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a0=1a_0=1a0=1
an+1=an+sin(an)a_{n+1}=a_n+\sin{(a_n)}an+1=an+sin(an)
Explain why the following occurs:
a1=1+sin(1)≈1.841470985a_1=1+\sin{(1)}\approx 1.841470985a1=1+sin(1)≈1.841470985
a2=1+sin(1)+sin(1+sin(1))≈2.805061709a_2=1+\sin{(1)}+\sin{(1+\sin{(1)})}\approx 2.805061709a2=1+sin(1)+sin(1+sin(1))≈2.805061709
a3=1+sin(1)+sin(1+sin(1))+sin(1+sin(1)+sin(1+sin(1)))≈3.135276333a_3=1+\sin{(1)}+\sin{(1+\sin{(1)})}+\sin{(1+\sin{(1)}+\sin{(1+\sin{(1)})})}\approx 3.135276333a3=1+sin(1)+sin(1+sin(1))+sin(1+sin(1)+sin(1+sin(1)))≈3.135276333
a4≈3.141592612a_4\approx 3.141592612a4≈3.141592612
a5≈3.141592654≈πa_5\approx 3.141592654\approx\pia5≈3.141592654≈π
Note by Jack Han 5 years, 11 months ago
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Let's see here...
The interesting thing I was talking about is the fact that the given series always converges into a value ana_nan such that sin(an)=0sin(a_n)=0sin(an)=0 for different values of a0a_0a0. To put it in more exact terms, it always converges to a value of ana_nan such that cos(an)=−1⇒an=mπcos(a_n)=-1\Rightarrow a_n=m\picos(an)=−1⇒an=mπ, where mmm is an odd integer. (mmm 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(x)=cos(x)⇒df(x)dx=−sin(x)f\left( x \right) =\cos { (x) } \\ \Rightarrow \frac { df\left( x \right) }{ dx } =-\sin { (x) } f(x)=cos(x)⇒dxdf(x)=−sin(x)
Now see what happens when we take some arbitrary value of xxx (say x=1x=1x=1)and then do the following repeatedly:
x:=x−df(x)dx=x+sin(x)x:=x-\frac { df\left( x \right) }{ dx }=x+\sin { (x) } x:=x−dxdf(x)=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 xxx(=1=1=1). The other is brown, and is after one iteration of above step.
We can see that when we do x:=x+sin(x)x:=x+sin(x)x:=x+sin(x), what is actually happening is that xxx is surfing along the slope of the curve cos(x)cos(x)cos(x). We move the value of xxx down the tangent. Change xxx little by little, so that finally, after many iterations it moves closer and closer to the minima, i.e x=πx=\pix=π.
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 an=an−1−αdf(an−1)dan−1a_n=a_{n-1}-\alpha\frac { df\left(a_{n-1} \right) }{ da_{n-1} }an=an−1−αdan−1df(an−1)
Will converge to the nearest value of aaa (nearest to a0a_0a0) such that f(a)f(a)f(a) is minimum, provided the value of α\alphaα is not too large.
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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
Good work! Newton's method for estimating roots.
Pretty much seals the deal. Great solution +1.
Good work.
@Shashwat Shukla @Mvs Saketh
As usual, since the series converges.. this means that when n→∞n\to\inftyn→∞ ,
an+1=ana_{n+1}=a_{n}an+1=an. But an+1=an+sin(an)a_{n+1}=a_{n}+sin(a_{n})an+1=an+sin(an)
⇒sin(an)=0\Rightarrow sin(a_{n})=0⇒sin(an)=0
Now how do we know that an=πa_{n}=\pian=π? We know this since a0=1a_0=1a0=1 and the series is constantly increasing. Therefore, it converges onto the first value of x>1x>1x>1 such that sin(x)=0sin(x)=0sin(x)=0.
Bro , But It is not always true limn→∞(an)=limn→∞(an+1)\lim _{ n\rightarrow \infty }{ ({ a }_{ n }) } =\lim _{ n\rightarrow \infty }{ { (a }_{ n+1 }) } limn→∞(an)=limn→∞(an+1) .
Why not? Do you have a counter-example?
Which value converges to pi ??
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
This is turning out to be very interesting...
I want to know if we can find a general form for a function f(x)f(x)f(x) such that the series a1,a2,...a_1,a_2,...a1,a2,... defined by:
an+1=an−1+f(an−1)a_{n+1}=a_{n-1}+f(a_{n-1})an+1=an−1+f(an−1)
Converges for a given value of a0a_0a0.
Further, is it true that all of the values of such ana_nan as n→∞n\to \inftyn→∞ satisfy f(an)=0f(a_n)=0f(an)=0?
I think the answer to this is going to be extremely interesting.... I have a feeling... Is anyone else thinking what I'm thinking?
What are you thinking?
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Top NewestLet's see here...
The interesting thing I was talking about is the fact that the given series always converges into a value an such that sin(an)=0 for different values of a0. To put it in more exact terms, it always converges to a value of an such that cos(an)=−1⇒an=mπ, 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(x)=cos(x)⇒dxdf(x)=−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−dxdf(x)=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=π.
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 an=an−1−αdan−1df(an−1)
Will converge to the nearest value of a (nearest to a0) such that f(a) is minimum, provided the value of α is not too large.
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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
Log in to reply
Good work! Newton's method for estimating roots.
Pretty much seals the deal. Great solution +1.
Log in to reply
Good work.
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@Shashwat Shukla @Mvs Saketh
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As usual, since the series converges.. this means that when n→∞ ,
an+1=an. But an+1=an+sin(an)
⇒sin(an)=0
Now how do we know that an=π? We know this since a0=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 limn→∞(an)=limn→∞(an+1) .
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Why not? Do you have a counter-example?
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Which value converges to pi ??
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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
Log in to reply
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 a1,a2,... defined by:
an+1=an−1+f(an−1)
Converges for a given value of a0.
Further, is it true that all of the values of such an as n→∞ satisfy f(an)=0?
Log in to reply
I think the answer to this is going to be extremely interesting.... I have a feeling... Is anyone else thinking what I'm thinking?
Log in to reply
What are you thinking?
Log in to reply