Mixing
Baby Rudin (“Principles of Mathematical Analysis”), chapter 3 exercise 3 [duplicate]
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This question already has an answer here:
Prob. 3, Chap. 3 in Baby Rudin: If $s_1 = sqrt{2}$, and $s_{n+1} = sqrt{2 + sqrt{s_n}}$, what is the limit of this sequence?
3 answers
If $s_1=sqrt 2$,
and $$s_{n+1}=sqrt {2+sqrt{s_n}}$$
Prove that ${s_n}$ is converging, and that $s_n<2$ for all $n=1,2ldots$
My attempt to use mathematical induction: if $s_n<2$ then I use
$s_{n+1}=sqrt {2+sqrt s_n}$, to prove $s_{n+1}<sqrt {2+2}=2$.
I guess $s_n$ is an increasing sequence, I can use the $s_n$ is bounded and increasing, and then prove that $s_n$ converges.
real-analysis calculus sequences-and-series limits analysis
edited Feb 2 at 14:49
Peter Mortensen
561310
asked Feb 2 at 9:35
jacksonjackson
1499
$endgroup$
marked as duplicate by rtybase, stressed out, user21820, Trevor Gunn, StubbornAtom Feb 2 at 22:37
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
|
show 1 more comment
$begingroup$
This question already has an answer here:
Prob. 3, Chap. 3 in Baby Rudin: If $s_1 = sqrt{2}$, and $s_{n+1} = sqrt{2 + sqrt{s_n}}$, what is the limit of this sequence?
3 answers
If $s_1=sqrt 2$,
and $$s_{n+1}=sqrt {2+sqrt{s_n}}$$
Prove that ${s_n}$ is converging, and that $s_n<2$ for all $n=1,2ldots$
My attempt to use mathematical induction: if $s_n<2$ then I use
$s_{n+1}=sqrt {2+sqrt s_n}$, to prove $s_{n+1}<sqrt {2+2}=2$.
I guess $s_n$ is an increasing sequence, I can use the $s_n$ is bounded and increasing, and then prove that $s_n$ converges.
real-analysis calculus sequences-and-series limits analysis
edited Feb 2 at 14:49
Peter Mortensen
561310
asked Feb 2 at 9:35
jacksonjackson
1499
$endgroup$
marked as duplicate by rtybase, stressed out, user21820, Trevor Gunn, StubbornAtom Feb 2 at 22:37
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
2
$begingroup$
Jackson.All is in place.Just work it out clearly.+
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– Peter Szilas
Feb 2 at 10:05
1
$begingroup$
Thanks a lot for you answer my question ,I figure it out
$endgroup$
– jackson
Feb 2 at 10:06
1
$begingroup$
also covered here
$endgroup$
– rtybase
Feb 2 at 10:28
$begingroup$
@rtybase how you search for that I have searched before I asked but I still don’t find the answer....
$endgroup$
– jackson
Feb 2 at 10:28
$begingroup$
I have dealt with this problem before ... just from the top of my head.
$endgroup$
– rtybase
Feb 2 at 10:29
|
show 1 more comment
$begingroup$
This question already has an answer here:
Prob. 3, Chap. 3 in Baby Rudin: If $s_1 = sqrt{2}$, and $s_{n+1} = sqrt{2 + sqrt{s_n}}$, what is the limit of this sequence?
3 answers
If $s_1=sqrt 2$,
and $$s_{n+1}=sqrt {2+sqrt{s_n}}$$
Prove that ${s_n}$ is converging, and that $s_n<2$ for all $n=1,2ldots$
My attempt to use mathematical induction: if $s_n<2$ then I use
$s_{n+1}=sqrt {2+sqrt s_n}$, to prove $s_{n+1}<sqrt {2+2}=2$.
I guess $s_n$ is an increasing sequence, I can use the $s_n$ is bounded and increasing, and then prove that $s_n$ converges.
real-analysis calculus sequences-and-series limits analysis
edited Feb 2 at 14:49
Peter Mortensen
561310
asked Feb 2 at 9:35
jacksonjackson
1499
$endgroup$
This question already has an answer here:
Prob. 3, Chap. 3 in Baby Rudin: If $s_1 = sqrt{2}$, and $s_{n+1} = sqrt{2 + sqrt{s_n}}$, what is the limit of this sequence?
3 answers
If $s_1=sqrt 2$,
and $$s_{n+1}=sqrt {2+sqrt{s_n}}$$
Prove that ${s_n}$ is converging, and that $s_n<2$ for all $n=1,2ldots$
My attempt to use mathematical induction: if $s_n<2$ then I use
$s_{n+1}=sqrt {2+sqrt s_n}$, to prove $s_{n+1}<sqrt {2+2}=2$.
I guess $s_n$ is an increasing sequence, I can use the $s_n$ is bounded and increasing, and then prove that $s_n$ converges.
This question already has an answer here:
Prob. 3, Chap. 3 in Baby Rudin: If $s_1 = sqrt{2}$, and $s_{n+1} = sqrt{2 + sqrt{s_n}}$, what is the limit of this sequence?
3 answers
real-analysis calculus sequences-and-series limits analysis
real-analysis calculus sequences-and-series limits analysis
edited Feb 2 at 14:49
Peter Mortensen
561310
asked Feb 2 at 9:35
jacksonjackson
1499
edited Feb 2 at 14:49
Peter Mortensen
561310
asked Feb 2 at 9:35
jacksonjackson
1499
edited Feb 2 at 14:49
Peter Mortensen
561310
edited Feb 2 at 14:49
Peter Mortensen
561310
edited Feb 2 at 14:49
Peter Mortensen
561310
561310
asked Feb 2 at 9:35
jacksonjackson
1499
asked Feb 2 at 9:35
jacksonjackson
1499
asked Feb 2 at 9:35
jacksonjackson
1499
1499
marked as duplicate by rtybase, stressed out, user21820, Trevor Gunn, StubbornAtom Feb 2 at 22:37
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by rtybase, stressed out, user21820, Trevor Gunn, StubbornAtom Feb 2 at 22:37
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
2
$begingroup$
Jackson.All is in place.Just work it out clearly.+
$endgroup$
– Peter Szilas
Feb 2 at 10:05
1
$begingroup$
Thanks a lot for you answer my question ,I figure it out
$endgroup$
– jackson
Feb 2 at 10:06
1
$begingroup$
also covered here
$endgroup$
– rtybase
Feb 2 at 10:28
$begingroup$
@rtybase how you search for that I have searched before I asked but I still don’t find the answer....
$endgroup$
– jackson
Feb 2 at 10:28
$begingroup$
I have dealt with this problem before ... just from the top of my head.
$endgroup$
– rtybase
Feb 2 at 10:29
|
show 1 more comment
2
$begingroup$
Jackson.All is in place.Just work it out clearly.+
$endgroup$
– Peter Szilas
Feb 2 at 10:05
1
$begingroup$
Thanks a lot for you answer my question ,I figure it out
$endgroup$
– jackson
Feb 2 at 10:06
1
$begingroup$
also covered here
$endgroup$
– rtybase
Feb 2 at 10:28
$begingroup$
@rtybase how you search for that I have searched before I asked but I still don’t find the answer....
$endgroup$
– jackson
Feb 2 at 10:28
$begingroup$
I have dealt with this problem before ... just from the top of my head.
$endgroup$
– rtybase
Feb 2 at 10:29
2
2
$begingroup$
Jackson.All is in place.Just work it out clearly.+
$endgroup$
– Peter Szilas
Feb 2 at 10:05
$begingroup$
Jackson.All is in place.Just work it out clearly.+
$endgroup$
– Peter Szilas
Feb 2 at 10:05
1
1
$begingroup$
Thanks a lot for you answer my question ,I figure it out
$endgroup$
– jackson
Feb 2 at 10:06
$begingroup$
Thanks a lot for you answer my question ,I figure it out
$endgroup$
– jackson
Feb 2 at 10:06
1
1
$begingroup$
also covered here
$endgroup$
– rtybase
Feb 2 at 10:28
$begingroup$
also covered here
$endgroup$
– rtybase
Feb 2 at 10:28
$begingroup$
@rtybase how you search for that I have searched before I asked but I still don’t find the answer....
$endgroup$
– jackson
Feb 2 at 10:28
$begingroup$
@rtybase how you search for that I have searched before I asked but I still don’t find the answer....
$endgroup$
– jackson
Feb 2 at 10:28
$begingroup$
I have dealt with this problem before ... just from the top of my head.
$endgroup$
– rtybase
Feb 2 at 10:29
$begingroup$
I have dealt with this problem before ... just from the top of my head.
$endgroup$
– rtybase
Feb 2 at 10:29
|
show 1 more comment
4 Answers
4
active
oldest
votes
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To give a name to your idea, you want to use the monotone convergence theorem; a bounded and increasing sequence converges to the supremum of that set, indeed a standard approach would be to use induction.
Lemma: $s_n$ is bounded
Base case:
$n=1$
$$s_1= sqrt{2} <2$$
Suppose now for arbitrary $k$
$$ s_k< 2$$
Now observe as the inductive step:
$$ s_{k+1}=sqrt{2 + sqrt{s_k}}< sqrt{2+2}=2$$
By the principle of mathematical induction we have found that $2$ is a bound for all $nin
mathbb N$.
Lemma: $s_n$ is increasing
Base case:
$$s_2-s_1>0$$
$$sqrt{2 + sqrt{sqrt{2}}}- sqrt{2}>sqrt{2 + 0}- sqrt{2}=0 $$
Now suppose for arbitrary $k$ we have:
$$ s_{k}-s_{k-1}>0 implies s_k > s_{k-1}implies sqrt{s_k}> sqrt{s_{k-1}}$$
We now make our inductive step:
$$ s_{k+1}-s_k=sqrt{2 + sqrt{s_k}}-sqrt{2 + sqrt{s_{k-1}}}>sqrt{2 + sqrt{s_{k-1}}}-sqrt{2 + sqrt{s_{k-1}}}=0$$
Hence we have an increasing sequence for all $n in mathbb N$ this completes the proof and our sequence is indeed convergent $square$.
answered Feb 2 at 9:55
Wesley StrikWesley Strik
2,189424
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add a comment |
$begingroup$
Your idea is perfect. The monotone convergence theorem guarantees convergence of a sequence if you can show that the sequence is increasing and bounded above (or decreasing and bounded below). To get that your sequence is increasing, the trick is to use strong induction:
The base case is easy, so I'll leave that to you. Next, suppose that each subsequent term is larger than the previous for each of the first $k$ terms. We want to show $s_{k+1} > s_k$, and because we used strong induction, we can assume $s_{k} > s_{k-1}$, which gives the inequality $s_{k+1} = sqrt{2 + sqrt{s_k}} > sqrt{2+sqrt{s_{k-1}}}$. And what is $sqrt{2 + sqrt{s_{k-1}}} $?
This argument is implicitly using the fact that $sqrt{cdot}$ is an increasing function, but this is fairly easy to prove.
If you wanted to actually find the limit of the sequence, first note that ${s_{n+1} }$ is a subsequence of ${s_n}$. If $ {s_n }$ converges, any subsequence also converges to the same value. Taking a limit of both sides of that equality, we get:
$$lim(s_{n+1}) = lim left( sqrt{2 + sqrt{s_n}} right)$$
Applying limit laws, and recognizing that $sqrt{cdot}$ is a continuous function$^dagger$, we'll have:
$$lim(s_{n+1}) = sqrt{ 2 + sqrt{ lim(s_n)}}$$
We know ${s_n}$ converges to some $L in mathbb{R}^+$ and ${s_{n+1} }$ to the same, so the above reduces to:
$$L = sqrt{2 + sqrt{L}}$$
Solving for $L$, we find that it is a root of the quartic $L^4 - 4L^2 - L + 4$. Notice that $1$ is a root, so we can factor out $(L-1)$, reducing the problem to finding roots of the cubic $L^3 + L^2 - 3L - 4$. You can find its three roots using the cubic formula (two of them will have nonzero imaginary part, so our $L$ will be the single real root). The cubic formula is a mess, so you can also arrive at an arbitrarily good approximation for $L$ using Newton's method. Either way, we get $L approx 1.8312$.
$^dagger$We're applying the fact that, if $f$ is a continuous function and ${a_n}$ a convergent sequence in the domain of $f$, then $a_n to M implies f(a_n) to f(M)$.
answered Feb 2 at 9:43
Kaj HansenKaj Hansen
27.8k43980
$endgroup$
$begingroup$
in fact I want to know how to prove $s_n$ is increasing😂😅
$endgroup$
– jackson
Feb 2 at 9:48
1
$begingroup$
I know the fact you add,but I still don’t understand why $s_n$ is increasing
$endgroup$
– jackson
Feb 2 at 9:53
$begingroup$
Added! And ugh, I have no idea why my TeX isn't rendering in the bit I added. Anyone have any ideas?
$endgroup$
– Kaj Hansen
Feb 2 at 9:57
$begingroup$
There we go, lol
$endgroup$
– Kaj Hansen
Feb 2 at 9:58
1
$begingroup$
Oh, just realized you weren't actually asked to find the value to which it converges. Well, that's how you'd do it! Best of luck.
$endgroup$
– Kaj Hansen
Feb 2 at 10:04
add a comment |
$begingroup$
It is a fixed point iteration of the form $s_{n+1}=g(s_n)$. You can easily check that $g([0,2]) subset [0,2]$ an so, if you start with $s_1 in [0,2]$, the sequence will remain in $[0,2]$, which proves part of the question. Regarding convergence, you can use again the properties of $g$ to check that the sequence in monotone, like you guessed. Note that you can look at $g$ as a real function and use differential calculus...
answered Feb 2 at 9:47
PierreCarrePierreCarre
2,178215
$endgroup$
$begingroup$
It's kind of cheating to use tools of differential calculus when you are still building it up and technically "it does not exist yet"
$endgroup$
– Wesley Strik
Feb 2 at 10:02
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But great answer :)
$endgroup$
– Wesley Strik
Feb 2 at 10:03
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Sure @WesleyStrik, it is a little bit cheating :), but it is also a bit more general in the sense that you could prove convergence (Banach's fixed point) even without monotonicity.
$endgroup$
– PierreCarre
Feb 3 at 12:26
add a comment |
$begingroup$
Wesley's lemma 1 nicely shows that $s_n$ is bounded above.
Rephrasing:
Show that $s_n$ is increasing:
1) $n=1$: $s_2 > s_1√$
2) Hypothesis : $s_{n+1} ge s_n.$
Step $n+1:$
$s_{n+2} =sqrt{2+√s_{n+1}} {Hyp. atop ge} sqrt{2+√s_n} =s_{n+1}.$
Used: $f(x)=√x$ is an increasing function.
answered Feb 2 at 10:22
Peter SzilasPeter Szilas
12k2822
$endgroup$
1
$begingroup$
You give a nice proof!
$endgroup$
– jackson
Feb 2 at 10:24
1
$begingroup$
Jackson.Thanks, once I got your lead, not so very difficult:))
$endgroup$
– Peter Szilas
Feb 2 at 10:25
add a comment |
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
To give a name to your idea, you want to use the monotone convergence theorem; a bounded and increasing sequence converges to the supremum of that set, indeed a standard approach would be to use induction.
Lemma: $s_n$ is bounded
Base case:
$n=1$
$$s_1= sqrt{2} <2$$
Suppose now for arbitrary $k$
$$ s_k< 2$$
Now observe as the inductive step:
$$ s_{k+1}=sqrt{2 + sqrt{s_k}}< sqrt{2+2}=2$$
By the principle of mathematical induction we have found that $2$ is a bound for all $nin
mathbb N$.
Lemma: $s_n$ is increasing
Base case:
$$s_2-s_1>0$$
$$sqrt{2 + sqrt{sqrt{2}}}- sqrt{2}>sqrt{2 + 0}- sqrt{2}=0 $$
Now suppose for arbitrary $k$ we have:
$$ s_{k}-s_{k-1}>0 implies s_k > s_{k-1}implies sqrt{s_k}> sqrt{s_{k-1}}$$
We now make our inductive step:
$$ s_{k+1}-s_k=sqrt{2 + sqrt{s_k}}-sqrt{2 + sqrt{s_{k-1}}}>sqrt{2 + sqrt{s_{k-1}}}-sqrt{2 + sqrt{s_{k-1}}}=0$$
Hence we have an increasing sequence for all $n in mathbb N$ this completes the proof and our sequence is indeed convergent $square$.
answered Feb 2 at 9:55
Wesley StrikWesley Strik
2,189424
$endgroup$
add a comment |
$begingroup$
To give a name to your idea, you want to use the monotone convergence theorem; a bounded and increasing sequence converges to the supremum of that set, indeed a standard approach would be to use induction.
Lemma: $s_n$ is bounded
Base case:
$n=1$
$$s_1= sqrt{2} <2$$
Suppose now for arbitrary $k$
$$ s_k< 2$$
Now observe as the inductive step:
$$ s_{k+1}=sqrt{2 + sqrt{s_k}}< sqrt{2+2}=2$$
By the principle of mathematical induction we have found that $2$ is a bound for all $nin
mathbb N$.
Lemma: $s_n$ is increasing
Base case:
$$s_2-s_1>0$$
$$sqrt{2 + sqrt{sqrt{2}}}- sqrt{2}>sqrt{2 + 0}- sqrt{2}=0 $$
Now suppose for arbitrary $k$ we have:
$$ s_{k}-s_{k-1}>0 implies s_k > s_{k-1}implies sqrt{s_k}> sqrt{s_{k-1}}$$
We now make our inductive step:
$$ s_{k+1}-s_k=sqrt{2 + sqrt{s_k}}-sqrt{2 + sqrt{s_{k-1}}}>sqrt{2 + sqrt{s_{k-1}}}-sqrt{2 + sqrt{s_{k-1}}}=0$$
Hence we have an increasing sequence for all $n in mathbb N$ this completes the proof and our sequence is indeed convergent $square$.
answered Feb 2 at 9:55
Wesley StrikWesley Strik
2,189424
$endgroup$
add a comment |
$begingroup$
To give a name to your idea, you want to use the monotone convergence theorem; a bounded and increasing sequence converges to the supremum of that set, indeed a standard approach would be to use induction.
Lemma: $s_n$ is bounded
Base case:
$n=1$
$$s_1= sqrt{2} <2$$
Suppose now for arbitrary $k$
$$ s_k< 2$$
Now observe as the inductive step:
$$ s_{k+1}=sqrt{2 + sqrt{s_k}}< sqrt{2+2}=2$$
By the principle of mathematical induction we have found that $2$ is a bound for all $nin
mathbb N$.
Lemma: $s_n$ is increasing
Base case:
$$s_2-s_1>0$$
$$sqrt{2 + sqrt{sqrt{2}}}- sqrt{2}>sqrt{2 + 0}- sqrt{2}=0 $$
Now suppose for arbitrary $k$ we have:
$$ s_{k}-s_{k-1}>0 implies s_k > s_{k-1}implies sqrt{s_k}> sqrt{s_{k-1}}$$
We now make our inductive step:
$$ s_{k+1}-s_k=sqrt{2 + sqrt{s_k}}-sqrt{2 + sqrt{s_{k-1}}}>sqrt{2 + sqrt{s_{k-1}}}-sqrt{2 + sqrt{s_{k-1}}}=0$$
Hence we have an increasing sequence for all $n in mathbb N$ this completes the proof and our sequence is indeed convergent $square$.
answered Feb 2 at 9:55
Wesley StrikWesley Strik
2,189424
$endgroup$
To give a name to your idea, you want to use the monotone convergence theorem; a bounded and increasing sequence converges to the supremum of that set, indeed a standard approach would be to use induction.
Lemma: $s_n$ is bounded
Base case:
$n=1$
$$s_1= sqrt{2} <2$$
Suppose now for arbitrary $k$
$$ s_k< 2$$
Now observe as the inductive step:
$$ s_{k+1}=sqrt{2 + sqrt{s_k}}< sqrt{2+2}=2$$
By the principle of mathematical induction we have found that $2$ is a bound for all $nin
mathbb N$.
Lemma: $s_n$ is increasing
Base case:
$$s_2-s_1>0$$
$$sqrt{2 + sqrt{sqrt{2}}}- sqrt{2}>sqrt{2 + 0}- sqrt{2}=0 $$
Now suppose for arbitrary $k$ we have:
$$ s_{k}-s_{k-1}>0 implies s_k > s_{k-1}implies sqrt{s_k}> sqrt{s_{k-1}}$$
We now make our inductive step:
$$ s_{k+1}-s_k=sqrt{2 + sqrt{s_k}}-sqrt{2 + sqrt{s_{k-1}}}>sqrt{2 + sqrt{s_{k-1}}}-sqrt{2 + sqrt{s_{k-1}}}=0$$
Hence we have an increasing sequence for all $n in mathbb N$ this completes the proof and our sequence is indeed convergent $square$.
answered Feb 2 at 9:55
Wesley StrikWesley Strik
2,189424
edited Feb 2 at 17:09
answered Feb 2 at 9:55
Wesley StrikWesley Strik
2,189424
answered Feb 2 at 9:55
Wesley StrikWesley Strik
2,189424
answered Feb 2 at 9:55
Wesley StrikWesley Strik
2,189424
2,189424
add a comment |
add a comment |
$begingroup$
Your idea is perfect. The monotone convergence theorem guarantees convergence of a sequence if you can show that the sequence is increasing and bounded above (or decreasing and bounded below). To get that your sequence is increasing, the trick is to use strong induction:
The base case is easy, so I'll leave that to you. Next, suppose that each subsequent term is larger than the previous for each of the first $k$ terms. We want to show $s_{k+1} > s_k$, and because we used strong induction, we can assume $s_{k} > s_{k-1}$, which gives the inequality $s_{k+1} = sqrt{2 + sqrt{s_k}} > sqrt{2+sqrt{s_{k-1}}}$. And what is $sqrt{2 + sqrt{s_{k-1}}} $?
This argument is implicitly using the fact that $sqrt{cdot}$ is an increasing function, but this is fairly easy to prove.
If you wanted to actually find the limit of the sequence, first note that ${s_{n+1} }$ is a subsequence of ${s_n}$. If $ {s_n }$ converges, any subsequence also converges to the same value. Taking a limit of both sides of that equality, we get:
$$lim(s_{n+1}) = lim left( sqrt{2 + sqrt{s_n}} right)$$
Applying limit laws, and recognizing that $sqrt{cdot}$ is a continuous function$^dagger$, we'll have:
$$lim(s_{n+1}) = sqrt{ 2 + sqrt{ lim(s_n)}}$$
We know ${s_n}$ converges to some $L in mathbb{R}^+$ and ${s_{n+1} }$ to the same, so the above reduces to:
$$L = sqrt{2 + sqrt{L}}$$
Solving for $L$, we find that it is a root of the quartic $L^4 - 4L^2 - L + 4$. Notice that $1$ is a root, so we can factor out $(L-1)$, reducing the problem to finding roots of the cubic $L^3 + L^2 - 3L - 4$. You can find its three roots using the cubic formula (two of them will have nonzero imaginary part, so our $L$ will be the single real root). The cubic formula is a mess, so you can also arrive at an arbitrarily good approximation for $L$ using Newton's method. Either way, we get $L approx 1.8312$.
$^dagger$We're applying the fact that, if $f$ is a continuous function and ${a_n}$ a convergent sequence in the domain of $f$, then $a_n to M implies f(a_n) to f(M)$.
answered Feb 2 at 9:43
Kaj HansenKaj Hansen
27.8k43980
$endgroup$
$begingroup$
in fact I want to know how to prove $s_n$ is increasing😂😅
$endgroup$
– jackson
Feb 2 at 9:48
1
$begingroup$
I know the fact you add,but I still don’t understand why $s_n$ is increasing
$endgroup$
– jackson
Feb 2 at 9:53
$begingroup$
Added! And ugh, I have no idea why my TeX isn't rendering in the bit I added. Anyone have any ideas?
$endgroup$
– Kaj Hansen
Feb 2 at 9:57
$begingroup$
There we go, lol
$endgroup$
– Kaj Hansen
Feb 2 at 9:58
1
$begingroup$
Oh, just realized you weren't actually asked to find the value to which it converges. Well, that's how you'd do it! Best of luck.
$endgroup$
– Kaj Hansen
Feb 2 at 10:04
add a comment |
$begingroup$
Your idea is perfect. The monotone convergence theorem guarantees convergence of a sequence if you can show that the sequence is increasing and bounded above (or decreasing and bounded below). To get that your sequence is increasing, the trick is to use strong induction:
The base case is easy, so I'll leave that to you. Next, suppose that each subsequent term is larger than the previous for each of the first $k$ terms. We want to show $s_{k+1} > s_k$, and because we used strong induction, we can assume $s_{k} > s_{k-1}$, which gives the inequality $s_{k+1} = sqrt{2 + sqrt{s_k}} > sqrt{2+sqrt{s_{k-1}}}$. And what is $sqrt{2 + sqrt{s_{k-1}}} $?
This argument is implicitly using the fact that $sqrt{cdot}$ is an increasing function, but this is fairly easy to prove.
If you wanted to actually find the limit of the sequence, first note that ${s_{n+1} }$ is a subsequence of ${s_n}$. If $ {s_n }$ converges, any subsequence also converges to the same value. Taking a limit of both sides of that equality, we get:
$$lim(s_{n+1}) = lim left( sqrt{2 + sqrt{s_n}} right)$$
Applying limit laws, and recognizing that $sqrt{cdot}$ is a continuous function$^dagger$, we'll have:
$$lim(s_{n+1}) = sqrt{ 2 + sqrt{ lim(s_n)}}$$
We know ${s_n}$ converges to some $L in mathbb{R}^+$ and ${s_{n+1} }$ to the same, so the above reduces to:
$$L = sqrt{2 + sqrt{L}}$$
Solving for $L$, we find that it is a root of the quartic $L^4 - 4L^2 - L + 4$. Notice that $1$ is a root, so we can factor out $(L-1)$, reducing the problem to finding roots of the cubic $L^3 + L^2 - 3L - 4$. You can find its three roots using the cubic formula (two of them will have nonzero imaginary part, so our $L$ will be the single real root). The cubic formula is a mess, so you can also arrive at an arbitrarily good approximation for $L$ using Newton's method. Either way, we get $L approx 1.8312$.
$^dagger$We're applying the fact that, if $f$ is a continuous function and ${a_n}$ a convergent sequence in the domain of $f$, then $a_n to M implies f(a_n) to f(M)$.
answered Feb 2 at 9:43
Kaj HansenKaj Hansen
27.8k43980
$endgroup$
$begingroup$
in fact I want to know how to prove $s_n$ is increasing😂😅
$endgroup$
– jackson
Feb 2 at 9:48
1
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I know the fact you add,but I still don’t understand why $s_n$ is increasing
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– jackson
Feb 2 at 9:53
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Added! And ugh, I have no idea why my TeX isn't rendering in the bit I added. Anyone have any ideas?
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– Kaj Hansen
Feb 2 at 9:57
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There we go, lol
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– Kaj Hansen
Feb 2 at 9:58
1
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Oh, just realized you weren't actually asked to find the value to which it converges. Well, that's how you'd do it! Best of luck.
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– Kaj Hansen
Feb 2 at 10:04
add a comment |
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Your idea is perfect. The monotone convergence theorem guarantees convergence of a sequence if you can show that the sequence is increasing and bounded above (or decreasing and bounded below). To get that your sequence is increasing, the trick is to use strong induction:
The base case is easy, so I'll leave that to you. Next, suppose that each subsequent term is larger than the previous for each of the first $k$ terms. We want to show $s_{k+1} > s_k$, and because we used strong induction, we can assume $s_{k} > s_{k-1}$, which gives the inequality $s_{k+1} = sqrt{2 + sqrt{s_k}} > sqrt{2+sqrt{s_{k-1}}}$. And what is $sqrt{2 + sqrt{s_{k-1}}} $?
This argument is implicitly using the fact that $sqrt{cdot}$ is an increasing function, but this is fairly easy to prove.
If you wanted to actually find the limit of the sequence, first note that ${s_{n+1} }$ is a subsequence of ${s_n}$. If $ {s_n }$ converges, any subsequence also converges to the same value. Taking a limit of both sides of that equality, we get:
$$lim(s_{n+1}) = lim left( sqrt{2 + sqrt{s_n}} right)$$
Applying limit laws, and recognizing that $sqrt{cdot}$ is a continuous function$^dagger$, we'll have:
$$lim(s_{n+1}) = sqrt{ 2 + sqrt{ lim(s_n)}}$$
We know ${s_n}$ converges to some $L in mathbb{R}^+$ and ${s_{n+1} }$ to the same, so the above reduces to:
$$L = sqrt{2 + sqrt{L}}$$
Solving for $L$, we find that it is a root of the quartic $L^4 - 4L^2 - L + 4$. Notice that $1$ is a root, so we can factor out $(L-1)$, reducing the problem to finding roots of the cubic $L^3 + L^2 - 3L - 4$. You can find its three roots using the cubic formula (two of them will have nonzero imaginary part, so our $L$ will be the single real root). The cubic formula is a mess, so you can also arrive at an arbitrarily good approximation for $L$ using Newton's method. Either way, we get $L approx 1.8312$.
$^dagger$We're applying the fact that, if $f$ is a continuous function and ${a_n}$ a convergent sequence in the domain of $f$, then $a_n to M implies f(a_n) to f(M)$.
answered Feb 2 at 9:43
Kaj HansenKaj Hansen
27.8k43980
$endgroup$
Your idea is perfect. The monotone convergence theorem guarantees convergence of a sequence if you can show that the sequence is increasing and bounded above (or decreasing and bounded below). To get that your sequence is increasing, the trick is to use strong induction:
The base case is easy, so I'll leave that to you. Next, suppose that each subsequent term is larger than the previous for each of the first $k$ terms. We want to show $s_{k+1} > s_k$, and because we used strong induction, we can assume $s_{k} > s_{k-1}$, which gives the inequality $s_{k+1} = sqrt{2 + sqrt{s_k}} > sqrt{2+sqrt{s_{k-1}}}$. And what is $sqrt{2 + sqrt{s_{k-1}}} $?
This argument is implicitly using the fact that $sqrt{cdot}$ is an increasing function, but this is fairly easy to prove.
If you wanted to actually find the limit of the sequence, first note that ${s_{n+1} }$ is a subsequence of ${s_n}$. If $ {s_n }$ converges, any subsequence also converges to the same value. Taking a limit of both sides of that equality, we get:
$$lim(s_{n+1}) = lim left( sqrt{2 + sqrt{s_n}} right)$$
Applying limit laws, and recognizing that $sqrt{cdot}$ is a continuous function$^dagger$, we'll have:
$$lim(s_{n+1}) = sqrt{ 2 + sqrt{ lim(s_n)}}$$
We know ${s_n}$ converges to some $L in mathbb{R}^+$ and ${s_{n+1} }$ to the same, so the above reduces to:
$$L = sqrt{2 + sqrt{L}}$$
Solving for $L$, we find that it is a root of the quartic $L^4 - 4L^2 - L + 4$. Notice that $1$ is a root, so we can factor out $(L-1)$, reducing the problem to finding roots of the cubic $L^3 + L^2 - 3L - 4$. You can find its three roots using the cubic formula (two of them will have nonzero imaginary part, so our $L$ will be the single real root). The cubic formula is a mess, so you can also arrive at an arbitrarily good approximation for $L$ using Newton's method. Either way, we get $L approx 1.8312$.
$^dagger$We're applying the fact that, if $f$ is a continuous function and ${a_n}$ a convergent sequence in the domain of $f$, then $a_n to M implies f(a_n) to f(M)$.
answered Feb 2 at 9:43
Kaj HansenKaj Hansen
27.8k43980
edited Feb 2 at 22:24
answered Feb 2 at 9:43
Kaj HansenKaj Hansen
27.8k43980
answered Feb 2 at 9:43
Kaj HansenKaj Hansen
27.8k43980
answered Feb 2 at 9:43
Kaj HansenKaj Hansen
27.8k43980
27.8k43980
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in fact I want to know how to prove $s_n$ is increasing😂😅
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– jackson
Feb 2 at 9:48
1
$begingroup$
I know the fact you add,but I still don’t understand why $s_n$ is increasing
$endgroup$
– jackson
Feb 2 at 9:53
$begingroup$
Added! And ugh, I have no idea why my TeX isn't rendering in the bit I added. Anyone have any ideas?
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– Kaj Hansen
Feb 2 at 9:57
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There we go, lol
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– Kaj Hansen
Feb 2 at 9:58
1
$begingroup$
Oh, just realized you weren't actually asked to find the value to which it converges. Well, that's how you'd do it! Best of luck.
$endgroup$
– Kaj Hansen
Feb 2 at 10:04
add a comment |
$begingroup$
in fact I want to know how to prove $s_n$ is increasing😂😅
$endgroup$
– jackson
Feb 2 at 9:48
1
$begingroup$
I know the fact you add,but I still don’t understand why $s_n$ is increasing
$endgroup$
– jackson
Feb 2 at 9:53
$begingroup$
Added! And ugh, I have no idea why my TeX isn't rendering in the bit I added. Anyone have any ideas?
$endgroup$
– Kaj Hansen
Feb 2 at 9:57
$begingroup$
There we go, lol
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– Kaj Hansen
Feb 2 at 9:58
1
$begingroup$
Oh, just realized you weren't actually asked to find the value to which it converges. Well, that's how you'd do it! Best of luck.
$endgroup$
– Kaj Hansen
Feb 2 at 10:04
$begingroup$
in fact I want to know how to prove $s_n$ is increasing😂😅
$endgroup$
– jackson
Feb 2 at 9:48
$begingroup$
in fact I want to know how to prove $s_n$ is increasing😂😅
$endgroup$
– jackson
Feb 2 at 9:48
1
1
$begingroup$
I know the fact you add,but I still don’t understand why $s_n$ is increasing
$endgroup$
– jackson
Feb 2 at 9:53
$begingroup$
I know the fact you add,but I still don’t understand why $s_n$ is increasing
$endgroup$
– jackson
Feb 2 at 9:53
$begingroup$
Added! And ugh, I have no idea why my TeX isn't rendering in the bit I added. Anyone have any ideas?
$endgroup$
– Kaj Hansen
Feb 2 at 9:57
$begingroup$
Added! And ugh, I have no idea why my TeX isn't rendering in the bit I added. Anyone have any ideas?
$endgroup$
– Kaj Hansen
Feb 2 at 9:57
$begingroup$
There we go, lol
$endgroup$
– Kaj Hansen
Feb 2 at 9:58
$begingroup$
There we go, lol
$endgroup$
– Kaj Hansen
Feb 2 at 9:58
1
1
$begingroup$
Oh, just realized you weren't actually asked to find the value to which it converges. Well, that's how you'd do it! Best of luck.
$endgroup$
– Kaj Hansen
Feb 2 at 10:04
$begingroup$
Oh, just realized you weren't actually asked to find the value to which it converges. Well, that's how you'd do it! Best of luck.
$endgroup$
– Kaj Hansen
Feb 2 at 10:04
add a comment |
$begingroup$
It is a fixed point iteration of the form $s_{n+1}=g(s_n)$. You can easily check that $g([0,2]) subset [0,2]$ an so, if you start with $s_1 in [0,2]$, the sequence will remain in $[0,2]$, which proves part of the question. Regarding convergence, you can use again the properties of $g$ to check that the sequence in monotone, like you guessed. Note that you can look at $g$ as a real function and use differential calculus...
answered Feb 2 at 9:47
PierreCarrePierreCarre
2,178215
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It's kind of cheating to use tools of differential calculus when you are still building it up and technically "it does not exist yet"
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– Wesley Strik
Feb 2 at 10:02
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But great answer :)
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– Wesley Strik
Feb 2 at 10:03
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Sure @WesleyStrik, it is a little bit cheating :), but it is also a bit more general in the sense that you could prove convergence (Banach's fixed point) even without monotonicity.
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– PierreCarre
Feb 3 at 12:26
add a comment |
$begingroup$
It is a fixed point iteration of the form $s_{n+1}=g(s_n)$. You can easily check that $g([0,2]) subset [0,2]$ an so, if you start with $s_1 in [0,2]$, the sequence will remain in $[0,2]$, which proves part of the question. Regarding convergence, you can use again the properties of $g$ to check that the sequence in monotone, like you guessed. Note that you can look at $g$ as a real function and use differential calculus...
answered Feb 2 at 9:47
PierreCarrePierreCarre
2,178215
$endgroup$
$begingroup$
It's kind of cheating to use tools of differential calculus when you are still building it up and technically "it does not exist yet"
$endgroup$
– Wesley Strik
Feb 2 at 10:02
$begingroup$
But great answer :)
$endgroup$
– Wesley Strik
Feb 2 at 10:03
$begingroup$
Sure @WesleyStrik, it is a little bit cheating :), but it is also a bit more general in the sense that you could prove convergence (Banach's fixed point) even without monotonicity.
$endgroup$
– PierreCarre
Feb 3 at 12:26
add a comment |
$begingroup$
It is a fixed point iteration of the form $s_{n+1}=g(s_n)$. You can easily check that $g([0,2]) subset [0,2]$ an so, if you start with $s_1 in [0,2]$, the sequence will remain in $[0,2]$, which proves part of the question. Regarding convergence, you can use again the properties of $g$ to check that the sequence in monotone, like you guessed. Note that you can look at $g$ as a real function and use differential calculus...
answered Feb 2 at 9:47
PierreCarrePierreCarre
2,178215
$endgroup$
It is a fixed point iteration of the form $s_{n+1}=g(s_n)$. You can easily check that $g([0,2]) subset [0,2]$ an so, if you start with $s_1 in [0,2]$, the sequence will remain in $[0,2]$, which proves part of the question. Regarding convergence, you can use again the properties of $g$ to check that the sequence in monotone, like you guessed. Note that you can look at $g$ as a real function and use differential calculus...
answered Feb 2 at 9:47
PierreCarrePierreCarre
2,178215
answered Feb 2 at 9:47
PierreCarrePierreCarre
2,178215
answered Feb 2 at 9:47
PierreCarrePierreCarre
2,178215
answered Feb 2 at 9:47
PierreCarrePierreCarre
2,178215
2,178215
$begingroup$
It's kind of cheating to use tools of differential calculus when you are still building it up and technically "it does not exist yet"
$endgroup$
– Wesley Strik
Feb 2 at 10:02
$begingroup$
But great answer :)
$endgroup$
– Wesley Strik
Feb 2 at 10:03
$begingroup$
Sure @WesleyStrik, it is a little bit cheating :), but it is also a bit more general in the sense that you could prove convergence (Banach's fixed point) even without monotonicity.
$endgroup$
– PierreCarre
Feb 3 at 12:26
add a comment |
$begingroup$
It's kind of cheating to use tools of differential calculus when you are still building it up and technically "it does not exist yet"
$endgroup$
– Wesley Strik
Feb 2 at 10:02
$begingroup$
But great answer :)
$endgroup$
– Wesley Strik
Feb 2 at 10:03
$begingroup$
Sure @WesleyStrik, it is a little bit cheating :), but it is also a bit more general in the sense that you could prove convergence (Banach's fixed point) even without monotonicity.
$endgroup$
– PierreCarre
Feb 3 at 12:26
$begingroup$
It's kind of cheating to use tools of differential calculus when you are still building it up and technically "it does not exist yet"
$endgroup$
– Wesley Strik
Feb 2 at 10:02
$begingroup$
It's kind of cheating to use tools of differential calculus when you are still building it up and technically "it does not exist yet"
$endgroup$
– Wesley Strik
Feb 2 at 10:02
$begingroup$
But great answer :)
$endgroup$
– Wesley Strik
Feb 2 at 10:03
$begingroup$
But great answer :)
$endgroup$
– Wesley Strik
Feb 2 at 10:03
$begingroup$
Sure @WesleyStrik, it is a little bit cheating :), but it is also a bit more general in the sense that you could prove convergence (Banach's fixed point) even without monotonicity.
$endgroup$
– PierreCarre
Feb 3 at 12:26
$begingroup$
Sure @WesleyStrik, it is a little bit cheating :), but it is also a bit more general in the sense that you could prove convergence (Banach's fixed point) even without monotonicity.
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– PierreCarre
Feb 3 at 12:26
add a comment |
$begingroup$
Wesley's lemma 1 nicely shows that $s_n$ is bounded above.
Rephrasing:
Show that $s_n$ is increasing:
1) $n=1$: $s_2 > s_1√$
2) Hypothesis : $s_{n+1} ge s_n.$
Step $n+1:$
$s_{n+2} =sqrt{2+√s_{n+1}} {Hyp. atop ge} sqrt{2+√s_n} =s_{n+1}.$
Used: $f(x)=√x$ is an increasing function.
answered Feb 2 at 10:22
Peter SzilasPeter Szilas
12k2822
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1
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You give a nice proof!
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– jackson
Feb 2 at 10:24
1
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Jackson.Thanks, once I got your lead, not so very difficult:))
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– Peter Szilas
Feb 2 at 10:25
add a comment |
$begingroup$
Wesley's lemma 1 nicely shows that $s_n$ is bounded above.
Rephrasing:
Show that $s_n$ is increasing:
1) $n=1$: $s_2 > s_1√$
2) Hypothesis : $s_{n+1} ge s_n.$
Step $n+1:$
$s_{n+2} =sqrt{2+√s_{n+1}} {Hyp. atop ge} sqrt{2+√s_n} =s_{n+1}.$
Used: $f(x)=√x$ is an increasing function.
answered Feb 2 at 10:22
Peter SzilasPeter Szilas
12k2822
$endgroup$
1
$begingroup$
You give a nice proof!
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– jackson
Feb 2 at 10:24
1
$begingroup$
Jackson.Thanks, once I got your lead, not so very difficult:))
$endgroup$
– Peter Szilas
Feb 2 at 10:25
add a comment |
$begingroup$
Wesley's lemma 1 nicely shows that $s_n$ is bounded above.
Rephrasing:
Show that $s_n$ is increasing:
1) $n=1$: $s_2 > s_1√$
2) Hypothesis : $s_{n+1} ge s_n.$
Step $n+1:$
$s_{n+2} =sqrt{2+√s_{n+1}} {Hyp. atop ge} sqrt{2+√s_n} =s_{n+1}.$
Used: $f(x)=√x$ is an increasing function.
answered Feb 2 at 10:22
Peter SzilasPeter Szilas
12k2822
$endgroup$
Wesley's lemma 1 nicely shows that $s_n$ is bounded above.
Rephrasing:
Show that $s_n$ is increasing:
1) $n=1$: $s_2 > s_1√$
2) Hypothesis : $s_{n+1} ge s_n.$
Step $n+1:$
$s_{n+2} =sqrt{2+√s_{n+1}} {Hyp. atop ge} sqrt{2+√s_n} =s_{n+1}.$
Used: $f(x)=√x$ is an increasing function.
answered Feb 2 at 10:22
Peter SzilasPeter Szilas
12k2822
edited Feb 3 at 9:37
answered Feb 2 at 10:22
Peter SzilasPeter Szilas
12k2822
answered Feb 2 at 10:22
Peter SzilasPeter Szilas
12k2822
answered Feb 2 at 10:22
Peter SzilasPeter Szilas
12k2822
12k2822
1
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You give a nice proof!
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– jackson
Feb 2 at 10:24
1
$begingroup$
Jackson.Thanks, once I got your lead, not so very difficult:))
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– Peter Szilas
Feb 2 at 10:25
add a comment |
1
$begingroup$
You give a nice proof!
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– jackson
Feb 2 at 10:24
1
$begingroup$
Jackson.Thanks, once I got your lead, not so very difficult:))
$endgroup$
– Peter Szilas
Feb 2 at 10:25
1
1
$begingroup$
You give a nice proof!
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– jackson
Feb 2 at 10:24
$begingroup$
You give a nice proof!
$endgroup$
– jackson
Feb 2 at 10:24
1
1
$begingroup$
Jackson.Thanks, once I got your lead, not so very difficult:))
$endgroup$
– Peter Szilas
Feb 2 at 10:25
$begingroup$
Jackson.Thanks, once I got your lead, not so very difficult:))
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– Peter Szilas
Feb 2 at 10:25
add a comment |
2
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Jackson.All is in place.Just work it out clearly.+
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– Peter Szilas
Feb 2 at 10:05
1
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Thanks a lot for you answer my question ,I figure it out
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– jackson
Feb 2 at 10:06
1
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also covered here
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– rtybase
Feb 2 at 10:28
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@rtybase how you search for that I have searched before I asked but I still don’t find the answer....
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– jackson
Feb 2 at 10:28
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I have dealt with this problem before ... just from the top of my head.
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– rtybase
Feb 2 at 10:29