How did the guy in the comments on the fifth stage got this result? [on hold]
-4
Derivative of Riemann zeta, is this inequality true?
In the fifth stage, he got this:
lim h→0 {π/2*tan(π2(2n+1+h))−ζ′(−2n−h)/ζ(−2n−h)}
And shifted the equation to that:
lim h→0 {−1/h+O(h)+1/h−ζ′′(−2n)/2ζ′(−2n)+O(h)}
How did he do that?
limits derivatives asymptotics zeta-functions
asked Jan 2 at 16:26
Hercules III
1
New contributor
Hercules III is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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put on hold as off-topic by Adrian Keister, amWhy, Davide Giraudo, A. Pongrácz, José Carlos Santos Jan 3 at 0:03
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Adrian Keister, amWhy, Davide Giraudo, A. Pongrácz, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
-4
Derivative of Riemann zeta, is this inequality true?
In the fifth stage, he got this:
lim h→0 {π/2*tan(π2(2n+1+h))−ζ′(−2n−h)/ζ(−2n−h)}
And shifted the equation to that:
lim h→0 {−1/h+O(h)+1/h−ζ′′(−2n)/2ζ′(−2n)+O(h)}
How did he do that?
limits derivatives asymptotics zeta-functions
asked Jan 2 at 16:26
Hercules III
1
New contributor
Hercules III is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by Adrian Keister, amWhy, Davide Giraudo, A. Pongrácz, José Carlos Santos Jan 3 at 0:03
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Adrian Keister, amWhy, Davide Giraudo, A. Pongrácz, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
1
This will be much more readable if you use MathJax formatting. See here: math.meta.stackexchange.com/questions/5020/…
– postmortes
Jan 2 at 16:31
I don't know how to write with this web's language, enter the link I have attached
– Hercules III
Jan 2 at 17:04
Do you know how to evaluate $lim_{hto 0} fracpi2tanleft(frac{pi}2 (2n+1+h)right) + frac{1}{h}=A$ ? en.wikipedia.org/wiki/… And $lim_{hto 0} -frac{1}{h}-frac{zeta'(-2n-h)}{zeta(-2n-h)}=B $. Then $fracpi2tanleft(frac{pi}2 (2n+1+h)right)-frac{zeta'(-2n-h)}{zeta(-2n-h)} = A+B +O(h)$.
– reuns
Jan 2 at 19:21
@reuns why did you add +1/h and -1/h afterwards? Why did you add O(h) at the end?
– Hercules III
Jan 3 at 5:53
??????????????????????????
– Hercules III
Jan 3 at 19:30
add a comment |
-4
-4
-4
Derivative of Riemann zeta, is this inequality true?
In the fifth stage, he got this:
lim h→0 {π/2*tan(π2(2n+1+h))−ζ′(−2n−h)/ζ(−2n−h)}
And shifted the equation to that:
lim h→0 {−1/h+O(h)+1/h−ζ′′(−2n)/2ζ′(−2n)+O(h)}
How did he do that?
limits derivatives asymptotics zeta-functions
asked Jan 2 at 16:26
Hercules III
1
New contributor
Hercules III is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Derivative of Riemann zeta, is this inequality true?
In the fifth stage, he got this:
lim h→0 {π/2*tan(π2(2n+1+h))−ζ′(−2n−h)/ζ(−2n−h)}
And shifted the equation to that:
lim h→0 {−1/h+O(h)+1/h−ζ′′(−2n)/2ζ′(−2n)+O(h)}
How did he do that?
limits derivatives asymptotics zeta-functions
limits derivatives asymptotics zeta-functions
asked Jan 2 at 16:26
Hercules III
1
New contributor
Hercules III is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Jan 2 at 16:26
Hercules III
1
New contributor
Hercules III is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Jan 2 at 16:26
Hercules III
1
New contributor
Hercules III is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Jan 2 at 16:26
Hercules III
1
asked Jan 2 at 16:26
Hercules III
1
1
New contributor
Hercules III is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Hercules III is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Hercules III is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by Adrian Keister, amWhy, Davide Giraudo, A. Pongrácz, José Carlos Santos Jan 3 at 0:03
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Adrian Keister, amWhy, Davide Giraudo, A. Pongrácz, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as off-topic by Adrian Keister, amWhy, Davide Giraudo, A. Pongrácz, José Carlos Santos Jan 3 at 0:03
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Adrian Keister, amWhy, Davide Giraudo, A. Pongrácz, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
1
This will be much more readable if you use MathJax formatting. See here: math.meta.stackexchange.com/questions/5020/…
– postmortes
Jan 2 at 16:31
I don't know how to write with this web's language, enter the link I have attached
– Hercules III
Jan 2 at 17:04
Do you know how to evaluate $lim_{hto 0} fracpi2tanleft(frac{pi}2 (2n+1+h)right) + frac{1}{h}=A$ ? en.wikipedia.org/wiki/… And $lim_{hto 0} -frac{1}{h}-frac{zeta'(-2n-h)}{zeta(-2n-h)}=B $. Then $fracpi2tanleft(frac{pi}2 (2n+1+h)right)-frac{zeta'(-2n-h)}{zeta(-2n-h)} = A+B +O(h)$.
– reuns
Jan 2 at 19:21
@reuns why did you add +1/h and -1/h afterwards? Why did you add O(h) at the end?
– Hercules III
Jan 3 at 5:53
??????????????????????????
– Hercules III
Jan 3 at 19:30
add a comment |
1
This will be much more readable if you use MathJax formatting. See here: math.meta.stackexchange.com/questions/5020/…
– postmortes
Jan 2 at 16:31
I don't know how to write with this web's language, enter the link I have attached
– Hercules III
Jan 2 at 17:04
Do you know how to evaluate $lim_{hto 0} fracpi2tanleft(frac{pi}2 (2n+1+h)right) + frac{1}{h}=A$ ? en.wikipedia.org/wiki/… And $lim_{hto 0} -frac{1}{h}-frac{zeta'(-2n-h)}{zeta(-2n-h)}=B $. Then $fracpi2tanleft(frac{pi}2 (2n+1+h)right)-frac{zeta'(-2n-h)}{zeta(-2n-h)} = A+B +O(h)$.
– reuns
Jan 2 at 19:21
@reuns why did you add +1/h and -1/h afterwards? Why did you add O(h) at the end?
– Hercules III
Jan 3 at 5:53
??????????????????????????
– Hercules III
Jan 3 at 19:30
1
1
This will be much more readable if you use MathJax formatting. See here: math.meta.stackexchange.com/questions/5020/…
– postmortes
Jan 2 at 16:31
This will be much more readable if you use MathJax formatting. See here: math.meta.stackexchange.com/questions/5020/…
– postmortes
Jan 2 at 16:31
I don't know how to write with this web's language, enter the link I have attached
– Hercules III
Jan 2 at 17:04
I don't know how to write with this web's language, enter the link I have attached
– Hercules III
Jan 2 at 17:04
Do you know how to evaluate $lim_{hto 0} fracpi2tanleft(frac{pi}2 (2n+1+h)right) + frac{1}{h}=A$ ? en.wikipedia.org/wiki/… And $lim_{hto 0} -frac{1}{h}-frac{zeta'(-2n-h)}{zeta(-2n-h)}=B $. Then $fracpi2tanleft(frac{pi}2 (2n+1+h)right)-frac{zeta'(-2n-h)}{zeta(-2n-h)} = A+B +O(h)$.
– reuns
Jan 2 at 19:21
Do you know how to evaluate $lim_{hto 0} fracpi2tanleft(frac{pi}2 (2n+1+h)right) + frac{1}{h}=A$ ? en.wikipedia.org/wiki/… And $lim_{hto 0} -frac{1}{h}-frac{zeta'(-2n-h)}{zeta(-2n-h)}=B $. Then $fracpi2tanleft(frac{pi}2 (2n+1+h)right)-frac{zeta'(-2n-h)}{zeta(-2n-h)} = A+B +O(h)$.
– reuns
Jan 2 at 19:21
@reuns why did you add +1/h and -1/h afterwards? Why did you add O(h) at the end?
– Hercules III
Jan 3 at 5:53
@reuns why did you add +1/h and -1/h afterwards? Why did you add O(h) at the end?
– Hercules III
Jan 3 at 5:53
??????????????????????????
– Hercules III
Jan 3 at 19:30
??????????????????????????
– Hercules III
Jan 3 at 19:30
add a comment |
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1
This will be much more readable if you use MathJax formatting. See here: math.meta.stackexchange.com/questions/5020/…
– postmortes
Jan 2 at 16:31
I don't know how to write with this web's language, enter the link I have attached
– Hercules III
Jan 2 at 17:04
Do you know how to evaluate $lim_{hto 0} fracpi2tanleft(frac{pi}2 (2n+1+h)right) + frac{1}{h}=A$ ? en.wikipedia.org/wiki/… And $lim_{hto 0} -frac{1}{h}-frac{zeta'(-2n-h)}{zeta(-2n-h)}=B $. Then $fracpi2tanleft(frac{pi}2 (2n+1+h)right)-frac{zeta'(-2n-h)}{zeta(-2n-h)} = A+B +O(h)$.
– reuns
Jan 2 at 19:21
@reuns why did you add +1/h and -1/h afterwards? Why did you add O(h) at the end?
– Hercules III
Jan 3 at 5:53
??????????????????????????
– Hercules III
Jan 3 at 19:30