For sufficiently large $n$, Which number is bigger, $2^n$ or $n^{1000}$? [duplicate]
$begingroup$
This question already has an answer here:
How to prove that exponential grows faster than polynomial?
13 answers
Proof of a binomial theorem based inequality?
3 answers
How do I determine which number is bigger as $n$ gets sufficiently large, $2^n$ or $n^ {1000}$?
It seems to me it is a limit problem so I tried to tackle it that way.
$$lim_{nto infty} frac{2^n}{n^{1000}}$$
My thoughts are that, after some $n$, the numerator terms will be more than the terms in the denominator so we'll have something like $$frac{overbrace{2times 2timescdots times 2}^{1000text{ factors}}}{ntimes n times cdots times n} times 2^{n-1000}$$
At this point, I was thinking of using the fact that $2^n$ grows faster slower than $n!$ as $n$ gets larger so the limit, in this case, will be greater than $1$, meaning $2^n$ is bigger than $n^{1000}$ for sufficiently large $n$. This conclusion is really just a surmise based on a non-concrete formulation. Therefore, I'd appreciate any input on how to tackle this problem.
real-analysis sequences-and-series number-theory inequality asymptotics
asked Dec 6 '18 at 15:53
E.NoleE.Nole
139114
$endgroup$
marked as duplicate by Jyrki Lahtonen, user10354138, GNUSupporter 8964民主女神 地下教會, Fabian, jgon Dec 9 '18 at 15:08
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
How to prove that exponential grows faster than polynomial?
13 answers
Proof of a binomial theorem based inequality?
3 answers
How do I determine which number is bigger as $n$ gets sufficiently large, $2^n$ or $n^ {1000}$?
It seems to me it is a limit problem so I tried to tackle it that way.
$$lim_{nto infty} frac{2^n}{n^{1000}}$$
My thoughts are that, after some $n$, the numerator terms will be more than the terms in the denominator so we'll have something like $$frac{overbrace{2times 2timescdots times 2}^{1000text{ factors}}}{ntimes n times cdots times n} times 2^{n-1000}$$
At this point, I was thinking of using the fact that $2^n$ grows faster slower than $n!$ as $n$ gets larger so the limit, in this case, will be greater than $1$, meaning $2^n$ is bigger than $n^{1000}$ for sufficiently large $n$. This conclusion is really just a surmise based on a non-concrete formulation. Therefore, I'd appreciate any input on how to tackle this problem.
real-analysis sequences-and-series number-theory inequality asymptotics
asked Dec 6 '18 at 15:53
E.NoleE.Nole
139114
$endgroup$
marked as duplicate by Jyrki Lahtonen, user10354138, GNUSupporter 8964民主女神 地下教會, Fabian, jgon Dec 9 '18 at 15:08
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
2
$begingroup$
Exponential terms grow faster than polynomial terms for sufficiently large $n$. Also, $2^n$ does not grow faster than $n!$.
$endgroup$
– greelious
Dec 6 '18 at 15:58
$begingroup$
I mixed up the two, thanks. I'll edit it.
$endgroup$
– E.Nole
Dec 6 '18 at 16:00
2
$begingroup$
See related question/answers
$endgroup$
– farruhota
Dec 6 '18 at 16:56
add a comment |
$begingroup$
This question already has an answer here:
How to prove that exponential grows faster than polynomial?
13 answers
Proof of a binomial theorem based inequality?
3 answers
How do I determine which number is bigger as $n$ gets sufficiently large, $2^n$ or $n^ {1000}$?
It seems to me it is a limit problem so I tried to tackle it that way.
$$lim_{nto infty} frac{2^n}{n^{1000}}$$
My thoughts are that, after some $n$, the numerator terms will be more than the terms in the denominator so we'll have something like $$frac{overbrace{2times 2timescdots times 2}^{1000text{ factors}}}{ntimes n times cdots times n} times 2^{n-1000}$$
At this point, I was thinking of using the fact that $2^n$ grows faster slower than $n!$ as $n$ gets larger so the limit, in this case, will be greater than $1$, meaning $2^n$ is bigger than $n^{1000}$ for sufficiently large $n$. This conclusion is really just a surmise based on a non-concrete formulation. Therefore, I'd appreciate any input on how to tackle this problem.
real-analysis sequences-and-series number-theory inequality asymptotics
asked Dec 6 '18 at 15:53
E.NoleE.Nole
139114
$endgroup$
This question already has an answer here:
How to prove that exponential grows faster than polynomial?
13 answers
Proof of a binomial theorem based inequality?
3 answers
How do I determine which number is bigger as $n$ gets sufficiently large, $2^n$ or $n^ {1000}$?
It seems to me it is a limit problem so I tried to tackle it that way.
$$lim_{nto infty} frac{2^n}{n^{1000}}$$
My thoughts are that, after some $n$, the numerator terms will be more than the terms in the denominator so we'll have something like $$frac{overbrace{2times 2timescdots times 2}^{1000text{ factors}}}{ntimes n times cdots times n} times 2^{n-1000}$$
At this point, I was thinking of using the fact that $2^n$ grows faster slower than $n!$ as $n$ gets larger so the limit, in this case, will be greater than $1$, meaning $2^n$ is bigger than $n^{1000}$ for sufficiently large $n$. This conclusion is really just a surmise based on a non-concrete formulation. Therefore, I'd appreciate any input on how to tackle this problem.
This question already has an answer here:
How to prove that exponential grows faster than polynomial?
13 answers
Proof of a binomial theorem based inequality?
3 answers
real-analysis sequences-and-series number-theory inequality asymptotics
real-analysis sequences-and-series number-theory inequality asymptotics
asked Dec 6 '18 at 15:53
E.NoleE.Nole
139114
asked Dec 6 '18 at 15:53
E.NoleE.Nole
139114
edited Dec 6 '18 at 16:10
E.Nole
asked Dec 6 '18 at 15:53
E.NoleE.Nole
139114
asked Dec 6 '18 at 15:53
E.NoleE.Nole
139114
asked Dec 6 '18 at 15:53
E.NoleE.Nole
139114
139114
marked as duplicate by Jyrki Lahtonen, user10354138, GNUSupporter 8964民主女神 地下教會, Fabian, jgon Dec 9 '18 at 15:08
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Jyrki Lahtonen, user10354138, GNUSupporter 8964民主女神 地下教會, Fabian, jgon Dec 9 '18 at 15:08
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
2
$begingroup$
Exponential terms grow faster than polynomial terms for sufficiently large $n$. Also, $2^n$ does not grow faster than $n!$.
$endgroup$
– greelious
Dec 6 '18 at 15:58
$begingroup$
I mixed up the two, thanks. I'll edit it.
$endgroup$
– E.Nole
Dec 6 '18 at 16:00
2
$begingroup$
See related question/answers
$endgroup$
– farruhota
Dec 6 '18 at 16:56
add a comment |
2
$begingroup$
Exponential terms grow faster than polynomial terms for sufficiently large $n$. Also, $2^n$ does not grow faster than $n!$.
$endgroup$
– greelious
Dec 6 '18 at 15:58
$begingroup$
I mixed up the two, thanks. I'll edit it.
$endgroup$
– E.Nole
Dec 6 '18 at 16:00
2
$begingroup$
See related question/answers
$endgroup$
– farruhota
Dec 6 '18 at 16:56
2
2
$begingroup$
Exponential terms grow faster than polynomial terms for sufficiently large $n$. Also, $2^n$ does not grow faster than $n!$.
$endgroup$
– greelious
Dec 6 '18 at 15:58
$begingroup$
Exponential terms grow faster than polynomial terms for sufficiently large $n$. Also, $2^n$ does not grow faster than $n!$.
$endgroup$
– greelious
Dec 6 '18 at 15:58
$begingroup$
I mixed up the two, thanks. I'll edit it.
$endgroup$
– E.Nole
Dec 6 '18 at 16:00
$begingroup$
I mixed up the two, thanks. I'll edit it.
$endgroup$
– E.Nole
Dec 6 '18 at 16:00
2
2
$begingroup$
See related question/answers
$endgroup$
– farruhota
Dec 6 '18 at 16:56
$begingroup$
See related question/answers
$endgroup$
– farruhota
Dec 6 '18 at 16:56
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Note that$$frac{2^{n+1}}{2^n}=2text{ whereas }frac{(n+1)^{1,000}}{n^{1,000}}=left(1+frac1nright)^{1,000}.$$Since$$lim_{ntoinfty}left(1+frac1nright)^{1,000}=1<2,$$you have that$$left(1+frac1nright)^{1,000}<frac32$$if $n$ is large enough.It's not hard to deduce from this that $2^n>n^{1,000}$ if $n$ is large enough.
answered Dec 6 '18 at 15:58
José Carlos SantosJosé Carlos Santos
157k22126227
$endgroup$
$begingroup$
I'd like to clarify what the first line says about the two expressions. What I can deduce is that the purpose of the first line is to show that $2^n$ is increasing at a constant rate whereas $n^{1000}$ starts to slow down in terms of increase rate. Is that so?
$endgroup$
– E.Nole
Dec 6 '18 at 16:20
$begingroup$
It is more precise than that but, yes, it includes that.
$endgroup$
– José Carlos Santos
Dec 6 '18 at 16:23
$begingroup$
Well, what more does it say? I'm trying to understand your motivation for that approach.
$endgroup$
– E.Nole
Dec 6 '18 at 16:27
1
$begingroup$
It says that, after a certain point, the sequence $(n^{1,000})_{ninmathbb N}$ grows slower than the sequence $(2^n)_{ninmathbb N}$.
$endgroup$
– José Carlos Santos
Dec 6 '18 at 16:29
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Note that$$frac{2^{n+1}}{2^n}=2text{ whereas }frac{(n+1)^{1,000}}{n^{1,000}}=left(1+frac1nright)^{1,000}.$$Since$$lim_{ntoinfty}left(1+frac1nright)^{1,000}=1<2,$$you have that$$left(1+frac1nright)^{1,000}<frac32$$if $n$ is large enough.It's not hard to deduce from this that $2^n>n^{1,000}$ if $n$ is large enough.
answered Dec 6 '18 at 15:58
José Carlos SantosJosé Carlos Santos
157k22126227
$endgroup$
$begingroup$
I'd like to clarify what the first line says about the two expressions. What I can deduce is that the purpose of the first line is to show that $2^n$ is increasing at a constant rate whereas $n^{1000}$ starts to slow down in terms of increase rate. Is that so?
$endgroup$
– E.Nole
Dec 6 '18 at 16:20
$begingroup$
It is more precise than that but, yes, it includes that.
$endgroup$
– José Carlos Santos
Dec 6 '18 at 16:23
$begingroup$
Well, what more does it say? I'm trying to understand your motivation for that approach.
$endgroup$
– E.Nole
Dec 6 '18 at 16:27
1
$begingroup$
It says that, after a certain point, the sequence $(n^{1,000})_{ninmathbb N}$ grows slower than the sequence $(2^n)_{ninmathbb N}$.
$endgroup$
– José Carlos Santos
Dec 6 '18 at 16:29
add a comment |
$begingroup$
Note that$$frac{2^{n+1}}{2^n}=2text{ whereas }frac{(n+1)^{1,000}}{n^{1,000}}=left(1+frac1nright)^{1,000}.$$Since$$lim_{ntoinfty}left(1+frac1nright)^{1,000}=1<2,$$you have that$$left(1+frac1nright)^{1,000}<frac32$$if $n$ is large enough.It's not hard to deduce from this that $2^n>n^{1,000}$ if $n$ is large enough.
answered Dec 6 '18 at 15:58
José Carlos SantosJosé Carlos Santos
157k22126227
$endgroup$
$begingroup$
I'd like to clarify what the first line says about the two expressions. What I can deduce is that the purpose of the first line is to show that $2^n$ is increasing at a constant rate whereas $n^{1000}$ starts to slow down in terms of increase rate. Is that so?
$endgroup$
– E.Nole
Dec 6 '18 at 16:20
$begingroup$
It is more precise than that but, yes, it includes that.
$endgroup$
– José Carlos Santos
Dec 6 '18 at 16:23
$begingroup$
Well, what more does it say? I'm trying to understand your motivation for that approach.
$endgroup$
– E.Nole
Dec 6 '18 at 16:27
1
$begingroup$
It says that, after a certain point, the sequence $(n^{1,000})_{ninmathbb N}$ grows slower than the sequence $(2^n)_{ninmathbb N}$.
$endgroup$
– José Carlos Santos
Dec 6 '18 at 16:29
add a comment |
$begingroup$
Note that$$frac{2^{n+1}}{2^n}=2text{ whereas }frac{(n+1)^{1,000}}{n^{1,000}}=left(1+frac1nright)^{1,000}.$$Since$$lim_{ntoinfty}left(1+frac1nright)^{1,000}=1<2,$$you have that$$left(1+frac1nright)^{1,000}<frac32$$if $n$ is large enough.It's not hard to deduce from this that $2^n>n^{1,000}$ if $n$ is large enough.
answered Dec 6 '18 at 15:58
José Carlos SantosJosé Carlos Santos
157k22126227
$endgroup$
Note that$$frac{2^{n+1}}{2^n}=2text{ whereas }frac{(n+1)^{1,000}}{n^{1,000}}=left(1+frac1nright)^{1,000}.$$Since$$lim_{ntoinfty}left(1+frac1nright)^{1,000}=1<2,$$you have that$$left(1+frac1nright)^{1,000}<frac32$$if $n$ is large enough.It's not hard to deduce from this that $2^n>n^{1,000}$ if $n$ is large enough.
answered Dec 6 '18 at 15:58
José Carlos SantosJosé Carlos Santos
157k22126227
edited Dec 6 '18 at 16:16
answered Dec 6 '18 at 15:58
José Carlos SantosJosé Carlos Santos
157k22126227
answered Dec 6 '18 at 15:58
José Carlos SantosJosé Carlos Santos
157k22126227
answered Dec 6 '18 at 15:58
José Carlos SantosJosé Carlos Santos
157k22126227
157k22126227
$begingroup$
I'd like to clarify what the first line says about the two expressions. What I can deduce is that the purpose of the first line is to show that $2^n$ is increasing at a constant rate whereas $n^{1000}$ starts to slow down in terms of increase rate. Is that so?
$endgroup$
– E.Nole
Dec 6 '18 at 16:20
$begingroup$
It is more precise than that but, yes, it includes that.
$endgroup$
– José Carlos Santos
Dec 6 '18 at 16:23
$begingroup$
Well, what more does it say? I'm trying to understand your motivation for that approach.
$endgroup$
– E.Nole
Dec 6 '18 at 16:27
1
$begingroup$
It says that, after a certain point, the sequence $(n^{1,000})_{ninmathbb N}$ grows slower than the sequence $(2^n)_{ninmathbb N}$.
$endgroup$
– José Carlos Santos
Dec 6 '18 at 16:29
add a comment |
$begingroup$
I'd like to clarify what the first line says about the two expressions. What I can deduce is that the purpose of the first line is to show that $2^n$ is increasing at a constant rate whereas $n^{1000}$ starts to slow down in terms of increase rate. Is that so?
$endgroup$
– E.Nole
Dec 6 '18 at 16:20
$begingroup$
It is more precise than that but, yes, it includes that.
$endgroup$
– José Carlos Santos
Dec 6 '18 at 16:23
$begingroup$
Well, what more does it say? I'm trying to understand your motivation for that approach.
$endgroup$
– E.Nole
Dec 6 '18 at 16:27
1
$begingroup$
It says that, after a certain point, the sequence $(n^{1,000})_{ninmathbb N}$ grows slower than the sequence $(2^n)_{ninmathbb N}$.
$endgroup$
– José Carlos Santos
Dec 6 '18 at 16:29
$begingroup$
I'd like to clarify what the first line says about the two expressions. What I can deduce is that the purpose of the first line is to show that $2^n$ is increasing at a constant rate whereas $n^{1000}$ starts to slow down in terms of increase rate. Is that so?
$endgroup$
– E.Nole
Dec 6 '18 at 16:20
$begingroup$
I'd like to clarify what the first line says about the two expressions. What I can deduce is that the purpose of the first line is to show that $2^n$ is increasing at a constant rate whereas $n^{1000}$ starts to slow down in terms of increase rate. Is that so?
$endgroup$
– E.Nole
Dec 6 '18 at 16:20
$begingroup$
It is more precise than that but, yes, it includes that.
$endgroup$
– José Carlos Santos
Dec 6 '18 at 16:23
$begingroup$
It is more precise than that but, yes, it includes that.
$endgroup$
– José Carlos Santos
Dec 6 '18 at 16:23
$begingroup$
Well, what more does it say? I'm trying to understand your motivation for that approach.
$endgroup$
– E.Nole
Dec 6 '18 at 16:27
$begingroup$
Well, what more does it say? I'm trying to understand your motivation for that approach.
$endgroup$
– E.Nole
Dec 6 '18 at 16:27
1
1
$begingroup$
It says that, after a certain point, the sequence $(n^{1,000})_{ninmathbb N}$ grows slower than the sequence $(2^n)_{ninmathbb N}$.
$endgroup$
– José Carlos Santos
Dec 6 '18 at 16:29
$begingroup$
It says that, after a certain point, the sequence $(n^{1,000})_{ninmathbb N}$ grows slower than the sequence $(2^n)_{ninmathbb N}$.
$endgroup$
– José Carlos Santos
Dec 6 '18 at 16:29
add a comment |
2
$begingroup$
Exponential terms grow faster than polynomial terms for sufficiently large $n$. Also, $2^n$ does not grow faster than $n!$.
$endgroup$
– greelious
Dec 6 '18 at 15:58
$begingroup$
I mixed up the two, thanks. I'll edit it.
$endgroup$
– E.Nole
Dec 6 '18 at 16:00
2
$begingroup$
See related question/answers
$endgroup$
– farruhota
Dec 6 '18 at 16:56