Chapter V: Titchmarsh's book “The theory of the Riemann Zeta function”
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Through chapter V of Titchmarsh's book "The theory of the Riemann Zeta function" it is used a "counting technique" that I am not understanding. In particular, Theorem 5.12, p 106, uses something like: if $bleq 2a$, with $0<a<t$ and $sum_{a<nleq b}n^{-s-it}=O(f(t))$ then $sum_{nleq t}n^{-s-it}=O(f(t)log(t))$, as $ttoinfty$. My question is: how this $log(t)$ appears? That is how is it related to $bleq 2a$ and $0<a<t$?
real-analysis sequences-and-series complex-analysis summation
asked Dec 4 '18 at 19:39
GoldSoundzGoldSoundz
1098
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add a comment |
$begingroup$
Through chapter V of Titchmarsh's book "The theory of the Riemann Zeta function" it is used a "counting technique" that I am not understanding. In particular, Theorem 5.12, p 106, uses something like: if $bleq 2a$, with $0<a<t$ and $sum_{a<nleq b}n^{-s-it}=O(f(t))$ then $sum_{nleq t}n^{-s-it}=O(f(t)log(t))$, as $ttoinfty$. My question is: how this $log(t)$ appears? That is how is it related to $bleq 2a$ and $0<a<t$?
real-analysis sequences-and-series complex-analysis summation
asked Dec 4 '18 at 19:39
GoldSoundzGoldSoundz
1098
$endgroup$
$begingroup$
Here's a link to the book if anyone wants it.
$endgroup$
– Mason
Dec 4 '18 at 19:53
add a comment |
$begingroup$
Through chapter V of Titchmarsh's book "The theory of the Riemann Zeta function" it is used a "counting technique" that I am not understanding. In particular, Theorem 5.12, p 106, uses something like: if $bleq 2a$, with $0<a<t$ and $sum_{a<nleq b}n^{-s-it}=O(f(t))$ then $sum_{nleq t}n^{-s-it}=O(f(t)log(t))$, as $ttoinfty$. My question is: how this $log(t)$ appears? That is how is it related to $bleq 2a$ and $0<a<t$?
real-analysis sequences-and-series complex-analysis summation
asked Dec 4 '18 at 19:39
GoldSoundzGoldSoundz
1098
$endgroup$
Through chapter V of Titchmarsh's book "The theory of the Riemann Zeta function" it is used a "counting technique" that I am not understanding. In particular, Theorem 5.12, p 106, uses something like: if $bleq 2a$, with $0<a<t$ and $sum_{a<nleq b}n^{-s-it}=O(f(t))$ then $sum_{nleq t}n^{-s-it}=O(f(t)log(t))$, as $ttoinfty$. My question is: how this $log(t)$ appears? That is how is it related to $bleq 2a$ and $0<a<t$?
real-analysis sequences-and-series complex-analysis summation
real-analysis sequences-and-series complex-analysis summation
asked Dec 4 '18 at 19:39
GoldSoundzGoldSoundz
1098
asked Dec 4 '18 at 19:39
GoldSoundzGoldSoundz
1098
asked Dec 4 '18 at 19:39
GoldSoundzGoldSoundz
1098
asked Dec 4 '18 at 19:39
GoldSoundzGoldSoundz
1098
asked Dec 4 '18 at 19:39
GoldSoundzGoldSoundz
1098
1098
$begingroup$
Here's a link to the book if anyone wants it.
$endgroup$
– Mason
Dec 4 '18 at 19:53
add a comment |
$begingroup$
Here's a link to the book if anyone wants it.
$endgroup$
– Mason
Dec 4 '18 at 19:53
$begingroup$
Here's a link to the book if anyone wants it.
$endgroup$
– Mason
Dec 4 '18 at 19:53
$begingroup$
Here's a link to the book if anyone wants it.
$endgroup$
– Mason
Dec 4 '18 at 19:53
add a comment |
1 Answer
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You need to apply the result, as also stated in the book, multiple times: Take $b_n=t 2^{-n}$ and $a_n = t 2^{-(n+1)}$ (e.g. $b_0 =t$, $a_0 =t/2$, $b_1 =t/2$, $a_1 =t/4$, and so on) and apply this until $a_n le 1$. This happens exactly when $n le log(t)/log(2)$ and $log(t)/log(2) le (n+1)$. Thus, we only need to apply the result $ log(t)$-times.
answered Dec 4 '18 at 20:46
p4schp4sch
4,995217
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$begingroup$
Thanks! I think I got it.
$endgroup$
– GoldSoundz
Dec 4 '18 at 20:59
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
You need to apply the result, as also stated in the book, multiple times: Take $b_n=t 2^{-n}$ and $a_n = t 2^{-(n+1)}$ (e.g. $b_0 =t$, $a_0 =t/2$, $b_1 =t/2$, $a_1 =t/4$, and so on) and apply this until $a_n le 1$. This happens exactly when $n le log(t)/log(2)$ and $log(t)/log(2) le (n+1)$. Thus, we only need to apply the result $ log(t)$-times.
answered Dec 4 '18 at 20:46
p4schp4sch
4,995217
$endgroup$
$begingroup$
Thanks! I think I got it.
$endgroup$
– GoldSoundz
Dec 4 '18 at 20:59
add a comment |
$begingroup$
You need to apply the result, as also stated in the book, multiple times: Take $b_n=t 2^{-n}$ and $a_n = t 2^{-(n+1)}$ (e.g. $b_0 =t$, $a_0 =t/2$, $b_1 =t/2$, $a_1 =t/4$, and so on) and apply this until $a_n le 1$. This happens exactly when $n le log(t)/log(2)$ and $log(t)/log(2) le (n+1)$. Thus, we only need to apply the result $ log(t)$-times.
answered Dec 4 '18 at 20:46
p4schp4sch
4,995217
$endgroup$
$begingroup$
Thanks! I think I got it.
$endgroup$
– GoldSoundz
Dec 4 '18 at 20:59
add a comment |
$begingroup$
You need to apply the result, as also stated in the book, multiple times: Take $b_n=t 2^{-n}$ and $a_n = t 2^{-(n+1)}$ (e.g. $b_0 =t$, $a_0 =t/2$, $b_1 =t/2$, $a_1 =t/4$, and so on) and apply this until $a_n le 1$. This happens exactly when $n le log(t)/log(2)$ and $log(t)/log(2) le (n+1)$. Thus, we only need to apply the result $ log(t)$-times.
answered Dec 4 '18 at 20:46
p4schp4sch
4,995217
$endgroup$
You need to apply the result, as also stated in the book, multiple times: Take $b_n=t 2^{-n}$ and $a_n = t 2^{-(n+1)}$ (e.g. $b_0 =t$, $a_0 =t/2$, $b_1 =t/2$, $a_1 =t/4$, and so on) and apply this until $a_n le 1$. This happens exactly when $n le log(t)/log(2)$ and $log(t)/log(2) le (n+1)$. Thus, we only need to apply the result $ log(t)$-times.
answered Dec 4 '18 at 20:46
p4schp4sch
4,995217
answered Dec 4 '18 at 20:46
p4schp4sch
4,995217
answered Dec 4 '18 at 20:46
p4schp4sch
4,995217
answered Dec 4 '18 at 20:46
p4schp4sch
4,995217
4,995217
$begingroup$
Thanks! I think I got it.
$endgroup$
– GoldSoundz
Dec 4 '18 at 20:59
add a comment |
$begingroup$
Thanks! I think I got it.
$endgroup$
– GoldSoundz
Dec 4 '18 at 20:59
$begingroup$
Thanks! I think I got it.
$endgroup$
– GoldSoundz
Dec 4 '18 at 20:59
$begingroup$
Thanks! I think I got it.
$endgroup$
– GoldSoundz
Dec 4 '18 at 20:59
add a comment |
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$begingroup$
Here's a link to the book if anyone wants it.
$endgroup$
– Mason
Dec 4 '18 at 19:53