How can $O(fg)=fO(g)$
$begingroup$
So it is known that $mathcal{O}(fg)=fcdot mathcal{O}(g)$. But what if $f$ is always negative, say a constant? Then $h=mathcal{O}(fg)$ implies there exist $C>0, x_0$ such that $|h(x)|le Cg(x)f(x)$ for $xge x_0$, and $g(x)f(x)$ are strictly positive for $xge x_0$. But this means that, in this case, $g(x)$ is always negative for $xge x_0$, since $f$ is always negative. Now what this means is that if some function is $mathcal{O}(g)$ then $g$ must take positive values too (by definition of how big-oh works), otherwise there can't be $mathcal{O}(g)$. Thus if $h=fmathcal{O}(g)$ then $|h(x)|le f(x)cdot Dcdot g(x)$ with $D>0$ and $g>0$ for these values of $x$, so that $|h(x)|$ is negative, which cannot be.
So what is it that I'm misunderstanding here?
real-analysis asymptotics computational-mathematics
asked Jan 24 at 3:41
sequencesequence
4,25731437
$endgroup$
add a comment |
$begingroup$
So it is known that $mathcal{O}(fg)=fcdot mathcal{O}(g)$. But what if $f$ is always negative, say a constant? Then $h=mathcal{O}(fg)$ implies there exist $C>0, x_0$ such that $|h(x)|le Cg(x)f(x)$ for $xge x_0$, and $g(x)f(x)$ are strictly positive for $xge x_0$. But this means that, in this case, $g(x)$ is always negative for $xge x_0$, since $f$ is always negative. Now what this means is that if some function is $mathcal{O}(g)$ then $g$ must take positive values too (by definition of how big-oh works), otherwise there can't be $mathcal{O}(g)$. Thus if $h=fmathcal{O}(g)$ then $|h(x)|le f(x)cdot Dcdot g(x)$ with $D>0$ and $g>0$ for these values of $x$, so that $|h(x)|$ is negative, which cannot be.
So what is it that I'm misunderstanding here?
real-analysis asymptotics computational-mathematics
asked Jan 24 at 3:41
sequencesequence
4,25731437
$endgroup$
add a comment |
$begingroup$
So it is known that $mathcal{O}(fg)=fcdot mathcal{O}(g)$. But what if $f$ is always negative, say a constant? Then $h=mathcal{O}(fg)$ implies there exist $C>0, x_0$ such that $|h(x)|le Cg(x)f(x)$ for $xge x_0$, and $g(x)f(x)$ are strictly positive for $xge x_0$. But this means that, in this case, $g(x)$ is always negative for $xge x_0$, since $f$ is always negative. Now what this means is that if some function is $mathcal{O}(g)$ then $g$ must take positive values too (by definition of how big-oh works), otherwise there can't be $mathcal{O}(g)$. Thus if $h=fmathcal{O}(g)$ then $|h(x)|le f(x)cdot Dcdot g(x)$ with $D>0$ and $g>0$ for these values of $x$, so that $|h(x)|$ is negative, which cannot be.
So what is it that I'm misunderstanding here?
real-analysis asymptotics computational-mathematics
asked Jan 24 at 3:41
sequencesequence
4,25731437
$endgroup$
So it is known that $mathcal{O}(fg)=fcdot mathcal{O}(g)$. But what if $f$ is always negative, say a constant? Then $h=mathcal{O}(fg)$ implies there exist $C>0, x_0$ such that $|h(x)|le Cg(x)f(x)$ for $xge x_0$, and $g(x)f(x)$ are strictly positive for $xge x_0$. But this means that, in this case, $g(x)$ is always negative for $xge x_0$, since $f$ is always negative. Now what this means is that if some function is $mathcal{O}(g)$ then $g$ must take positive values too (by definition of how big-oh works), otherwise there can't be $mathcal{O}(g)$. Thus if $h=fmathcal{O}(g)$ then $|h(x)|le f(x)cdot Dcdot g(x)$ with $D>0$ and $g>0$ for these values of $x$, so that $|h(x)|$ is negative, which cannot be.
So what is it that I'm misunderstanding here?
real-analysis asymptotics computational-mathematics
real-analysis asymptotics computational-mathematics
asked Jan 24 at 3:41
sequencesequence
4,25731437
asked Jan 24 at 3:41
sequencesequence
4,25731437
asked Jan 24 at 3:41
sequencesequence
4,25731437
asked Jan 24 at 3:41
sequencesequence
4,25731437
asked Jan 24 at 3:41
sequencesequence
4,25731437
4,25731437
add a comment |
add a comment |
1 Answer
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$begingroup$
$mathcal{O}(g)$ only makes sense if $g(x)$ is strictly positive for sufficiently large $x$, and similarly $mathcal{O}(fg)$ only makes sense if $f(x)g(x)$ is positive for sufficiently large $x$. So the statement $fmathcal{O}(g)=mathcal{O}(fg)$ only makes sense if both $f(x)$ and $g(x)$ are strictly positive for sufficiently large $x$.
answered Jan 24 at 4:08
kccukccu
10.6k11228
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$begingroup$
$mathcal{O}(g)$ only makes sense if $g(x)$ is strictly positive for sufficiently large $x$, and similarly $mathcal{O}(fg)$ only makes sense if $f(x)g(x)$ is positive for sufficiently large $x$. So the statement $fmathcal{O}(g)=mathcal{O}(fg)$ only makes sense if both $f(x)$ and $g(x)$ are strictly positive for sufficiently large $x$.
answered Jan 24 at 4:08
kccukccu
10.6k11228
$endgroup$
add a comment |
$begingroup$
$mathcal{O}(g)$ only makes sense if $g(x)$ is strictly positive for sufficiently large $x$, and similarly $mathcal{O}(fg)$ only makes sense if $f(x)g(x)$ is positive for sufficiently large $x$. So the statement $fmathcal{O}(g)=mathcal{O}(fg)$ only makes sense if both $f(x)$ and $g(x)$ are strictly positive for sufficiently large $x$.
answered Jan 24 at 4:08
kccukccu
10.6k11228
$endgroup$
add a comment |
$begingroup$
$mathcal{O}(g)$ only makes sense if $g(x)$ is strictly positive for sufficiently large $x$, and similarly $mathcal{O}(fg)$ only makes sense if $f(x)g(x)$ is positive for sufficiently large $x$. So the statement $fmathcal{O}(g)=mathcal{O}(fg)$ only makes sense if both $f(x)$ and $g(x)$ are strictly positive for sufficiently large $x$.
answered Jan 24 at 4:08
kccukccu
10.6k11228
$endgroup$
$mathcal{O}(g)$ only makes sense if $g(x)$ is strictly positive for sufficiently large $x$, and similarly $mathcal{O}(fg)$ only makes sense if $f(x)g(x)$ is positive for sufficiently large $x$. So the statement $fmathcal{O}(g)=mathcal{O}(fg)$ only makes sense if both $f(x)$ and $g(x)$ are strictly positive for sufficiently large $x$.
answered Jan 24 at 4:08
kccukccu
10.6k11228
answered Jan 24 at 4:08
kccukccu
10.6k11228
answered Jan 24 at 4:08
kccukccu
10.6k11228
answered Jan 24 at 4:08
kccukccu
10.6k11228
10.6k11228
add a comment |
add a comment |
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