Prove $sum_kmid nmu(k)d(k)=(-1)^omega(n)$ The Next CEO of Stack OverflowShowing $sum_dmid n mu(d)tau(n/d)=1$ and $sum_dmid n mu(d)tau(d)=(-1)^r$Möbius function verificationProve $sum_k = 1^n mu(k)left[ frac nk right] = 1$Convolution identity involving the Möbius function $sum_n,d>0 |mu(d)| = 2^omega(n)$Prove that $sum_t vert n d^3(t) = (sum_t vert nd(t))^2$ for all $n in mathbbN$Proof of inequality involving multiplicative function?Bound for the sum of the divisors of a numberProve that $sum_n, d geq 1 = 2^omega(n)$Formula for unique distribution of colored balls into boxesUpper bound for the divisor counting function?How to invert an arithmetic function where Möbius inversion may not apply?
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Prove $sum_kmid nmu(k)d(k)=(-1)^omega(n)$
The Next CEO of Stack OverflowShowing $sum_dmid n mu(d)tau(n/d)=1$ and $sum_dmid n mu(d)tau(d)=(-1)^r$Möbius function verificationProve $sum_k = 1^n mu(k)left[ frac nk right] = 1$Convolution identity involving the Möbius function $sum_n,d>0 |mu(d)| = 2^omega(n)$Prove that $sum_t vert n d^3(t) = (sum_t vert nd(t))^2$ for all $n in mathbbN$Proof of inequality involving multiplicative function?Bound for the sum of the divisors of a numberProve that $sum_n, d geq 1 = 2^omega(n)$Formula for unique distribution of colored balls into boxesUpper bound for the divisor counting function?How to invert an arithmetic function where Möbius inversion may not apply?
$begingroup$
I have the following exercise.
Show that for all natural numbers $n$, the following equality holds
$$sum_dmu(d)d(d)=(-1)^omega(n)$$
Here, $mu$ is the Möbius function, $d$ counts the number of divisors of $n$, and $omega$ counts the number of distinct prime divisors of $n$.
I tried looking at a number like $n=10$ just to see what it looks like expanded. So since the divisors of $n$ are $1,2,5,$ and $10$, I can show that
$$sum_10mu(d)d(d)=(-1)^omega(10)=mu(1)d(1)+mu(2)d(2)+mu(5)d(5)+mu(10)d(10)=1$$
This gives me
$$1cdot 1+-1cdot 2+-1cdot 2 +1cdot 4 $$
I'm thinking somehow since both $mu$ and $d$ are multiplicative that we can rewrite though as
$$mu(1)d(1)+mu(2)d(2)+mu(5)d(5)+mu(2)d(2)mu(5)d(5)$$
$$=mu(1)d(1)+mu(5)d(5)+mu(2)d(2)+mu(2)d(2)mu(5)d(5)$$
$$=mu(1)d(1)+mu(5)d(5)+mu(2)d(2)(1+mu(5)d(5))$$
$$=mu(1)d(1)+mu(5)d(5)+mu(2)d(2)(mu(1)d(1)+mu(5)d(5))$$
$$=(1+mu(2)d(2))(1+mu(5)d(5))$$
$$=sum_dmu(d)d(d)sum_dmu(d)d(d)$$
So this original summation function is multiplicative. But this isn't helping me see the how to move forward. I know that the $mu$ function is defined as $mu(n)=(-1)^omega(n)$ if $n$ is square free and $0$ if divisible by a square, so I think this plays a role somehow, but again, I'm feeling lost.
EDIT: Would looking at $n=prod_i=1^kp_i^alpha_i$ be a more useful approach to the problem? Knowing the larger summed function is multiplicative means I can focus my approach on the $p_i^alpha_i$...
number-theory multiplicative-function divisor-counting-function mobius-inversion dirichlet-convolution
edited yesterday
Eric Wofsey
191k14216349
asked Jul 19 '14 at 18:19
LalaloopsyLalaloopsy
89111221
$endgroup$
add a comment |
$begingroup$
I have the following exercise.
Show that for all natural numbers $n$, the following equality holds
$$sum_dmu(d)d(d)=(-1)^omega(n)$$
Here, $mu$ is the Möbius function, $d$ counts the number of divisors of $n$, and $omega$ counts the number of distinct prime divisors of $n$.
I tried looking at a number like $n=10$ just to see what it looks like expanded. So since the divisors of $n$ are $1,2,5,$ and $10$, I can show that
$$sum_10mu(d)d(d)=(-1)^omega(10)=mu(1)d(1)+mu(2)d(2)+mu(5)d(5)+mu(10)d(10)=1$$
This gives me
$$1cdot 1+-1cdot 2+-1cdot 2 +1cdot 4 $$
I'm thinking somehow since both $mu$ and $d$ are multiplicative that we can rewrite though as
$$mu(1)d(1)+mu(2)d(2)+mu(5)d(5)+mu(2)d(2)mu(5)d(5)$$
$$=mu(1)d(1)+mu(5)d(5)+mu(2)d(2)+mu(2)d(2)mu(5)d(5)$$
$$=mu(1)d(1)+mu(5)d(5)+mu(2)d(2)(1+mu(5)d(5))$$
$$=mu(1)d(1)+mu(5)d(5)+mu(2)d(2)(mu(1)d(1)+mu(5)d(5))$$
$$=(1+mu(2)d(2))(1+mu(5)d(5))$$
$$=sum_dmu(d)d(d)sum_dmu(d)d(d)$$
So this original summation function is multiplicative. But this isn't helping me see the how to move forward. I know that the $mu$ function is defined as $mu(n)=(-1)^omega(n)$ if $n$ is square free and $0$ if divisible by a square, so I think this plays a role somehow, but again, I'm feeling lost.
EDIT: Would looking at $n=prod_i=1^kp_i^alpha_i$ be a more useful approach to the problem? Knowing the larger summed function is multiplicative means I can focus my approach on the $p_i^alpha_i$...
number-theory multiplicative-function divisor-counting-function mobius-inversion dirichlet-convolution
edited yesterday
Eric Wofsey
191k14216349
asked Jul 19 '14 at 18:19
LalaloopsyLalaloopsy
89111221
$endgroup$
$begingroup$
The book I am working out of is Niven's Introduction to the Theory of Numbers and they use $d$, so I've just gotten in the habit of using it. I know it's confusing. My professor did reveal that $tau$ was also used, but I guess I'm just used to the idea of $d$=divisor function.
$endgroup$
– Lalaloopsy
Jul 19 '14 at 18:32
5
$begingroup$
$d(d)$ is such a confusing notation...
$endgroup$
– CuriousGuest
Jul 20 '14 at 6:54
add a comment |
$begingroup$
I have the following exercise.
Show that for all natural numbers $n$, the following equality holds
$$sum_dmu(d)d(d)=(-1)^omega(n)$$
Here, $mu$ is the Möbius function, $d$ counts the number of divisors of $n$, and $omega$ counts the number of distinct prime divisors of $n$.
I tried looking at a number like $n=10$ just to see what it looks like expanded. So since the divisors of $n$ are $1,2,5,$ and $10$, I can show that
$$sum_10mu(d)d(d)=(-1)^omega(10)=mu(1)d(1)+mu(2)d(2)+mu(5)d(5)+mu(10)d(10)=1$$
This gives me
$$1cdot 1+-1cdot 2+-1cdot 2 +1cdot 4 $$
I'm thinking somehow since both $mu$ and $d$ are multiplicative that we can rewrite though as
$$mu(1)d(1)+mu(2)d(2)+mu(5)d(5)+mu(2)d(2)mu(5)d(5)$$
$$=mu(1)d(1)+mu(5)d(5)+mu(2)d(2)+mu(2)d(2)mu(5)d(5)$$
$$=mu(1)d(1)+mu(5)d(5)+mu(2)d(2)(1+mu(5)d(5))$$
$$=mu(1)d(1)+mu(5)d(5)+mu(2)d(2)(mu(1)d(1)+mu(5)d(5))$$
$$=(1+mu(2)d(2))(1+mu(5)d(5))$$
$$=sum_dmu(d)d(d)sum_dmu(d)d(d)$$
So this original summation function is multiplicative. But this isn't helping me see the how to move forward. I know that the $mu$ function is defined as $mu(n)=(-1)^omega(n)$ if $n$ is square free and $0$ if divisible by a square, so I think this plays a role somehow, but again, I'm feeling lost.
EDIT: Would looking at $n=prod_i=1^kp_i^alpha_i$ be a more useful approach to the problem? Knowing the larger summed function is multiplicative means I can focus my approach on the $p_i^alpha_i$...
number-theory multiplicative-function divisor-counting-function mobius-inversion dirichlet-convolution
edited yesterday
Eric Wofsey
191k14216349
asked Jul 19 '14 at 18:19
LalaloopsyLalaloopsy
89111221
$endgroup$
I have the following exercise.
Show that for all natural numbers $n$, the following equality holds
$$sum_dmu(d)d(d)=(-1)^omega(n)$$
Here, $mu$ is the Möbius function, $d$ counts the number of divisors of $n$, and $omega$ counts the number of distinct prime divisors of $n$.
I tried looking at a number like $n=10$ just to see what it looks like expanded. So since the divisors of $n$ are $1,2,5,$ and $10$, I can show that
$$sum_10mu(d)d(d)=(-1)^omega(10)=mu(1)d(1)+mu(2)d(2)+mu(5)d(5)+mu(10)d(10)=1$$
This gives me
$$1cdot 1+-1cdot 2+-1cdot 2 +1cdot 4 $$
I'm thinking somehow since both $mu$ and $d$ are multiplicative that we can rewrite though as
$$mu(1)d(1)+mu(2)d(2)+mu(5)d(5)+mu(2)d(2)mu(5)d(5)$$
$$=mu(1)d(1)+mu(5)d(5)+mu(2)d(2)+mu(2)d(2)mu(5)d(5)$$
$$=mu(1)d(1)+mu(5)d(5)+mu(2)d(2)(1+mu(5)d(5))$$
$$=mu(1)d(1)+mu(5)d(5)+mu(2)d(2)(mu(1)d(1)+mu(5)d(5))$$
$$=(1+mu(2)d(2))(1+mu(5)d(5))$$
$$=sum_dmu(d)d(d)sum_dmu(d)d(d)$$
So this original summation function is multiplicative. But this isn't helping me see the how to move forward. I know that the $mu$ function is defined as $mu(n)=(-1)^omega(n)$ if $n$ is square free and $0$ if divisible by a square, so I think this plays a role somehow, but again, I'm feeling lost.
EDIT: Would looking at $n=prod_i=1^kp_i^alpha_i$ be a more useful approach to the problem? Knowing the larger summed function is multiplicative means I can focus my approach on the $p_i^alpha_i$...
number-theory multiplicative-function divisor-counting-function mobius-inversion dirichlet-convolution
number-theory multiplicative-function divisor-counting-function mobius-inversion dirichlet-convolution
edited yesterday
Eric Wofsey
191k14216349
asked Jul 19 '14 at 18:19
LalaloopsyLalaloopsy
89111221
edited yesterday
Eric Wofsey
191k14216349
asked Jul 19 '14 at 18:19
LalaloopsyLalaloopsy
89111221
edited yesterday
Eric Wofsey
191k14216349
edited yesterday
Eric Wofsey
191k14216349
edited yesterday
Eric Wofsey
191k14216349
191k14216349
asked Jul 19 '14 at 18:19
LalaloopsyLalaloopsy
89111221
asked Jul 19 '14 at 18:19
LalaloopsyLalaloopsy
89111221
asked Jul 19 '14 at 18:19
LalaloopsyLalaloopsy
89111221
89111221
$begingroup$
The book I am working out of is Niven's Introduction to the Theory of Numbers and they use $d$, so I've just gotten in the habit of using it. I know it's confusing. My professor did reveal that $tau$ was also used, but I guess I'm just used to the idea of $d$=divisor function.
$endgroup$
– Lalaloopsy
Jul 19 '14 at 18:32
5
$begingroup$
$d(d)$ is such a confusing notation...
$endgroup$
– CuriousGuest
Jul 20 '14 at 6:54
add a comment |
$begingroup$
The book I am working out of is Niven's Introduction to the Theory of Numbers and they use $d$, so I've just gotten in the habit of using it. I know it's confusing. My professor did reveal that $tau$ was also used, but I guess I'm just used to the idea of $d$=divisor function.
$endgroup$
– Lalaloopsy
Jul 19 '14 at 18:32
5
$begingroup$
$d(d)$ is such a confusing notation...
$endgroup$
– CuriousGuest
Jul 20 '14 at 6:54
$begingroup$
The book I am working out of is Niven's Introduction to the Theory of Numbers and they use $d$, so I've just gotten in the habit of using it. I know it's confusing. My professor did reveal that $tau$ was also used, but I guess I'm just used to the idea of $d$=divisor function.
$endgroup$
– Lalaloopsy
Jul 19 '14 at 18:32
$begingroup$
The book I am working out of is Niven's Introduction to the Theory of Numbers and they use $d$, so I've just gotten in the habit of using it. I know it's confusing. My professor did reveal that $tau$ was also used, but I guess I'm just used to the idea of $d$=divisor function.
$endgroup$
– Lalaloopsy
Jul 19 '14 at 18:32
5
5
$begingroup$
$d(d)$ is such a confusing notation...
$endgroup$
– CuriousGuest
Jul 20 '14 at 6:54
$begingroup$
$d(d)$ is such a confusing notation...
$endgroup$
– CuriousGuest
Jul 20 '14 at 6:54
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Look at the sum for a prime power $p^k$. The divisors of $p^k$ are $1,p,p^2,...,p^k-1$. All of them contain a square except $1$ and $p$. That means $mu(p^a)=0$. So the sum is
$$mu(1)d(1)+mu(p)d(p)\=1times 1+(-1)times 2=1-2=-1$$
So each different prime, or its power, contributes a factor (-1).
answered Jul 19 '14 at 18:26
Empy2Empy2
33.6k12462
$endgroup$
1
$begingroup$
In particular, since $mu(n)d(n)$ is a multiplicative function, so is $sum_n'mid n mu(n')d(n')$...
$endgroup$
– Thomas Andrews
Jul 19 '14 at 18:32
add a comment |
$begingroup$
Another, Combinatorial, way would be like $$sum _d mu (d) d(d)=sum _i=0^w(n)sum _p_1<p_2 ldots < p_imu (p_1 dots p_i)d (p_1 dots p_i)=sum _i=0^w(n)binomw(n)i(-1)^i2^i=(1-2)^w(n)$$ where the $p_j$ are primes in the descomposition of $n$.
answered Jul 21 '14 at 2:23
PhicarPhicar
2,6601915
$endgroup$
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
votes
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active
oldest
votes
$begingroup$
Look at the sum for a prime power $p^k$. The divisors of $p^k$ are $1,p,p^2,...,p^k-1$. All of them contain a square except $1$ and $p$. That means $mu(p^a)=0$. So the sum is
$$mu(1)d(1)+mu(p)d(p)\=1times 1+(-1)times 2=1-2=-1$$
So each different prime, or its power, contributes a factor (-1).
answered Jul 19 '14 at 18:26
Empy2Empy2
33.6k12462
$endgroup$
1
$begingroup$
In particular, since $mu(n)d(n)$ is a multiplicative function, so is $sum_n'mid n mu(n')d(n')$...
$endgroup$
– Thomas Andrews
Jul 19 '14 at 18:32
add a comment |
$begingroup$
Look at the sum for a prime power $p^k$. The divisors of $p^k$ are $1,p,p^2,...,p^k-1$. All of them contain a square except $1$ and $p$. That means $mu(p^a)=0$. So the sum is
$$mu(1)d(1)+mu(p)d(p)\=1times 1+(-1)times 2=1-2=-1$$
So each different prime, or its power, contributes a factor (-1).
answered Jul 19 '14 at 18:26
Empy2Empy2
33.6k12462
$endgroup$
1
$begingroup$
In particular, since $mu(n)d(n)$ is a multiplicative function, so is $sum_n'mid n mu(n')d(n')$...
$endgroup$
– Thomas Andrews
Jul 19 '14 at 18:32
add a comment |
$begingroup$
Look at the sum for a prime power $p^k$. The divisors of $p^k$ are $1,p,p^2,...,p^k-1$. All of them contain a square except $1$ and $p$. That means $mu(p^a)=0$. So the sum is
$$mu(1)d(1)+mu(p)d(p)\=1times 1+(-1)times 2=1-2=-1$$
So each different prime, or its power, contributes a factor (-1).
answered Jul 19 '14 at 18:26
Empy2Empy2
33.6k12462
$endgroup$
Look at the sum for a prime power $p^k$. The divisors of $p^k$ are $1,p,p^2,...,p^k-1$. All of them contain a square except $1$ and $p$. That means $mu(p^a)=0$. So the sum is
$$mu(1)d(1)+mu(p)d(p)\=1times 1+(-1)times 2=1-2=-1$$
So each different prime, or its power, contributes a factor (-1).
answered Jul 19 '14 at 18:26
Empy2Empy2
33.6k12462
answered Jul 19 '14 at 18:26
Empy2Empy2
33.6k12462
answered Jul 19 '14 at 18:26
Empy2Empy2
33.6k12462
answered Jul 19 '14 at 18:26
Empy2Empy2
33.6k12462
33.6k12462
1
$begingroup$
In particular, since $mu(n)d(n)$ is a multiplicative function, so is $sum_n'mid n mu(n')d(n')$...
$endgroup$
– Thomas Andrews
Jul 19 '14 at 18:32
add a comment |
1
$begingroup$
In particular, since $mu(n)d(n)$ is a multiplicative function, so is $sum_n'mid n mu(n')d(n')$...
$endgroup$
– Thomas Andrews
Jul 19 '14 at 18:32
1
1
$begingroup$
In particular, since $mu(n)d(n)$ is a multiplicative function, so is $sum_n'mid n mu(n')d(n')$...
$endgroup$
– Thomas Andrews
Jul 19 '14 at 18:32
$begingroup$
In particular, since $mu(n)d(n)$ is a multiplicative function, so is $sum_n'mid n mu(n')d(n')$...
$endgroup$
– Thomas Andrews
Jul 19 '14 at 18:32
add a comment |
$begingroup$
Another, Combinatorial, way would be like $$sum _d mu (d) d(d)=sum _i=0^w(n)sum _p_1<p_2 ldots < p_imu (p_1 dots p_i)d (p_1 dots p_i)=sum _i=0^w(n)binomw(n)i(-1)^i2^i=(1-2)^w(n)$$ where the $p_j$ are primes in the descomposition of $n$.
answered Jul 21 '14 at 2:23
PhicarPhicar
2,6601915
$endgroup$
add a comment |
$begingroup$
Another, Combinatorial, way would be like $$sum _d mu (d) d(d)=sum _i=0^w(n)sum _p_1<p_2 ldots < p_imu (p_1 dots p_i)d (p_1 dots p_i)=sum _i=0^w(n)binomw(n)i(-1)^i2^i=(1-2)^w(n)$$ where the $p_j$ are primes in the descomposition of $n$.
answered Jul 21 '14 at 2:23
PhicarPhicar
2,6601915
$endgroup$
add a comment |
$begingroup$
Another, Combinatorial, way would be like $$sum _d mu (d) d(d)=sum _i=0^w(n)sum _p_1<p_2 ldots < p_imu (p_1 dots p_i)d (p_1 dots p_i)=sum _i=0^w(n)binomw(n)i(-1)^i2^i=(1-2)^w(n)$$ where the $p_j$ are primes in the descomposition of $n$.
answered Jul 21 '14 at 2:23
PhicarPhicar
2,6601915
$endgroup$
Another, Combinatorial, way would be like $$sum _d mu (d) d(d)=sum _i=0^w(n)sum _p_1<p_2 ldots < p_imu (p_1 dots p_i)d (p_1 dots p_i)=sum _i=0^w(n)binomw(n)i(-1)^i2^i=(1-2)^w(n)$$ where the $p_j$ are primes in the descomposition of $n$.
answered Jul 21 '14 at 2:23
PhicarPhicar
2,6601915
answered Jul 21 '14 at 2:23
PhicarPhicar
2,6601915
answered Jul 21 '14 at 2:23
PhicarPhicar
2,6601915
answered Jul 21 '14 at 2:23
PhicarPhicar
2,6601915
2,6601915
add a comment |
add a comment |
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The book I am working out of is Niven's Introduction to the Theory of Numbers and they use $d$, so I've just gotten in the habit of using it. I know it's confusing. My professor did reveal that $tau$ was also used, but I guess I'm just used to the idea of $d$=divisor function.
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– Lalaloopsy
Jul 19 '14 at 18:32
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$d(d)$ is such a confusing notation...
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– CuriousGuest
Jul 20 '14 at 6:54