Now this baffles me.

I have a simple question. Does there exists a finite constant α0>0 {\alpha}_{0}>0 , Such that the sequence {an}n=0 \lbrace {a}_{n} \rbrace_{n=0} , defined by the iteration :

an=an1+αsin(an1) {a}_{n} = {a}_{n-1}+\alpha\sin({a}_{n-1})

Not converges to any single finite value for a0=1,α>α0 {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 α \alpha , I calculated nn such that an{a}_{n} gets within 0.01%0.01 \% the value of π\pi . I got this :

α=1.99,n=622\alpha = 1.99 , n=622

α=1.991,n=686\alpha = 1.991 , n=686

α=1.992,n=766\alpha = 1.992 , n=766

α=1.993,n=867\alpha = 1.993 , n=867

α=1.994,n=1000\alpha = 1.994 , n=1000

α=1.995,n=1183\alpha = 1.995 , n=1183

α=1.996,n=1453\alpha = 1.996 , n=1453

α=1.997,n=1893\alpha = 1.997 , n=1893

α=1.998,n=2743\alpha = 1.998 , n=2743

α=1.999,n=5150\alpha = 1.999 , n=5150

And the last one is :

α=2,n=15198162 \alpha = 2 , n=15198162

Also when I tried for α=2.001 \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 α \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
4 years, 3 months ago

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Here are my thoughts,

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

an1y(an1y(an1){ a }_{ n-1 }-\frac { y(a_{n-1} }{ y'(a_{n-1}) } =an)a_{n})

then comparing we require

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

y=(cot(x)+cosec(x))1αy=(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 xπx \rightarrow \pi

1+cos(x)sin(x)=2cos(x2)2sin(x2)0forxclosetoπ\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 α0\alpha \geq 0

hence i think it should converge for all values of α\alpha given a0=1a_{0} = 1 to π\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 α=1\alpha =1

Mvs Saketh - 4 years, 3 months ago

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

Ronak Agarwal - 4 years, 3 months ago

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(1tan(x)+1sin(x))1t\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

Mvs Saketh - 4 years, 3 months ago

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@Mvs Saketh 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)

Ronak Agarwal - 4 years, 3 months ago

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@Ronak Agarwal 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

Mvs Saketh - 4 years, 3 months ago

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

Ronak Agarwal - 4 years, 3 months ago

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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 α\alpha became >2>2. And I think I know what's happening.

From what I saw in my program, the value of ana_n when α>2\alpha>2 as nn\to \infty seems to be shuttling between two values. What may be happening is that the value overshoots π\pi then becomes less than π\pi and so on. It might be possible that for some value a>π,b<πa>\pi,b<\pi, b=a+αsin(a)b=a+\alpha sin(a) and a=b+αsin(b)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))x=x+2.001sin(x)+2.001sin(x+2.001sin(x))

Raghav Vaidyanathan - 4 years, 3 months ago

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@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.

Ronak Agarwal - 4 years, 3 months ago

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Great! Notice that the value might have converged to pi if the initial value of a0a_0 was more than the infinite loop root value. (or so I think)

Raghav Vaidyanathan - 4 years, 3 months ago

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

Ronak Agarwal - 4 years, 3 months ago

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

For example for α=3 \alpha=3 , it begin to oscillate between 33 values, and so on.

Ronak Agarwal - 4 years, 3 months ago

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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 π\pi) for x=x+αsin(x)+αsin(x+αsin(x))x=x+\alpha sin(x)+\alpha sin(x+\alpha sin(x)). After simplification:

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

Now we get to the interesting part. This curve is symmetric about the point (π,0)(\pi,0)similar to the way x3x^3 is symmetric about (0,0)(0,0).

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

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

Now if we increase α/2\alpha /2 to more than 11.

f(π)<0\Rightarrow f^{'}(\pi)<0 and alsof(π)=0f(\pi)=0

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

Check it out:

It's like a cubic.

Raghav Vaidyanathan - 4 years, 3 months ago

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

Ronak Agarwal - 4 years, 3 months ago

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Thanks, the analysis was an interesting read.

Calvin Lin Staff - 4 years, 3 months ago

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

Azhaghu Roopesh M - 4 years, 3 months ago

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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 ;

                                                             }

Ronak Agarwal - 4 years, 3 months ago

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Well, The only thing I can add is that your conjecture isn't true. For example, take α=kπa0sin(a0) \alpha = \dfrac{k\pi - a_0}{sin(a_0)} . By taking a large enough k k , we will have kπa0sin(a0)>2 \dfrac{k\pi - a_0}{sin(a_0)} > 2

Siddhartha Srivastava - 4 years, 3 months ago

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

Ronak Agarwal - 4 years, 3 months ago

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Sorry. I just noticed you had a value for a0 a_0 aswell. Take α=π1sin(1) \alpha = \dfrac{\pi - 1}{sin(1)} . You can check to see that this is greater than 2 2 . You can find more values by taking α=kπ1sin(1) \alpha = \dfrac{k\pi - 1}{sin(1)} for a natural number k k

Siddhartha Srivastava - 4 years, 3 months ago

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

Ronak Agarwal - 4 years, 3 months ago

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

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

Now, I argue that if if α>2\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=by=b line. If ana_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 α\alpha becomes too large as the point escapes from this basin of cos(x)cos(x) itself, then I am assuming it will diverge(don't have a solid proof for this).

Raghav Vaidyanathan - 4 years, 3 months ago

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