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Asymptotic notations and inequalities
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Suppose that we have two implicit functions
$$f(t)=logfract2+Oleft(frac1tright),,
g(t)=log t+Oleft(frac1tright).$$
Let $z(t)$ be a function such that $f(t)< z(t) < g(t)$, when $t$ is sufficiently large.
Can the last inequalities be replaced by any known asymptotic notations?
A bit more general version of my question could be "Let $g(t)-f(t)=constant+o(1)$ and $f(t)< z(t) < g(t)$, as $ttoinfty$. Can the last inequalities be expressed by any known asymptotic notations?"
The idea of my question is that the constants implied by $O$ might be different and it is confusing to write $logfract2+Oleft(frac1tright)<z(t)<log t+Oleft(frac1tright)$ or $logfract2<z(t)+Oleft(frac1tright)<log t$, as $t to infty$.
inequality asymptotics
asked Jul 15 at 21:18
Fancier of Mathematica
212
add a comment |Â
up vote
2
down vote
favorite
Suppose that we have two implicit functions
$$f(t)=logfract2+Oleft(frac1tright),,
g(t)=log t+Oleft(frac1tright).$$
Let $z(t)$ be a function such that $f(t)< z(t) < g(t)$, when $t$ is sufficiently large.
Can the last inequalities be replaced by any known asymptotic notations?
A bit more general version of my question could be "Let $g(t)-f(t)=constant+o(1)$ and $f(t)< z(t) < g(t)$, as $ttoinfty$. Can the last inequalities be expressed by any known asymptotic notations?"
The idea of my question is that the constants implied by $O$ might be different and it is confusing to write $logfract2+Oleft(frac1tright)<z(t)<log t+Oleft(frac1tright)$ or $logfract2<z(t)+Oleft(frac1tright)<log t$, as $t to infty$.
inequality asymptotics
asked Jul 15 at 21:18
Fancier of Mathematica
212
The closest I can think of is big theta. It is not exactly what you want but for $f=mathcal O(g)$ you can work with it
– Holo
Jul 15 at 21:26
From the inequalities $f < z < g$ you can deduce that $z(t) = log t + O(1)$, but that isn't equivalent to the inequalities.
– Antonio Vargas
Jul 15 at 23:19
@AntonioVargas I don't see how you can deduce $O(1)$ there.
– user58697
Jul 16 at 0:16
1
There exist constants $A, B, T > 0$ such that $$f(t) > log fract2 - A = log t - log 2 - A,$$ $$g(t) < log t + B,$$ and $f(t) < z(t) < g(t)$ for all $t > T$. Thus $$log t - log 2 - A < z(t) < log t + B$$ and hence $$-log 2 - A < z(t) - log t < B$$ for all $t > T$, which implies that $z(t) = log t + O(1)$. @user58697
– Antonio Vargas
Jul 16 at 2:36
You should write $Omega(frac1t)$ on the LHS, and this also solves the problem of confusing the two implied constants.
– Aravind
Jul 16 at 5:14
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Suppose that we have two implicit functions
$$f(t)=logfract2+Oleft(frac1tright),,
g(t)=log t+Oleft(frac1tright).$$
Let $z(t)$ be a function such that $f(t)< z(t) < g(t)$, when $t$ is sufficiently large.
Can the last inequalities be replaced by any known asymptotic notations?
A bit more general version of my question could be "Let $g(t)-f(t)=constant+o(1)$ and $f(t)< z(t) < g(t)$, as $ttoinfty$. Can the last inequalities be expressed by any known asymptotic notations?"
The idea of my question is that the constants implied by $O$ might be different and it is confusing to write $logfract2+Oleft(frac1tright)<z(t)<log t+Oleft(frac1tright)$ or $logfract2<z(t)+Oleft(frac1tright)<log t$, as $t to infty$.
inequality asymptotics
asked Jul 15 at 21:18
Fancier of Mathematica
212
Suppose that we have two implicit functions
$$f(t)=logfract2+Oleft(frac1tright),,
g(t)=log t+Oleft(frac1tright).$$
Let $z(t)$ be a function such that $f(t)< z(t) < g(t)$, when $t$ is sufficiently large.
Can the last inequalities be replaced by any known asymptotic notations?
A bit more general version of my question could be "Let $g(t)-f(t)=constant+o(1)$ and $f(t)< z(t) < g(t)$, as $ttoinfty$. Can the last inequalities be expressed by any known asymptotic notations?"
The idea of my question is that the constants implied by $O$ might be different and it is confusing to write $logfract2+Oleft(frac1tright)<z(t)<log t+Oleft(frac1tright)$ or $logfract2<z(t)+Oleft(frac1tright)<log t$, as $t to infty$.
inequality asymptotics
asked Jul 15 at 21:18
Fancier of Mathematica
212
asked Jul 15 at 21:18
Fancier of Mathematica
212
asked Jul 15 at 21:18
Fancier of Mathematica
212
asked Jul 15 at 21:18
Fancier of Mathematica
212
212
The closest I can think of is big theta. It is not exactly what you want but for $f=mathcal O(g)$ you can work with it
– Holo
Jul 15 at 21:26
From the inequalities $f < z < g$ you can deduce that $z(t) = log t + O(1)$, but that isn't equivalent to the inequalities.
– Antonio Vargas
Jul 15 at 23:19
@AntonioVargas I don't see how you can deduce $O(1)$ there.
– user58697
Jul 16 at 0:16
1
There exist constants $A, B, T > 0$ such that $$f(t) > log fract2 - A = log t - log 2 - A,$$ $$g(t) < log t + B,$$ and $f(t) < z(t) < g(t)$ for all $t > T$. Thus $$log t - log 2 - A < z(t) < log t + B$$ and hence $$-log 2 - A < z(t) - log t < B$$ for all $t > T$, which implies that $z(t) = log t + O(1)$. @user58697
– Antonio Vargas
Jul 16 at 2:36
You should write $Omega(frac1t)$ on the LHS, and this also solves the problem of confusing the two implied constants.
– Aravind
Jul 16 at 5:14
add a comment |Â
The closest I can think of is big theta. It is not exactly what you want but for $f=mathcal O(g)$ you can work with it
– Holo
Jul 15 at 21:26
From the inequalities $f < z < g$ you can deduce that $z(t) = log t + O(1)$, but that isn't equivalent to the inequalities.
– Antonio Vargas
Jul 15 at 23:19
@AntonioVargas I don't see how you can deduce $O(1)$ there.
– user58697
Jul 16 at 0:16
1
There exist constants $A, B, T > 0$ such that $$f(t) > log fract2 - A = log t - log 2 - A,$$ $$g(t) < log t + B,$$ and $f(t) < z(t) < g(t)$ for all $t > T$. Thus $$log t - log 2 - A < z(t) < log t + B$$ and hence $$-log 2 - A < z(t) - log t < B$$ for all $t > T$, which implies that $z(t) = log t + O(1)$. @user58697
– Antonio Vargas
Jul 16 at 2:36
You should write $Omega(frac1t)$ on the LHS, and this also solves the problem of confusing the two implied constants.
– Aravind
Jul 16 at 5:14
The closest I can think of is big theta. It is not exactly what you want but for $f=mathcal O(g)$ you can work with it
– Holo
Jul 15 at 21:26
The closest I can think of is big theta. It is not exactly what you want but for $f=mathcal O(g)$ you can work with it
– Holo
Jul 15 at 21:26
From the inequalities $f < z < g$ you can deduce that $z(t) = log t + O(1)$, but that isn't equivalent to the inequalities.
– Antonio Vargas
Jul 15 at 23:19
From the inequalities $f < z < g$ you can deduce that $z(t) = log t + O(1)$, but that isn't equivalent to the inequalities.
– Antonio Vargas
Jul 15 at 23:19
@AntonioVargas I don't see how you can deduce $O(1)$ there.
– user58697
Jul 16 at 0:16
@AntonioVargas I don't see how you can deduce $O(1)$ there.
– user58697
Jul 16 at 0:16
1
1
There exist constants $A, B, T > 0$ such that $$f(t) > log fract2 - A = log t - log 2 - A,$$ $$g(t) < log t + B,$$ and $f(t) < z(t) < g(t)$ for all $t > T$. Thus $$log t - log 2 - A < z(t) < log t + B$$ and hence $$-log 2 - A < z(t) - log t < B$$ for all $t > T$, which implies that $z(t) = log t + O(1)$. @user58697
– Antonio Vargas
Jul 16 at 2:36
There exist constants $A, B, T > 0$ such that $$f(t) > log fract2 - A = log t - log 2 - A,$$ $$g(t) < log t + B,$$ and $f(t) < z(t) < g(t)$ for all $t > T$. Thus $$log t - log 2 - A < z(t) < log t + B$$ and hence $$-log 2 - A < z(t) - log t < B$$ for all $t > T$, which implies that $z(t) = log t + O(1)$. @user58697
– Antonio Vargas
Jul 16 at 2:36
You should write $Omega(frac1t)$ on the LHS, and this also solves the problem of confusing the two implied constants.
– Aravind
Jul 16 at 5:14
You should write $Omega(frac1t)$ on the LHS, and this also solves the problem of confusing the two implied constants.
– Aravind
Jul 16 at 5:14
add a comment |Â
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The closest I can think of is big theta. It is not exactly what you want but for $f=mathcal O(g)$ you can work with it
– Holo
Jul 15 at 21:26
From the inequalities $f < z < g$ you can deduce that $z(t) = log t + O(1)$, but that isn't equivalent to the inequalities.
– Antonio Vargas
Jul 15 at 23:19
@AntonioVargas I don't see how you can deduce $O(1)$ there.
– user58697
Jul 16 at 0:16
1
There exist constants $A, B, T > 0$ such that $$f(t) > log fract2 - A = log t - log 2 - A,$$ $$g(t) < log t + B,$$ and $f(t) < z(t) < g(t)$ for all $t > T$. Thus $$log t - log 2 - A < z(t) < log t + B$$ and hence $$-log 2 - A < z(t) - log t < B$$ for all $t > T$, which implies that $z(t) = log t + O(1)$. @user58697
– Antonio Vargas
Jul 16 at 2:36
You should write $Omega(frac1t)$ on the LHS, and this also solves the problem of confusing the two implied constants.
– Aravind
Jul 16 at 5:14