For $epsilon > 0$, is there always $n,m in mathbbN$ so that $e^n$ is $epsilon$-close to $m$? [closed] The 2019 Stack Overflow Developer Survey Results Are InIs there a proof that there is no general method to solve transcendental equations?Does the “no gaps”-property with transcendental numbers mean that there is only “one number”?Is there a basis for the continuous functions space?Help with Spivak's Calculus: Chapter 1 problem 21Show that $partial A$ is always a closed setProving that an increasing sequence with no upper bound is positive at a certain $n$Are there any other fields other than $mathbbR,mathbbC$, rich enough to have analysis built on them?Are there transcendental numbers that cannot be reached?Prove that for each $alpha in [0,1]$, there exists $E in M$ with $mu(E)=alpha$Show that $sup_t in mathbbR sum_j=1^infty frac(it)^j(k+j)!$ is bounded for any fixed $k>2$
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For $epsilon > 0$, is there always $n,m in mathbbN$ so that $e^n$ is $epsilon$-close to $m$? [closed]
The 2019 Stack Overflow Developer Survey Results Are InIs there a proof that there is no general method to solve transcendental equations?Does the “no gaps”-property with transcendental numbers mean that there is only “one number”?Is there a basis for the continuous functions space?Help with Spivak's Calculus: Chapter 1 problem 21Show that $partial A$ is always a closed setProving that an increasing sequence with no upper bound is positive at a certain $n$Are there any other fields other than $mathbbR,mathbbC$, rich enough to have analysis built on them?Are there transcendental numbers that cannot be reached?Prove that for each $alpha in [0,1]$, there exists $E in M$ with $mu(E)=alpha$Show that $sup_t in mathbbR sum_j=1^infty frac(it)^j(k+j)!$ is bounded for any fixed $k>2$
$begingroup$
I don't have much to go off of, so I can't demonstrate any attempts here. I just want to know if there has been any answer or partial answer to this question.
analysis transcendental-numbers
asked Mar 23 at 0:52
FredFred
62659
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closed as off-topic by Saad, Eevee Trainer, Lee David Chung Lin, max_zorn, Javi Mar 23 at 10:41
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Eevee Trainer, Lee David Chung Lin, max_zorn, Javi
add a comment |
$begingroup$
I don't have much to go off of, so I can't demonstrate any attempts here. I just want to know if there has been any answer or partial answer to this question.
analysis transcendental-numbers
asked Mar 23 at 0:52
FredFred
62659
$endgroup$
closed as off-topic by Saad, Eevee Trainer, Lee David Chung Lin, max_zorn, Javi Mar 23 at 10:41
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Eevee Trainer, Lee David Chung Lin, max_zorn, Javi
add a comment |
$begingroup$
I don't have much to go off of, so I can't demonstrate any attempts here. I just want to know if there has been any answer or partial answer to this question.
analysis transcendental-numbers
asked Mar 23 at 0:52
FredFred
62659
$endgroup$
I don't have much to go off of, so I can't demonstrate any attempts here. I just want to know if there has been any answer or partial answer to this question.
analysis transcendental-numbers
analysis transcendental-numbers
asked Mar 23 at 0:52
FredFred
62659
asked Mar 23 at 0:52
FredFred
62659
edited Mar 23 at 2:15
Fred
asked Mar 23 at 0:52
FredFred
62659
asked Mar 23 at 0:52
FredFred
62659
asked Mar 23 at 0:52
FredFred
62659
62659
closed as off-topic by Saad, Eevee Trainer, Lee David Chung Lin, max_zorn, Javi Mar 23 at 10:41
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Eevee Trainer, Lee David Chung Lin, max_zorn, Javi
closed as off-topic by Saad, Eevee Trainer, Lee David Chung Lin, max_zorn, Javi Mar 23 at 10:41
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Eevee Trainer, Lee David Chung Lin, max_zorn, Javi
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You can make $e^a$ exactly equal to any positive number $y$ (integer or not) by setting
$a = ln y$. You can get as close as you like with rational $a$.
I suspect that $e^n$ comes arbitrarily close to an integer for integral $n$, but have not been able to find a reference. Weyl's theorem implies that the integral multiples of $e$ (or any other irrational number) are equidistributed modulo $1$.
The powers of $e$ probably are, but caution is called for. There are irrational numbers $gamma$ whose powers are not equidistributed.
(https://math.unm.edu/~crisp/courses/wavelets/fall13/wavelet-weyl-report2.pdf)
answered Mar 23 at 0:58
Ethan BolkerEthan Bolker
45.9k553120
$endgroup$
$begingroup$
Ah now I feel silly. Of course... the integer values of $a$ are interesting. Thank you.
$endgroup$
– Fred
Mar 23 at 1:11
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You can make $e^a$ exactly equal to any positive number $y$ (integer or not) by setting
$a = ln y$. You can get as close as you like with rational $a$.
I suspect that $e^n$ comes arbitrarily close to an integer for integral $n$, but have not been able to find a reference. Weyl's theorem implies that the integral multiples of $e$ (or any other irrational number) are equidistributed modulo $1$.
The powers of $e$ probably are, but caution is called for. There are irrational numbers $gamma$ whose powers are not equidistributed.
(https://math.unm.edu/~crisp/courses/wavelets/fall13/wavelet-weyl-report2.pdf)
answered Mar 23 at 0:58
Ethan BolkerEthan Bolker
45.9k553120
$endgroup$
$begingroup$
Ah now I feel silly. Of course... the integer values of $a$ are interesting. Thank you.
$endgroup$
– Fred
Mar 23 at 1:11
add a comment |
$begingroup$
You can make $e^a$ exactly equal to any positive number $y$ (integer or not) by setting
$a = ln y$. You can get as close as you like with rational $a$.
I suspect that $e^n$ comes arbitrarily close to an integer for integral $n$, but have not been able to find a reference. Weyl's theorem implies that the integral multiples of $e$ (or any other irrational number) are equidistributed modulo $1$.
The powers of $e$ probably are, but caution is called for. There are irrational numbers $gamma$ whose powers are not equidistributed.
(https://math.unm.edu/~crisp/courses/wavelets/fall13/wavelet-weyl-report2.pdf)
answered Mar 23 at 0:58
Ethan BolkerEthan Bolker
45.9k553120
$endgroup$
$begingroup$
Ah now I feel silly. Of course... the integer values of $a$ are interesting. Thank you.
$endgroup$
– Fred
Mar 23 at 1:11
add a comment |
$begingroup$
You can make $e^a$ exactly equal to any positive number $y$ (integer or not) by setting
$a = ln y$. You can get as close as you like with rational $a$.
I suspect that $e^n$ comes arbitrarily close to an integer for integral $n$, but have not been able to find a reference. Weyl's theorem implies that the integral multiples of $e$ (or any other irrational number) are equidistributed modulo $1$.
The powers of $e$ probably are, but caution is called for. There are irrational numbers $gamma$ whose powers are not equidistributed.
(https://math.unm.edu/~crisp/courses/wavelets/fall13/wavelet-weyl-report2.pdf)
answered Mar 23 at 0:58
Ethan BolkerEthan Bolker
45.9k553120
$endgroup$
You can make $e^a$ exactly equal to any positive number $y$ (integer or not) by setting
$a = ln y$. You can get as close as you like with rational $a$.
I suspect that $e^n$ comes arbitrarily close to an integer for integral $n$, but have not been able to find a reference. Weyl's theorem implies that the integral multiples of $e$ (or any other irrational number) are equidistributed modulo $1$.
The powers of $e$ probably are, but caution is called for. There are irrational numbers $gamma$ whose powers are not equidistributed.
(https://math.unm.edu/~crisp/courses/wavelets/fall13/wavelet-weyl-report2.pdf)
answered Mar 23 at 0:58
Ethan BolkerEthan Bolker
45.9k553120
edited Mar 23 at 2:32
answered Mar 23 at 0:58
Ethan BolkerEthan Bolker
45.9k553120
answered Mar 23 at 0:58
Ethan BolkerEthan Bolker
45.9k553120
answered Mar 23 at 0:58
Ethan BolkerEthan Bolker
45.9k553120
45.9k553120
$begingroup$
Ah now I feel silly. Of course... the integer values of $a$ are interesting. Thank you.
$endgroup$
– Fred
Mar 23 at 1:11
add a comment |
$begingroup$
Ah now I feel silly. Of course... the integer values of $a$ are interesting. Thank you.
$endgroup$
– Fred
Mar 23 at 1:11
$begingroup$
Ah now I feel silly. Of course... the integer values of $a$ are interesting. Thank you.
$endgroup$
– Fred
Mar 23 at 1:11
$begingroup$
Ah now I feel silly. Of course... the integer values of $a$ are interesting. Thank you.
$endgroup$
– Fred
Mar 23 at 1:11
add a comment |
