What is an asymptotically nonnegative function?
$begingroup$
I was reading up on the definition of theta- notation and came across this,
The definition of $Theta(g(n))$ requires that every member $f(n) = Theta(g(n))$ be
asymptotically nonnegative, that is, that $f(n)$ be nonnegative whenever $n$ is sufficiently large.
I'm having a little difficulty understanding what this means.
notation asymptotics
edited Jan 10 at 16:48
Clement C.
50.8k33992
asked Jan 10 at 16:19
Frantz PaulFrantz Paul
62
$endgroup$
add a comment |
$begingroup$
I was reading up on the definition of theta- notation and came across this,
The definition of $Theta(g(n))$ requires that every member $f(n) = Theta(g(n))$ be
asymptotically nonnegative, that is, that $f(n)$ be nonnegative whenever $n$ is sufficiently large.
I'm having a little difficulty understanding what this means.
notation asymptotics
edited Jan 10 at 16:48
Clement C.
50.8k33992
asked Jan 10 at 16:19
Frantz PaulFrantz Paul
62
$endgroup$
add a comment |
$begingroup$
I was reading up on the definition of theta- notation and came across this,
The definition of $Theta(g(n))$ requires that every member $f(n) = Theta(g(n))$ be
asymptotically nonnegative, that is, that $f(n)$ be nonnegative whenever $n$ is sufficiently large.
I'm having a little difficulty understanding what this means.
notation asymptotics
edited Jan 10 at 16:48
Clement C.
50.8k33992
asked Jan 10 at 16:19
Frantz PaulFrantz Paul
62
$endgroup$
I was reading up on the definition of theta- notation and came across this,
The definition of $Theta(g(n))$ requires that every member $f(n) = Theta(g(n))$ be
asymptotically nonnegative, that is, that $f(n)$ be nonnegative whenever $n$ is sufficiently large.
I'm having a little difficulty understanding what this means.
notation asymptotics
notation asymptotics
edited Jan 10 at 16:48
Clement C.
50.8k33992
asked Jan 10 at 16:19
Frantz PaulFrantz Paul
62
edited Jan 10 at 16:48
Clement C.
50.8k33992
asked Jan 10 at 16:19
Frantz PaulFrantz Paul
62
edited Jan 10 at 16:48
Clement C.
50.8k33992
edited Jan 10 at 16:48
Clement C.
50.8k33992
edited Jan 10 at 16:48
Clement C.
50.8k33992
50.8k33992
asked Jan 10 at 16:19
Frantz PaulFrantz Paul
62
asked Jan 10 at 16:19
Frantz PaulFrantz Paul
62
asked Jan 10 at 16:19
Frantz PaulFrantz Paul
62
62
add a comment |
add a comment |
1 Answer
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$begingroup$
It means that there is some positive $N$ such that for all $n > N$, we have $f(n) geq 0$.
answered Jan 10 at 16:22
user3482749user3482749
4,296919
$endgroup$
add a comment |
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$begingroup$
It means that there is some positive $N$ such that for all $n > N$, we have $f(n) geq 0$.
answered Jan 10 at 16:22
user3482749user3482749
4,296919
$endgroup$
add a comment |
$begingroup$
It means that there is some positive $N$ such that for all $n > N$, we have $f(n) geq 0$.
answered Jan 10 at 16:22
user3482749user3482749
4,296919
$endgroup$
add a comment |
$begingroup$
It means that there is some positive $N$ such that for all $n > N$, we have $f(n) geq 0$.
answered Jan 10 at 16:22
user3482749user3482749
4,296919
$endgroup$
It means that there is some positive $N$ such that for all $n > N$, we have $f(n) geq 0$.
answered Jan 10 at 16:22
user3482749user3482749
4,296919
answered Jan 10 at 16:22
user3482749user3482749
4,296919
answered Jan 10 at 16:22
user3482749user3482749
4,296919
answered Jan 10 at 16:22
user3482749user3482749
4,296919
4,296919
add a comment |
add a comment |
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