\(a_0=1\)
\(a_{n+1}=a_n+\sin{(a_n)}\)
Explain why the following occurs:
\(a_0=1\)
\(a_1=1+\sin{(1)}\approx 1.841470985\)
\(a_2=1+\sin{(1)}+\sin{(1+\sin{(1)})}\approx 2.805061709\)
\(a_3=1+\sin{(1)}+\sin{(1+\sin{(1)})}+\sin{(1+\sin{(1)}+\sin{(1+\sin{(1)})})}\approx 3.135276333\)
\(a_4\approx 3.141592612\)
\(a_5\approx 3.141592654\approx\pi\)
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Jack Han
3 years, 1 month ago
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The interesting thing I was talking about is the fact that the given series always converges into a value \(a_n\) such that \(sin(a_n)=0\) for different values of \(a_0\). To put it in more exact terms, it always converges to a value of \(a_n\) such that \(cos(a_n)=-1\Rightarrow a_n=m\pi\), where \(m\) is an odd integer. (\(m\) can also be even, but that is a degenerate case where all terms are same)
I'm going to use something here that I actually learned from gradient descent. If you don't know what it is, you can google it. But, the mathematics used below is an extremely tame form and is easy to understand with little knowledge of calculus.
Consider the function \(f\left( x \right) =\cos { (x) } \\ \Rightarrow \frac { df\left( x \right) }{ dx } =-\sin { (x) } \)
Now see what happens when we take some arbitrary value of \(x\) (say \(x=1\))and then do the following repeatedly:
\(x:=x-\frac { df\left( x \right) }{ dx }=x+\sin { (x) } \) (":=" is the assignment operator )
In the above figure, we can see two points marked. One is red, which represents the first value of \(x\)(\(=1\)). The other is brown, and is after one iteration of above step.
We can see that when we do \(x:=x+sin(x)\), what is actually happening is that \(x\) is surfing along the slope of the curve \(cos(x)\). We move the value of \(x\) down the tangent. Change \(x\) little by little, so that finally, after many iterations it moves closer and closer to the minima, i.e \(x=\pi\).
I know this is not a definitive proof of what happens... I'm sure you will realize the importance of this once you understand what is happening.
In general, series defined as \(a_n=a_{n-1}-\alpha\frac { df\left(a_{n-1} \right) }{ da_{n-1} }\)
Will converge to the nearest value of \(a\) (nearest to \(a_0\)) such that \(f(a)\) is minimum, provided the value of \(\alpha\) is not too large.
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Good work.
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Good work! Newton's method for estimating roots.
Pretty much seals the deal. Great solution +1.
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Yes, that was awesome, that is basically the newton Rhapsody method of estimation of roots, doing the following iteration for any curve will eventually lead us to the nearest root, that is great, actually i think this is pretty much the solution +1
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@Shashwat Shukla @Mvs Saketh
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As usual, since the series converges.. this means that when \(n\to\infty\) ,
\(a_{n+1}=a_{n}\). But \(a_{n+1}=a_{n}+sin(a_{n})\)
\(\Rightarrow sin(a_{n})=0\)
Now how do we know that \(a_{n}=\pi\)? We know this since \(a_0=1\) and the series is constantly increasing. Therefore, it converges onto the first value of \(x>1\) such that \(sin(x)=0\).
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Bro , But It is not always true \(\lim _{ n\rightarrow \infty }{ ({ a }_{ n }) } =\lim _{ n\rightarrow \infty }{ { (a }_{ n+1 }) } \) .
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Why not? Do you have a counter-example?
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Comment deleted Mar 13, 2015
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But can't the value of zero be discarded? Because here, \(a_{n}=1\) for all n.
There is no question of convergence at all as all the terms have a fixed value.
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My Bad , I wrongly co-relate two diffrent situations , Actually this example is given when I got two answwer to an single recursion , which is something Like that \({ a }_{ n+1 }={ \left( \sqrt { 2 } \right) }^{ { a }_{ n } }\) , I misinterpreate this current situation, So I deleted that irrelevant comment , But Yes this is good recurance , and I'am trying to prove convergence of this recurance. If I got correct than I report you.
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Thanks for joining the party :D
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Comment deleted Mar 13, 2015
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No, the sequence only increases, only that the rate of increase decreases till it reaches 0 at \(a_n= \pi\)
you can easily see it from the graph if x+sin(x), it is a non decreasing function and reaches inflection point at x= pi
but yes convergence remains unproven, i am trying
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My bad. Sorry I jumped to a wrong conclusion. Thanks for replying.
But the rate at which the value of \(a_{n}\) increases isn't the same as the rate at which \(x+sinx\) is increasing. So how do you directly conclude from the graph?
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it is not the increasing nature that confirms, it is the fact that the graph lies above x=y as well
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I sort of understand what you're saying...But isn't it easier to say this: The increment in the variable in every iteration is \(sina_{n}\) and as \(a_{n}\) tends to \(\pi\), \(\quad sina_{n}\) becomes smaller and smaller. (I think that is what you were saying initially about rate of change slowing down).
I have deleted my previous comment as it might only muddle things up for someone else.
Also, please check out my response to your post on this problem of mine.
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Indeed it is ,simply that sin(an) is positive, in the range and hence increment is positive or sequence rises :)
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Can someone prove the convergence of this recursion?
@Ronak Agarwal @Mvs Saketh @Azhaghu Roopesh M @Pratik Shastri @Raghav Vaidyanathan @Deepanshu Gupta or anyone else who's interested.
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This is turning out to be very interesting...
I want to know if we can find a general form for a function \(f(x)\) such that the series \(a_1,a_2,...\) defined by:
\(a_{n+1}=a_{n-1}+f(a_{n-1})\)
Converges for a given value of \(a_0\).
Further, is it true that all of the values of such \(a_n\) as \(n\to \infty\) satisfy \(f(a_n)=0\)?
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I think the answer to this is going to be extremely interesting.... I have a feeling... Is anyone else thinking what I'm thinking?
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What are you thinking?
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Comment deleted Mar 12, 2015
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are you sure you have differentiated it correctly ? there should be atleast one \(x\) in your \(f'(x)\) i think
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If you call L the value of the limit you obtain sin(L)=0. Now L can be pi or zero but zero is impossible because of the initial condition. More precisely you can say that the value of the sequence is LOW bounded
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Which value converges to pi ??
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