Mixing
Is $O(2^{n/2})$ the same as $O(2^n)$?
Why or why not? It seems like the answer should be no, but on the other hand, it's weird that you'd reach the same value in a constant multiple of n.
computational-complexity
edited Nov 20 '18 at 9:53
J.-E. Pin
18.3k21754
asked Nov 18 '18 at 0:40
Henry Bigelow
486
add a comment |
Why or why not? It seems like the answer should be no, but on the other hand, it's weird that you'd reach the same value in a constant multiple of n.
computational-complexity
edited Nov 20 '18 at 9:53
J.-E. Pin
18.3k21754
asked Nov 18 '18 at 0:40
Henry Bigelow
486
3
One may observe that, as $n to infty$, $displaystyle frac{2^n}{2^{n/2}} to infty.$
– Olivier Oloa
Nov 18 '18 at 0:45
3
a constant multiple of $n$ has big consequences when $n$ is in the exponent
– mathworker21
Nov 18 '18 at 0:48
add a comment |
Why or why not? It seems like the answer should be no, but on the other hand, it's weird that you'd reach the same value in a constant multiple of n.
computational-complexity
edited Nov 20 '18 at 9:53
J.-E. Pin
18.3k21754
asked Nov 18 '18 at 0:40
Henry Bigelow
486
Why or why not? It seems like the answer should be no, but on the other hand, it's weird that you'd reach the same value in a constant multiple of n.
computational-complexity
computational-complexity
edited Nov 20 '18 at 9:53
J.-E. Pin
18.3k21754
asked Nov 18 '18 at 0:40
Henry Bigelow
486
edited Nov 20 '18 at 9:53
J.-E. Pin
18.3k21754
asked Nov 18 '18 at 0:40
Henry Bigelow
486
edited Nov 20 '18 at 9:53
J.-E. Pin
18.3k21754
edited Nov 20 '18 at 9:53
J.-E. Pin
18.3k21754
edited Nov 20 '18 at 9:53
J.-E. Pin
18.3k21754
18.3k21754
asked Nov 18 '18 at 0:40
Henry Bigelow
486
asked Nov 18 '18 at 0:40
Henry Bigelow
486
asked Nov 18 '18 at 0:40
Henry Bigelow
486
486
3
One may observe that, as $n to infty$, $displaystyle frac{2^n}{2^{n/2}} to infty.$
– Olivier Oloa
Nov 18 '18 at 0:45
3
a constant multiple of $n$ has big consequences when $n$ is in the exponent
– mathworker21
Nov 18 '18 at 0:48
add a comment |
3
One may observe that, as $n to infty$, $displaystyle frac{2^n}{2^{n/2}} to infty.$
– Olivier Oloa
Nov 18 '18 at 0:45
3
a constant multiple of $n$ has big consequences when $n$ is in the exponent
– mathworker21
Nov 18 '18 at 0:48
3
3
One may observe that, as $n to infty$, $displaystyle frac{2^n}{2^{n/2}} to infty.$
– Olivier Oloa
Nov 18 '18 at 0:45
One may observe that, as $n to infty$, $displaystyle frac{2^n}{2^{n/2}} to infty.$
– Olivier Oloa
Nov 18 '18 at 0:45
3
3
a constant multiple of $n$ has big consequences when $n$ is in the exponent
– mathworker21
Nov 18 '18 at 0:48
a constant multiple of $n$ has big consequences when $n$ is in the exponent
– mathworker21
Nov 18 '18 at 0:48
add a comment |
3 Answers
3
active
oldest
votes
$2^{n}$ is $O(2^{n})$ but it is not $O(2^{n/2})$.
answered Nov 18 '18 at 0:42
Kavi Rama Murthy
50.4k31854
add a comment |
I made the same mistake once but if you write $x=2^n$, then you would have $O(x) = O(x^2)$ which of course doesn't make sense.
answered Nov 18 '18 at 0:52
Stan Tendijck
1,411210
add a comment |
By exponent rules, $O(2^{n/2}) = O((2^{1/2})^n) approx O(1.414^n)$ which clearly differs from $O(2^n)$ by more than a constant factor (consider the ratio).
answered Nov 18 '18 at 3:00
qwr
6,54042654
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$2^{n}$ is $O(2^{n})$ but it is not $O(2^{n/2})$.
answered Nov 18 '18 at 0:42
Kavi Rama Murthy
50.4k31854
add a comment |
$2^{n}$ is $O(2^{n})$ but it is not $O(2^{n/2})$.
answered Nov 18 '18 at 0:42
Kavi Rama Murthy
50.4k31854
add a comment |
$2^{n}$ is $O(2^{n})$ but it is not $O(2^{n/2})$.
answered Nov 18 '18 at 0:42
Kavi Rama Murthy
50.4k31854
$2^{n}$ is $O(2^{n})$ but it is not $O(2^{n/2})$.
answered Nov 18 '18 at 0:42
Kavi Rama Murthy
50.4k31854
answered Nov 18 '18 at 0:42
Kavi Rama Murthy
50.4k31854
answered Nov 18 '18 at 0:42
Kavi Rama Murthy
50.4k31854
answered Nov 18 '18 at 0:42
Kavi Rama Murthy
50.4k31854
50.4k31854
add a comment |
add a comment |
I made the same mistake once but if you write $x=2^n$, then you would have $O(x) = O(x^2)$ which of course doesn't make sense.
answered Nov 18 '18 at 0:52
Stan Tendijck
1,411210
add a comment |
I made the same mistake once but if you write $x=2^n$, then you would have $O(x) = O(x^2)$ which of course doesn't make sense.
answered Nov 18 '18 at 0:52
Stan Tendijck
1,411210
add a comment |
I made the same mistake once but if you write $x=2^n$, then you would have $O(x) = O(x^2)$ which of course doesn't make sense.
answered Nov 18 '18 at 0:52
Stan Tendijck
1,411210
I made the same mistake once but if you write $x=2^n$, then you would have $O(x) = O(x^2)$ which of course doesn't make sense.
answered Nov 18 '18 at 0:52
Stan Tendijck
1,411210
answered Nov 18 '18 at 0:52
Stan Tendijck
1,411210
answered Nov 18 '18 at 0:52
Stan Tendijck
1,411210
answered Nov 18 '18 at 0:52
Stan Tendijck
1,411210
1,411210
add a comment |
add a comment |
By exponent rules, $O(2^{n/2}) = O((2^{1/2})^n) approx O(1.414^n)$ which clearly differs from $O(2^n)$ by more than a constant factor (consider the ratio).
answered Nov 18 '18 at 3:00
qwr
6,54042654
add a comment |
By exponent rules, $O(2^{n/2}) = O((2^{1/2})^n) approx O(1.414^n)$ which clearly differs from $O(2^n)$ by more than a constant factor (consider the ratio).
answered Nov 18 '18 at 3:00
qwr
6,54042654
add a comment |
By exponent rules, $O(2^{n/2}) = O((2^{1/2})^n) approx O(1.414^n)$ which clearly differs from $O(2^n)$ by more than a constant factor (consider the ratio).
answered Nov 18 '18 at 3:00
qwr
6,54042654
By exponent rules, $O(2^{n/2}) = O((2^{1/2})^n) approx O(1.414^n)$ which clearly differs from $O(2^n)$ by more than a constant factor (consider the ratio).
answered Nov 18 '18 at 3:00
qwr
6,54042654
answered Nov 18 '18 at 3:00
qwr
6,54042654
answered Nov 18 '18 at 3:00
qwr
6,54042654
answered Nov 18 '18 at 3:00
qwr
6,54042654
6,54042654
add a comment |
add a comment |
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3
One may observe that, as $n to infty$, $displaystyle frac{2^n}{2^{n/2}} to infty.$
– Olivier Oloa
Nov 18 '18 at 0:45
3
a constant multiple of $n$ has big consequences when $n$ is in the exponent
– mathworker21
Nov 18 '18 at 0:48