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.

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TopNewestThanks, the analysis was an interesting read. – Calvin Lin Staff · 2 years 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 \(\alpha\) 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 \(x=x+2.001sin(x)+2.001sin(x+2.001sin(x))\) – Raghav Vaidyanathan · 2 years 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 · 2 years ago

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– Raghav Vaidyanathan · 2 years 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)Log in to reply

Wolfram Alpha Query – Ronak Agarwal · 2 years ago

Guess what, it again sticks to it's two-value cycle , check thisLog in to reply

For example for \( \alpha=3 \), it begin to oscillate between \(3\) values, and so on. – Ronak Agarwal · 2 years ago

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

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. – Raghav Vaidyanathan · 2 years ago

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– Ronak Agarwal · 2 years ago

I did this analysis too , well it's very interesting.Log in to reply

@Mvs Saketh @Azhaghu Roopesh M @Ronak Agarwal – Raghav Vaidyanathan · 2 years 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

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

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 \(\alpha =1\) – Mvs Saketh · 2 years ago

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– Ronak Agarwal · 2 years ago

I think we should check the conditions where newton's raphson's methos fails.Log in to reply

– Ronak Agarwal · 2 years ago

But this is just amazing at \( \alpha=2\) , the convergence rate goes awfully slow.Log in to reply

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 · 2 years ago

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– Ronak Agarwal · 2 years 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)Log in to reply

and yes i know and i am very afraid, really afraid, just hope to qualify – Mvs Saketh · 2 years 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 \(cos(x)\) "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). – Raghav Vaidyanathan · 2 years ago

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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 \( \dfrac{k\pi - a_0}{sin(a_0)} > 2 \) – Siddhartha Srivastava · 2 years ago

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– Ronak Agarwal · 2 years ago

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

– Siddhartha Srivastava · 2 years 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 \( k \)Log in to reply

– Ronak Agarwal · 2 years ago

Oh, now I realize I have mistaken, I didn't realize this.Log in to reply

Can you post here the Java code that you've used ? – Azhaghu Roopesh M · 2 years ago

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