How to prove $sum_{dmid q}frac{mu(d)log d}{d}=-frac{phi(q)}{q}sum_{pmid q}frac{log p}{p-1}$?
Prove that $$sum_{dmid q}frac{mu(d)log d}{d}=-frac{phi(q)}{q}sum_{pmid q}frac{log p}{p-1},$$
where $mu$ is Möbius function, $phi$ is Euler's totient function, and $q$ is a positive integer.
I can get
begin{align}
sum_{dmid q} frac{mu(d)log d}{d}& = sum_{dmid q}frac{mu(d)}{d}sum_{pmid d}log p \
& = sum_{pmid q} log p sum_{substack{dmid q \ pmid d}} frac{mu(d)}{d}
= sum_{pmid q} log p sum_{substack{d \ pmid d mid q}} frac{mu(d)}{d},
end{align}
Let $d=pr$, then $mu(d)=mu(p)mu(r)=-mu(r)$,
$$ sum_{pmid q} log p sum_{substack{d \ pmid d mid q}} frac{mu(d)}{d}= - sum_{pmid q} frac{log p}{p} sum_{substack{rmid q \ p nmid r}} frac{mu(r)}{r}.$$
But I don't know why
$$- sum_{pmid q} frac{log p}{p} sum_{substack{rmid q \ p nmid r}} frac{mu(r)}{r}=-frac{phi(q)}{q} sum_{pmid q} frac{log p}{p-1}?$$
Can you help me?
elementary-number-theory arithmetic-functions
add a comment |
Prove that $$sum_{dmid q}frac{mu(d)log d}{d}=-frac{phi(q)}{q}sum_{pmid q}frac{log p}{p-1},$$
where $mu$ is Möbius function, $phi$ is Euler's totient function, and $q$ is a positive integer.
I can get
begin{align}
sum_{dmid q} frac{mu(d)log d}{d}& = sum_{dmid q}frac{mu(d)}{d}sum_{pmid d}log p \
& = sum_{pmid q} log p sum_{substack{dmid q \ pmid d}} frac{mu(d)}{d}
= sum_{pmid q} log p sum_{substack{d \ pmid d mid q}} frac{mu(d)}{d},
end{align}
Let $d=pr$, then $mu(d)=mu(p)mu(r)=-mu(r)$,
$$ sum_{pmid q} log p sum_{substack{d \ pmid d mid q}} frac{mu(d)}{d}= - sum_{pmid q} frac{log p}{p} sum_{substack{rmid q \ p nmid r}} frac{mu(r)}{r}.$$
But I don't know why
$$- sum_{pmid q} frac{log p}{p} sum_{substack{rmid q \ p nmid r}} frac{mu(r)}{r}=-frac{phi(q)}{q} sum_{pmid q} frac{log p}{p-1}?$$
Can you help me?
elementary-number-theory arithmetic-functions
Can't you just prove the original statement when $q$ is a prime power and appeal to multiplicativity?
– mathworker21
Nov 12 at 5:40
1
@mathworker21 To obtain something multiplicative it would help to replace $log(d)$ by $d^{-s}$ and differentiate (at $s=0$) at the end. The RHS is the derivative of $prod_{p | q} (1-p^{-(s+1)})$
– reuns
Nov 12 at 6:15
Just to make sure, does the sum over $p$ only involve prime values of $p$?
– Batominovski
Nov 12 at 10:28
add a comment |
Prove that $$sum_{dmid q}frac{mu(d)log d}{d}=-frac{phi(q)}{q}sum_{pmid q}frac{log p}{p-1},$$
where $mu$ is Möbius function, $phi$ is Euler's totient function, and $q$ is a positive integer.
I can get
begin{align}
sum_{dmid q} frac{mu(d)log d}{d}& = sum_{dmid q}frac{mu(d)}{d}sum_{pmid d}log p \
& = sum_{pmid q} log p sum_{substack{dmid q \ pmid d}} frac{mu(d)}{d}
= sum_{pmid q} log p sum_{substack{d \ pmid d mid q}} frac{mu(d)}{d},
end{align}
Let $d=pr$, then $mu(d)=mu(p)mu(r)=-mu(r)$,
$$ sum_{pmid q} log p sum_{substack{d \ pmid d mid q}} frac{mu(d)}{d}= - sum_{pmid q} frac{log p}{p} sum_{substack{rmid q \ p nmid r}} frac{mu(r)}{r}.$$
But I don't know why
$$- sum_{pmid q} frac{log p}{p} sum_{substack{rmid q \ p nmid r}} frac{mu(r)}{r}=-frac{phi(q)}{q} sum_{pmid q} frac{log p}{p-1}?$$
Can you help me?
elementary-number-theory arithmetic-functions
Prove that $$sum_{dmid q}frac{mu(d)log d}{d}=-frac{phi(q)}{q}sum_{pmid q}frac{log p}{p-1},$$
where $mu$ is Möbius function, $phi$ is Euler's totient function, and $q$ is a positive integer.
I can get
begin{align}
sum_{dmid q} frac{mu(d)log d}{d}& = sum_{dmid q}frac{mu(d)}{d}sum_{pmid d}log p \
& = sum_{pmid q} log p sum_{substack{dmid q \ pmid d}} frac{mu(d)}{d}
= sum_{pmid q} log p sum_{substack{d \ pmid d mid q}} frac{mu(d)}{d},
end{align}
Let $d=pr$, then $mu(d)=mu(p)mu(r)=-mu(r)$,
$$ sum_{pmid q} log p sum_{substack{d \ pmid d mid q}} frac{mu(d)}{d}= - sum_{pmid q} frac{log p}{p} sum_{substack{rmid q \ p nmid r}} frac{mu(r)}{r}.$$
But I don't know why
$$- sum_{pmid q} frac{log p}{p} sum_{substack{rmid q \ p nmid r}} frac{mu(r)}{r}=-frac{phi(q)}{q} sum_{pmid q} frac{log p}{p-1}?$$
Can you help me?
elementary-number-theory arithmetic-functions
elementary-number-theory arithmetic-functions
edited Nov 12 at 5:45
asked Nov 12 at 5:23
arithmetic1
456
456
Can't you just prove the original statement when $q$ is a prime power and appeal to multiplicativity?
– mathworker21
Nov 12 at 5:40
1
@mathworker21 To obtain something multiplicative it would help to replace $log(d)$ by $d^{-s}$ and differentiate (at $s=0$) at the end. The RHS is the derivative of $prod_{p | q} (1-p^{-(s+1)})$
– reuns
Nov 12 at 6:15
Just to make sure, does the sum over $p$ only involve prime values of $p$?
– Batominovski
Nov 12 at 10:28
add a comment |
Can't you just prove the original statement when $q$ is a prime power and appeal to multiplicativity?
– mathworker21
Nov 12 at 5:40
1
@mathworker21 To obtain something multiplicative it would help to replace $log(d)$ by $d^{-s}$ and differentiate (at $s=0$) at the end. The RHS is the derivative of $prod_{p | q} (1-p^{-(s+1)})$
– reuns
Nov 12 at 6:15
Just to make sure, does the sum over $p$ only involve prime values of $p$?
– Batominovski
Nov 12 at 10:28
Can't you just prove the original statement when $q$ is a prime power and appeal to multiplicativity?
– mathworker21
Nov 12 at 5:40
Can't you just prove the original statement when $q$ is a prime power and appeal to multiplicativity?
– mathworker21
Nov 12 at 5:40
1
1
@mathworker21 To obtain something multiplicative it would help to replace $log(d)$ by $d^{-s}$ and differentiate (at $s=0$) at the end. The RHS is the derivative of $prod_{p | q} (1-p^{-(s+1)})$
– reuns
Nov 12 at 6:15
@mathworker21 To obtain something multiplicative it would help to replace $log(d)$ by $d^{-s}$ and differentiate (at $s=0$) at the end. The RHS is the derivative of $prod_{p | q} (1-p^{-(s+1)})$
– reuns
Nov 12 at 6:15
Just to make sure, does the sum over $p$ only involve prime values of $p$?
– Batominovski
Nov 12 at 10:28
Just to make sure, does the sum over $p$ only involve prime values of $p$?
– Batominovski
Nov 12 at 10:28
add a comment |
2 Answers
2
active
oldest
votes
Let me write $n$ instead of $q$.
We have
begin{align}
sum_{d|n}frac{mu(d)log(d)}d
&=sum_{d|n}frac{mu(d)}dsum_{p|d}log(p)\
&=sum_{p|n}log(p)sum_{p|d|n}frac{mu(d)}d\
&=frac 1nsum_{p|n}log(p)sum_{p|d|n}mu(d)frac nd
end{align}
Write $n=p^em$ with $pnmid m$.
Then $varphi(n)=p^{e-1}(p-1)varphi(m)$ and
begin{align}
sum_{p|d|n}mu(d)frac nd
&=sum_{dmid m}sum_{i=1}^emu(p^id)frac{p^em}{p^id}\
&=sum_{dmid m}mu(pd)frac{p^em}{pd}\
&=-sum_{dmid m}mu(d)frac{p^em}{pd}\
&=-p^{e-1}varphi(m)\
&=-frac{varphi(n)}{p-1}
end{align}
Thank you very much!
– arithmetic1
Nov 28 at 8:28
add a comment |
I find a paper "On some identities in multiplicative number theory", Olivier Bordellès and Benoit Cloitre, arXiv:1804.05332v2 https://arxiv.org/abs/1804.05332v2
Using Dirichlet convolution
begin{eqnarray*}
- frac{varphi(n)}{n} sum_{p mid n} frac{log p}{p-1} &=& - frac{1}{n} sum_{p mid n} varphi left( frac{n}{p} right) log p \
&=& - frac{1}{n} left( Lambda ast varphi right) (n) \
&=& - frac{1}{n} left( - mu log ast mathbf{1} ast mu ast mathrm{id} right) (n) \
&=& frac{1}{n} left( mu log ast mathrm{id} right) (n) \
&=& sum_{d mid n} frac{mu(d) log d}{d}.
end{eqnarray*}
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2994900%2fhow-to-prove-sum-d-mid-q-frac-mud-log-dd-frac-phiqq-sum-p-mid%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Let me write $n$ instead of $q$.
We have
begin{align}
sum_{d|n}frac{mu(d)log(d)}d
&=sum_{d|n}frac{mu(d)}dsum_{p|d}log(p)\
&=sum_{p|n}log(p)sum_{p|d|n}frac{mu(d)}d\
&=frac 1nsum_{p|n}log(p)sum_{p|d|n}mu(d)frac nd
end{align}
Write $n=p^em$ with $pnmid m$.
Then $varphi(n)=p^{e-1}(p-1)varphi(m)$ and
begin{align}
sum_{p|d|n}mu(d)frac nd
&=sum_{dmid m}sum_{i=1}^emu(p^id)frac{p^em}{p^id}\
&=sum_{dmid m}mu(pd)frac{p^em}{pd}\
&=-sum_{dmid m}mu(d)frac{p^em}{pd}\
&=-p^{e-1}varphi(m)\
&=-frac{varphi(n)}{p-1}
end{align}
Thank you very much!
– arithmetic1
Nov 28 at 8:28
add a comment |
Let me write $n$ instead of $q$.
We have
begin{align}
sum_{d|n}frac{mu(d)log(d)}d
&=sum_{d|n}frac{mu(d)}dsum_{p|d}log(p)\
&=sum_{p|n}log(p)sum_{p|d|n}frac{mu(d)}d\
&=frac 1nsum_{p|n}log(p)sum_{p|d|n}mu(d)frac nd
end{align}
Write $n=p^em$ with $pnmid m$.
Then $varphi(n)=p^{e-1}(p-1)varphi(m)$ and
begin{align}
sum_{p|d|n}mu(d)frac nd
&=sum_{dmid m}sum_{i=1}^emu(p^id)frac{p^em}{p^id}\
&=sum_{dmid m}mu(pd)frac{p^em}{pd}\
&=-sum_{dmid m}mu(d)frac{p^em}{pd}\
&=-p^{e-1}varphi(m)\
&=-frac{varphi(n)}{p-1}
end{align}
Thank you very much!
– arithmetic1
Nov 28 at 8:28
add a comment |
Let me write $n$ instead of $q$.
We have
begin{align}
sum_{d|n}frac{mu(d)log(d)}d
&=sum_{d|n}frac{mu(d)}dsum_{p|d}log(p)\
&=sum_{p|n}log(p)sum_{p|d|n}frac{mu(d)}d\
&=frac 1nsum_{p|n}log(p)sum_{p|d|n}mu(d)frac nd
end{align}
Write $n=p^em$ with $pnmid m$.
Then $varphi(n)=p^{e-1}(p-1)varphi(m)$ and
begin{align}
sum_{p|d|n}mu(d)frac nd
&=sum_{dmid m}sum_{i=1}^emu(p^id)frac{p^em}{p^id}\
&=sum_{dmid m}mu(pd)frac{p^em}{pd}\
&=-sum_{dmid m}mu(d)frac{p^em}{pd}\
&=-p^{e-1}varphi(m)\
&=-frac{varphi(n)}{p-1}
end{align}
Let me write $n$ instead of $q$.
We have
begin{align}
sum_{d|n}frac{mu(d)log(d)}d
&=sum_{d|n}frac{mu(d)}dsum_{p|d}log(p)\
&=sum_{p|n}log(p)sum_{p|d|n}frac{mu(d)}d\
&=frac 1nsum_{p|n}log(p)sum_{p|d|n}mu(d)frac nd
end{align}
Write $n=p^em$ with $pnmid m$.
Then $varphi(n)=p^{e-1}(p-1)varphi(m)$ and
begin{align}
sum_{p|d|n}mu(d)frac nd
&=sum_{dmid m}sum_{i=1}^emu(p^id)frac{p^em}{p^id}\
&=sum_{dmid m}mu(pd)frac{p^em}{pd}\
&=-sum_{dmid m}mu(d)frac{p^em}{pd}\
&=-p^{e-1}varphi(m)\
&=-frac{varphi(n)}{p-1}
end{align}
edited Nov 12 at 9:24
answered Nov 12 at 9:06
Fabio Lucchini
7,82311326
7,82311326
Thank you very much!
– arithmetic1
Nov 28 at 8:28
add a comment |
Thank you very much!
– arithmetic1
Nov 28 at 8:28
Thank you very much!
– arithmetic1
Nov 28 at 8:28
Thank you very much!
– arithmetic1
Nov 28 at 8:28
add a comment |
I find a paper "On some identities in multiplicative number theory", Olivier Bordellès and Benoit Cloitre, arXiv:1804.05332v2 https://arxiv.org/abs/1804.05332v2
Using Dirichlet convolution
begin{eqnarray*}
- frac{varphi(n)}{n} sum_{p mid n} frac{log p}{p-1} &=& - frac{1}{n} sum_{p mid n} varphi left( frac{n}{p} right) log p \
&=& - frac{1}{n} left( Lambda ast varphi right) (n) \
&=& - frac{1}{n} left( - mu log ast mathbf{1} ast mu ast mathrm{id} right) (n) \
&=& frac{1}{n} left( mu log ast mathrm{id} right) (n) \
&=& sum_{d mid n} frac{mu(d) log d}{d}.
end{eqnarray*}
add a comment |
I find a paper "On some identities in multiplicative number theory", Olivier Bordellès and Benoit Cloitre, arXiv:1804.05332v2 https://arxiv.org/abs/1804.05332v2
Using Dirichlet convolution
begin{eqnarray*}
- frac{varphi(n)}{n} sum_{p mid n} frac{log p}{p-1} &=& - frac{1}{n} sum_{p mid n} varphi left( frac{n}{p} right) log p \
&=& - frac{1}{n} left( Lambda ast varphi right) (n) \
&=& - frac{1}{n} left( - mu log ast mathbf{1} ast mu ast mathrm{id} right) (n) \
&=& frac{1}{n} left( mu log ast mathrm{id} right) (n) \
&=& sum_{d mid n} frac{mu(d) log d}{d}.
end{eqnarray*}
add a comment |
I find a paper "On some identities in multiplicative number theory", Olivier Bordellès and Benoit Cloitre, arXiv:1804.05332v2 https://arxiv.org/abs/1804.05332v2
Using Dirichlet convolution
begin{eqnarray*}
- frac{varphi(n)}{n} sum_{p mid n} frac{log p}{p-1} &=& - frac{1}{n} sum_{p mid n} varphi left( frac{n}{p} right) log p \
&=& - frac{1}{n} left( Lambda ast varphi right) (n) \
&=& - frac{1}{n} left( - mu log ast mathbf{1} ast mu ast mathrm{id} right) (n) \
&=& frac{1}{n} left( mu log ast mathrm{id} right) (n) \
&=& sum_{d mid n} frac{mu(d) log d}{d}.
end{eqnarray*}
I find a paper "On some identities in multiplicative number theory", Olivier Bordellès and Benoit Cloitre, arXiv:1804.05332v2 https://arxiv.org/abs/1804.05332v2
Using Dirichlet convolution
begin{eqnarray*}
- frac{varphi(n)}{n} sum_{p mid n} frac{log p}{p-1} &=& - frac{1}{n} sum_{p mid n} varphi left( frac{n}{p} right) log p \
&=& - frac{1}{n} left( Lambda ast varphi right) (n) \
&=& - frac{1}{n} left( - mu log ast mathbf{1} ast mu ast mathrm{id} right) (n) \
&=& frac{1}{n} left( mu log ast mathrm{id} right) (n) \
&=& sum_{d mid n} frac{mu(d) log d}{d}.
end{eqnarray*}
answered Nov 28 at 8:26
arithmetic1
456
456
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2994900%2fhow-to-prove-sum-d-mid-q-frac-mud-log-dd-frac-phiqq-sum-p-mid%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Can't you just prove the original statement when $q$ is a prime power and appeal to multiplicativity?
– mathworker21
Nov 12 at 5:40
1
@mathworker21 To obtain something multiplicative it would help to replace $log(d)$ by $d^{-s}$ and differentiate (at $s=0$) at the end. The RHS is the derivative of $prod_{p | q} (1-p^{-(s+1)})$
– reuns
Nov 12 at 6:15
Just to make sure, does the sum over $p$ only involve prime values of $p$?
– Batominovski
Nov 12 at 10:28