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Given is the iteration $x_{k+1}=frac{1}{11}(1-cos(x_{k}))$ with $x_{0}in (-frac{pi }{2},frac{pi }{2})$ without $0$. Check if the sequence converges to $x^{*}=0$ and find its convergence rate.

Well, i am stuck in this obviously very easy first part of the problem. Wolfram Alpha shows that $lim_{xrightarrow infty }frac{1}{11}(1-cos(x))=0$, so the sequence converges to the given $x^{*}=0$. Can anyone give me a hint how to solve this limit, please?

Second, the convergence rate. I know the definitions of sublinear convergence, superlinear convergence etc.
I need to calculate the $lim_{xrightarrow infty }frac{left | x_{k+1} -x^{*}right |}{left | x_{k}-x^{*} right |}$, so $lim _{xrightarrow infty } frac{left | frac{1}{11}(1-cos(x)) right |}{left | x right |}=lim_{xrightarrow infty }frac{1-cos(x)}{11x}= 0$. So, the sequnce converges superlinearly. Is the idea correct at all? Can anybody help me with this problem, please?
Thank you in advance!

analysis convergence numerical-methods

share|cite|improve this question

edited Mar 31 '14 at 20:22

Amir Hossein

2,38111750

asked Mar 31 '14 at 20:15

LullabyLullaby

912718

$endgroup$

add a comment | 

0

1

$begingroup$

Given is the iteration $x_{k+1}=frac{1}{11}(1-cos(x_{k}))$ with $x_{0}in (-frac{pi }{2},frac{pi }{2})$ without $0$. Check if the sequence converges to $x^{*}=0$ and find its convergence rate.

Well, i am stuck in this obviously very easy first part of the problem. Wolfram Alpha shows that $lim_{xrightarrow infty }frac{1}{11}(1-cos(x))=0$, so the sequence converges to the given $x^{*}=0$. Can anyone give me a hint how to solve this limit, please?

Second, the convergence rate. I know the definitions of sublinear convergence, superlinear convergence etc.
I need to calculate the $lim_{xrightarrow infty }frac{left | x_{k+1} -x^{*}right |}{left | x_{k}-x^{*} right |}$, so $lim _{xrightarrow infty } frac{left | frac{1}{11}(1-cos(x)) right |}{left | x right |}=lim_{xrightarrow infty }frac{1-cos(x)}{11x}= 0$. So, the sequnce converges superlinearly. Is the idea correct at all? Can anybody help me with this problem, please?
Thank you in advance!

analysis convergence numerical-methods

share|cite|improve this question

edited Mar 31 '14 at 20:22

Amir Hossein

2,38111750

asked Mar 31 '14 at 20:15

LullabyLullaby

912718

$endgroup$

add a comment | 

$begingroup$

Given is the iteration $x_{k+1}=frac{1}{11}(1-cos(x_{k}))$ with $x_{0}in (-frac{pi }{2},frac{pi }{2})$ without $0$. Check if the sequence converges to $x^{*}=0$ and find its convergence rate.

Well, i am stuck in this obviously very easy first part of the problem. Wolfram Alpha shows that $lim_{xrightarrow infty }frac{1}{11}(1-cos(x))=0$, so the sequence converges to the given $x^{*}=0$. Can anyone give me a hint how to solve this limit, please?

Second, the convergence rate. I know the definitions of sublinear convergence, superlinear convergence etc.
I need to calculate the $lim_{xrightarrow infty }frac{left | x_{k+1} -x^{*}right |}{left | x_{k}-x^{*} right |}$, so $lim _{xrightarrow infty } frac{left | frac{1}{11}(1-cos(x)) right |}{left | x right |}=lim_{xrightarrow infty }frac{1-cos(x)}{11x}= 0$. So, the sequnce converges superlinearly. Is the idea correct at all? Can anybody help me with this problem, please?
Thank you in advance!

analysis convergence numerical-methods

share|cite|improve this question

edited Mar 31 '14 at 20:22

Amir Hossein

2,38111750

asked Mar 31 '14 at 20:15

LullabyLullaby

912718

$endgroup$

share|cite|improve this question

edited Mar 31 '14 at 20:22

Amir Hossein

2,38111750

asked Mar 31 '14 at 20:15

LullabyLullaby

912718

share|cite|improve this question

edited Mar 31 '14 at 20:22

Amir Hossein

2,38111750

asked Mar 31 '14 at 20:15

LullabyLullaby

912718

edited Mar 31 '14 at 20:22

Amir Hossein

2,38111750

edited Mar 31 '14 at 20:22

Amir Hossein

2,38111750

asked Mar 31 '14 at 20:15

LullabyLullaby

912718

asked Mar 31 '14 at 20:15

LullabyLullaby

912718

1 Answer
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$begingroup$

I'm not so sure how you managed to obtain that $lim_{x→∞} (1-cos(x))/11 = 0$ because this limit doesn't exist. The graph is forever waving about between $1/11$ and $0$:

But more importantly, the behaviour of this function at infinity is not really relevant when we are talking about starting with $|x_0| < pi/2$ and trying to get a null sequence.

Hint: you can try showing that $1-frac{x^2}{2} < cos x $ and use this to get a bound on $|x_k|$ which should tend to $0$ irrespective of the starting point $x_0 ∈ (-π/2,π/2)setminus {0}$.

This will also give you a bound up to a constant on the convergence rate, so if a reasonable Big-O is all you want, you're done at this point. If you want to show
that there is a quadratic lower bound as well, try scaling $p(x):=-(x-π/2)(x+π/2)$ to have the same maximum height in $(-pi/2 , pi/2)$ as $cos(x)$ and use this to bound $cos(x)$ from above now.

share|cite|improve this answer

answered Mar 31 '14 at 20:54

Calvin KhorCalvin Khor

12.4k21439

$endgroup$

add a comment | 

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1

$begingroup$

I'm not so sure how you managed to obtain that $lim_{x→∞} (1-cos(x))/11 = 0$ because this limit doesn't exist. The graph is forever waving about between $1/11$ and $0$:

But more importantly, the behaviour of this function at infinity is not really relevant when we are talking about starting with $|x_0| < pi/2$ and trying to get a null sequence.

Hint: you can try showing that $1-frac{x^2}{2} < cos x $ and use this to get a bound on $|x_k|$ which should tend to $0$ irrespective of the starting point $x_0 ∈ (-π/2,π/2)setminus {0}$.

This will also give you a bound up to a constant on the convergence rate, so if a reasonable Big-O is all you want, you're done at this point. If you want to show
that there is a quadratic lower bound as well, try scaling $p(x):=-(x-π/2)(x+π/2)$ to have the same maximum height in $(-pi/2 , pi/2)$ as $cos(x)$ and use this to bound $cos(x)$ from above now.

share|cite|improve this answer

answered Mar 31 '14 at 20:54

Calvin KhorCalvin Khor

12.4k21439

$endgroup$

add a comment | 

1

$begingroup$

I'm not so sure how you managed to obtain that $lim_{x→∞} (1-cos(x))/11 = 0$ because this limit doesn't exist. The graph is forever waving about between $1/11$ and $0$:

But more importantly, the behaviour of this function at infinity is not really relevant when we are talking about starting with $|x_0| < pi/2$ and trying to get a null sequence.

Hint: you can try showing that $1-frac{x^2}{2} < cos x $ and use this to get a bound on $|x_k|$ which should tend to $0$ irrespective of the starting point $x_0 ∈ (-π/2,π/2)setminus {0}$.

This will also give you a bound up to a constant on the convergence rate, so if a reasonable Big-O is all you want, you're done at this point. If you want to show
that there is a quadratic lower bound as well, try scaling $p(x):=-(x-π/2)(x+π/2)$ to have the same maximum height in $(-pi/2 , pi/2)$ as $cos(x)$ and use this to bound $cos(x)$ from above now.

share|cite|improve this answer

answered Mar 31 '14 at 20:54

Calvin KhorCalvin Khor

12.4k21439

$endgroup$

add a comment | 

$begingroup$

I'm not so sure how you managed to obtain that $lim_{x→∞} (1-cos(x))/11 = 0$ because this limit doesn't exist. The graph is forever waving about between $1/11$ and $0$:

But more importantly, the behaviour of this function at infinity is not really relevant when we are talking about starting with $|x_0| < pi/2$ and trying to get a null sequence.

Hint: you can try showing that $1-frac{x^2}{2} < cos x $ and use this to get a bound on $|x_k|$ which should tend to $0$ irrespective of the starting point $x_0 ∈ (-π/2,π/2)setminus {0}$.

This will also give you a bound up to a constant on the convergence rate, so if a reasonable Big-O is all you want, you're done at this point. If you want to show
that there is a quadratic lower bound as well, try scaling $p(x):=-(x-π/2)(x+π/2)$ to have the same maximum height in $(-pi/2 , pi/2)$ as $cos(x)$ and use this to bound $cos(x)$ from above now.

share|cite|improve this answer

answered Mar 31 '14 at 20:54

Calvin KhorCalvin Khor

12.4k21439

$endgroup$

share|cite|improve this answer

answered Mar 31 '14 at 20:54

Calvin KhorCalvin Khor

12.4k21439

answered Mar 31 '14 at 20:54

Calvin KhorCalvin Khor

12.4k21439

answered Mar 31 '14 at 20:54

Calvin KhorCalvin Khor

12.4k21439