Mixing
Why does infinitely repeating $sqrt[n]{x}$ converge to 1? [closed]
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A coworker showed me an interesting fact of math this morning. He stated that if you get the square root of $x$ and then get the square root of the result, and so on, and so forth, to infinity, then the result will converge to $1$ as your iteration approaches infinity. So essentially, the following statement should always be true:
$$1 < sqrt[n]{x} < x$$
No matter how many times you iterate.
Why does this hold true, even if $x = 1.0...1$? For example:
$$1 < sqrt[1000]{sqrt[9]{sqrt[52]{sqrt[99]{sqrt[3]{sqrt[563]{1.000001}}}}}} < 1.000001$$
nested-radicals
asked Jan 23 at 17:38
PerpetualJPerpetualJ
1767
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closed as unclear what you're asking by Xander Henderson, Mark Viola, Holo, Namaste, José Carlos Santos Jan 23 at 18:54
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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show 5 more comments
$begingroup$
A coworker showed me an interesting fact of math this morning. He stated that if you get the square root of $x$ and then get the square root of the result, and so on, and so forth, to infinity, then the result will converge to $1$ as your iteration approaches infinity. So essentially, the following statement should always be true:
$$1 < sqrt[n]{x} < x$$
No matter how many times you iterate.
Why does this hold true, even if $x = 1.0...1$? For example:
$$1 < sqrt[1000]{sqrt[9]{sqrt[52]{sqrt[99]{sqrt[3]{sqrt[563]{1.000001}}}}}} < 1.000001$$
nested-radicals
asked Jan 23 at 17:38
PerpetualJPerpetualJ
1767
$endgroup$
closed as unclear what you're asking by Xander Henderson, Mark Viola, Holo, Namaste, José Carlos Santos Jan 23 at 18:54
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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What is the question? Is it the title? Or is it in the body of the text?
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– lightxbulb
Jan 23 at 17:42
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The Banach fixed point theorem is mildly relevant. It might also be relevant to note that $$ sqrt[n]{sqrt[n]{x}} = (x^{1/n})^{1/n} = x^{1/n^2}. $$ As $nto infty$, the exponent $1/n^2 to 0$.
$endgroup$
– Xander Henderson
Jan 23 at 17:42
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@XanderHenderson Does the OP mean the $n$'th root or $n$ applications of the square root, or yet something else?
$endgroup$
– Mark Viola
Jan 23 at 17:46
2
$begingroup$
@XanderHenderson I asked you because your comment focused on the $n$'th root of $x$, not $n$ applications of the square root. So, one could infer that you had made an interpretation.
$endgroup$
– Mark Viola
Jan 23 at 17:51
2
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@amWhy What point?
$endgroup$
– Mark Viola
Jan 23 at 17:56
|
show 5 more comments
$begingroup$
A coworker showed me an interesting fact of math this morning. He stated that if you get the square root of $x$ and then get the square root of the result, and so on, and so forth, to infinity, then the result will converge to $1$ as your iteration approaches infinity. So essentially, the following statement should always be true:
$$1 < sqrt[n]{x} < x$$
No matter how many times you iterate.
Why does this hold true, even if $x = 1.0...1$? For example:
$$1 < sqrt[1000]{sqrt[9]{sqrt[52]{sqrt[99]{sqrt[3]{sqrt[563]{1.000001}}}}}} < 1.000001$$
nested-radicals
asked Jan 23 at 17:38
PerpetualJPerpetualJ
1767
$endgroup$
A coworker showed me an interesting fact of math this morning. He stated that if you get the square root of $x$ and then get the square root of the result, and so on, and so forth, to infinity, then the result will converge to $1$ as your iteration approaches infinity. So essentially, the following statement should always be true:
$$1 < sqrt[n]{x} < x$$
No matter how many times you iterate.
Why does this hold true, even if $x = 1.0...1$? For example:
$$1 < sqrt[1000]{sqrt[9]{sqrt[52]{sqrt[99]{sqrt[3]{sqrt[563]{1.000001}}}}}} < 1.000001$$
nested-radicals
nested-radicals
asked Jan 23 at 17:38
PerpetualJPerpetualJ
1767
asked Jan 23 at 17:38
PerpetualJPerpetualJ
1767
asked Jan 23 at 17:38
PerpetualJPerpetualJ
1767
asked Jan 23 at 17:38
PerpetualJPerpetualJ
1767
asked Jan 23 at 17:38
PerpetualJPerpetualJ
1767
1767
closed as unclear what you're asking by Xander Henderson, Mark Viola, Holo, Namaste, José Carlos Santos Jan 23 at 18:54
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by Xander Henderson, Mark Viola, Holo, Namaste, José Carlos Santos Jan 23 at 18:54
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
What is the question? Is it the title? Or is it in the body of the text?
$endgroup$
– lightxbulb
Jan 23 at 17:42
$begingroup$
The Banach fixed point theorem is mildly relevant. It might also be relevant to note that $$ sqrt[n]{sqrt[n]{x}} = (x^{1/n})^{1/n} = x^{1/n^2}. $$ As $nto infty$, the exponent $1/n^2 to 0$.
$endgroup$
– Xander Henderson
Jan 23 at 17:42
$begingroup$
@XanderHenderson Does the OP mean the $n$'th root or $n$ applications of the square root, or yet something else?
$endgroup$
– Mark Viola
Jan 23 at 17:46
2
$begingroup$
@XanderHenderson I asked you because your comment focused on the $n$'th root of $x$, not $n$ applications of the square root. So, one could infer that you had made an interpretation.
$endgroup$
– Mark Viola
Jan 23 at 17:51
2
$begingroup$
@amWhy What point?
$endgroup$
– Mark Viola
Jan 23 at 17:56
|
show 5 more comments
$begingroup$
What is the question? Is it the title? Or is it in the body of the text?
$endgroup$
– lightxbulb
Jan 23 at 17:42
$begingroup$
The Banach fixed point theorem is mildly relevant. It might also be relevant to note that $$ sqrt[n]{sqrt[n]{x}} = (x^{1/n})^{1/n} = x^{1/n^2}. $$ As $nto infty$, the exponent $1/n^2 to 0$.
$endgroup$
– Xander Henderson
Jan 23 at 17:42
$begingroup$
@XanderHenderson Does the OP mean the $n$'th root or $n$ applications of the square root, or yet something else?
$endgroup$
– Mark Viola
Jan 23 at 17:46
2
$begingroup$
@XanderHenderson I asked you because your comment focused on the $n$'th root of $x$, not $n$ applications of the square root. So, one could infer that you had made an interpretation.
$endgroup$
– Mark Viola
Jan 23 at 17:51
2
$begingroup$
@amWhy What point?
$endgroup$
– Mark Viola
Jan 23 at 17:56
$begingroup$
What is the question? Is it the title? Or is it in the body of the text?
$endgroup$
– lightxbulb
Jan 23 at 17:42
$begingroup$
What is the question? Is it the title? Or is it in the body of the text?
$endgroup$
– lightxbulb
Jan 23 at 17:42
$begingroup$
The Banach fixed point theorem is mildly relevant. It might also be relevant to note that $$ sqrt[n]{sqrt[n]{x}} = (x^{1/n})^{1/n} = x^{1/n^2}. $$ As $nto infty$, the exponent $1/n^2 to 0$.
$endgroup$
– Xander Henderson
Jan 23 at 17:42
$begingroup$
The Banach fixed point theorem is mildly relevant. It might also be relevant to note that $$ sqrt[n]{sqrt[n]{x}} = (x^{1/n})^{1/n} = x^{1/n^2}. $$ As $nto infty$, the exponent $1/n^2 to 0$.
$endgroup$
– Xander Henderson
Jan 23 at 17:42
$begingroup$
@XanderHenderson Does the OP mean the $n$'th root or $n$ applications of the square root, or yet something else?
$endgroup$
– Mark Viola
Jan 23 at 17:46
$begingroup$
@XanderHenderson Does the OP mean the $n$'th root or $n$ applications of the square root, or yet something else?
$endgroup$
– Mark Viola
Jan 23 at 17:46
2
2
$begingroup$
@XanderHenderson I asked you because your comment focused on the $n$'th root of $x$, not $n$ applications of the square root. So, one could infer that you had made an interpretation.
$endgroup$
– Mark Viola
Jan 23 at 17:51
$begingroup$
@XanderHenderson I asked you because your comment focused on the $n$'th root of $x$, not $n$ applications of the square root. So, one could infer that you had made an interpretation.
$endgroup$
– Mark Viola
Jan 23 at 17:51
2
2
$begingroup$
@amWhy What point?
$endgroup$
– Mark Viola
Jan 23 at 17:56
$begingroup$
@amWhy What point?
$endgroup$
– Mark Viola
Jan 23 at 17:56
|
show 5 more comments
3 Answers
3
active
oldest
votes
$begingroup$
Yes, if you start with $x>1$. The square root is monotone: that is, it preserves inequalities between positive numbers. So, if $1<x$, then $1=sqrt 1<sqrt x$. And you also have $sqrt x<x$, since this is equivalent to $sqrt x>1$. So, yes,
$$
1<x^{1/2^n}<x
$$
for all $n$. Note that applying the square root successively is to take the power $1/2$ $n$ times. So what you wrote $sqrt[n]{x}$ is $x^{1/2^n}$. The notation $sqrt[n]{x}$ usually means $x^{1/n}$.
As for the limit, you have
$$
x^{1/2^n}=e^{tfrac1{2^n}log x}xrightarrow[ntoinfty]{}e^0=1.
$$
answered Jan 23 at 17:51
Martin ArgeramiMartin Argerami
128k1184184
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add a comment |
$begingroup$
First, the notation $sqrt[n]{x}$ means the $n$'th root of $x$ and not $n$ applications of the square root.
Second, note that
$$underbrace{sqrt{dotssqrt{sqrt{sqrt{x}}}}}_{n,,text{applications}}=x^{1/2^n}$$
Hence, as $n$ increases, $1/2^n$ decreases and as $nto infty$, $1/2^nto0$. Therefore, no matter the value of $x>0$, $x^0=1$.
Aside, note that for $x>1$, $1<sqrt{x}<x$ while for $0<x<1$, $1>sqrt{x}>x$.
answered Jan 23 at 17:49
Mark ViolaMark Viola
133k1278176
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add a comment |
$begingroup$
$sqrt[n]{x} = exp(log(sqrt[n]{x})) = expleft(frac{1}{n}log(x)right)$. Can you solve your question now?
answered Jan 23 at 17:48
KlausKlaus
2,30712
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add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Yes, if you start with $x>1$. The square root is monotone: that is, it preserves inequalities between positive numbers. So, if $1<x$, then $1=sqrt 1<sqrt x$. And you also have $sqrt x<x$, since this is equivalent to $sqrt x>1$. So, yes,
$$
1<x^{1/2^n}<x
$$
for all $n$. Note that applying the square root successively is to take the power $1/2$ $n$ times. So what you wrote $sqrt[n]{x}$ is $x^{1/2^n}$. The notation $sqrt[n]{x}$ usually means $x^{1/n}$.
As for the limit, you have
$$
x^{1/2^n}=e^{tfrac1{2^n}log x}xrightarrow[ntoinfty]{}e^0=1.
$$
answered Jan 23 at 17:51
Martin ArgeramiMartin Argerami
128k1184184
$endgroup$
add a comment |
$begingroup$
Yes, if you start with $x>1$. The square root is monotone: that is, it preserves inequalities between positive numbers. So, if $1<x$, then $1=sqrt 1<sqrt x$. And you also have $sqrt x<x$, since this is equivalent to $sqrt x>1$. So, yes,
$$
1<x^{1/2^n}<x
$$
for all $n$. Note that applying the square root successively is to take the power $1/2$ $n$ times. So what you wrote $sqrt[n]{x}$ is $x^{1/2^n}$. The notation $sqrt[n]{x}$ usually means $x^{1/n}$.
As for the limit, you have
$$
x^{1/2^n}=e^{tfrac1{2^n}log x}xrightarrow[ntoinfty]{}e^0=1.
$$
answered Jan 23 at 17:51
Martin ArgeramiMartin Argerami
128k1184184
$endgroup$
add a comment |
$begingroup$
Yes, if you start with $x>1$. The square root is monotone: that is, it preserves inequalities between positive numbers. So, if $1<x$, then $1=sqrt 1<sqrt x$. And you also have $sqrt x<x$, since this is equivalent to $sqrt x>1$. So, yes,
$$
1<x^{1/2^n}<x
$$
for all $n$. Note that applying the square root successively is to take the power $1/2$ $n$ times. So what you wrote $sqrt[n]{x}$ is $x^{1/2^n}$. The notation $sqrt[n]{x}$ usually means $x^{1/n}$.
As for the limit, you have
$$
x^{1/2^n}=e^{tfrac1{2^n}log x}xrightarrow[ntoinfty]{}e^0=1.
$$
answered Jan 23 at 17:51
Martin ArgeramiMartin Argerami
128k1184184
$endgroup$
Yes, if you start with $x>1$. The square root is monotone: that is, it preserves inequalities between positive numbers. So, if $1<x$, then $1=sqrt 1<sqrt x$. And you also have $sqrt x<x$, since this is equivalent to $sqrt x>1$. So, yes,
$$
1<x^{1/2^n}<x
$$
for all $n$. Note that applying the square root successively is to take the power $1/2$ $n$ times. So what you wrote $sqrt[n]{x}$ is $x^{1/2^n}$. The notation $sqrt[n]{x}$ usually means $x^{1/n}$.
As for the limit, you have
$$
x^{1/2^n}=e^{tfrac1{2^n}log x}xrightarrow[ntoinfty]{}e^0=1.
$$
answered Jan 23 at 17:51
Martin ArgeramiMartin Argerami
128k1184184
answered Jan 23 at 17:51
Martin ArgeramiMartin Argerami
128k1184184
answered Jan 23 at 17:51
Martin ArgeramiMartin Argerami
128k1184184
answered Jan 23 at 17:51
Martin ArgeramiMartin Argerami
128k1184184
128k1184184
add a comment |
add a comment |
$begingroup$
First, the notation $sqrt[n]{x}$ means the $n$'th root of $x$ and not $n$ applications of the square root.
Second, note that
$$underbrace{sqrt{dotssqrt{sqrt{sqrt{x}}}}}_{n,,text{applications}}=x^{1/2^n}$$
Hence, as $n$ increases, $1/2^n$ decreases and as $nto infty$, $1/2^nto0$. Therefore, no matter the value of $x>0$, $x^0=1$.
Aside, note that for $x>1$, $1<sqrt{x}<x$ while for $0<x<1$, $1>sqrt{x}>x$.
answered Jan 23 at 17:49
Mark ViolaMark Viola
133k1278176
$endgroup$
add a comment |
$begingroup$
First, the notation $sqrt[n]{x}$ means the $n$'th root of $x$ and not $n$ applications of the square root.
Second, note that
$$underbrace{sqrt{dotssqrt{sqrt{sqrt{x}}}}}_{n,,text{applications}}=x^{1/2^n}$$
Hence, as $n$ increases, $1/2^n$ decreases and as $nto infty$, $1/2^nto0$. Therefore, no matter the value of $x>0$, $x^0=1$.
Aside, note that for $x>1$, $1<sqrt{x}<x$ while for $0<x<1$, $1>sqrt{x}>x$.
answered Jan 23 at 17:49
Mark ViolaMark Viola
133k1278176
$endgroup$
add a comment |
$begingroup$
First, the notation $sqrt[n]{x}$ means the $n$'th root of $x$ and not $n$ applications of the square root.
Second, note that
$$underbrace{sqrt{dotssqrt{sqrt{sqrt{x}}}}}_{n,,text{applications}}=x^{1/2^n}$$
Hence, as $n$ increases, $1/2^n$ decreases and as $nto infty$, $1/2^nto0$. Therefore, no matter the value of $x>0$, $x^0=1$.
Aside, note that for $x>1$, $1<sqrt{x}<x$ while for $0<x<1$, $1>sqrt{x}>x$.
answered Jan 23 at 17:49
Mark ViolaMark Viola
133k1278176
$endgroup$
First, the notation $sqrt[n]{x}$ means the $n$'th root of $x$ and not $n$ applications of the square root.
Second, note that
$$underbrace{sqrt{dotssqrt{sqrt{sqrt{x}}}}}_{n,,text{applications}}=x^{1/2^n}$$
Hence, as $n$ increases, $1/2^n$ decreases and as $nto infty$, $1/2^nto0$. Therefore, no matter the value of $x>0$, $x^0=1$.
Aside, note that for $x>1$, $1<sqrt{x}<x$ while for $0<x<1$, $1>sqrt{x}>x$.
answered Jan 23 at 17:49
Mark ViolaMark Viola
133k1278176
edited Jan 23 at 17:54
answered Jan 23 at 17:49
Mark ViolaMark Viola
133k1278176
answered Jan 23 at 17:49
Mark ViolaMark Viola
133k1278176
answered Jan 23 at 17:49
Mark ViolaMark Viola
133k1278176
133k1278176
add a comment |
add a comment |
$begingroup$
$sqrt[n]{x} = exp(log(sqrt[n]{x})) = expleft(frac{1}{n}log(x)right)$. Can you solve your question now?
answered Jan 23 at 17:48
KlausKlaus
2,30712
$endgroup$
add a comment |
$begingroup$
$sqrt[n]{x} = exp(log(sqrt[n]{x})) = expleft(frac{1}{n}log(x)right)$. Can you solve your question now?
answered Jan 23 at 17:48
KlausKlaus
2,30712
$endgroup$
add a comment |
$begingroup$
$sqrt[n]{x} = exp(log(sqrt[n]{x})) = expleft(frac{1}{n}log(x)right)$. Can you solve your question now?
answered Jan 23 at 17:48
KlausKlaus
2,30712
$endgroup$
$sqrt[n]{x} = exp(log(sqrt[n]{x})) = expleft(frac{1}{n}log(x)right)$. Can you solve your question now?
answered Jan 23 at 17:48
KlausKlaus
2,30712
answered Jan 23 at 17:48
KlausKlaus
2,30712
answered Jan 23 at 17:48
KlausKlaus
2,30712
answered Jan 23 at 17:48
KlausKlaus
2,30712
2,30712
add a comment |
add a comment |
$begingroup$
What is the question? Is it the title? Or is it in the body of the text?
$endgroup$
– lightxbulb
Jan 23 at 17:42
$begingroup$
The Banach fixed point theorem is mildly relevant. It might also be relevant to note that $$ sqrt[n]{sqrt[n]{x}} = (x^{1/n})^{1/n} = x^{1/n^2}. $$ As $nto infty$, the exponent $1/n^2 to 0$.
$endgroup$
– Xander Henderson
Jan 23 at 17:42
$begingroup$
@XanderHenderson Does the OP mean the $n$'th root or $n$ applications of the square root, or yet something else?
$endgroup$
– Mark Viola
Jan 23 at 17:46
2
$begingroup$
@XanderHenderson I asked you because your comment focused on the $n$'th root of $x$, not $n$ applications of the square root. So, one could infer that you had made an interpretation.
$endgroup$
– Mark Viola
Jan 23 at 17:51
2
$begingroup$
@amWhy What point?
$endgroup$
– Mark Viola
Jan 23 at 17:56