How can I have x-axis ticks that show ticks scaled in powers of ten?
4
$begingroup$
I am having trouble achieving this effect with on the x-axis. Does anybody have an idea how I can achieve this effect?
plotting
edited yesterday
m_goldberg
87.4k872198
asked yesterday
Kanye WestKanye West
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4
$begingroup$
I am having trouble achieving this effect with on the x-axis. Does anybody have an idea how I can achieve this effect?
plotting
edited yesterday
m_goldberg
87.4k872198
asked yesterday
Kanye WestKanye West
292
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Kanye West is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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2
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As a slight correction to kglr's comment: the "number theory" version of the logarithmic integral is expressed asLogIntegral[x] - LogIntegral[2].
$endgroup$
– J. M. is computer-less♦
yesterday
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Welcome to Mathematica.SE, Kanye! I suggest the following: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Take the tour and check the faqs! 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign!
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– Chris K
19 hours ago
add a comment |
4
4
4
$begingroup$
I am having trouble achieving this effect with on the x-axis. Does anybody have an idea how I can achieve this effect?
plotting
edited yesterday
m_goldberg
87.4k872198
asked yesterday
Kanye WestKanye West
292
New contributor
Kanye West is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I am having trouble achieving this effect with on the x-axis. Does anybody have an idea how I can achieve this effect?
plotting
plotting
edited yesterday
m_goldberg
87.4k872198
asked yesterday
Kanye WestKanye West
292
New contributor
Kanye West is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
m_goldberg
87.4k872198
asked yesterday
Kanye WestKanye West
292
New contributor
Kanye West is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited yesterday
m_goldberg
87.4k872198
edited yesterday
m_goldberg
87.4k872198
edited yesterday
m_goldberg
87.4k872198
87.4k872198
asked yesterday
Kanye WestKanye West
292
New contributor
Kanye West is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
Kanye WestKanye West
292
asked yesterday
Kanye WestKanye West
292
292
New contributor
Kanye West is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Kanye West is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Kanye West is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
2
$begingroup$
As a slight correction to kglr's comment: the "number theory" version of the logarithmic integral is expressed asLogIntegral[x] - LogIntegral[2].
$endgroup$
– J. M. is computer-less♦
yesterday
$begingroup$
Welcome to Mathematica.SE, Kanye! I suggest the following: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Take the tour and check the faqs! 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign!
$endgroup$
– Chris K
19 hours ago
add a comment |
2
$begingroup$
As a slight correction to kglr's comment: the "number theory" version of the logarithmic integral is expressed asLogIntegral[x] - LogIntegral[2].
$endgroup$
– J. M. is computer-less♦
yesterday
$begingroup$
Welcome to Mathematica.SE, Kanye! I suggest the following: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Take the tour and check the faqs! 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign!
$endgroup$
– Chris K
19 hours ago
2
2
$begingroup$
As a slight correction to kglr's comment: the "number theory" version of the logarithmic integral is expressed as
LogIntegral[x] - LogIntegral[2].$endgroup$
– J. M. is computer-less♦
yesterday
$begingroup$
As a slight correction to kglr's comment: the "number theory" version of the logarithmic integral is expressed as
LogIntegral[x] - LogIntegral[2].$endgroup$
– J. M. is computer-less♦
yesterday
$begingroup$
Welcome to Mathematica.SE, Kanye! I suggest the following: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Take the tour and check the faqs! 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign!
$endgroup$
– Chris K
19 hours ago
$begingroup$
Welcome to Mathematica.SE, Kanye! I suggest the following: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Take the tour and check the faqs! 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign!
$endgroup$
– Chris K
19 hours ago
add a comment |
1 Answer
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LogLinearPlot[{PrimePi[x]/(LogIntegral[x]-LogIntegral[2]),
PrimePi[x]/(x/Log[x])}, {x, 2, 10000000}, GridLines -> {None, {1}}]
Thanks: @J.M.iscomputer-less for the LogIntegral[2] correction.
answered yesterday
kglrkglr
188k10203421
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1 Answer
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1 Answer
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9
$begingroup$
LogLinearPlot[{PrimePi[x]/(LogIntegral[x]-LogIntegral[2]),
PrimePi[x]/(x/Log[x])}, {x, 2, 10000000}, GridLines -> {None, {1}}]
Thanks: @J.M.iscomputer-less for the LogIntegral[2] correction.
answered yesterday
kglrkglr
188k10203421
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add a comment |
9
$begingroup$
LogLinearPlot[{PrimePi[x]/(LogIntegral[x]-LogIntegral[2]),
PrimePi[x]/(x/Log[x])}, {x, 2, 10000000}, GridLines -> {None, {1}}]
Thanks: @J.M.iscomputer-less for the LogIntegral[2] correction.
answered yesterday
kglrkglr
188k10203421
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9
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$begingroup$
LogLinearPlot[{PrimePi[x]/(LogIntegral[x]-LogIntegral[2]),
PrimePi[x]/(x/Log[x])}, {x, 2, 10000000}, GridLines -> {None, {1}}]
Thanks: @J.M.iscomputer-less for the LogIntegral[2] correction.
answered yesterday
kglrkglr
188k10203421
$endgroup$
LogLinearPlot[{PrimePi[x]/(LogIntegral[x]-LogIntegral[2]),
PrimePi[x]/(x/Log[x])}, {x, 2, 10000000}, GridLines -> {None, {1}}]
Thanks: @J.M.iscomputer-less for the LogIntegral[2] correction.
answered yesterday
kglrkglr
188k10203421
edited yesterday
answered yesterday
kglrkglr
188k10203421
answered yesterday
kglrkglr
188k10203421
answered yesterday
kglrkglr
188k10203421
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2
$begingroup$
As a slight correction to kglr's comment: the "number theory" version of the logarithmic integral is expressed as
LogIntegral[x] - LogIntegral[2].$endgroup$
– J. M. is computer-less♦
yesterday
$begingroup$
Welcome to Mathematica.SE, Kanye! I suggest the following: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Take the tour and check the faqs! 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign!
$endgroup$
– Chris K
19 hours ago