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Justify: if $xgt 0$, $;lim_ntoinfty sqrtncdotoverbracesinsincdotssin^nspacetextsines(x)=sqrt3$
Clash Royale CLAN TAG#URR8PPP
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I believe that I have managed to show that (if $xgt 0$)
$$lim_ntoinfty sqrtncdotoverbracesinsincdotssin^nspacetextsines(x)=sqrt3$$
I did this by defining a sequence as $a_0=x$ and the recursion
$$a_n+1=sin a_n$$
I then approximated the recursion with the first two nonzero terms of the Maclaurin series for sine, giving me
$$Delta a_n=-fracx^36$$
I then approximated this with a differential equation
$$y'=-fracy^36$$
Which I then easily solved... the answer follows from here.
Question: How can this be made more rigorous? I don't know how to justify that my approximations are good enough for th error to vanish under the limit. What theorems are generally used to justify approximations of discrete recursions with differential equations? I think I know how to justify the approximation of some with its Maclaurin series using the Lagrange error bound.
limits reference-request recurrence-relations approximation
edited Jul 30 at 22:56
amWhy
189k25219431
asked Jul 30 at 15:24
Frpzzd
16.9k63489
 |Â
show 2 more comments
up vote
7
down vote
favorite
I believe that I have managed to show that (if $xgt 0$)
$$lim_ntoinfty sqrtncdotoverbracesinsincdotssin^nspacetextsines(x)=sqrt3$$
I did this by defining a sequence as $a_0=x$ and the recursion
$$a_n+1=sin a_n$$
I then approximated the recursion with the first two nonzero terms of the Maclaurin series for sine, giving me
$$Delta a_n=-fracx^36$$
I then approximated this with a differential equation
$$y'=-fracy^36$$
Which I then easily solved... the answer follows from here.
Question: How can this be made more rigorous? I don't know how to justify that my approximations are good enough for th error to vanish under the limit. What theorems are generally used to justify approximations of discrete recursions with differential equations? I think I know how to justify the approximation of some with its Maclaurin series using the Lagrange error bound.
limits reference-request recurrence-relations approximation
edited Jul 30 at 22:56
amWhy
189k25219431
asked Jul 30 at 15:24
Frpzzd
16.9k63489
@Dzoooks It is a finite difference. $$Delta a_n:=a_n+1-a_n$$
– Frpzzd
Jul 30 at 15:48
1
How is this possible? $$|overbracesin dots sin^n textsines(x)| leq 1,$$ so your limit is 0 by the Squeeze Theorem. Should you be dividing by $sqrtx$?
– Dzoooks
Jul 30 at 15:51
This is true for $x in (0, pi/2): lim a_nsqrt n = sqrt 3$, which could be justified by applying Stolz formula.
– xbh
Jul 30 at 15:53
@xbh and Dzoooks: sorry, thanks for pointing that out. The $sqrt n$ should be multiplied.
– Frpzzd
Jul 30 at 15:56
@xbh Can you explain why this only works for x in that interval? And if you walked me through the application of Stolz and how to carefully justify this, that would be the kind of answer I am looking for. In particular, I would like to justify the fact that the error is small enough to be negligible as n goes to infinity.
– Frpzzd
Jul 30 at 16:00
 |Â
show 2 more comments
up vote
7
down vote
favorite
up vote
7
down vote
favorite
I believe that I have managed to show that (if $xgt 0$)
$$lim_ntoinfty sqrtncdotoverbracesinsincdotssin^nspacetextsines(x)=sqrt3$$
I did this by defining a sequence as $a_0=x$ and the recursion
$$a_n+1=sin a_n$$
I then approximated the recursion with the first two nonzero terms of the Maclaurin series for sine, giving me
$$Delta a_n=-fracx^36$$
I then approximated this with a differential equation
$$y'=-fracy^36$$
Which I then easily solved... the answer follows from here.
Question: How can this be made more rigorous? I don't know how to justify that my approximations are good enough for th error to vanish under the limit. What theorems are generally used to justify approximations of discrete recursions with differential equations? I think I know how to justify the approximation of some with its Maclaurin series using the Lagrange error bound.
limits reference-request recurrence-relations approximation
edited Jul 30 at 22:56
amWhy
189k25219431
asked Jul 30 at 15:24
Frpzzd
16.9k63489
I believe that I have managed to show that (if $xgt 0$)
$$lim_ntoinfty sqrtncdotoverbracesinsincdotssin^nspacetextsines(x)=sqrt3$$
I did this by defining a sequence as $a_0=x$ and the recursion
$$a_n+1=sin a_n$$
I then approximated the recursion with the first two nonzero terms of the Maclaurin series for sine, giving me
$$Delta a_n=-fracx^36$$
I then approximated this with a differential equation
$$y'=-fracy^36$$
Which I then easily solved... the answer follows from here.
Question: How can this be made more rigorous? I don't know how to justify that my approximations are good enough for th error to vanish under the limit. What theorems are generally used to justify approximations of discrete recursions with differential equations? I think I know how to justify the approximation of some with its Maclaurin series using the Lagrange error bound.
limits reference-request recurrence-relations approximation
edited Jul 30 at 22:56
amWhy
189k25219431
asked Jul 30 at 15:24
Frpzzd
16.9k63489
edited Jul 30 at 22:56
amWhy
189k25219431
edited Jul 30 at 22:56
amWhy
189k25219431
edited Jul 30 at 22:56
amWhy
189k25219431
189k25219431
asked Jul 30 at 15:24
Frpzzd
16.9k63489
asked Jul 30 at 15:24
Frpzzd
16.9k63489
asked Jul 30 at 15:24
Frpzzd
16.9k63489
16.9k63489
@Dzoooks It is a finite difference. $$Delta a_n:=a_n+1-a_n$$
– Frpzzd
Jul 30 at 15:48
1
How is this possible? $$|overbracesin dots sin^n textsines(x)| leq 1,$$ so your limit is 0 by the Squeeze Theorem. Should you be dividing by $sqrtx$?
– Dzoooks
Jul 30 at 15:51
This is true for $x in (0, pi/2): lim a_nsqrt n = sqrt 3$, which could be justified by applying Stolz formula.
– xbh
Jul 30 at 15:53
@xbh and Dzoooks: sorry, thanks for pointing that out. The $sqrt n$ should be multiplied.
– Frpzzd
Jul 30 at 15:56
@xbh Can you explain why this only works for x in that interval? And if you walked me through the application of Stolz and how to carefully justify this, that would be the kind of answer I am looking for. In particular, I would like to justify the fact that the error is small enough to be negligible as n goes to infinity.
– Frpzzd
Jul 30 at 16:00
 |Â
show 2 more comments
@Dzoooks It is a finite difference. $$Delta a_n:=a_n+1-a_n$$
– Frpzzd
Jul 30 at 15:48
1
How is this possible? $$|overbracesin dots sin^n textsines(x)| leq 1,$$ so your limit is 0 by the Squeeze Theorem. Should you be dividing by $sqrtx$?
– Dzoooks
Jul 30 at 15:51
This is true for $x in (0, pi/2): lim a_nsqrt n = sqrt 3$, which could be justified by applying Stolz formula.
– xbh
Jul 30 at 15:53
@xbh and Dzoooks: sorry, thanks for pointing that out. The $sqrt n$ should be multiplied.
– Frpzzd
Jul 30 at 15:56
@xbh Can you explain why this only works for x in that interval? And if you walked me through the application of Stolz and how to carefully justify this, that would be the kind of answer I am looking for. In particular, I would like to justify the fact that the error is small enough to be negligible as n goes to infinity.
– Frpzzd
Jul 30 at 16:00
@Dzoooks It is a finite difference. $$Delta a_n:=a_n+1-a_n$$
– Frpzzd
Jul 30 at 15:48
@Dzoooks It is a finite difference. $$Delta a_n:=a_n+1-a_n$$
– Frpzzd
Jul 30 at 15:48
1
1
How is this possible? $$|overbracesin dots sin^n textsines(x)| leq 1,$$ so your limit is 0 by the Squeeze Theorem. Should you be dividing by $sqrtx$?
– Dzoooks
Jul 30 at 15:51
How is this possible? $$|overbracesin dots sin^n textsines(x)| leq 1,$$ so your limit is 0 by the Squeeze Theorem. Should you be dividing by $sqrtx$?
– Dzoooks
Jul 30 at 15:51
This is true for $x in (0, pi/2): lim a_nsqrt n = sqrt 3$, which could be justified by applying Stolz formula.
– xbh
Jul 30 at 15:53
This is true for $x in (0, pi/2): lim a_nsqrt n = sqrt 3$, which could be justified by applying Stolz formula.
– xbh
Jul 30 at 15:53
@xbh and Dzoooks: sorry, thanks for pointing that out. The $sqrt n$ should be multiplied.
– Frpzzd
Jul 30 at 15:56
@xbh and Dzoooks: sorry, thanks for pointing that out. The $sqrt n$ should be multiplied.
– Frpzzd
Jul 30 at 15:56
@xbh Can you explain why this only works for x in that interval? And if you walked me through the application of Stolz and how to carefully justify this, that would be the kind of answer I am looking for. In particular, I would like to justify the fact that the error is small enough to be negligible as n goes to infinity.
– Frpzzd
Jul 30 at 16:00
@xbh Can you explain why this only works for x in that interval? And if you walked me through the application of Stolz and how to carefully justify this, that would be the kind of answer I am looking for. In particular, I would like to justify the fact that the error is small enough to be negligible as n goes to infinity.
– Frpzzd
Jul 30 at 16:00
 |Â
show 2 more comments
4 Answers
4
active
oldest
votes
up vote
7
down vote
This may not be the answer you wanted, cuz it may not be generalized.
Claim: $lim n a_n^2 = 3$.
Proof. $blacktriangleleft$ Note that not matter which value $x$ is, $a_1 = sin (x) in (-1,1)$, so it suffice to consider if $x in (0, pi/2)$. Now $a_1 > 0$ and $a_2 < a_1$. By induction it would be easy to verify that $0 < a_n< a_n-1$. Hence $a_n searrow$ and $a_n > 0$. Therefore the limit $A = lim a_n$ exists. Solve $sin (A) = A$ in $[0, pi/2]$ gives that $A=0$, i.e. $a_n = o(1)$.
Now compute the limit:
beginalign*
lim n a_n^2 &= lim frac n a_n^-2 \
&= lim frac 1 a_n+1^-2 - a_n^-2quad [text Stolz formula]\
&= lim frac a_n^2 a_n+1^2 a_n^2 - a_n+1^2 \
&= lim frac a_n^2 sin(a_n^2) (a_n - sin(a_n) )(a_n + sin (a_n)) \
&= lim frac a_n^4 a_n^3/6 cdot 2a_n quad [sin(a_n) sim a_n; a_n - sin (a_n) sim a_n^3/6]\
&= 3.
endalign*
Hence $lim a_n sqrt n = sqrt 3. blacktriangleright$
answered Jul 30 at 16:13
xbh
1,0257
Even if you start with $x ge pi/2$, already the next iterate $sin x$ is in the interval $[0,pi/2)$, so there's no problem.
– Hans Lundmark
Jul 30 at 18:15
@HansLundmark Thanks, I forgot about this.
– xbh
Jul 30 at 18:30
Very clever use of Cesaro-Stolz. +1
– Paramanand Singh
Jul 31 at 0:43
@ParamanandSingh Thanks!
– xbh
Jul 31 at 2:07
add a comment |Â
up vote
6
down vote
I'll assume $0<x<1$, so that your sequence decreases to zero.
Write
$$b_n=a_n^2=sin^2(sin^[n-1](x))$$
where I'll use $sin^[m](x)$ for the $m$-fold composite of $sin$.
We'd like to prove $b_nsim 3/n$.
But $b_n+1=phi(b_n)$
where $$phi(t)=sin^2sqrt t=t-fract^23+fract^312+cdots.$$
Then
$$frac1b_n+1=frac1b_nleft(1+b_n+O(b_n^2)right)
=frac1b_n+frac13+O(b_n).$$
As $b_nto0$ then $1/b_n$ increases by at least $1/4$ eventually,
so that $1/b_n<n/4+C$ for some constant $C$, and so $b_n=O(1/n)$.
Then$$frac1b_n=frac n3+O(ln n)$$
which is good enough.
answered Jul 30 at 16:05
Lord Shark the Unknown
84.5k950111
Better than mine. Thanks for the elegant analysis!
– xbh
Jul 30 at 16:19
This is hard to understand. How did you obtain the expression for $1/b_n+1$?
– Frpzzd
Jul 30 at 16:47
@Frpzzd By writing $1/b_n+1=1/phi(b_n)$ and taking the first few terms of the Laurent series.
– Lord Shark the Unknown
Jul 30 at 16:49
add a comment |Â
up vote
1
down vote
Since you asked for a reference,
here is one that
I have enjoyed for many years:
Asymptotic Methods in Analysis
by N. G. de Bruijn,
available quite inexpensively at
Dover and Amazon.
Chapter 8 has an extensive discussion
of iterated functions
which includes most of the methods
in the answers to
this question.
answered Aug 2 at 17:15
marty cohen
69.1k446122
Thank you very much! I will have to add this to my "to read" list.
– Frpzzd
Aug 2 at 19:22
add a comment |Â
up vote
0
down vote
We have
$$
sqrt n sin(cdots sin x) = a_n\
sqrtn+1sin(sin(cdots sin x)) = a_n+1
$$
hence
$$
sinleft(fraca_nsqrt nright) = fraca_n+1sqrtn+1
$$
Analyzing now the iterative procedure
$$
u_n+1 = sin u_n
$$
with $u_0 = fracpi2$ the sequence
$$
sqrt k u_k
$$
is convergent and has a behavior shown in the attached plot
(in red $sqrt 3$ and in blue $sqrt k u_k$)
answered Aug 2 at 16:53
Cesareo
5,6012412
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
This may not be the answer you wanted, cuz it may not be generalized.
Claim: $lim n a_n^2 = 3$.
Proof. $blacktriangleleft$ Note that not matter which value $x$ is, $a_1 = sin (x) in (-1,1)$, so it suffice to consider if $x in (0, pi/2)$. Now $a_1 > 0$ and $a_2 < a_1$. By induction it would be easy to verify that $0 < a_n< a_n-1$. Hence $a_n searrow$ and $a_n > 0$. Therefore the limit $A = lim a_n$ exists. Solve $sin (A) = A$ in $[0, pi/2]$ gives that $A=0$, i.e. $a_n = o(1)$.
Now compute the limit:
beginalign*
lim n a_n^2 &= lim frac n a_n^-2 \
&= lim frac 1 a_n+1^-2 - a_n^-2quad [text Stolz formula]\
&= lim frac a_n^2 a_n+1^2 a_n^2 - a_n+1^2 \
&= lim frac a_n^2 sin(a_n^2) (a_n - sin(a_n) )(a_n + sin (a_n)) \
&= lim frac a_n^4 a_n^3/6 cdot 2a_n quad [sin(a_n) sim a_n; a_n - sin (a_n) sim a_n^3/6]\
&= 3.
endalign*
Hence $lim a_n sqrt n = sqrt 3. blacktriangleright$
answered Jul 30 at 16:13
xbh
1,0257
Even if you start with $x ge pi/2$, already the next iterate $sin x$ is in the interval $[0,pi/2)$, so there's no problem.
– Hans Lundmark
Jul 30 at 18:15
@HansLundmark Thanks, I forgot about this.
– xbh
Jul 30 at 18:30
Very clever use of Cesaro-Stolz. +1
– Paramanand Singh
Jul 31 at 0:43
@ParamanandSingh Thanks!
– xbh
Jul 31 at 2:07
add a comment |Â
up vote
7
down vote
This may not be the answer you wanted, cuz it may not be generalized.
Claim: $lim n a_n^2 = 3$.
Proof. $blacktriangleleft$ Note that not matter which value $x$ is, $a_1 = sin (x) in (-1,1)$, so it suffice to consider if $x in (0, pi/2)$. Now $a_1 > 0$ and $a_2 < a_1$. By induction it would be easy to verify that $0 < a_n< a_n-1$. Hence $a_n searrow$ and $a_n > 0$. Therefore the limit $A = lim a_n$ exists. Solve $sin (A) = A$ in $[0, pi/2]$ gives that $A=0$, i.e. $a_n = o(1)$.
Now compute the limit:
beginalign*
lim n a_n^2 &= lim frac n a_n^-2 \
&= lim frac 1 a_n+1^-2 - a_n^-2quad [text Stolz formula]\
&= lim frac a_n^2 a_n+1^2 a_n^2 - a_n+1^2 \
&= lim frac a_n^2 sin(a_n^2) (a_n - sin(a_n) )(a_n + sin (a_n)) \
&= lim frac a_n^4 a_n^3/6 cdot 2a_n quad [sin(a_n) sim a_n; a_n - sin (a_n) sim a_n^3/6]\
&= 3.
endalign*
Hence $lim a_n sqrt n = sqrt 3. blacktriangleright$
answered Jul 30 at 16:13
xbh
1,0257
Even if you start with $x ge pi/2$, already the next iterate $sin x$ is in the interval $[0,pi/2)$, so there's no problem.
– Hans Lundmark
Jul 30 at 18:15
@HansLundmark Thanks, I forgot about this.
– xbh
Jul 30 at 18:30
Very clever use of Cesaro-Stolz. +1
– Paramanand Singh
Jul 31 at 0:43
@ParamanandSingh Thanks!
– xbh
Jul 31 at 2:07
add a comment |Â
up vote
7
down vote
up vote
7
down vote
This may not be the answer you wanted, cuz it may not be generalized.
Claim: $lim n a_n^2 = 3$.
Proof. $blacktriangleleft$ Note that not matter which value $x$ is, $a_1 = sin (x) in (-1,1)$, so it suffice to consider if $x in (0, pi/2)$. Now $a_1 > 0$ and $a_2 < a_1$. By induction it would be easy to verify that $0 < a_n< a_n-1$. Hence $a_n searrow$ and $a_n > 0$. Therefore the limit $A = lim a_n$ exists. Solve $sin (A) = A$ in $[0, pi/2]$ gives that $A=0$, i.e. $a_n = o(1)$.
Now compute the limit:
beginalign*
lim n a_n^2 &= lim frac n a_n^-2 \
&= lim frac 1 a_n+1^-2 - a_n^-2quad [text Stolz formula]\
&= lim frac a_n^2 a_n+1^2 a_n^2 - a_n+1^2 \
&= lim frac a_n^2 sin(a_n^2) (a_n - sin(a_n) )(a_n + sin (a_n)) \
&= lim frac a_n^4 a_n^3/6 cdot 2a_n quad [sin(a_n) sim a_n; a_n - sin (a_n) sim a_n^3/6]\
&= 3.
endalign*
Hence $lim a_n sqrt n = sqrt 3. blacktriangleright$
answered Jul 30 at 16:13
xbh
1,0257
This may not be the answer you wanted, cuz it may not be generalized.
Claim: $lim n a_n^2 = 3$.
Proof. $blacktriangleleft$ Note that not matter which value $x$ is, $a_1 = sin (x) in (-1,1)$, so it suffice to consider if $x in (0, pi/2)$. Now $a_1 > 0$ and $a_2 < a_1$. By induction it would be easy to verify that $0 < a_n< a_n-1$. Hence $a_n searrow$ and $a_n > 0$. Therefore the limit $A = lim a_n$ exists. Solve $sin (A) = A$ in $[0, pi/2]$ gives that $A=0$, i.e. $a_n = o(1)$.
Now compute the limit:
beginalign*
lim n a_n^2 &= lim frac n a_n^-2 \
&= lim frac 1 a_n+1^-2 - a_n^-2quad [text Stolz formula]\
&= lim frac a_n^2 a_n+1^2 a_n^2 - a_n+1^2 \
&= lim frac a_n^2 sin(a_n^2) (a_n - sin(a_n) )(a_n + sin (a_n)) \
&= lim frac a_n^4 a_n^3/6 cdot 2a_n quad [sin(a_n) sim a_n; a_n - sin (a_n) sim a_n^3/6]\
&= 3.
endalign*
Hence $lim a_n sqrt n = sqrt 3. blacktriangleright$
answered Jul 30 at 16:13
xbh
1,0257
edited Jul 30 at 18:34
answered Jul 30 at 16:13
xbh
1,0257
answered Jul 30 at 16:13
xbh
1,0257
answered Jul 30 at 16:13
xbh
1,0257
1,0257
Even if you start with $x ge pi/2$, already the next iterate $sin x$ is in the interval $[0,pi/2)$, so there's no problem.
– Hans Lundmark
Jul 30 at 18:15
@HansLundmark Thanks, I forgot about this.
– xbh
Jul 30 at 18:30
Very clever use of Cesaro-Stolz. +1
– Paramanand Singh
Jul 31 at 0:43
@ParamanandSingh Thanks!
– xbh
Jul 31 at 2:07
add a comment |Â
Even if you start with $x ge pi/2$, already the next iterate $sin x$ is in the interval $[0,pi/2)$, so there's no problem.
– Hans Lundmark
Jul 30 at 18:15
@HansLundmark Thanks, I forgot about this.
– xbh
Jul 30 at 18:30
Very clever use of Cesaro-Stolz. +1
– Paramanand Singh
Jul 31 at 0:43
@ParamanandSingh Thanks!
– xbh
Jul 31 at 2:07
Even if you start with $x ge pi/2$, already the next iterate $sin x$ is in the interval $[0,pi/2)$, so there's no problem.
– Hans Lundmark
Jul 30 at 18:15
Even if you start with $x ge pi/2$, already the next iterate $sin x$ is in the interval $[0,pi/2)$, so there's no problem.
– Hans Lundmark
Jul 30 at 18:15
@HansLundmark Thanks, I forgot about this.
– xbh
Jul 30 at 18:30
@HansLundmark Thanks, I forgot about this.
– xbh
Jul 30 at 18:30
Very clever use of Cesaro-Stolz. +1
– Paramanand Singh
Jul 31 at 0:43
Very clever use of Cesaro-Stolz. +1
– Paramanand Singh
Jul 31 at 0:43
@ParamanandSingh Thanks!
– xbh
Jul 31 at 2:07
@ParamanandSingh Thanks!
– xbh
Jul 31 at 2:07
add a comment |Â
up vote
6
down vote
I'll assume $0<x<1$, so that your sequence decreases to zero.
Write
$$b_n=a_n^2=sin^2(sin^[n-1](x))$$
where I'll use $sin^[m](x)$ for the $m$-fold composite of $sin$.
We'd like to prove $b_nsim 3/n$.
But $b_n+1=phi(b_n)$
where $$phi(t)=sin^2sqrt t=t-fract^23+fract^312+cdots.$$
Then
$$frac1b_n+1=frac1b_nleft(1+b_n+O(b_n^2)right)
=frac1b_n+frac13+O(b_n).$$
As $b_nto0$ then $1/b_n$ increases by at least $1/4$ eventually,
so that $1/b_n<n/4+C$ for some constant $C$, and so $b_n=O(1/n)$.
Then$$frac1b_n=frac n3+O(ln n)$$
which is good enough.
answered Jul 30 at 16:05
Lord Shark the Unknown
84.5k950111
Better than mine. Thanks for the elegant analysis!
– xbh
Jul 30 at 16:19
This is hard to understand. How did you obtain the expression for $1/b_n+1$?
– Frpzzd
Jul 30 at 16:47
@Frpzzd By writing $1/b_n+1=1/phi(b_n)$ and taking the first few terms of the Laurent series.
– Lord Shark the Unknown
Jul 30 at 16:49
add a comment |Â
up vote
6
down vote
I'll assume $0<x<1$, so that your sequence decreases to zero.
Write
$$b_n=a_n^2=sin^2(sin^[n-1](x))$$
where I'll use $sin^[m](x)$ for the $m$-fold composite of $sin$.
We'd like to prove $b_nsim 3/n$.
But $b_n+1=phi(b_n)$
where $$phi(t)=sin^2sqrt t=t-fract^23+fract^312+cdots.$$
Then
$$frac1b_n+1=frac1b_nleft(1+b_n+O(b_n^2)right)
=frac1b_n+frac13+O(b_n).$$
As $b_nto0$ then $1/b_n$ increases by at least $1/4$ eventually,
so that $1/b_n<n/4+C$ for some constant $C$, and so $b_n=O(1/n)$.
Then$$frac1b_n=frac n3+O(ln n)$$
which is good enough.
answered Jul 30 at 16:05
Lord Shark the Unknown
84.5k950111
Better than mine. Thanks for the elegant analysis!
– xbh
Jul 30 at 16:19
This is hard to understand. How did you obtain the expression for $1/b_n+1$?
– Frpzzd
Jul 30 at 16:47
@Frpzzd By writing $1/b_n+1=1/phi(b_n)$ and taking the first few terms of the Laurent series.
– Lord Shark the Unknown
Jul 30 at 16:49
add a comment |Â
up vote
6
down vote
up vote
6
down vote
I'll assume $0<x<1$, so that your sequence decreases to zero.
Write
$$b_n=a_n^2=sin^2(sin^[n-1](x))$$
where I'll use $sin^[m](x)$ for the $m$-fold composite of $sin$.
We'd like to prove $b_nsim 3/n$.
But $b_n+1=phi(b_n)$
where $$phi(t)=sin^2sqrt t=t-fract^23+fract^312+cdots.$$
Then
$$frac1b_n+1=frac1b_nleft(1+b_n+O(b_n^2)right)
=frac1b_n+frac13+O(b_n).$$
As $b_nto0$ then $1/b_n$ increases by at least $1/4$ eventually,
so that $1/b_n<n/4+C$ for some constant $C$, and so $b_n=O(1/n)$.
Then$$frac1b_n=frac n3+O(ln n)$$
which is good enough.
answered Jul 30 at 16:05
Lord Shark the Unknown
84.5k950111
I'll assume $0<x<1$, so that your sequence decreases to zero.
Write
$$b_n=a_n^2=sin^2(sin^[n-1](x))$$
where I'll use $sin^[m](x)$ for the $m$-fold composite of $sin$.
We'd like to prove $b_nsim 3/n$.
But $b_n+1=phi(b_n)$
where $$phi(t)=sin^2sqrt t=t-fract^23+fract^312+cdots.$$
Then
$$frac1b_n+1=frac1b_nleft(1+b_n+O(b_n^2)right)
=frac1b_n+frac13+O(b_n).$$
As $b_nto0$ then $1/b_n$ increases by at least $1/4$ eventually,
so that $1/b_n<n/4+C$ for some constant $C$, and so $b_n=O(1/n)$.
Then$$frac1b_n=frac n3+O(ln n)$$
which is good enough.
answered Jul 30 at 16:05
Lord Shark the Unknown
84.5k950111
answered Jul 30 at 16:05
Lord Shark the Unknown
84.5k950111
answered Jul 30 at 16:05
Lord Shark the Unknown
84.5k950111
answered Jul 30 at 16:05
Lord Shark the Unknown
84.5k950111
84.5k950111
Better than mine. Thanks for the elegant analysis!
– xbh
Jul 30 at 16:19
This is hard to understand. How did you obtain the expression for $1/b_n+1$?
– Frpzzd
Jul 30 at 16:47
@Frpzzd By writing $1/b_n+1=1/phi(b_n)$ and taking the first few terms of the Laurent series.
– Lord Shark the Unknown
Jul 30 at 16:49
add a comment |Â
Better than mine. Thanks for the elegant analysis!
– xbh
Jul 30 at 16:19
This is hard to understand. How did you obtain the expression for $1/b_n+1$?
– Frpzzd
Jul 30 at 16:47
@Frpzzd By writing $1/b_n+1=1/phi(b_n)$ and taking the first few terms of the Laurent series.
– Lord Shark the Unknown
Jul 30 at 16:49
Better than mine. Thanks for the elegant analysis!
– xbh
Jul 30 at 16:19
Better than mine. Thanks for the elegant analysis!
– xbh
Jul 30 at 16:19
This is hard to understand. How did you obtain the expression for $1/b_n+1$?
– Frpzzd
Jul 30 at 16:47
This is hard to understand. How did you obtain the expression for $1/b_n+1$?
– Frpzzd
Jul 30 at 16:47
@Frpzzd By writing $1/b_n+1=1/phi(b_n)$ and taking the first few terms of the Laurent series.
– Lord Shark the Unknown
Jul 30 at 16:49
@Frpzzd By writing $1/b_n+1=1/phi(b_n)$ and taking the first few terms of the Laurent series.
– Lord Shark the Unknown
Jul 30 at 16:49
add a comment |Â
up vote
1
down vote
Since you asked for a reference,
here is one that
I have enjoyed for many years:
Asymptotic Methods in Analysis
by N. G. de Bruijn,
available quite inexpensively at
Dover and Amazon.
Chapter 8 has an extensive discussion
of iterated functions
which includes most of the methods
in the answers to
this question.
answered Aug 2 at 17:15
marty cohen
69.1k446122
Thank you very much! I will have to add this to my "to read" list.
– Frpzzd
Aug 2 at 19:22
add a comment |Â
up vote
1
down vote
Since you asked for a reference,
here is one that
I have enjoyed for many years:
Asymptotic Methods in Analysis
by N. G. de Bruijn,
available quite inexpensively at
Dover and Amazon.
Chapter 8 has an extensive discussion
of iterated functions
which includes most of the methods
in the answers to
this question.
answered Aug 2 at 17:15
marty cohen
69.1k446122
Thank you very much! I will have to add this to my "to read" list.
– Frpzzd
Aug 2 at 19:22
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Since you asked for a reference,
here is one that
I have enjoyed for many years:
Asymptotic Methods in Analysis
by N. G. de Bruijn,
available quite inexpensively at
Dover and Amazon.
Chapter 8 has an extensive discussion
of iterated functions
which includes most of the methods
in the answers to
this question.
answered Aug 2 at 17:15
marty cohen
69.1k446122
Since you asked for a reference,
here is one that
I have enjoyed for many years:
Asymptotic Methods in Analysis
by N. G. de Bruijn,
available quite inexpensively at
Dover and Amazon.
Chapter 8 has an extensive discussion
of iterated functions
which includes most of the methods
in the answers to
this question.
answered Aug 2 at 17:15
marty cohen
69.1k446122
answered Aug 2 at 17:15
marty cohen
69.1k446122
answered Aug 2 at 17:15
marty cohen
69.1k446122
answered Aug 2 at 17:15
marty cohen
69.1k446122
69.1k446122
Thank you very much! I will have to add this to my "to read" list.
– Frpzzd
Aug 2 at 19:22
add a comment |Â
Thank you very much! I will have to add this to my "to read" list.
– Frpzzd
Aug 2 at 19:22
Thank you very much! I will have to add this to my "to read" list.
– Frpzzd
Aug 2 at 19:22
Thank you very much! I will have to add this to my "to read" list.
– Frpzzd
Aug 2 at 19:22
add a comment |Â
up vote
0
down vote
We have
$$
sqrt n sin(cdots sin x) = a_n\
sqrtn+1sin(sin(cdots sin x)) = a_n+1
$$
hence
$$
sinleft(fraca_nsqrt nright) = fraca_n+1sqrtn+1
$$
Analyzing now the iterative procedure
$$
u_n+1 = sin u_n
$$
with $u_0 = fracpi2$ the sequence
$$
sqrt k u_k
$$
is convergent and has a behavior shown in the attached plot
(in red $sqrt 3$ and in blue $sqrt k u_k$)
answered Aug 2 at 16:53
Cesareo
5,6012412
add a comment |Â
up vote
0
down vote
We have
$$
sqrt n sin(cdots sin x) = a_n\
sqrtn+1sin(sin(cdots sin x)) = a_n+1
$$
hence
$$
sinleft(fraca_nsqrt nright) = fraca_n+1sqrtn+1
$$
Analyzing now the iterative procedure
$$
u_n+1 = sin u_n
$$
with $u_0 = fracpi2$ the sequence
$$
sqrt k u_k
$$
is convergent and has a behavior shown in the attached plot
(in red $sqrt 3$ and in blue $sqrt k u_k$)
answered Aug 2 at 16:53
Cesareo
5,6012412
add a comment |Â
up vote
0
down vote
up vote
0
down vote
We have
$$
sqrt n sin(cdots sin x) = a_n\
sqrtn+1sin(sin(cdots sin x)) = a_n+1
$$
hence
$$
sinleft(fraca_nsqrt nright) = fraca_n+1sqrtn+1
$$
Analyzing now the iterative procedure
$$
u_n+1 = sin u_n
$$
with $u_0 = fracpi2$ the sequence
$$
sqrt k u_k
$$
is convergent and has a behavior shown in the attached plot
(in red $sqrt 3$ and in blue $sqrt k u_k$)
answered Aug 2 at 16:53
Cesareo
5,6012412
We have
$$
sqrt n sin(cdots sin x) = a_n\
sqrtn+1sin(sin(cdots sin x)) = a_n+1
$$
hence
$$
sinleft(fraca_nsqrt nright) = fraca_n+1sqrtn+1
$$
Analyzing now the iterative procedure
$$
u_n+1 = sin u_n
$$
with $u_0 = fracpi2$ the sequence
$$
sqrt k u_k
$$
is convergent and has a behavior shown in the attached plot
(in red $sqrt 3$ and in blue $sqrt k u_k$)
answered Aug 2 at 16:53
Cesareo
5,6012412
edited Aug 2 at 16:59
answered Aug 2 at 16:53
Cesareo
5,6012412
answered Aug 2 at 16:53
Cesareo
5,6012412
answered Aug 2 at 16:53
Cesareo
5,6012412
5,6012412
add a comment |Â
add a comment |Â
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@Dzoooks It is a finite difference. $$Delta a_n:=a_n+1-a_n$$
– Frpzzd
Jul 30 at 15:48
1
How is this possible? $$|overbracesin dots sin^n textsines(x)| leq 1,$$ so your limit is 0 by the Squeeze Theorem. Should you be dividing by $sqrtx$?
– Dzoooks
Jul 30 at 15:51
This is true for $x in (0, pi/2): lim a_nsqrt n = sqrt 3$, which could be justified by applying Stolz formula.
– xbh
Jul 30 at 15:53
@xbh and Dzoooks: sorry, thanks for pointing that out. The $sqrt n$ should be multiplied.
– Frpzzd
Jul 30 at 15:56
@xbh Can you explain why this only works for x in that interval? And if you walked me through the application of Stolz and how to carefully justify this, that would be the kind of answer I am looking for. In particular, I would like to justify the fact that the error is small enough to be negligible as n goes to infinity.
– Frpzzd
Jul 30 at 16:00