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Big Oh of values of Riemann zeta function
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There is a equality in a proof in Apostol's Analytic Number Theory as follows:
$O(x^{alpha} zeta(alpha)) = O(x^{alpha})$ for arbitrary real number $alpha ge 0$.
How do we say that? Does anything changes when $alpha in [0,1]$?
number-theory analytic-number-theory riemann-zeta
asked Jan 17 at 20:09
NinjaNinja
1,105720
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add a comment |
$begingroup$
There is a equality in a proof in Apostol's Analytic Number Theory as follows:
$O(x^{alpha} zeta(alpha)) = O(x^{alpha})$ for arbitrary real number $alpha ge 0$.
How do we say that? Does anything changes when $alpha in [0,1]$?
number-theory analytic-number-theory riemann-zeta
asked Jan 17 at 20:09
NinjaNinja
1,105720
$endgroup$
1
$begingroup$
It is not correct. Check out $zeta(1)$ for $alpha=1$, e.g., here. What section/chapter is this in Apostol?
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– Dietrich Burde
Jan 17 at 20:12
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It is in Chapter 3, Theorem 3.5.
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– Ninja
Jan 18 at 10:45
$begingroup$
My bad, $alpha$ is any positive number except $1$. What happens now?
$endgroup$
– Ninja
Jan 18 at 10:46
add a comment |
$begingroup$
There is a equality in a proof in Apostol's Analytic Number Theory as follows:
$O(x^{alpha} zeta(alpha)) = O(x^{alpha})$ for arbitrary real number $alpha ge 0$.
How do we say that? Does anything changes when $alpha in [0,1]$?
number-theory analytic-number-theory riemann-zeta
asked Jan 17 at 20:09
NinjaNinja
1,105720
$endgroup$
There is a equality in a proof in Apostol's Analytic Number Theory as follows:
$O(x^{alpha} zeta(alpha)) = O(x^{alpha})$ for arbitrary real number $alpha ge 0$.
How do we say that? Does anything changes when $alpha in [0,1]$?
number-theory analytic-number-theory riemann-zeta
number-theory analytic-number-theory riemann-zeta
asked Jan 17 at 20:09
NinjaNinja
1,105720
asked Jan 17 at 20:09
NinjaNinja
1,105720
asked Jan 17 at 20:09
NinjaNinja
1,105720
asked Jan 17 at 20:09
NinjaNinja
1,105720
asked Jan 17 at 20:09
NinjaNinja
1,105720
1,105720
1
$begingroup$
It is not correct. Check out $zeta(1)$ for $alpha=1$, e.g., here. What section/chapter is this in Apostol?
$endgroup$
– Dietrich Burde
Jan 17 at 20:12
$begingroup$
It is in Chapter 3, Theorem 3.5.
$endgroup$
– Ninja
Jan 18 at 10:45
$begingroup$
My bad, $alpha$ is any positive number except $1$. What happens now?
$endgroup$
– Ninja
Jan 18 at 10:46
add a comment |
1
$begingroup$
It is not correct. Check out $zeta(1)$ for $alpha=1$, e.g., here. What section/chapter is this in Apostol?
$endgroup$
– Dietrich Burde
Jan 17 at 20:12
$begingroup$
It is in Chapter 3, Theorem 3.5.
$endgroup$
– Ninja
Jan 18 at 10:45
$begingroup$
My bad, $alpha$ is any positive number except $1$. What happens now?
$endgroup$
– Ninja
Jan 18 at 10:46
1
1
$begingroup$
It is not correct. Check out $zeta(1)$ for $alpha=1$, e.g., here. What section/chapter is this in Apostol?
$endgroup$
– Dietrich Burde
Jan 17 at 20:12
$begingroup$
It is not correct. Check out $zeta(1)$ for $alpha=1$, e.g., here. What section/chapter is this in Apostol?
$endgroup$
– Dietrich Burde
Jan 17 at 20:12
$begingroup$
It is in Chapter 3, Theorem 3.5.
$endgroup$
– Ninja
Jan 18 at 10:45
$begingroup$
It is in Chapter 3, Theorem 3.5.
$endgroup$
– Ninja
Jan 18 at 10:45
$begingroup$
My bad, $alpha$ is any positive number except $1$. What happens now?
$endgroup$
– Ninja
Jan 18 at 10:46
$begingroup$
My bad, $alpha$ is any positive number except $1$. What happens now?
$endgroup$
– Ninja
Jan 18 at 10:46
add a comment |
1 Answer
1
active
oldest
votes
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The thing is - as far as the implied limit there is concerned - for fixed $alpha$ as $x$ varies, $zeta(alpha)$ is just a number. $O(cf(x))=O(f(x))$ for any constant $c$ and function $f$.
As noted in the comments, $alpha=1$ is an exception; there, $O(x^{alpha}zeta(alpha))$ becomes $O(infty)$, which is vacuously true for any function of $x$.
In practice, the places you'll actually want to use this always come from values of $alpha$ for which the series for $zeta$ converges - namely, $alpha>1$. However, it's still true for $alpha<1$, using the definitionof $zeta$ by analytic continuation.
answered Jan 17 at 20:31
jmerryjmerry
10.6k1225
$endgroup$
$begingroup$
OK, $alpha=1$ is special. $O(infty)$ is a rather meaningless proposition.
$endgroup$
– jmerry
Jan 17 at 21:47
add a comment |
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1 Answer
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1 Answer
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$begingroup$
The thing is - as far as the implied limit there is concerned - for fixed $alpha$ as $x$ varies, $zeta(alpha)$ is just a number. $O(cf(x))=O(f(x))$ for any constant $c$ and function $f$.
As noted in the comments, $alpha=1$ is an exception; there, $O(x^{alpha}zeta(alpha))$ becomes $O(infty)$, which is vacuously true for any function of $x$.
In practice, the places you'll actually want to use this always come from values of $alpha$ for which the series for $zeta$ converges - namely, $alpha>1$. However, it's still true for $alpha<1$, using the definitionof $zeta$ by analytic continuation.
answered Jan 17 at 20:31
jmerryjmerry
10.6k1225
$endgroup$
$begingroup$
OK, $alpha=1$ is special. $O(infty)$ is a rather meaningless proposition.
$endgroup$
– jmerry
Jan 17 at 21:47
add a comment |
$begingroup$
The thing is - as far as the implied limit there is concerned - for fixed $alpha$ as $x$ varies, $zeta(alpha)$ is just a number. $O(cf(x))=O(f(x))$ for any constant $c$ and function $f$.
As noted in the comments, $alpha=1$ is an exception; there, $O(x^{alpha}zeta(alpha))$ becomes $O(infty)$, which is vacuously true for any function of $x$.
In practice, the places you'll actually want to use this always come from values of $alpha$ for which the series for $zeta$ converges - namely, $alpha>1$. However, it's still true for $alpha<1$, using the definitionof $zeta$ by analytic continuation.
answered Jan 17 at 20:31
jmerryjmerry
10.6k1225
$endgroup$
$begingroup$
OK, $alpha=1$ is special. $O(infty)$ is a rather meaningless proposition.
$endgroup$
– jmerry
Jan 17 at 21:47
add a comment |
$begingroup$
The thing is - as far as the implied limit there is concerned - for fixed $alpha$ as $x$ varies, $zeta(alpha)$ is just a number. $O(cf(x))=O(f(x))$ for any constant $c$ and function $f$.
As noted in the comments, $alpha=1$ is an exception; there, $O(x^{alpha}zeta(alpha))$ becomes $O(infty)$, which is vacuously true for any function of $x$.
In practice, the places you'll actually want to use this always come from values of $alpha$ for which the series for $zeta$ converges - namely, $alpha>1$. However, it's still true for $alpha<1$, using the definitionof $zeta$ by analytic continuation.
answered Jan 17 at 20:31
jmerryjmerry
10.6k1225
$endgroup$
The thing is - as far as the implied limit there is concerned - for fixed $alpha$ as $x$ varies, $zeta(alpha)$ is just a number. $O(cf(x))=O(f(x))$ for any constant $c$ and function $f$.
As noted in the comments, $alpha=1$ is an exception; there, $O(x^{alpha}zeta(alpha))$ becomes $O(infty)$, which is vacuously true for any function of $x$.
In practice, the places you'll actually want to use this always come from values of $alpha$ for which the series for $zeta$ converges - namely, $alpha>1$. However, it's still true for $alpha<1$, using the definitionof $zeta$ by analytic continuation.
answered Jan 17 at 20:31
jmerryjmerry
10.6k1225
edited Jan 17 at 21:50
answered Jan 17 at 20:31
jmerryjmerry
10.6k1225
answered Jan 17 at 20:31
jmerryjmerry
10.6k1225
answered Jan 17 at 20:31
jmerryjmerry
10.6k1225
10.6k1225
$begingroup$
OK, $alpha=1$ is special. $O(infty)$ is a rather meaningless proposition.
$endgroup$
– jmerry
Jan 17 at 21:47
add a comment |
$begingroup$
OK, $alpha=1$ is special. $O(infty)$ is a rather meaningless proposition.
$endgroup$
– jmerry
Jan 17 at 21:47
$begingroup$
OK, $alpha=1$ is special. $O(infty)$ is a rather meaningless proposition.
$endgroup$
– jmerry
Jan 17 at 21:47
$begingroup$
OK, $alpha=1$ is special. $O(infty)$ is a rather meaningless proposition.
$endgroup$
– jmerry
Jan 17 at 21:47
add a comment |
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It is not correct. Check out $zeta(1)$ for $alpha=1$, e.g., here. What section/chapter is this in Apostol?
$endgroup$
– Dietrich Burde
Jan 17 at 20:12
$begingroup$
It is in Chapter 3, Theorem 3.5.
$endgroup$
– Ninja
Jan 18 at 10:45
$begingroup$
My bad, $alpha$ is any positive number except $1$. What happens now?
$endgroup$
– Ninja
Jan 18 at 10:46