Asymptotic order
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Below is a question I faced from an online test for preparation of exam and I had doubt in solution provided so I wanted to discuss my approach and ask about it.
$$frac{e^{nlog n}}n(A),n^{sqrt n}(B),2^{nlog n}(C)$$
I want to order these functions based on increasing asymptotic order.
I rewrite A as $frac{n^{nlog e}}{n}=frac{n^{1.44n}}{n}=n^{1.44n-1}$
C is re-written as $n^{nlog2}=n^n$
So, finally, my order came to be A, B, C in increasing asymptotic growth.
Am I correct?
asymptotics
edited Nov 17 at 9:27
Parcly Taxel
41k137198
asked Nov 17 at 7:18
user3767495
1518
add a comment |
up vote
2
down vote
favorite
Below is a question I faced from an online test for preparation of exam and I had doubt in solution provided so I wanted to discuss my approach and ask about it.
$$frac{e^{nlog n}}n(A),n^{sqrt n}(B),2^{nlog n}(C)$$
I want to order these functions based on increasing asymptotic order.
I rewrite A as $frac{n^{nlog e}}{n}=frac{n^{1.44n}}{n}=n^{1.44n-1}$
C is re-written as $n^{nlog2}=n^n$
So, finally, my order came to be A, B, C in increasing asymptotic growth.
Am I correct?
asymptotics
edited Nov 17 at 9:27
Parcly Taxel
41k137198
asked Nov 17 at 7:18
user3767495
1518
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Below is a question I faced from an online test for preparation of exam and I had doubt in solution provided so I wanted to discuss my approach and ask about it.
$$frac{e^{nlog n}}n(A),n^{sqrt n}(B),2^{nlog n}(C)$$
I want to order these functions based on increasing asymptotic order.
I rewrite A as $frac{n^{nlog e}}{n}=frac{n^{1.44n}}{n}=n^{1.44n-1}$
C is re-written as $n^{nlog2}=n^n$
So, finally, my order came to be A, B, C in increasing asymptotic growth.
Am I correct?
asymptotics
edited Nov 17 at 9:27
Parcly Taxel
41k137198
asked Nov 17 at 7:18
user3767495
1518
Below is a question I faced from an online test for preparation of exam and I had doubt in solution provided so I wanted to discuss my approach and ask about it.
$$frac{e^{nlog n}}n(A),n^{sqrt n}(B),2^{nlog n}(C)$$
I want to order these functions based on increasing asymptotic order.
I rewrite A as $frac{n^{nlog e}}{n}=frac{n^{1.44n}}{n}=n^{1.44n-1}$
C is re-written as $n^{nlog2}=n^n$
So, finally, my order came to be A, B, C in increasing asymptotic growth.
Am I correct?
asymptotics
asymptotics
edited Nov 17 at 9:27
Parcly Taxel
41k137198
asked Nov 17 at 7:18
user3767495
1518
edited Nov 17 at 9:27
Parcly Taxel
41k137198
asked Nov 17 at 7:18
user3767495
1518
edited Nov 17 at 9:27
Parcly Taxel
41k137198
edited Nov 17 at 9:27
Parcly Taxel
41k137198
edited Nov 17 at 9:27
Parcly Taxel
41k137198
41k137198
asked Nov 17 at 7:18
user3767495
1518
asked Nov 17 at 7:18
user3767495
1518
asked Nov 17 at 7:18
user3767495
1518
1518
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
$$frac{e^{nlog_2n}}n=frac{e^{n(ln n)/(ln 2)}}n=frac{n^{n/ln 2}}n=n^{n/ln 2-1}$$
$$2^{nlog_2 n}=n^n$$
We now compare powers: $sqrt n<frac n{ln 2}-1<n$. Thus $B<C<A$, not $A<B<C$ as you claimed.
answered Nov 17 at 7:36
Parcly Taxel
41k137198
@Parcly-I think ln means log to the base e, but log is taken to be of base 2.
– user3767495
Nov 17 at 7:43
1
@user3767495 My analysis would still hold in that case.
– Parcly Taxel
Nov 17 at 7:44
@Parcly-Thanks.I realised I was on the correct path, but somehow ordering in wrong way.
– user3767495
Nov 17 at 9:51
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
$$frac{e^{nlog_2n}}n=frac{e^{n(ln n)/(ln 2)}}n=frac{n^{n/ln 2}}n=n^{n/ln 2-1}$$
$$2^{nlog_2 n}=n^n$$
We now compare powers: $sqrt n<frac n{ln 2}-1<n$. Thus $B<C<A$, not $A<B<C$ as you claimed.
answered Nov 17 at 7:36
Parcly Taxel
41k137198
@Parcly-I think ln means log to the base e, but log is taken to be of base 2.
– user3767495
Nov 17 at 7:43
1
@user3767495 My analysis would still hold in that case.
– Parcly Taxel
Nov 17 at 7:44
@Parcly-Thanks.I realised I was on the correct path, but somehow ordering in wrong way.
– user3767495
Nov 17 at 9:51
add a comment |
up vote
1
down vote
accepted
$$frac{e^{nlog_2n}}n=frac{e^{n(ln n)/(ln 2)}}n=frac{n^{n/ln 2}}n=n^{n/ln 2-1}$$
$$2^{nlog_2 n}=n^n$$
We now compare powers: $sqrt n<frac n{ln 2}-1<n$. Thus $B<C<A$, not $A<B<C$ as you claimed.
answered Nov 17 at 7:36
Parcly Taxel
41k137198
@Parcly-I think ln means log to the base e, but log is taken to be of base 2.
– user3767495
Nov 17 at 7:43
1
@user3767495 My analysis would still hold in that case.
– Parcly Taxel
Nov 17 at 7:44
@Parcly-Thanks.I realised I was on the correct path, but somehow ordering in wrong way.
– user3767495
Nov 17 at 9:51
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
$$frac{e^{nlog_2n}}n=frac{e^{n(ln n)/(ln 2)}}n=frac{n^{n/ln 2}}n=n^{n/ln 2-1}$$
$$2^{nlog_2 n}=n^n$$
We now compare powers: $sqrt n<frac n{ln 2}-1<n$. Thus $B<C<A$, not $A<B<C$ as you claimed.
answered Nov 17 at 7:36
Parcly Taxel
41k137198
$$frac{e^{nlog_2n}}n=frac{e^{n(ln n)/(ln 2)}}n=frac{n^{n/ln 2}}n=n^{n/ln 2-1}$$
$$2^{nlog_2 n}=n^n$$
We now compare powers: $sqrt n<frac n{ln 2}-1<n$. Thus $B<C<A$, not $A<B<C$ as you claimed.
answered Nov 17 at 7:36
Parcly Taxel
41k137198
edited Nov 17 at 7:46
answered Nov 17 at 7:36
Parcly Taxel
41k137198
answered Nov 17 at 7:36
Parcly Taxel
41k137198
answered Nov 17 at 7:36
Parcly Taxel
41k137198
41k137198
@Parcly-I think ln means log to the base e, but log is taken to be of base 2.
– user3767495
Nov 17 at 7:43
1
@user3767495 My analysis would still hold in that case.
– Parcly Taxel
Nov 17 at 7:44
@Parcly-Thanks.I realised I was on the correct path, but somehow ordering in wrong way.
– user3767495
Nov 17 at 9:51
add a comment |
@Parcly-I think ln means log to the base e, but log is taken to be of base 2.
– user3767495
Nov 17 at 7:43
1
@user3767495 My analysis would still hold in that case.
– Parcly Taxel
Nov 17 at 7:44
@Parcly-Thanks.I realised I was on the correct path, but somehow ordering in wrong way.
– user3767495
Nov 17 at 9:51
@Parcly-I think ln means log to the base e, but log is taken to be of base 2.
– user3767495
Nov 17 at 7:43
@Parcly-I think ln means log to the base e, but log is taken to be of base 2.
– user3767495
Nov 17 at 7:43
1
1
@user3767495 My analysis would still hold in that case.
– Parcly Taxel
Nov 17 at 7:44
@user3767495 My analysis would still hold in that case.
– Parcly Taxel
Nov 17 at 7:44
@Parcly-Thanks.I realised I was on the correct path, but somehow ordering in wrong way.
– user3767495
Nov 17 at 9:51
@Parcly-Thanks.I realised I was on the correct path, but somehow ordering in wrong way.
– user3767495
Nov 17 at 9:51
add a comment |
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