Mixing
Proof Verification of Baby Rudin 3.3
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I am currently studying Rudin's Foundations of Real Analysis. I wrote a solution to Rudin's problem 3.3, and wanted to get some input on my proof. I appreciate any suggestions on the technicalities, style, and clarity.
Problem:
If $s_1=sqrt2$, and
$s_{n+1} = sqrt{2+sqrt{s_n}}$ where $n = 1, 2, 3, ...$ prove that {${s_n}$} converges, and that $s_n<2$ for $n = 1, 2, 3, ...$
My proof:
We first show $s_n<2$ for $n = 1, 2, 3, ...$ by induction.
When $n = 1$, it is clear that $sqrt2 < 2$.
If $s_{n-1}$ < 2, then
$0<sqrt{2+sqrt{s_{n-1}}} < 2$
$leftrightarrow 2+sqrt{s_{n-1}}<4$
$leftrightarrow sqrt{s_{n-1}} < 2$
$leftrightarrow s_{n-1} < 4$
But $s_{n-1} < 2$ by induction hypothesis. Thus $s_n<2$ for $n = 1, 2, 3, ...$ as desired.
We next show that {$s_{n}$} is monotonically increasing by induction.
It is clear that $s_1leq s_2$.
We need to show $s_n leq s_{n+1}$ for all $n geq 2$. Assume $s_{n-1} leq s_{n}$. Then,
$s_n leq s_{n+1}$
$leftrightarrow 0 < sqrt{2+sqrt{s_{n-1}}} leq sqrt{2+sqrt{s_{n}}}$
$leftrightarrow 2+sqrt{s_{n-1}} leq 2+sqrt{s_{n}}$
$leftrightarrow sqrt{s_{n-1}} leq sqrt{s_{n}}$
$leftrightarrow s_{n-1} leq s_{n}$
But $s_{n-1} leq s_{n}$ by the induction hypothesis. Thus the sequence is monotonically increasing.
Since {$s_n$} is bounded above and monotonically increasing, it converges.
sequences-and-series
asked Nov 5 at 5:20
b_chao
143
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I am currently studying Rudin's Foundations of Real Analysis. I wrote a solution to Rudin's problem 3.3, and wanted to get some input on my proof. I appreciate any suggestions on the technicalities, style, and clarity.
Problem:
If $s_1=sqrt2$, and
$s_{n+1} = sqrt{2+sqrt{s_n}}$ where $n = 1, 2, 3, ...$ prove that {${s_n}$} converges, and that $s_n<2$ for $n = 1, 2, 3, ...$
My proof:
We first show $s_n<2$ for $n = 1, 2, 3, ...$ by induction.
When $n = 1$, it is clear that $sqrt2 < 2$.
If $s_{n-1}$ < 2, then
$0<sqrt{2+sqrt{s_{n-1}}} < 2$
$leftrightarrow 2+sqrt{s_{n-1}}<4$
$leftrightarrow sqrt{s_{n-1}} < 2$
$leftrightarrow s_{n-1} < 4$
But $s_{n-1} < 2$ by induction hypothesis. Thus $s_n<2$ for $n = 1, 2, 3, ...$ as desired.
We next show that {$s_{n}$} is monotonically increasing by induction.
It is clear that $s_1leq s_2$.
We need to show $s_n leq s_{n+1}$ for all $n geq 2$. Assume $s_{n-1} leq s_{n}$. Then,
$s_n leq s_{n+1}$
$leftrightarrow 0 < sqrt{2+sqrt{s_{n-1}}} leq sqrt{2+sqrt{s_{n}}}$
$leftrightarrow 2+sqrt{s_{n-1}} leq 2+sqrt{s_{n}}$
$leftrightarrow sqrt{s_{n-1}} leq sqrt{s_{n}}$
$leftrightarrow s_{n-1} leq s_{n}$
But $s_{n-1} leq s_{n}$ by the induction hypothesis. Thus the sequence is monotonically increasing.
Since {$s_n$} is bounded above and monotonically increasing, it converges.
sequences-and-series
asked Nov 5 at 5:20
b_chao
143
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am currently studying Rudin's Foundations of Real Analysis. I wrote a solution to Rudin's problem 3.3, and wanted to get some input on my proof. I appreciate any suggestions on the technicalities, style, and clarity.
Problem:
If $s_1=sqrt2$, and
$s_{n+1} = sqrt{2+sqrt{s_n}}$ where $n = 1, 2, 3, ...$ prove that {${s_n}$} converges, and that $s_n<2$ for $n = 1, 2, 3, ...$
My proof:
We first show $s_n<2$ for $n = 1, 2, 3, ...$ by induction.
When $n = 1$, it is clear that $sqrt2 < 2$.
If $s_{n-1}$ < 2, then
$0<sqrt{2+sqrt{s_{n-1}}} < 2$
$leftrightarrow 2+sqrt{s_{n-1}}<4$
$leftrightarrow sqrt{s_{n-1}} < 2$
$leftrightarrow s_{n-1} < 4$
But $s_{n-1} < 2$ by induction hypothesis. Thus $s_n<2$ for $n = 1, 2, 3, ...$ as desired.
We next show that {$s_{n}$} is monotonically increasing by induction.
It is clear that $s_1leq s_2$.
We need to show $s_n leq s_{n+1}$ for all $n geq 2$. Assume $s_{n-1} leq s_{n}$. Then,
$s_n leq s_{n+1}$
$leftrightarrow 0 < sqrt{2+sqrt{s_{n-1}}} leq sqrt{2+sqrt{s_{n}}}$
$leftrightarrow 2+sqrt{s_{n-1}} leq 2+sqrt{s_{n}}$
$leftrightarrow sqrt{s_{n-1}} leq sqrt{s_{n}}$
$leftrightarrow s_{n-1} leq s_{n}$
But $s_{n-1} leq s_{n}$ by the induction hypothesis. Thus the sequence is monotonically increasing.
Since {$s_n$} is bounded above and monotonically increasing, it converges.
sequences-and-series
asked Nov 5 at 5:20
b_chao
143
I am currently studying Rudin's Foundations of Real Analysis. I wrote a solution to Rudin's problem 3.3, and wanted to get some input on my proof. I appreciate any suggestions on the technicalities, style, and clarity.
Problem:
If $s_1=sqrt2$, and
$s_{n+1} = sqrt{2+sqrt{s_n}}$ where $n = 1, 2, 3, ...$ prove that {${s_n}$} converges, and that $s_n<2$ for $n = 1, 2, 3, ...$
My proof:
We first show $s_n<2$ for $n = 1, 2, 3, ...$ by induction.
When $n = 1$, it is clear that $sqrt2 < 2$.
If $s_{n-1}$ < 2, then
$0<sqrt{2+sqrt{s_{n-1}}} < 2$
$leftrightarrow 2+sqrt{s_{n-1}}<4$
$leftrightarrow sqrt{s_{n-1}} < 2$
$leftrightarrow s_{n-1} < 4$
But $s_{n-1} < 2$ by induction hypothesis. Thus $s_n<2$ for $n = 1, 2, 3, ...$ as desired.
We next show that {$s_{n}$} is monotonically increasing by induction.
It is clear that $s_1leq s_2$.
We need to show $s_n leq s_{n+1}$ for all $n geq 2$. Assume $s_{n-1} leq s_{n}$. Then,
$s_n leq s_{n+1}$
$leftrightarrow 0 < sqrt{2+sqrt{s_{n-1}}} leq sqrt{2+sqrt{s_{n}}}$
$leftrightarrow 2+sqrt{s_{n-1}} leq 2+sqrt{s_{n}}$
$leftrightarrow sqrt{s_{n-1}} leq sqrt{s_{n}}$
$leftrightarrow s_{n-1} leq s_{n}$
But $s_{n-1} leq s_{n}$ by the induction hypothesis. Thus the sequence is monotonically increasing.
Since {$s_n$} is bounded above and monotonically increasing, it converges.
sequences-and-series
sequences-and-series
asked Nov 5 at 5:20
b_chao
143
asked Nov 5 at 5:20
b_chao
143
edited 18 hours ago
asked Nov 5 at 5:20
b_chao
143
asked Nov 5 at 5:20
b_chao
143
asked Nov 5 at 5:20
b_chao
143
143
add a comment |
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1 Answer
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0
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- For the first proof, Also, I would show a series of forwards implications rather than the string if "if and only if" statements you are using, since your proof doesn't require both directions.
In this spirit, your first inductive proof, upper-bounding $s_{n}$ by $2$, could be re-phrased as follows:
We claim $s_{n} < 2$ for all $n$ by induction.
For $n = 1$ this is clear. Assume $s_{n-1} < 2$. Then since the square root function is monotonically increasing on the non-negative reals,
begin{eqnarray}
sqrt{s_{n-1}} < 2 \
2 + sqrt{s_{n-1}} < 4 \
sqrt{2 + sqrt{s_{n-1}}} < 2
end{eqnarray}
Since $s_n = sqrt{2 + sqrt{s_{n-1}}}$ this completes the inductive step.
I made a point of saying the square-root function is monotonic. This is fairly pedantic, and I don't think is really needed, but given that your argument implicitly uses this fact, it doesn't hurt.
- Your second proof has what I believe to be a typo.
We need show $s_n leq s_{n−1}$ for all $n geq 2$. Assume $s_n leq s_{n-1}$.
At first glance, I wasn't sure if this was a typo or if you meant to do a proof by contradiction. I think what you meant to show was $s_n leq s_{n+1}$. In that case, replacing $s_{n-1}$ with $s_{n+1}$ where appropriate should be fine, and stylistically I would say the same things as regarding your first proof.
answered yesterday
Akhil Jalan
465
New contributor
Akhil Jalan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Thanks for the review! I corrected the typos you mentioned.
– b_chao
18 hours ago
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
- For the first proof, Also, I would show a series of forwards implications rather than the string if "if and only if" statements you are using, since your proof doesn't require both directions.
In this spirit, your first inductive proof, upper-bounding $s_{n}$ by $2$, could be re-phrased as follows:
We claim $s_{n} < 2$ for all $n$ by induction.
For $n = 1$ this is clear. Assume $s_{n-1} < 2$. Then since the square root function is monotonically increasing on the non-negative reals,
begin{eqnarray}
sqrt{s_{n-1}} < 2 \
2 + sqrt{s_{n-1}} < 4 \
sqrt{2 + sqrt{s_{n-1}}} < 2
end{eqnarray}
Since $s_n = sqrt{2 + sqrt{s_{n-1}}}$ this completes the inductive step.
I made a point of saying the square-root function is monotonic. This is fairly pedantic, and I don't think is really needed, but given that your argument implicitly uses this fact, it doesn't hurt.
- Your second proof has what I believe to be a typo.
We need show $s_n leq s_{n−1}$ for all $n geq 2$. Assume $s_n leq s_{n-1}$.
At first glance, I wasn't sure if this was a typo or if you meant to do a proof by contradiction. I think what you meant to show was $s_n leq s_{n+1}$. In that case, replacing $s_{n-1}$ with $s_{n+1}$ where appropriate should be fine, and stylistically I would say the same things as regarding your first proof.
answered yesterday
Akhil Jalan
465
New contributor
Akhil Jalan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Thanks for the review! I corrected the typos you mentioned.
– b_chao
18 hours ago
add a comment |
up vote
0
down vote
- For the first proof, Also, I would show a series of forwards implications rather than the string if "if and only if" statements you are using, since your proof doesn't require both directions.
In this spirit, your first inductive proof, upper-bounding $s_{n}$ by $2$, could be re-phrased as follows:
We claim $s_{n} < 2$ for all $n$ by induction.
For $n = 1$ this is clear. Assume $s_{n-1} < 2$. Then since the square root function is monotonically increasing on the non-negative reals,
begin{eqnarray}
sqrt{s_{n-1}} < 2 \
2 + sqrt{s_{n-1}} < 4 \
sqrt{2 + sqrt{s_{n-1}}} < 2
end{eqnarray}
Since $s_n = sqrt{2 + sqrt{s_{n-1}}}$ this completes the inductive step.
I made a point of saying the square-root function is monotonic. This is fairly pedantic, and I don't think is really needed, but given that your argument implicitly uses this fact, it doesn't hurt.
- Your second proof has what I believe to be a typo.
We need show $s_n leq s_{n−1}$ for all $n geq 2$. Assume $s_n leq s_{n-1}$.
At first glance, I wasn't sure if this was a typo or if you meant to do a proof by contradiction. I think what you meant to show was $s_n leq s_{n+1}$. In that case, replacing $s_{n-1}$ with $s_{n+1}$ where appropriate should be fine, and stylistically I would say the same things as regarding your first proof.
answered yesterday
Akhil Jalan
465
New contributor
Akhil Jalan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Thanks for the review! I corrected the typos you mentioned.
– b_chao
18 hours ago
add a comment |
up vote
0
down vote
up vote
0
down vote
- For the first proof, Also, I would show a series of forwards implications rather than the string if "if and only if" statements you are using, since your proof doesn't require both directions.
In this spirit, your first inductive proof, upper-bounding $s_{n}$ by $2$, could be re-phrased as follows:
We claim $s_{n} < 2$ for all $n$ by induction.
For $n = 1$ this is clear. Assume $s_{n-1} < 2$. Then since the square root function is monotonically increasing on the non-negative reals,
begin{eqnarray}
sqrt{s_{n-1}} < 2 \
2 + sqrt{s_{n-1}} < 4 \
sqrt{2 + sqrt{s_{n-1}}} < 2
end{eqnarray}
Since $s_n = sqrt{2 + sqrt{s_{n-1}}}$ this completes the inductive step.
I made a point of saying the square-root function is monotonic. This is fairly pedantic, and I don't think is really needed, but given that your argument implicitly uses this fact, it doesn't hurt.
- Your second proof has what I believe to be a typo.
We need show $s_n leq s_{n−1}$ for all $n geq 2$. Assume $s_n leq s_{n-1}$.
At first glance, I wasn't sure if this was a typo or if you meant to do a proof by contradiction. I think what you meant to show was $s_n leq s_{n+1}$. In that case, replacing $s_{n-1}$ with $s_{n+1}$ where appropriate should be fine, and stylistically I would say the same things as regarding your first proof.
answered yesterday
Akhil Jalan
465
New contributor
Akhil Jalan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
- For the first proof, Also, I would show a series of forwards implications rather than the string if "if and only if" statements you are using, since your proof doesn't require both directions.
In this spirit, your first inductive proof, upper-bounding $s_{n}$ by $2$, could be re-phrased as follows:
We claim $s_{n} < 2$ for all $n$ by induction.
For $n = 1$ this is clear. Assume $s_{n-1} < 2$. Then since the square root function is monotonically increasing on the non-negative reals,
begin{eqnarray}
sqrt{s_{n-1}} < 2 \
2 + sqrt{s_{n-1}} < 4 \
sqrt{2 + sqrt{s_{n-1}}} < 2
end{eqnarray}
Since $s_n = sqrt{2 + sqrt{s_{n-1}}}$ this completes the inductive step.
I made a point of saying the square-root function is monotonic. This is fairly pedantic, and I don't think is really needed, but given that your argument implicitly uses this fact, it doesn't hurt.
- Your second proof has what I believe to be a typo.
We need show $s_n leq s_{n−1}$ for all $n geq 2$. Assume $s_n leq s_{n-1}$.
At first glance, I wasn't sure if this was a typo or if you meant to do a proof by contradiction. I think what you meant to show was $s_n leq s_{n+1}$. In that case, replacing $s_{n-1}$ with $s_{n+1}$ where appropriate should be fine, and stylistically I would say the same things as regarding your first proof.
answered yesterday
Akhil Jalan
465
New contributor
Akhil Jalan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered yesterday
Akhil Jalan
465
New contributor
Akhil Jalan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered yesterday
Akhil Jalan
465
answered yesterday
Akhil Jalan
465
465
New contributor
Akhil Jalan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Akhil Jalan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Akhil Jalan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Thanks for the review! I corrected the typos you mentioned.
– b_chao
18 hours ago
add a comment |
Thanks for the review! I corrected the typos you mentioned.
– b_chao
18 hours ago
Thanks for the review! I corrected the typos you mentioned.
– b_chao
18 hours ago
Thanks for the review! I corrected the typos you mentioned.
– b_chao
18 hours ago
add a comment |
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