Mixing
Put the $sqrt[n]{n}$ numbers in to an increasing order
I had a test today and there was an extra problem I couldn’t solve.
Put the $sqrt[2]{2}, sqrt[3]{3}, dots, sqrt[100]{100}$ numbers in to an increasing order.
I just have no idea. I can handle the numbers $sqrt[2^k]{2^k}$ numbers, but I can’t solve the problem.
radicals
edited Nov 20 '18 at 14:39
Lehs
6,94431662
asked Nov 20 '18 at 14:33
Ti Tu Lea
314
add a comment |
I had a test today and there was an extra problem I couldn’t solve.
Put the $sqrt[2]{2}, sqrt[3]{3}, dots, sqrt[100]{100}$ numbers in to an increasing order.
I just have no idea. I can handle the numbers $sqrt[2^k]{2^k}$ numbers, but I can’t solve the problem.
radicals
edited Nov 20 '18 at 14:39
Lehs
6,94431662
asked Nov 20 '18 at 14:33
Ti Tu Lea
314
1
What class are you taking? We need to know what tools are available to you.
– saulspatz
Nov 20 '18 at 14:37
This is already solved here.
– Dietrich Burde
Nov 20 '18 at 14:43
1
Possible duplicate of Show that $n^{(1/n)}$ is eventually decreasing
– Dietrich Burde
Nov 20 '18 at 14:44
add a comment |
I had a test today and there was an extra problem I couldn’t solve.
Put the $sqrt[2]{2}, sqrt[3]{3}, dots, sqrt[100]{100}$ numbers in to an increasing order.
I just have no idea. I can handle the numbers $sqrt[2^k]{2^k}$ numbers, but I can’t solve the problem.
radicals
edited Nov 20 '18 at 14:39
Lehs
6,94431662
asked Nov 20 '18 at 14:33
Ti Tu Lea
314
I had a test today and there was an extra problem I couldn’t solve.
Put the $sqrt[2]{2}, sqrt[3]{3}, dots, sqrt[100]{100}$ numbers in to an increasing order.
I just have no idea. I can handle the numbers $sqrt[2^k]{2^k}$ numbers, but I can’t solve the problem.
radicals
radicals
edited Nov 20 '18 at 14:39
Lehs
6,94431662
asked Nov 20 '18 at 14:33
Ti Tu Lea
314
edited Nov 20 '18 at 14:39
Lehs
6,94431662
asked Nov 20 '18 at 14:33
Ti Tu Lea
314
edited Nov 20 '18 at 14:39
Lehs
6,94431662
edited Nov 20 '18 at 14:39
Lehs
6,94431662
edited Nov 20 '18 at 14:39
Lehs
6,94431662
6,94431662
asked Nov 20 '18 at 14:33
Ti Tu Lea
314
asked Nov 20 '18 at 14:33
Ti Tu Lea
314
asked Nov 20 '18 at 14:33
Ti Tu Lea
314
314
1
What class are you taking? We need to know what tools are available to you.
– saulspatz
Nov 20 '18 at 14:37
This is already solved here.
– Dietrich Burde
Nov 20 '18 at 14:43
1
Possible duplicate of Show that $n^{(1/n)}$ is eventually decreasing
– Dietrich Burde
Nov 20 '18 at 14:44
add a comment |
1
What class are you taking? We need to know what tools are available to you.
– saulspatz
Nov 20 '18 at 14:37
This is already solved here.
– Dietrich Burde
Nov 20 '18 at 14:43
1
Possible duplicate of Show that $n^{(1/n)}$ is eventually decreasing
– Dietrich Burde
Nov 20 '18 at 14:44
1
1
What class are you taking? We need to know what tools are available to you.
– saulspatz
Nov 20 '18 at 14:37
What class are you taking? We need to know what tools are available to you.
– saulspatz
Nov 20 '18 at 14:37
This is already solved here.
– Dietrich Burde
Nov 20 '18 at 14:43
This is already solved here.
– Dietrich Burde
Nov 20 '18 at 14:43
1
1
Possible duplicate of Show that $n^{(1/n)}$ is eventually decreasing
– Dietrich Burde
Nov 20 '18 at 14:44
Possible duplicate of Show that $n^{(1/n)}$ is eventually decreasing
– Dietrich Burde
Nov 20 '18 at 14:44
add a comment |
4 Answers
4
active
oldest
votes
Consider the function $f(x)=x^{1/x}$ defined for $x>0$.
Then $f(x)=exp(frac{log x}{x})$ and so
$$
f'(x)=f(x)frac{1-log x}{x^2}
$$
which is positive over $(0,e)$ and negative over $(e,infty)$. So we know that
$$
sqrt[n]{n}>sqrt[n+1]{n+1}
$$
for $nge 3$.
Also $(sqrt{2})^6=8$, $(sqrt[3]{3})^6=9$, hence $sqrt{2}<sqrt[3]{3}$. Moreover $sqrt{2}=sqrt[4]{4}$.
answered Nov 20 '18 at 14:41
egreg
178k1484201
add a comment |
Hint only.
I suppose you know how to study functions, you know power functions and also that you know how to de the derivative a combination of functions.
Then, can you find a function that goes through all the points you are describing?
answered Nov 20 '18 at 14:41
Martigan
5,215917
add a comment |
Hint 1 If you have differential calculus available: Differentiate $x mapsto x^{1 / x}$ (or, since $log$ is increasing, $x mapsto log(x^{1 / x}) = frac{log x}{x}$) to find a single maximum that lies between $2$ and $3$.
Then, it remains only to find where in the list $sqrt{2}$ fits, but you've already written that you can handle the cases $sqrt[2^k]{2^k}$.
Hint 2 Alternatively, we can compare the ratio $frac{sqrt[n + 1]{n + 1}}{sqrt[n]{n}}$ of successive values; raising that ratio to the $n(n + 1)$ power gives $$frac{(n + 1)^n}{n^{n + 1}} = frac{1}{n} left(1 + frac{1}{n}right)^n .$$ But $left(1 + frac{1}{n}right)^n$ increases monotonically to $e < 3$.
answered Nov 20 '18 at 14:44
Travis
59.7k767146
add a comment |
Note that $ln n^{1/n}=frac {ln n}{n}$
Upon differentiating the function $$f(x)=frac {ln x}{x}$$
we get $$f'(x)=frac {1-ln x}{x^2}<0$$ for $xge 3$
Thus the sequence $n^{1/n}$ is decreasing for $nge 3$
The ordering is $$ 1<2^{1/2}<3^{1/3}$$ and $3^{1/3}>4^{1/4}>5^{1/5}>.....$
answered Nov 20 '18 at 15:03
Mohammad Riazi-Kermani
40.9k42059
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
Consider the function $f(x)=x^{1/x}$ defined for $x>0$.
Then $f(x)=exp(frac{log x}{x})$ and so
$$
f'(x)=f(x)frac{1-log x}{x^2}
$$
which is positive over $(0,e)$ and negative over $(e,infty)$. So we know that
$$
sqrt[n]{n}>sqrt[n+1]{n+1}
$$
for $nge 3$.
Also $(sqrt{2})^6=8$, $(sqrt[3]{3})^6=9$, hence $sqrt{2}<sqrt[3]{3}$. Moreover $sqrt{2}=sqrt[4]{4}$.
answered Nov 20 '18 at 14:41
egreg
178k1484201
add a comment |
Consider the function $f(x)=x^{1/x}$ defined for $x>0$.
Then $f(x)=exp(frac{log x}{x})$ and so
$$
f'(x)=f(x)frac{1-log x}{x^2}
$$
which is positive over $(0,e)$ and negative over $(e,infty)$. So we know that
$$
sqrt[n]{n}>sqrt[n+1]{n+1}
$$
for $nge 3$.
Also $(sqrt{2})^6=8$, $(sqrt[3]{3})^6=9$, hence $sqrt{2}<sqrt[3]{3}$. Moreover $sqrt{2}=sqrt[4]{4}$.
answered Nov 20 '18 at 14:41
egreg
178k1484201
add a comment |
Consider the function $f(x)=x^{1/x}$ defined for $x>0$.
Then $f(x)=exp(frac{log x}{x})$ and so
$$
f'(x)=f(x)frac{1-log x}{x^2}
$$
which is positive over $(0,e)$ and negative over $(e,infty)$. So we know that
$$
sqrt[n]{n}>sqrt[n+1]{n+1}
$$
for $nge 3$.
Also $(sqrt{2})^6=8$, $(sqrt[3]{3})^6=9$, hence $sqrt{2}<sqrt[3]{3}$. Moreover $sqrt{2}=sqrt[4]{4}$.
answered Nov 20 '18 at 14:41
egreg
178k1484201
Consider the function $f(x)=x^{1/x}$ defined for $x>0$.
Then $f(x)=exp(frac{log x}{x})$ and so
$$
f'(x)=f(x)frac{1-log x}{x^2}
$$
which is positive over $(0,e)$ and negative over $(e,infty)$. So we know that
$$
sqrt[n]{n}>sqrt[n+1]{n+1}
$$
for $nge 3$.
Also $(sqrt{2})^6=8$, $(sqrt[3]{3})^6=9$, hence $sqrt{2}<sqrt[3]{3}$. Moreover $sqrt{2}=sqrt[4]{4}$.
answered Nov 20 '18 at 14:41
egreg
178k1484201
answered Nov 20 '18 at 14:41
egreg
178k1484201
answered Nov 20 '18 at 14:41
egreg
178k1484201
answered Nov 20 '18 at 14:41
egreg
178k1484201
178k1484201
add a comment |
add a comment |
Hint only.
I suppose you know how to study functions, you know power functions and also that you know how to de the derivative a combination of functions.
Then, can you find a function that goes through all the points you are describing?
answered Nov 20 '18 at 14:41
Martigan
5,215917
add a comment |
Hint only.
I suppose you know how to study functions, you know power functions and also that you know how to de the derivative a combination of functions.
Then, can you find a function that goes through all the points you are describing?
answered Nov 20 '18 at 14:41
Martigan
5,215917
add a comment |
Hint only.
I suppose you know how to study functions, you know power functions and also that you know how to de the derivative a combination of functions.
Then, can you find a function that goes through all the points you are describing?
answered Nov 20 '18 at 14:41
Martigan
5,215917
Hint only.
I suppose you know how to study functions, you know power functions and also that you know how to de the derivative a combination of functions.
Then, can you find a function that goes through all the points you are describing?
answered Nov 20 '18 at 14:41
Martigan
5,215917
answered Nov 20 '18 at 14:41
Martigan
5,215917
answered Nov 20 '18 at 14:41
Martigan
5,215917
answered Nov 20 '18 at 14:41
Martigan
5,215917
5,215917
add a comment |
add a comment |
Hint 1 If you have differential calculus available: Differentiate $x mapsto x^{1 / x}$ (or, since $log$ is increasing, $x mapsto log(x^{1 / x}) = frac{log x}{x}$) to find a single maximum that lies between $2$ and $3$.
Then, it remains only to find where in the list $sqrt{2}$ fits, but you've already written that you can handle the cases $sqrt[2^k]{2^k}$.
Hint 2 Alternatively, we can compare the ratio $frac{sqrt[n + 1]{n + 1}}{sqrt[n]{n}}$ of successive values; raising that ratio to the $n(n + 1)$ power gives $$frac{(n + 1)^n}{n^{n + 1}} = frac{1}{n} left(1 + frac{1}{n}right)^n .$$ But $left(1 + frac{1}{n}right)^n$ increases monotonically to $e < 3$.
answered Nov 20 '18 at 14:44
Travis
59.7k767146
add a comment |
Hint 1 If you have differential calculus available: Differentiate $x mapsto x^{1 / x}$ (or, since $log$ is increasing, $x mapsto log(x^{1 / x}) = frac{log x}{x}$) to find a single maximum that lies between $2$ and $3$.
Then, it remains only to find where in the list $sqrt{2}$ fits, but you've already written that you can handle the cases $sqrt[2^k]{2^k}$.
Hint 2 Alternatively, we can compare the ratio $frac{sqrt[n + 1]{n + 1}}{sqrt[n]{n}}$ of successive values; raising that ratio to the $n(n + 1)$ power gives $$frac{(n + 1)^n}{n^{n + 1}} = frac{1}{n} left(1 + frac{1}{n}right)^n .$$ But $left(1 + frac{1}{n}right)^n$ increases monotonically to $e < 3$.
answered Nov 20 '18 at 14:44
Travis
59.7k767146
add a comment |
Hint 1 If you have differential calculus available: Differentiate $x mapsto x^{1 / x}$ (or, since $log$ is increasing, $x mapsto log(x^{1 / x}) = frac{log x}{x}$) to find a single maximum that lies between $2$ and $3$.
Then, it remains only to find where in the list $sqrt{2}$ fits, but you've already written that you can handle the cases $sqrt[2^k]{2^k}$.
Hint 2 Alternatively, we can compare the ratio $frac{sqrt[n + 1]{n + 1}}{sqrt[n]{n}}$ of successive values; raising that ratio to the $n(n + 1)$ power gives $$frac{(n + 1)^n}{n^{n + 1}} = frac{1}{n} left(1 + frac{1}{n}right)^n .$$ But $left(1 + frac{1}{n}right)^n$ increases monotonically to $e < 3$.
answered Nov 20 '18 at 14:44
Travis
59.7k767146
Hint 1 If you have differential calculus available: Differentiate $x mapsto x^{1 / x}$ (or, since $log$ is increasing, $x mapsto log(x^{1 / x}) = frac{log x}{x}$) to find a single maximum that lies between $2$ and $3$.
Then, it remains only to find where in the list $sqrt{2}$ fits, but you've already written that you can handle the cases $sqrt[2^k]{2^k}$.
Hint 2 Alternatively, we can compare the ratio $frac{sqrt[n + 1]{n + 1}}{sqrt[n]{n}}$ of successive values; raising that ratio to the $n(n + 1)$ power gives $$frac{(n + 1)^n}{n^{n + 1}} = frac{1}{n} left(1 + frac{1}{n}right)^n .$$ But $left(1 + frac{1}{n}right)^n$ increases monotonically to $e < 3$.
answered Nov 20 '18 at 14:44
Travis
59.7k767146
answered Nov 20 '18 at 14:44
Travis
59.7k767146
answered Nov 20 '18 at 14:44
Travis
59.7k767146
answered Nov 20 '18 at 14:44
Travis
59.7k767146
59.7k767146
add a comment |
add a comment |
Note that $ln n^{1/n}=frac {ln n}{n}$
Upon differentiating the function $$f(x)=frac {ln x}{x}$$
we get $$f'(x)=frac {1-ln x}{x^2}<0$$ for $xge 3$
Thus the sequence $n^{1/n}$ is decreasing for $nge 3$
The ordering is $$ 1<2^{1/2}<3^{1/3}$$ and $3^{1/3}>4^{1/4}>5^{1/5}>.....$
answered Nov 20 '18 at 15:03
Mohammad Riazi-Kermani
40.9k42059
add a comment |
Note that $ln n^{1/n}=frac {ln n}{n}$
Upon differentiating the function $$f(x)=frac {ln x}{x}$$
we get $$f'(x)=frac {1-ln x}{x^2}<0$$ for $xge 3$
Thus the sequence $n^{1/n}$ is decreasing for $nge 3$
The ordering is $$ 1<2^{1/2}<3^{1/3}$$ and $3^{1/3}>4^{1/4}>5^{1/5}>.....$
answered Nov 20 '18 at 15:03
Mohammad Riazi-Kermani
40.9k42059
add a comment |
Note that $ln n^{1/n}=frac {ln n}{n}$
Upon differentiating the function $$f(x)=frac {ln x}{x}$$
we get $$f'(x)=frac {1-ln x}{x^2}<0$$ for $xge 3$
Thus the sequence $n^{1/n}$ is decreasing for $nge 3$
The ordering is $$ 1<2^{1/2}<3^{1/3}$$ and $3^{1/3}>4^{1/4}>5^{1/5}>.....$
answered Nov 20 '18 at 15:03
Mohammad Riazi-Kermani
40.9k42059
Note that $ln n^{1/n}=frac {ln n}{n}$
Upon differentiating the function $$f(x)=frac {ln x}{x}$$
we get $$f'(x)=frac {1-ln x}{x^2}<0$$ for $xge 3$
Thus the sequence $n^{1/n}$ is decreasing for $nge 3$
The ordering is $$ 1<2^{1/2}<3^{1/3}$$ and $3^{1/3}>4^{1/4}>5^{1/5}>.....$
answered Nov 20 '18 at 15:03
Mohammad Riazi-Kermani
40.9k42059
answered Nov 20 '18 at 15:03
Mohammad Riazi-Kermani
40.9k42059
answered Nov 20 '18 at 15:03
Mohammad Riazi-Kermani
40.9k42059
answered Nov 20 '18 at 15:03
Mohammad Riazi-Kermani
40.9k42059
40.9k42059
add a comment |
add a comment |
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1
What class are you taking? We need to know what tools are available to you.
– saulspatz
Nov 20 '18 at 14:37
This is already solved here.
– Dietrich Burde
Nov 20 '18 at 14:43
1
Possible duplicate of Show that $n^{(1/n)}$ is eventually decreasing
– Dietrich Burde
Nov 20 '18 at 14:44