situation. By the way today OCSC-2015 list will be released(wow)

- 3 years, 4 months ago

no bro , see you will observe that when t=0.01, the graph has an almost horizontal tangent near pi and hence falls very fast to 0, at t=2, you can see that it has a near vertical tangent at x=pi meaning it doesnt fall so slow and just falls to 0 at x=pi

and yes i know and i am very afraid, really afraid, just hope to qualify

- 3 years, 4 months ago

@Ronak Agarwal I think we can now kill this problem.

I'm going to use gradient descent inspired, mostly qualitative approach for proving my arguments. Consider this:

This is the $$cos(x)$$ "attraction basin" that is relevant to this problem. If $$\alpha$$ is small enough, then, as we know, we end up at $$\pi$$.

But, if $$\alpha>2$$, we see that there exists a value $$b$$ such that $$\pi-b=\pi+b+\alpha sin(\pi+b)$$ and $$\pi+b=\pi-b+\alpha sin(\pi-b)$$. (refer comments for proof).

Now, I argue that if if $$\alpha>2$$, the iterations will start shuttling between the two sides of the basin. Further, a small investigation into the iteration function can show us that it now "attracts" the iterations to the $$y=b$$ line. If $$a_n$$ is above it, it is pulled down

If it is below, it is pulled up.

The red line represents the motion of the point through successive iterations.

Going even deeper, if $$\alpha$$ becomes too large as the point escapes from this basin of $$cos(x)$$ itself, then I am assuming it will diverge(don't have a solid proof for this).

- 3 years, 4 months ago

Well, The only thing I can add is that your conjecture isn't true. For example, take $$\alpha = \dfrac{k\pi - a_0}{sin(a_0)}$$. By taking a large enough $$k$$, we will have $$\dfrac{k\pi - a_0}{sin(a_0)} > 2$$

- 3 years, 4 months ago

Well you must give me a counter example to my conjecture, and I can't understand what you have written,kindly elaborate.

- 3 years, 4 months ago

Sorry. I just noticed you had a value for $$a_0$$ aswell. Take $$\alpha = \dfrac{\pi - 1}{sin(1)}$$. You can check to see that this is greater than $$2$$. You can find more values by taking $$\alpha = \dfrac{k\pi - 1}{sin(1)}$$ for a natural number $$k$$

- 3 years, 4 months ago

Oh, now I realize I have mistaken, I didn't realize this.

- 3 years, 4 months ago

Can you post here the Java code that you've used ?

- 3 years, 4 months ago

  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 double d=1 ; int i=1 ; for(int j=990 ; j<1001 ; j=j+1) { for( ; ; i=i+1){ d = d + (1.00+(j/1000.00))*Math.sin(d) ; if(((10000*(d-Math.PI))/(Math.PI))<1 && -1 < ((10000*(d-Math.PI))/(Math.PI)) ) break ; } System.out.print(j+"-"+i+",") ; d=1 ; i=1 ; } 

- 3 years, 4 months ago