Limit of $a_n^2$ when the sequence is recursive
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
The problem states:
Given the sequence:
$$a_0 in left(0,1right)$$
$$a_n+1= 1 - sqrt 1-a_n^2$$
First part.
Check whether it converges, and if so find its limit.
Second part.
Find:
$$lim_n to infty frac a_n^2a_n+1$$
The first part is clear to me, and I've done it. I've proven that the sequence is bounded (0 from below and 1 from above), and also that it is monotone decreasing, so the sequence converges to 0.
The second part is what confuses me. If the sequence is given recursively, how do I know what $a_n$ even is? Furthermore, how do I put it in a limit that I'm supposed to evaluate?
Any ideas?
Thanks.
real-analysis sequences-and-series convergence
edited Aug 28 at 13:50
amWhy
190k26221433
asked Aug 28 at 13:28
Koy
965
add a comment |Â
up vote
1
down vote
favorite
The problem states:
Given the sequence:
$$a_0 in left(0,1right)$$
$$a_n+1= 1 - sqrt 1-a_n^2$$
First part.
Check whether it converges, and if so find its limit.
Second part.
Find:
$$lim_n to infty frac a_n^2a_n+1$$
The first part is clear to me, and I've done it. I've proven that the sequence is bounded (0 from below and 1 from above), and also that it is monotone decreasing, so the sequence converges to 0.
The second part is what confuses me. If the sequence is given recursively, how do I know what $a_n$ even is? Furthermore, how do I put it in a limit that I'm supposed to evaluate?
Any ideas?
Thanks.
real-analysis sequences-and-series convergence
edited Aug 28 at 13:50
amWhy
190k26221433
asked Aug 28 at 13:28
Koy
965
3
Do not use embedded images please. Instead, please write your question with mathjax (in both the title and in the body of the question).
– Lee Mosher
Aug 28 at 13:37
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
The problem states:
Given the sequence:
$$a_0 in left(0,1right)$$
$$a_n+1= 1 - sqrt 1-a_n^2$$
First part.
Check whether it converges, and if so find its limit.
Second part.
Find:
$$lim_n to infty frac a_n^2a_n+1$$
The first part is clear to me, and I've done it. I've proven that the sequence is bounded (0 from below and 1 from above), and also that it is monotone decreasing, so the sequence converges to 0.
The second part is what confuses me. If the sequence is given recursively, how do I know what $a_n$ even is? Furthermore, how do I put it in a limit that I'm supposed to evaluate?
Any ideas?
Thanks.
real-analysis sequences-and-series convergence
edited Aug 28 at 13:50
amWhy
190k26221433
asked Aug 28 at 13:28
Koy
965
The problem states:
Given the sequence:
$$a_0 in left(0,1right)$$
$$a_n+1= 1 - sqrt 1-a_n^2$$
First part.
Check whether it converges, and if so find its limit.
Second part.
Find:
$$lim_n to infty frac a_n^2a_n+1$$
The first part is clear to me, and I've done it. I've proven that the sequence is bounded (0 from below and 1 from above), and also that it is monotone decreasing, so the sequence converges to 0.
The second part is what confuses me. If the sequence is given recursively, how do I know what $a_n$ even is? Furthermore, how do I put it in a limit that I'm supposed to evaluate?
Any ideas?
Thanks.
real-analysis sequences-and-series convergence
edited Aug 28 at 13:50
amWhy
190k26221433
asked Aug 28 at 13:28
Koy
965
edited Aug 28 at 13:50
amWhy
190k26221433
edited Aug 28 at 13:50
amWhy
190k26221433
edited Aug 28 at 13:50
amWhy
190k26221433
190k26221433
asked Aug 28 at 13:28
Koy
965
asked Aug 28 at 13:28
Koy
965
asked Aug 28 at 13:28
Koy
965
965
3
Do not use embedded images please. Instead, please write your question with mathjax (in both the title and in the body of the question).
– Lee Mosher
Aug 28 at 13:37
add a comment |Â
3
Do not use embedded images please. Instead, please write your question with mathjax (in both the title and in the body of the question).
– Lee Mosher
Aug 28 at 13:37
3
3
Do not use embedded images please. Instead, please write your question with mathjax (in both the title and in the body of the question).
– Lee Mosher
Aug 28 at 13:37
Do not use embedded images please. Instead, please write your question with mathjax (in both the title and in the body of the question).
– Lee Mosher
Aug 28 at 13:37
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
5
down vote
accepted
I would like to point out that the fact that $a_n$ is bounded below by $0$ and decreasing only implies that it converges and that its limit $L$ satisfies $Lgeqslant 0$.
In fact, in this case $L$ must be $0$, but this must be deduced, for instance via
beginalign
L = 1 - sqrt1-L^2
&implies (L-1) = -sqrt1-L^2
\
&implies (L-1)^2 = 1-L^2
\
&implies 2L^2-2L = 0
\&implies L(L-1) = 0 implies L=0 text or L=1.
endalign
As for the second part, consider that your recurrence relation implies
beginalign
&(a_n+1-1) = -sqrt1-a_n^2
\
implies &(a_n+1-1)^2 = 1-a_n^2
\
implies &a_n^2 = 1 - (a_n+1^2-2a_n+1+1) = 2a_n+1 - a_n+1^2.
endalign
It follows that
beginalign
lim_ntoinftyfraca_n^2a_n+1
&=
lim_ntoinftyfrac2a_n+1 - a_n+1^2a_n+1
\&=
lim_ntoinfty 2 - a_n+1 = 2,
endalign
where the last step follows from your previous calculation.
answered Aug 28 at 13:43
Fimpellizieri
16.9k11835
@OscarLanzi If $a_n in A$ for all $n$ and $lim_ntoinfty a_n$ exists, then $lim_ntoinfty a_n in overlineA$ $($the closure of $A)$. In our case, $A = (0,1)$, so $L in [0,1]$, ie, $0leqslant L leqslant 1$.
– Fimpellizieri
Aug 28 at 15:51
Got my signs mixed up, but partly because I am led down the garden path. If $Lge 0$ and the choices are $0$ and $1$, we miss a step that eliminates the latter. Put $0le L le a_1$ instead of just $Lge 0$ and things go smoothly.
– Oscar Lanzi
Aug 28 at 15:55
If $a_n in (0,1)$ for all $n$, $(a_n)$ is decreasing and $L=0$ or $1$, then the choice is obvious.
– Fimpellizieri
Aug 28 at 16:15
add a comment |Â
up vote
2
down vote
Since $a_n to 0$ we have
$$fraca_n^2a_n+1=fraca_n^21-sqrt1-a_n^2=fraca_n^21-sqrt1-a_n^2,frac1+sqrt1-a_n^21+sqrt1-a_n^2=$$$$=fraca_n^2left(1+sqrt1-a_n^2right)1-1+a_n^2=1+sqrt1-a_n^2to 2$$
answered Aug 28 at 13:37
gimusi
71k73786
I would argue that the $O$-notation is not reasonable for a task obviously given to someone who recently started doing actual math. While what you said is true, it doesn't really help the understanding imho.
– Sellerie
Aug 28 at 13:48
@Sellerie I'm using little-o notation. If the OP is not aware about that we can avoid that using the limit $frac1-sqrt1-xx to frac 12$.
– gimusi
Aug 28 at 13:52
@Sellerie I've simplified all the derivation without little-o notation.
– gimusi
Aug 28 at 14:00
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
I would like to point out that the fact that $a_n$ is bounded below by $0$ and decreasing only implies that it converges and that its limit $L$ satisfies $Lgeqslant 0$.
In fact, in this case $L$ must be $0$, but this must be deduced, for instance via
beginalign
L = 1 - sqrt1-L^2
&implies (L-1) = -sqrt1-L^2
\
&implies (L-1)^2 = 1-L^2
\
&implies 2L^2-2L = 0
\&implies L(L-1) = 0 implies L=0 text or L=1.
endalign
As for the second part, consider that your recurrence relation implies
beginalign
&(a_n+1-1) = -sqrt1-a_n^2
\
implies &(a_n+1-1)^2 = 1-a_n^2
\
implies &a_n^2 = 1 - (a_n+1^2-2a_n+1+1) = 2a_n+1 - a_n+1^2.
endalign
It follows that
beginalign
lim_ntoinftyfraca_n^2a_n+1
&=
lim_ntoinftyfrac2a_n+1 - a_n+1^2a_n+1
\&=
lim_ntoinfty 2 - a_n+1 = 2,
endalign
where the last step follows from your previous calculation.
answered Aug 28 at 13:43
Fimpellizieri
16.9k11835
@OscarLanzi If $a_n in A$ for all $n$ and $lim_ntoinfty a_n$ exists, then $lim_ntoinfty a_n in overlineA$ $($the closure of $A)$. In our case, $A = (0,1)$, so $L in [0,1]$, ie, $0leqslant L leqslant 1$.
– Fimpellizieri
Aug 28 at 15:51
Got my signs mixed up, but partly because I am led down the garden path. If $Lge 0$ and the choices are $0$ and $1$, we miss a step that eliminates the latter. Put $0le L le a_1$ instead of just $Lge 0$ and things go smoothly.
– Oscar Lanzi
Aug 28 at 15:55
If $a_n in (0,1)$ for all $n$, $(a_n)$ is decreasing and $L=0$ or $1$, then the choice is obvious.
– Fimpellizieri
Aug 28 at 16:15
add a comment |Â
up vote
5
down vote
accepted
I would like to point out that the fact that $a_n$ is bounded below by $0$ and decreasing only implies that it converges and that its limit $L$ satisfies $Lgeqslant 0$.
In fact, in this case $L$ must be $0$, but this must be deduced, for instance via
beginalign
L = 1 - sqrt1-L^2
&implies (L-1) = -sqrt1-L^2
\
&implies (L-1)^2 = 1-L^2
\
&implies 2L^2-2L = 0
\&implies L(L-1) = 0 implies L=0 text or L=1.
endalign
As for the second part, consider that your recurrence relation implies
beginalign
&(a_n+1-1) = -sqrt1-a_n^2
\
implies &(a_n+1-1)^2 = 1-a_n^2
\
implies &a_n^2 = 1 - (a_n+1^2-2a_n+1+1) = 2a_n+1 - a_n+1^2.
endalign
It follows that
beginalign
lim_ntoinftyfraca_n^2a_n+1
&=
lim_ntoinftyfrac2a_n+1 - a_n+1^2a_n+1
\&=
lim_ntoinfty 2 - a_n+1 = 2,
endalign
where the last step follows from your previous calculation.
answered Aug 28 at 13:43
Fimpellizieri
16.9k11835
@OscarLanzi If $a_n in A$ for all $n$ and $lim_ntoinfty a_n$ exists, then $lim_ntoinfty a_n in overlineA$ $($the closure of $A)$. In our case, $A = (0,1)$, so $L in [0,1]$, ie, $0leqslant L leqslant 1$.
– Fimpellizieri
Aug 28 at 15:51
Got my signs mixed up, but partly because I am led down the garden path. If $Lge 0$ and the choices are $0$ and $1$, we miss a step that eliminates the latter. Put $0le L le a_1$ instead of just $Lge 0$ and things go smoothly.
– Oscar Lanzi
Aug 28 at 15:55
If $a_n in (0,1)$ for all $n$, $(a_n)$ is decreasing and $L=0$ or $1$, then the choice is obvious.
– Fimpellizieri
Aug 28 at 16:15
add a comment |Â
up vote
5
down vote
accepted
up vote
5
down vote
accepted
I would like to point out that the fact that $a_n$ is bounded below by $0$ and decreasing only implies that it converges and that its limit $L$ satisfies $Lgeqslant 0$.
In fact, in this case $L$ must be $0$, but this must be deduced, for instance via
beginalign
L = 1 - sqrt1-L^2
&implies (L-1) = -sqrt1-L^2
\
&implies (L-1)^2 = 1-L^2
\
&implies 2L^2-2L = 0
\&implies L(L-1) = 0 implies L=0 text or L=1.
endalign
As for the second part, consider that your recurrence relation implies
beginalign
&(a_n+1-1) = -sqrt1-a_n^2
\
implies &(a_n+1-1)^2 = 1-a_n^2
\
implies &a_n^2 = 1 - (a_n+1^2-2a_n+1+1) = 2a_n+1 - a_n+1^2.
endalign
It follows that
beginalign
lim_ntoinftyfraca_n^2a_n+1
&=
lim_ntoinftyfrac2a_n+1 - a_n+1^2a_n+1
\&=
lim_ntoinfty 2 - a_n+1 = 2,
endalign
where the last step follows from your previous calculation.
answered Aug 28 at 13:43
Fimpellizieri
16.9k11835
I would like to point out that the fact that $a_n$ is bounded below by $0$ and decreasing only implies that it converges and that its limit $L$ satisfies $Lgeqslant 0$.
In fact, in this case $L$ must be $0$, but this must be deduced, for instance via
beginalign
L = 1 - sqrt1-L^2
&implies (L-1) = -sqrt1-L^2
\
&implies (L-1)^2 = 1-L^2
\
&implies 2L^2-2L = 0
\&implies L(L-1) = 0 implies L=0 text or L=1.
endalign
As for the second part, consider that your recurrence relation implies
beginalign
&(a_n+1-1) = -sqrt1-a_n^2
\
implies &(a_n+1-1)^2 = 1-a_n^2
\
implies &a_n^2 = 1 - (a_n+1^2-2a_n+1+1) = 2a_n+1 - a_n+1^2.
endalign
It follows that
beginalign
lim_ntoinftyfraca_n^2a_n+1
&=
lim_ntoinftyfrac2a_n+1 - a_n+1^2a_n+1
\&=
lim_ntoinfty 2 - a_n+1 = 2,
endalign
where the last step follows from your previous calculation.
answered Aug 28 at 13:43
Fimpellizieri
16.9k11835
answered Aug 28 at 13:43
Fimpellizieri
16.9k11835
answered Aug 28 at 13:43
Fimpellizieri
16.9k11835
answered Aug 28 at 13:43
Fimpellizieri
16.9k11835
16.9k11835
@OscarLanzi If $a_n in A$ for all $n$ and $lim_ntoinfty a_n$ exists, then $lim_ntoinfty a_n in overlineA$ $($the closure of $A)$. In our case, $A = (0,1)$, so $L in [0,1]$, ie, $0leqslant L leqslant 1$.
– Fimpellizieri
Aug 28 at 15:51
Got my signs mixed up, but partly because I am led down the garden path. If $Lge 0$ and the choices are $0$ and $1$, we miss a step that eliminates the latter. Put $0le L le a_1$ instead of just $Lge 0$ and things go smoothly.
– Oscar Lanzi
Aug 28 at 15:55
If $a_n in (0,1)$ for all $n$, $(a_n)$ is decreasing and $L=0$ or $1$, then the choice is obvious.
– Fimpellizieri
Aug 28 at 16:15
add a comment |Â
@OscarLanzi If $a_n in A$ for all $n$ and $lim_ntoinfty a_n$ exists, then $lim_ntoinfty a_n in overlineA$ $($the closure of $A)$. In our case, $A = (0,1)$, so $L in [0,1]$, ie, $0leqslant L leqslant 1$.
– Fimpellizieri
Aug 28 at 15:51
Got my signs mixed up, but partly because I am led down the garden path. If $Lge 0$ and the choices are $0$ and $1$, we miss a step that eliminates the latter. Put $0le L le a_1$ instead of just $Lge 0$ and things go smoothly.
– Oscar Lanzi
Aug 28 at 15:55
If $a_n in (0,1)$ for all $n$, $(a_n)$ is decreasing and $L=0$ or $1$, then the choice is obvious.
– Fimpellizieri
Aug 28 at 16:15
@OscarLanzi If $a_n in A$ for all $n$ and $lim_ntoinfty a_n$ exists, then $lim_ntoinfty a_n in overlineA$ $($the closure of $A)$. In our case, $A = (0,1)$, so $L in [0,1]$, ie, $0leqslant L leqslant 1$.
– Fimpellizieri
Aug 28 at 15:51
@OscarLanzi If $a_n in A$ for all $n$ and $lim_ntoinfty a_n$ exists, then $lim_ntoinfty a_n in overlineA$ $($the closure of $A)$. In our case, $A = (0,1)$, so $L in [0,1]$, ie, $0leqslant L leqslant 1$.
– Fimpellizieri
Aug 28 at 15:51
Got my signs mixed up, but partly because I am led down the garden path. If $Lge 0$ and the choices are $0$ and $1$, we miss a step that eliminates the latter. Put $0le L le a_1$ instead of just $Lge 0$ and things go smoothly.
– Oscar Lanzi
Aug 28 at 15:55
Got my signs mixed up, but partly because I am led down the garden path. If $Lge 0$ and the choices are $0$ and $1$, we miss a step that eliminates the latter. Put $0le L le a_1$ instead of just $Lge 0$ and things go smoothly.
– Oscar Lanzi
Aug 28 at 15:55
If $a_n in (0,1)$ for all $n$, $(a_n)$ is decreasing and $L=0$ or $1$, then the choice is obvious.
– Fimpellizieri
Aug 28 at 16:15
If $a_n in (0,1)$ for all $n$, $(a_n)$ is decreasing and $L=0$ or $1$, then the choice is obvious.
– Fimpellizieri
Aug 28 at 16:15
add a comment |Â
up vote
2
down vote
Since $a_n to 0$ we have
$$fraca_n^2a_n+1=fraca_n^21-sqrt1-a_n^2=fraca_n^21-sqrt1-a_n^2,frac1+sqrt1-a_n^21+sqrt1-a_n^2=$$$$=fraca_n^2left(1+sqrt1-a_n^2right)1-1+a_n^2=1+sqrt1-a_n^2to 2$$
answered Aug 28 at 13:37
gimusi
71k73786
I would argue that the $O$-notation is not reasonable for a task obviously given to someone who recently started doing actual math. While what you said is true, it doesn't really help the understanding imho.
– Sellerie
Aug 28 at 13:48
@Sellerie I'm using little-o notation. If the OP is not aware about that we can avoid that using the limit $frac1-sqrt1-xx to frac 12$.
– gimusi
Aug 28 at 13:52
@Sellerie I've simplified all the derivation without little-o notation.
– gimusi
Aug 28 at 14:00
add a comment |Â
up vote
2
down vote
Since $a_n to 0$ we have
$$fraca_n^2a_n+1=fraca_n^21-sqrt1-a_n^2=fraca_n^21-sqrt1-a_n^2,frac1+sqrt1-a_n^21+sqrt1-a_n^2=$$$$=fraca_n^2left(1+sqrt1-a_n^2right)1-1+a_n^2=1+sqrt1-a_n^2to 2$$
answered Aug 28 at 13:37
gimusi
71k73786
I would argue that the $O$-notation is not reasonable for a task obviously given to someone who recently started doing actual math. While what you said is true, it doesn't really help the understanding imho.
– Sellerie
Aug 28 at 13:48
@Sellerie I'm using little-o notation. If the OP is not aware about that we can avoid that using the limit $frac1-sqrt1-xx to frac 12$.
– gimusi
Aug 28 at 13:52
@Sellerie I've simplified all the derivation without little-o notation.
– gimusi
Aug 28 at 14:00
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Since $a_n to 0$ we have
$$fraca_n^2a_n+1=fraca_n^21-sqrt1-a_n^2=fraca_n^21-sqrt1-a_n^2,frac1+sqrt1-a_n^21+sqrt1-a_n^2=$$$$=fraca_n^2left(1+sqrt1-a_n^2right)1-1+a_n^2=1+sqrt1-a_n^2to 2$$
answered Aug 28 at 13:37
gimusi
71k73786
Since $a_n to 0$ we have
$$fraca_n^2a_n+1=fraca_n^21-sqrt1-a_n^2=fraca_n^21-sqrt1-a_n^2,frac1+sqrt1-a_n^21+sqrt1-a_n^2=$$$$=fraca_n^2left(1+sqrt1-a_n^2right)1-1+a_n^2=1+sqrt1-a_n^2to 2$$
answered Aug 28 at 13:37
gimusi
71k73786
edited Aug 28 at 13:55
answered Aug 28 at 13:37
gimusi
71k73786
answered Aug 28 at 13:37
gimusi
71k73786
answered Aug 28 at 13:37
gimusi
71k73786
71k73786
I would argue that the $O$-notation is not reasonable for a task obviously given to someone who recently started doing actual math. While what you said is true, it doesn't really help the understanding imho.
– Sellerie
Aug 28 at 13:48
@Sellerie I'm using little-o notation. If the OP is not aware about that we can avoid that using the limit $frac1-sqrt1-xx to frac 12$.
– gimusi
Aug 28 at 13:52
@Sellerie I've simplified all the derivation without little-o notation.
– gimusi
Aug 28 at 14:00
add a comment |Â
I would argue that the $O$-notation is not reasonable for a task obviously given to someone who recently started doing actual math. While what you said is true, it doesn't really help the understanding imho.
– Sellerie
Aug 28 at 13:48
@Sellerie I'm using little-o notation. If the OP is not aware about that we can avoid that using the limit $frac1-sqrt1-xx to frac 12$.
– gimusi
Aug 28 at 13:52
@Sellerie I've simplified all the derivation without little-o notation.
– gimusi
Aug 28 at 14:00
I would argue that the $O$-notation is not reasonable for a task obviously given to someone who recently started doing actual math. While what you said is true, it doesn't really help the understanding imho.
– Sellerie
Aug 28 at 13:48
I would argue that the $O$-notation is not reasonable for a task obviously given to someone who recently started doing actual math. While what you said is true, it doesn't really help the understanding imho.
– Sellerie
Aug 28 at 13:48
@Sellerie I'm using little-o notation. If the OP is not aware about that we can avoid that using the limit $frac1-sqrt1-xx to frac 12$.
– gimusi
Aug 28 at 13:52
@Sellerie I'm using little-o notation. If the OP is not aware about that we can avoid that using the limit $frac1-sqrt1-xx to frac 12$.
– gimusi
Aug 28 at 13:52
@Sellerie I've simplified all the derivation without little-o notation.
– gimusi
Aug 28 at 14:00
@Sellerie I've simplified all the derivation without little-o notation.
– gimusi
Aug 28 at 14:00
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2897265%2flimit-of-a-n2-when-the-sequence-is-recursive%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
3
Do not use embedded images please. Instead, please write your question with mathjax (in both the title and in the body of the question).
– Lee Mosher
Aug 28 at 13:37