Showing $sum_dmid n mu(d)tau(n/d)=1$ and $sum_dmid n mu(d)tau(d)=(-1)^r$ [closed] Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Prove $sum_kmid nmu(k)d(k)=(-1)^omega(n)$Sum of Positive Divisors: $sum_n mu(n/d)nu(d)=1$ and $sum_n mu(n/d)sigma(d)=n$An application of Mobius Inversion $sum_d mid n mu(fracnd)nu(d) = 1$Number theory Exercise: $sum_d mid n mu(d) d(d) = (-1)^omega(n)$ and $sum_d mid n mu(d) sigma (d)$Prove that $sum_d = 2^omega(n)$Evaluating a sum with Mobius function $sum_ntau(d)mu(d)$Prove $sum_kmid nmu(k)d(k)=(-1)^omega(n)$Norm of Prime IdealDemostrate that: $ tau(n)varphi(n) ≥ n$$tau(n) = 1992$ and $phi(n) mid n$An application of Mobius Inversion $sum_d mid n mu(fracnd)nu(d) = 1$Writing as a Product of $k$ factorsDecomposing Dirichlet charcatersShow the following equality involving $tau$, the number of divisors of the input.Prove that the two sums involving $tau$ are equal.A bound on number of divisors of $n$
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Showing $sum_dmid n mu(d)tau(n/d)=1$ and $sum_dmid n mu(d)tau(d)=(-1)^r$ [closed]
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Prove $sum_kmid nmu(k)d(k)=(-1)^omega(n)$Sum of Positive Divisors: $sum_n mu(n/d)nu(d)=1$ and $sum_n mu(n/d)sigma(d)=n$An application of Mobius Inversion $sum_d mid n mu(fracnd)nu(d) = 1$Number theory Exercise: $sum_d mid n mu(d) d(d) = (-1)^omega(n)$ and $sum_d mid n mu(d) sigma (d)$Prove that $sum_d = 2^omega(n)$Evaluating a sum with Mobius function $sum_ntau(d)mu(d)$Prove $sum_kmid nmu(k)d(k)=(-1)^omega(n)$Norm of Prime IdealDemostrate that: $ tau(n)varphi(n) ≥ n$$tau(n) = 1992$ and $phi(n) mid n$An application of Mobius Inversion $sum_d mid n mu(fracnd)nu(d) = 1$Writing as a Product of $k$ factorsDecomposing Dirichlet charcatersShow the following equality involving $tau$, the number of divisors of the input.Prove that the two sums involving $tau$ are equal.A bound on number of divisors of $n$
$begingroup$
Need some help on this question from Victor Shoup
Let $tau(n)$ be the number of positive divisors of $n$. Show that:
$sum_dmid n mu(d)tau(n/d)=1$;
$sum_dmid n mu(d)tau(d)=(-1)^r$, where $n=p_1^e_1cdots p_r^e_r$ is the prime factorization of $n$.
I have tried both of them but cant find any solution!We have to use Mobius Function properties to prove this Question.
number-theory mobius-function multiplicative-function divisor-counting-function mobius-inversion
edited Mar 27 at 2:41
Saad
20.6k92452
asked Jan 15 '17 at 10:05
da6932da6932
143
$endgroup$
closed as off-topic by Saad, Eevee Trainer, Theo Bendit, mrtaurho, Parcly Taxel Mar 28 at 0:47
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." – Saad, Eevee Trainer, Theo Bendit, mrtaurho, Parcly Taxel
add a comment |
$begingroup$
Need some help on this question from Victor Shoup
Let $tau(n)$ be the number of positive divisors of $n$. Show that:
$sum_dmid n mu(d)tau(n/d)=1$;
$sum_dmid n mu(d)tau(d)=(-1)^r$, where $n=p_1^e_1cdots p_r^e_r$ is the prime factorization of $n$.
I have tried both of them but cant find any solution!We have to use Mobius Function properties to prove this Question.
number-theory mobius-function multiplicative-function divisor-counting-function mobius-inversion
edited Mar 27 at 2:41
Saad
20.6k92452
asked Jan 15 '17 at 10:05
da6932da6932
143
$endgroup$
closed as off-topic by Saad, Eevee Trainer, Theo Bendit, mrtaurho, Parcly Taxel Mar 28 at 0:47
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." – Saad, Eevee Trainer, Theo Bendit, mrtaurho, Parcly Taxel
3
$begingroup$
Trying to search for the first one, I found these questions: An application of Mobius Inversion $sum_d mid n mu(fracnd)nu(d) = 1$ and Sum of Positive Divisors: $sum_n mu(n/d)nu(d)=1$ and $sum_n mu(n/d)sigma(d)=n$. Perhaps with a bit of searching you can find more copies of that question.
$endgroup$
– Martin Sleziak
Jan 15 '17 at 11:17
3
$begingroup$
Similarly, with a bit of searching, you will probably be able to find several posts about the other one. Like here or here. In fact, you do not need to search at all, just look at the list of related questions in the sidebar on the right to find this one.
$endgroup$
– Martin Sleziak
Jan 15 '17 at 11:22
$begingroup$
The sad thing is, the fact that it's two questions in one makes it difficult to mark it as a duplicate.
$endgroup$
– rabota
Jan 15 '17 at 20:12
$begingroup$
Forgive me this time . Will try to search from the next time , newbie mistake . Also thnx alot for citing the answers . :)
$endgroup$
– da6932
Jan 16 '17 at 19:19
add a comment |
$begingroup$
Need some help on this question from Victor Shoup
Let $tau(n)$ be the number of positive divisors of $n$. Show that:
$sum_dmid n mu(d)tau(n/d)=1$;
$sum_dmid n mu(d)tau(d)=(-1)^r$, where $n=p_1^e_1cdots p_r^e_r$ is the prime factorization of $n$.
I have tried both of them but cant find any solution!We have to use Mobius Function properties to prove this Question.
number-theory mobius-function multiplicative-function divisor-counting-function mobius-inversion
edited Mar 27 at 2:41
Saad
20.6k92452
asked Jan 15 '17 at 10:05
da6932da6932
143
$endgroup$
Need some help on this question from Victor Shoup
Let $tau(n)$ be the number of positive divisors of $n$. Show that:
$sum_dmid n mu(d)tau(n/d)=1$;
$sum_dmid n mu(d)tau(d)=(-1)^r$, where $n=p_1^e_1cdots p_r^e_r$ is the prime factorization of $n$.
I have tried both of them but cant find any solution!We have to use Mobius Function properties to prove this Question.
number-theory mobius-function multiplicative-function divisor-counting-function mobius-inversion
number-theory mobius-function multiplicative-function divisor-counting-function mobius-inversion
edited Mar 27 at 2:41
Saad
20.6k92452
asked Jan 15 '17 at 10:05
da6932da6932
143
edited Mar 27 at 2:41
Saad
20.6k92452
asked Jan 15 '17 at 10:05
da6932da6932
143
edited Mar 27 at 2:41
Saad
20.6k92452
edited Mar 27 at 2:41
Saad
20.6k92452
edited Mar 27 at 2:41
Saad
20.6k92452
20.6k92452
asked Jan 15 '17 at 10:05
da6932da6932
143
asked Jan 15 '17 at 10:05
da6932da6932
143
asked Jan 15 '17 at 10:05
da6932da6932
143
143
closed as off-topic by Saad, Eevee Trainer, Theo Bendit, mrtaurho, Parcly Taxel Mar 28 at 0:47
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." – Saad, Eevee Trainer, Theo Bendit, mrtaurho, Parcly Taxel
closed as off-topic by Saad, Eevee Trainer, Theo Bendit, mrtaurho, Parcly Taxel Mar 28 at 0:47
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." – Saad, Eevee Trainer, Theo Bendit, mrtaurho, Parcly Taxel
3
$begingroup$
Trying to search for the first one, I found these questions: An application of Mobius Inversion $sum_d mid n mu(fracnd)nu(d) = 1$ and Sum of Positive Divisors: $sum_n mu(n/d)nu(d)=1$ and $sum_n mu(n/d)sigma(d)=n$. Perhaps with a bit of searching you can find more copies of that question.
$endgroup$
– Martin Sleziak
Jan 15 '17 at 11:17
3
$begingroup$
Similarly, with a bit of searching, you will probably be able to find several posts about the other one. Like here or here. In fact, you do not need to search at all, just look at the list of related questions in the sidebar on the right to find this one.
$endgroup$
– Martin Sleziak
Jan 15 '17 at 11:22
$begingroup$
The sad thing is, the fact that it's two questions in one makes it difficult to mark it as a duplicate.
$endgroup$
– rabota
Jan 15 '17 at 20:12
$begingroup$
Forgive me this time . Will try to search from the next time , newbie mistake . Also thnx alot for citing the answers . :)
$endgroup$
– da6932
Jan 16 '17 at 19:19
add a comment |
3
$begingroup$
Trying to search for the first one, I found these questions: An application of Mobius Inversion $sum_d mid n mu(fracnd)nu(d) = 1$ and Sum of Positive Divisors: $sum_n mu(n/d)nu(d)=1$ and $sum_n mu(n/d)sigma(d)=n$. Perhaps with a bit of searching you can find more copies of that question.
$endgroup$
– Martin Sleziak
Jan 15 '17 at 11:17
3
$begingroup$
Similarly, with a bit of searching, you will probably be able to find several posts about the other one. Like here or here. In fact, you do not need to search at all, just look at the list of related questions in the sidebar on the right to find this one.
$endgroup$
– Martin Sleziak
Jan 15 '17 at 11:22
$begingroup$
The sad thing is, the fact that it's two questions in one makes it difficult to mark it as a duplicate.
$endgroup$
– rabota
Jan 15 '17 at 20:12
$begingroup$
Forgive me this time . Will try to search from the next time , newbie mistake . Also thnx alot for citing the answers . :)
$endgroup$
– da6932
Jan 16 '17 at 19:19
3
3
$begingroup$
Trying to search for the first one, I found these questions: An application of Mobius Inversion $sum_d mid n mu(fracnd)nu(d) = 1$ and Sum of Positive Divisors: $sum_n mu(n/d)nu(d)=1$ and $sum_n mu(n/d)sigma(d)=n$. Perhaps with a bit of searching you can find more copies of that question.
$endgroup$
– Martin Sleziak
Jan 15 '17 at 11:17
$begingroup$
Trying to search for the first one, I found these questions: An application of Mobius Inversion $sum_d mid n mu(fracnd)nu(d) = 1$ and Sum of Positive Divisors: $sum_n mu(n/d)nu(d)=1$ and $sum_n mu(n/d)sigma(d)=n$. Perhaps with a bit of searching you can find more copies of that question.
$endgroup$
– Martin Sleziak
Jan 15 '17 at 11:17
3
3
$begingroup$
Similarly, with a bit of searching, you will probably be able to find several posts about the other one. Like here or here. In fact, you do not need to search at all, just look at the list of related questions in the sidebar on the right to find this one.
$endgroup$
– Martin Sleziak
Jan 15 '17 at 11:22
$begingroup$
Similarly, with a bit of searching, you will probably be able to find several posts about the other one. Like here or here. In fact, you do not need to search at all, just look at the list of related questions in the sidebar on the right to find this one.
$endgroup$
– Martin Sleziak
Jan 15 '17 at 11:22
$begingroup$
The sad thing is, the fact that it's two questions in one makes it difficult to mark it as a duplicate.
$endgroup$
– rabota
Jan 15 '17 at 20:12
$begingroup$
The sad thing is, the fact that it's two questions in one makes it difficult to mark it as a duplicate.
$endgroup$
– rabota
Jan 15 '17 at 20:12
$begingroup$
Forgive me this time . Will try to search from the next time , newbie mistake . Also thnx alot for citing the answers . :)
$endgroup$
– da6932
Jan 16 '17 at 19:19
$begingroup$
Forgive me this time . Will try to search from the next time , newbie mistake . Also thnx alot for citing the answers . :)
$endgroup$
– da6932
Jan 16 '17 at 19:19
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Using the fact that both $mu$ and $tau$ are multiplicative for the first one we get from first principles the value
$$tau(n) prod_q=1^r left(1+(-1)timesfrace_qe_q+1right)$$
which simplifies to
$$tau(n) prod_q=1^r frac1e_q+1 = 1.$$
For the second one we may write
$$prod_q=1^r (1+(-1)times 2) = (-1)^r.$$
Here we have used that for a subset $S$ of the set of prime factors $P$ corresponding to a squarefree divisor $d$ of $n$ we get $mu(d) = (-1)^$ and $tau(n/d) = tau(n) prod_p_qin S frace_qe_q+1.$
answered Jan 15 '17 at 23:29
Marko RiedelMarko Riedel
41.5k341112
$endgroup$
add a comment |
$begingroup$
1) $τ(n)= sum_n 1$ => According to the Moebius Inversion Formula $sum_n μ(d)τ(n/d) = 1$
2) If d has a squared prime factor, then μ(d)=0 => $sum_n μ(d)τ(d) = sum_rad(n) μ(d)τ(d)$.
If the number of prime factors of n is even, then μ(rad(n)/d)= μ(d).
If it is odd, then μ(rad(n)/d)= -μ(d) => $sum_n μ(d)τ(d) = sum_rad(n) μ(d)τ(d) = (-1)^r sum_rad(n) μ(rad(n)/d)τ(d)= (-1)^r sum_rad(n) μ(d)τ(rad(n)/d)=(-1)^r.$
answered Jan 15 '17 at 17:24
Yanior WegYanior Weg
2,85611547
$endgroup$
$begingroup$
Really nice observation for tau(n) , and indeed nice proof for first . Thnx .
$endgroup$
– da6932
Jan 16 '17 at 19:23
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Using the fact that both $mu$ and $tau$ are multiplicative for the first one we get from first principles the value
$$tau(n) prod_q=1^r left(1+(-1)timesfrace_qe_q+1right)$$
which simplifies to
$$tau(n) prod_q=1^r frac1e_q+1 = 1.$$
For the second one we may write
$$prod_q=1^r (1+(-1)times 2) = (-1)^r.$$
Here we have used that for a subset $S$ of the set of prime factors $P$ corresponding to a squarefree divisor $d$ of $n$ we get $mu(d) = (-1)^$ and $tau(n/d) = tau(n) prod_p_qin S frace_qe_q+1.$
answered Jan 15 '17 at 23:29
Marko RiedelMarko Riedel
41.5k341112
$endgroup$
add a comment |
$begingroup$
Using the fact that both $mu$ and $tau$ are multiplicative for the first one we get from first principles the value
$$tau(n) prod_q=1^r left(1+(-1)timesfrace_qe_q+1right)$$
which simplifies to
$$tau(n) prod_q=1^r frac1e_q+1 = 1.$$
For the second one we may write
$$prod_q=1^r (1+(-1)times 2) = (-1)^r.$$
Here we have used that for a subset $S$ of the set of prime factors $P$ corresponding to a squarefree divisor $d$ of $n$ we get $mu(d) = (-1)^$ and $tau(n/d) = tau(n) prod_p_qin S frace_qe_q+1.$
answered Jan 15 '17 at 23:29
Marko RiedelMarko Riedel
41.5k341112
$endgroup$
add a comment |
$begingroup$
Using the fact that both $mu$ and $tau$ are multiplicative for the first one we get from first principles the value
$$tau(n) prod_q=1^r left(1+(-1)timesfrace_qe_q+1right)$$
which simplifies to
$$tau(n) prod_q=1^r frac1e_q+1 = 1.$$
For the second one we may write
$$prod_q=1^r (1+(-1)times 2) = (-1)^r.$$
Here we have used that for a subset $S$ of the set of prime factors $P$ corresponding to a squarefree divisor $d$ of $n$ we get $mu(d) = (-1)^$ and $tau(n/d) = tau(n) prod_p_qin S frace_qe_q+1.$
answered Jan 15 '17 at 23:29
Marko RiedelMarko Riedel
41.5k341112
$endgroup$
Using the fact that both $mu$ and $tau$ are multiplicative for the first one we get from first principles the value
$$tau(n) prod_q=1^r left(1+(-1)timesfrace_qe_q+1right)$$
which simplifies to
$$tau(n) prod_q=1^r frac1e_q+1 = 1.$$
For the second one we may write
$$prod_q=1^r (1+(-1)times 2) = (-1)^r.$$
Here we have used that for a subset $S$ of the set of prime factors $P$ corresponding to a squarefree divisor $d$ of $n$ we get $mu(d) = (-1)^$ and $tau(n/d) = tau(n) prod_p_qin S frace_qe_q+1.$
answered Jan 15 '17 at 23:29
Marko RiedelMarko Riedel
41.5k341112
edited Jan 16 '17 at 21:08
answered Jan 15 '17 at 23:29
Marko RiedelMarko Riedel
41.5k341112
answered Jan 15 '17 at 23:29
Marko RiedelMarko Riedel
41.5k341112
answered Jan 15 '17 at 23:29
Marko RiedelMarko Riedel
41.5k341112
41.5k341112
add a comment |
add a comment |
$begingroup$
1) $τ(n)= sum_n 1$ => According to the Moebius Inversion Formula $sum_n μ(d)τ(n/d) = 1$
2) If d has a squared prime factor, then μ(d)=0 => $sum_n μ(d)τ(d) = sum_rad(n) μ(d)τ(d)$.
If the number of prime factors of n is even, then μ(rad(n)/d)= μ(d).
If it is odd, then μ(rad(n)/d)= -μ(d) => $sum_n μ(d)τ(d) = sum_rad(n) μ(d)τ(d) = (-1)^r sum_rad(n) μ(rad(n)/d)τ(d)= (-1)^r sum_rad(n) μ(d)τ(rad(n)/d)=(-1)^r.$
answered Jan 15 '17 at 17:24
Yanior WegYanior Weg
2,85611547
$endgroup$
$begingroup$
Really nice observation for tau(n) , and indeed nice proof for first . Thnx .
$endgroup$
– da6932
Jan 16 '17 at 19:23
add a comment |
$begingroup$
1) $τ(n)= sum_n 1$ => According to the Moebius Inversion Formula $sum_n μ(d)τ(n/d) = 1$
2) If d has a squared prime factor, then μ(d)=0 => $sum_n μ(d)τ(d) = sum_rad(n) μ(d)τ(d)$.
If the number of prime factors of n is even, then μ(rad(n)/d)= μ(d).
If it is odd, then μ(rad(n)/d)= -μ(d) => $sum_n μ(d)τ(d) = sum_rad(n) μ(d)τ(d) = (-1)^r sum_rad(n) μ(rad(n)/d)τ(d)= (-1)^r sum_rad(n) μ(d)τ(rad(n)/d)=(-1)^r.$
answered Jan 15 '17 at 17:24
Yanior WegYanior Weg
2,85611547
$endgroup$
$begingroup$
Really nice observation for tau(n) , and indeed nice proof for first . Thnx .
$endgroup$
– da6932
Jan 16 '17 at 19:23
add a comment |
$begingroup$
1) $τ(n)= sum_n 1$ => According to the Moebius Inversion Formula $sum_n μ(d)τ(n/d) = 1$
2) If d has a squared prime factor, then μ(d)=0 => $sum_n μ(d)τ(d) = sum_rad(n) μ(d)τ(d)$.
If the number of prime factors of n is even, then μ(rad(n)/d)= μ(d).
If it is odd, then μ(rad(n)/d)= -μ(d) => $sum_n μ(d)τ(d) = sum_rad(n) μ(d)τ(d) = (-1)^r sum_rad(n) μ(rad(n)/d)τ(d)= (-1)^r sum_rad(n) μ(d)τ(rad(n)/d)=(-1)^r.$
answered Jan 15 '17 at 17:24
Yanior WegYanior Weg
2,85611547
$endgroup$
1) $τ(n)= sum_n 1$ => According to the Moebius Inversion Formula $sum_n μ(d)τ(n/d) = 1$
2) If d has a squared prime factor, then μ(d)=0 => $sum_n μ(d)τ(d) = sum_rad(n) μ(d)τ(d)$.
If the number of prime factors of n is even, then μ(rad(n)/d)= μ(d).
If it is odd, then μ(rad(n)/d)= -μ(d) => $sum_n μ(d)τ(d) = sum_rad(n) μ(d)τ(d) = (-1)^r sum_rad(n) μ(rad(n)/d)τ(d)= (-1)^r sum_rad(n) μ(d)τ(rad(n)/d)=(-1)^r.$
answered Jan 15 '17 at 17:24
Yanior WegYanior Weg
2,85611547
edited Jan 15 '17 at 20:05
answered Jan 15 '17 at 17:24
Yanior WegYanior Weg
2,85611547
answered Jan 15 '17 at 17:24
Yanior WegYanior Weg
2,85611547
answered Jan 15 '17 at 17:24
Yanior WegYanior Weg
2,85611547
2,85611547
$begingroup$
Really nice observation for tau(n) , and indeed nice proof for first . Thnx .
$endgroup$
– da6932
Jan 16 '17 at 19:23
add a comment |
$begingroup$
Really nice observation for tau(n) , and indeed nice proof for first . Thnx .
$endgroup$
– da6932
Jan 16 '17 at 19:23
$begingroup$
Really nice observation for tau(n) , and indeed nice proof for first . Thnx .
$endgroup$
– da6932
Jan 16 '17 at 19:23
$begingroup$
Really nice observation for tau(n) , and indeed nice proof for first . Thnx .
$endgroup$
– da6932
Jan 16 '17 at 19:23
add a comment |
3
$begingroup$
Trying to search for the first one, I found these questions: An application of Mobius Inversion $sum_d mid n mu(fracnd)nu(d) = 1$ and Sum of Positive Divisors: $sum_n mu(n/d)nu(d)=1$ and $sum_n mu(n/d)sigma(d)=n$. Perhaps with a bit of searching you can find more copies of that question.
$endgroup$
– Martin Sleziak
Jan 15 '17 at 11:17
3
$begingroup$
Similarly, with a bit of searching, you will probably be able to find several posts about the other one. Like here or here. In fact, you do not need to search at all, just look at the list of related questions in the sidebar on the right to find this one.
$endgroup$
– Martin Sleziak
Jan 15 '17 at 11:22
$begingroup$
The sad thing is, the fact that it's two questions in one makes it difficult to mark it as a duplicate.
$endgroup$
– rabota
Jan 15 '17 at 20:12
$begingroup$
Forgive me this time . Will try to search from the next time , newbie mistake . Also thnx alot for citing the answers . :)
$endgroup$
– da6932
Jan 16 '17 at 19:19