Prove that $forall n, , exists N,x :lfloor{x^{N}}rfloor =n , land ,lfloor{x^{N+1}}rfloor =n+1$
The question is related to the interesting problem raised by following OP.
Notably I'm trying to prove the following fact
For any $nin mathbb{N} quad exists xin mathbb{R} quad x>1$ and $exists Nin mathbb{N}$ such that
$$lfloor{x^{N}}rfloor =n quad land quad lfloor{x^{N+1}}rfloor =n+1$$
My idea for the proof is to consider
- $x=sqrt[N]{n} implies lfloor{x^{N}}rfloor = lfloor{{(sqrt[N]{n})}^{N}}rfloor =lfloor nrfloor=n$
and then we need that
$x^{N+1}=xcdot n=n+1+alpha,$ for some $alpha in(0,1)$
therefore finally we need to show that for any $nin mathbb{N} quad exists alphain mathbb{R} quad alpha in(0,1)$ and $exists Nin mathbb{N}$ such that
$$sqrt[N]{n}=1+frac{1+alpha}{n}$$
but I'm totally stuck here and I can't find any method to prove that.
Thanks in advance for any idea or suggestion about that!
real-analysis
asked Nov 27 '18 at 21:57
gimusigimusi
1
add a comment |
The question is related to the interesting problem raised by following OP.
Notably I'm trying to prove the following fact
For any $nin mathbb{N} quad exists xin mathbb{R} quad x>1$ and $exists Nin mathbb{N}$ such that
$$lfloor{x^{N}}rfloor =n quad land quad lfloor{x^{N+1}}rfloor =n+1$$
My idea for the proof is to consider
- $x=sqrt[N]{n} implies lfloor{x^{N}}rfloor = lfloor{{(sqrt[N]{n})}^{N}}rfloor =lfloor nrfloor=n$
and then we need that
$x^{N+1}=xcdot n=n+1+alpha,$ for some $alpha in(0,1)$
therefore finally we need to show that for any $nin mathbb{N} quad exists alphain mathbb{R} quad alpha in(0,1)$ and $exists Nin mathbb{N}$ such that
$$sqrt[N]{n}=1+frac{1+alpha}{n}$$
but I'm totally stuck here and I can't find any method to prove that.
Thanks in advance for any idea or suggestion about that!
real-analysis
asked Nov 27 '18 at 21:57
gimusigimusi
1
add a comment |
The question is related to the interesting problem raised by following OP.
Notably I'm trying to prove the following fact
For any $nin mathbb{N} quad exists xin mathbb{R} quad x>1$ and $exists Nin mathbb{N}$ such that
$$lfloor{x^{N}}rfloor =n quad land quad lfloor{x^{N+1}}rfloor =n+1$$
My idea for the proof is to consider
- $x=sqrt[N]{n} implies lfloor{x^{N}}rfloor = lfloor{{(sqrt[N]{n})}^{N}}rfloor =lfloor nrfloor=n$
and then we need that
$x^{N+1}=xcdot n=n+1+alpha,$ for some $alpha in(0,1)$
therefore finally we need to show that for any $nin mathbb{N} quad exists alphain mathbb{R} quad alpha in(0,1)$ and $exists Nin mathbb{N}$ such that
$$sqrt[N]{n}=1+frac{1+alpha}{n}$$
but I'm totally stuck here and I can't find any method to prove that.
Thanks in advance for any idea or suggestion about that!
real-analysis
asked Nov 27 '18 at 21:57
gimusigimusi
1
The question is related to the interesting problem raised by following OP.
Notably I'm trying to prove the following fact
For any $nin mathbb{N} quad exists xin mathbb{R} quad x>1$ and $exists Nin mathbb{N}$ such that
$$lfloor{x^{N}}rfloor =n quad land quad lfloor{x^{N+1}}rfloor =n+1$$
My idea for the proof is to consider
- $x=sqrt[N]{n} implies lfloor{x^{N}}rfloor = lfloor{{(sqrt[N]{n})}^{N}}rfloor =lfloor nrfloor=n$
and then we need that
$x^{N+1}=xcdot n=n+1+alpha,$ for some $alpha in(0,1)$
therefore finally we need to show that for any $nin mathbb{N} quad exists alphain mathbb{R} quad alpha in(0,1)$ and $exists Nin mathbb{N}$ such that
$$sqrt[N]{n}=1+frac{1+alpha}{n}$$
but I'm totally stuck here and I can't find any method to prove that.
Thanks in advance for any idea or suggestion about that!
real-analysis
real-analysis
asked Nov 27 '18 at 21:57
gimusigimusi
1
asked Nov 27 '18 at 21:57
gimusigimusi
1
asked Nov 27 '18 at 21:57
gimusigimusi
1
asked Nov 27 '18 at 21:57
gimusigimusi
1
asked Nov 27 '18 at 21:57
gimusigimusi
1
1
add a comment |
add a comment |
1 Answer
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Start two sequences $(x_i)$ and $(N_i)$ with $x_0=n, N_0=1$ and define $x_{i+1}=sqrt{x_i}, N_{i+1}=2N_i, i=0,1,ldots$. Obviously we have $x_i^{N_i}=n$ and $lim_{i to infty} x_i=1$. So for some $k$, $x_k < frac{n+2}{n+1}$.
We consider $x_k^{N_k+0}, ldots, x_k^{N_k+i}, ldots$. Since $x_k > 1$, there is an index $r ge 0$ with $x_k^{N_k+r} < n+1$ and $x_k^{N_k+r+1} ge n+1$. Since $x_k < frac{n+2}{n+1}$, we also have $x_k^{N_k+r+1} = x_kx_k^{N_k+r} < (n+1)frac{n+2}{n+1} = n+2$ which finally leads to the desired
$$ lfloor x_k^{N_k+r}rfloor=n, quad lfloor x_k^{N_k+r+1}rfloor = n+1$$
answered Nov 27 '18 at 22:49
IngixIngix
3,354145
Thanks! I need to study that a little bit in order to digest all step! Bye
– gimusi
Nov 27 '18 at 23:00
Fell free to ask if anything is unclear.
– Ingix
Nov 27 '18 at 23:01
Yes all seems clear to me, it seems really a great proof! I'll await for some others answer or advice for other users but I think your one is a great job! Thanks again
– gimusi
Nov 27 '18 at 23:07
Just for curiosity, It is a standard way to proceed or you have just invented that for this specific case? Do you consider the result trivial?
– gimusi
Nov 27 '18 at 23:12
Not trivial, but rather easy. It's standard to consider the case where one power is below $n+1$ and the next above. The remainder is technique to make sure that the next power is not 'to big'.
– Ingix
Nov 27 '18 at 23:27
|
show 1 more comment
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1 Answer
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Start two sequences $(x_i)$ and $(N_i)$ with $x_0=n, N_0=1$ and define $x_{i+1}=sqrt{x_i}, N_{i+1}=2N_i, i=0,1,ldots$. Obviously we have $x_i^{N_i}=n$ and $lim_{i to infty} x_i=1$. So for some $k$, $x_k < frac{n+2}{n+1}$.
We consider $x_k^{N_k+0}, ldots, x_k^{N_k+i}, ldots$. Since $x_k > 1$, there is an index $r ge 0$ with $x_k^{N_k+r} < n+1$ and $x_k^{N_k+r+1} ge n+1$. Since $x_k < frac{n+2}{n+1}$, we also have $x_k^{N_k+r+1} = x_kx_k^{N_k+r} < (n+1)frac{n+2}{n+1} = n+2$ which finally leads to the desired
$$ lfloor x_k^{N_k+r}rfloor=n, quad lfloor x_k^{N_k+r+1}rfloor = n+1$$
answered Nov 27 '18 at 22:49
IngixIngix
3,354145
Thanks! I need to study that a little bit in order to digest all step! Bye
– gimusi
Nov 27 '18 at 23:00
Fell free to ask if anything is unclear.
– Ingix
Nov 27 '18 at 23:01
Yes all seems clear to me, it seems really a great proof! I'll await for some others answer or advice for other users but I think your one is a great job! Thanks again
– gimusi
Nov 27 '18 at 23:07
Just for curiosity, It is a standard way to proceed or you have just invented that for this specific case? Do you consider the result trivial?
– gimusi
Nov 27 '18 at 23:12
Not trivial, but rather easy. It's standard to consider the case where one power is below $n+1$ and the next above. The remainder is technique to make sure that the next power is not 'to big'.
– Ingix
Nov 27 '18 at 23:27
|
show 1 more comment
Start two sequences $(x_i)$ and $(N_i)$ with $x_0=n, N_0=1$ and define $x_{i+1}=sqrt{x_i}, N_{i+1}=2N_i, i=0,1,ldots$. Obviously we have $x_i^{N_i}=n$ and $lim_{i to infty} x_i=1$. So for some $k$, $x_k < frac{n+2}{n+1}$.
We consider $x_k^{N_k+0}, ldots, x_k^{N_k+i}, ldots$. Since $x_k > 1$, there is an index $r ge 0$ with $x_k^{N_k+r} < n+1$ and $x_k^{N_k+r+1} ge n+1$. Since $x_k < frac{n+2}{n+1}$, we also have $x_k^{N_k+r+1} = x_kx_k^{N_k+r} < (n+1)frac{n+2}{n+1} = n+2$ which finally leads to the desired
$$ lfloor x_k^{N_k+r}rfloor=n, quad lfloor x_k^{N_k+r+1}rfloor = n+1$$
answered Nov 27 '18 at 22:49
IngixIngix
3,354145
Thanks! I need to study that a little bit in order to digest all step! Bye
– gimusi
Nov 27 '18 at 23:00
Fell free to ask if anything is unclear.
– Ingix
Nov 27 '18 at 23:01
Yes all seems clear to me, it seems really a great proof! I'll await for some others answer or advice for other users but I think your one is a great job! Thanks again
– gimusi
Nov 27 '18 at 23:07
Just for curiosity, It is a standard way to proceed or you have just invented that for this specific case? Do you consider the result trivial?
– gimusi
Nov 27 '18 at 23:12
Not trivial, but rather easy. It's standard to consider the case where one power is below $n+1$ and the next above. The remainder is technique to make sure that the next power is not 'to big'.
– Ingix
Nov 27 '18 at 23:27
|
show 1 more comment
Start two sequences $(x_i)$ and $(N_i)$ with $x_0=n, N_0=1$ and define $x_{i+1}=sqrt{x_i}, N_{i+1}=2N_i, i=0,1,ldots$. Obviously we have $x_i^{N_i}=n$ and $lim_{i to infty} x_i=1$. So for some $k$, $x_k < frac{n+2}{n+1}$.
We consider $x_k^{N_k+0}, ldots, x_k^{N_k+i}, ldots$. Since $x_k > 1$, there is an index $r ge 0$ with $x_k^{N_k+r} < n+1$ and $x_k^{N_k+r+1} ge n+1$. Since $x_k < frac{n+2}{n+1}$, we also have $x_k^{N_k+r+1} = x_kx_k^{N_k+r} < (n+1)frac{n+2}{n+1} = n+2$ which finally leads to the desired
$$ lfloor x_k^{N_k+r}rfloor=n, quad lfloor x_k^{N_k+r+1}rfloor = n+1$$
answered Nov 27 '18 at 22:49
IngixIngix
3,354145
Start two sequences $(x_i)$ and $(N_i)$ with $x_0=n, N_0=1$ and define $x_{i+1}=sqrt{x_i}, N_{i+1}=2N_i, i=0,1,ldots$. Obviously we have $x_i^{N_i}=n$ and $lim_{i to infty} x_i=1$. So for some $k$, $x_k < frac{n+2}{n+1}$.
We consider $x_k^{N_k+0}, ldots, x_k^{N_k+i}, ldots$. Since $x_k > 1$, there is an index $r ge 0$ with $x_k^{N_k+r} < n+1$ and $x_k^{N_k+r+1} ge n+1$. Since $x_k < frac{n+2}{n+1}$, we also have $x_k^{N_k+r+1} = x_kx_k^{N_k+r} < (n+1)frac{n+2}{n+1} = n+2$ which finally leads to the desired
$$ lfloor x_k^{N_k+r}rfloor=n, quad lfloor x_k^{N_k+r+1}rfloor = n+1$$
answered Nov 27 '18 at 22:49
IngixIngix
3,354145
answered Nov 27 '18 at 22:49
IngixIngix
3,354145
answered Nov 27 '18 at 22:49
IngixIngix
3,354145
answered Nov 27 '18 at 22:49
IngixIngix
3,354145
3,354145
Thanks! I need to study that a little bit in order to digest all step! Bye
– gimusi
Nov 27 '18 at 23:00
Fell free to ask if anything is unclear.
– Ingix
Nov 27 '18 at 23:01
Yes all seems clear to me, it seems really a great proof! I'll await for some others answer or advice for other users but I think your one is a great job! Thanks again
– gimusi
Nov 27 '18 at 23:07
Just for curiosity, It is a standard way to proceed or you have just invented that for this specific case? Do you consider the result trivial?
– gimusi
Nov 27 '18 at 23:12
Not trivial, but rather easy. It's standard to consider the case where one power is below $n+1$ and the next above. The remainder is technique to make sure that the next power is not 'to big'.
– Ingix
Nov 27 '18 at 23:27
|
show 1 more comment
Thanks! I need to study that a little bit in order to digest all step! Bye
– gimusi
Nov 27 '18 at 23:00
Fell free to ask if anything is unclear.
– Ingix
Nov 27 '18 at 23:01
Yes all seems clear to me, it seems really a great proof! I'll await for some others answer or advice for other users but I think your one is a great job! Thanks again
– gimusi
Nov 27 '18 at 23:07
Just for curiosity, It is a standard way to proceed or you have just invented that for this specific case? Do you consider the result trivial?
– gimusi
Nov 27 '18 at 23:12
Not trivial, but rather easy. It's standard to consider the case where one power is below $n+1$ and the next above. The remainder is technique to make sure that the next power is not 'to big'.
– Ingix
Nov 27 '18 at 23:27
Thanks! I need to study that a little bit in order to digest all step! Bye
– gimusi
Nov 27 '18 at 23:00
Thanks! I need to study that a little bit in order to digest all step! Bye
– gimusi
Nov 27 '18 at 23:00
Fell free to ask if anything is unclear.
– Ingix
Nov 27 '18 at 23:01
Fell free to ask if anything is unclear.
– Ingix
Nov 27 '18 at 23:01
Yes all seems clear to me, it seems really a great proof! I'll await for some others answer or advice for other users but I think your one is a great job! Thanks again
– gimusi
Nov 27 '18 at 23:07
Yes all seems clear to me, it seems really a great proof! I'll await for some others answer or advice for other users but I think your one is a great job! Thanks again
– gimusi
Nov 27 '18 at 23:07
Just for curiosity, It is a standard way to proceed or you have just invented that for this specific case? Do you consider the result trivial?
– gimusi
Nov 27 '18 at 23:12
Just for curiosity, It is a standard way to proceed or you have just invented that for this specific case? Do you consider the result trivial?
– gimusi
Nov 27 '18 at 23:12
Not trivial, but rather easy. It's standard to consider the case where one power is below $n+1$ and the next above. The remainder is technique to make sure that the next power is not 'to big'.
– Ingix
Nov 27 '18 at 23:27
Not trivial, but rather easy. It's standard to consider the case where one power is below $n+1$ and the next above. The remainder is technique to make sure that the next power is not 'to big'.
– Ingix
Nov 27 '18 at 23:27
|
show 1 more comment
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