Mixing
A proof of $sumlimits_{d|n} sigma(d) = n sumlimits_{d|n} {tau(d) over d}$
$begingroup$
I'm trying to proof the following statement:
Let $n in mathbb{Z}$ and the $sum$ are on the divisors $d$ of $n$. Show that
$$sumlimits_{d|n} sigma(d) = n sumlimits_{d|n} {tau(d) over d}.$$
I arrive to a point where I have a product of polynomials and I can't simply find a way to reorder the factors in order to resemble some kind of useful structure for this problem. Any suggestions?
EDIT: due to unsuscribe of user593746, I can't ask delucidation, so I reopen the question because I wish to have detailed delucidation on the passages on the first row of his good proof (summatory index manipulations and t outside the first sum):
We have $$sum_{dmid n}sigma(d)=sum_{dmid n}sum_{tmid d}t=sum_{tmid n}sum_{kmid frac{n}{t}}t=sum_{tmid n}tsum_{kmid frac{n}t}1,$$
where $k=frac{d}{t}$. So
$$sum_{dmid n}sigma(d)=sum_{tmid n}ttauleft(frac{n}tright)=sum_{delta mid n}frac{n}{delta}tau(delta)=nsum_{deltamid n}frac{tau(delta)}{delta},$$
where $delta =frac{n}{t}$.
elementary-number-theory summation divisor-sum arithmetic-functions divisor-counting-function
asked Nov 27 '18 at 14:46
AlessarAlessar
313115
$endgroup$
add a comment |
$begingroup$
I'm trying to proof the following statement:
Let $n in mathbb{Z}$ and the $sum$ are on the divisors $d$ of $n$. Show that
$$sumlimits_{d|n} sigma(d) = n sumlimits_{d|n} {tau(d) over d}.$$
I arrive to a point where I have a product of polynomials and I can't simply find a way to reorder the factors in order to resemble some kind of useful structure for this problem. Any suggestions?
EDIT: due to unsuscribe of user593746, I can't ask delucidation, so I reopen the question because I wish to have detailed delucidation on the passages on the first row of his good proof (summatory index manipulations and t outside the first sum):
We have $$sum_{dmid n}sigma(d)=sum_{dmid n}sum_{tmid d}t=sum_{tmid n}sum_{kmid frac{n}{t}}t=sum_{tmid n}tsum_{kmid frac{n}t}1,$$
where $k=frac{d}{t}$. So
$$sum_{dmid n}sigma(d)=sum_{tmid n}ttauleft(frac{n}tright)=sum_{delta mid n}frac{n}{delta}tau(delta)=nsum_{deltamid n}frac{tau(delta)}{delta},$$
where $delta =frac{n}{t}$.
elementary-number-theory summation divisor-sum arithmetic-functions divisor-counting-function
asked Nov 27 '18 at 14:46
AlessarAlessar
313115
$endgroup$
$begingroup$
Use math.stackexchange.com/questions/1948422/…
$endgroup$
– lab bhattacharjee
Nov 27 '18 at 14:51
1
$begingroup$
The sums express two different ways to count the sum of how many divisors $cmid n$ each divisor $dmid n$ divides.
$endgroup$
– Servaes
Nov 27 '18 at 14:55
add a comment |
$begingroup$
I'm trying to proof the following statement:
Let $n in mathbb{Z}$ and the $sum$ are on the divisors $d$ of $n$. Show that
$$sumlimits_{d|n} sigma(d) = n sumlimits_{d|n} {tau(d) over d}.$$
I arrive to a point where I have a product of polynomials and I can't simply find a way to reorder the factors in order to resemble some kind of useful structure for this problem. Any suggestions?
EDIT: due to unsuscribe of user593746, I can't ask delucidation, so I reopen the question because I wish to have detailed delucidation on the passages on the first row of his good proof (summatory index manipulations and t outside the first sum):
We have $$sum_{dmid n}sigma(d)=sum_{dmid n}sum_{tmid d}t=sum_{tmid n}sum_{kmid frac{n}{t}}t=sum_{tmid n}tsum_{kmid frac{n}t}1,$$
where $k=frac{d}{t}$. So
$$sum_{dmid n}sigma(d)=sum_{tmid n}ttauleft(frac{n}tright)=sum_{delta mid n}frac{n}{delta}tau(delta)=nsum_{deltamid n}frac{tau(delta)}{delta},$$
where $delta =frac{n}{t}$.
elementary-number-theory summation divisor-sum arithmetic-functions divisor-counting-function
asked Nov 27 '18 at 14:46
AlessarAlessar
313115
$endgroup$
I'm trying to proof the following statement:
Let $n in mathbb{Z}$ and the $sum$ are on the divisors $d$ of $n$. Show that
$$sumlimits_{d|n} sigma(d) = n sumlimits_{d|n} {tau(d) over d}.$$
I arrive to a point where I have a product of polynomials and I can't simply find a way to reorder the factors in order to resemble some kind of useful structure for this problem. Any suggestions?
EDIT: due to unsuscribe of user593746, I can't ask delucidation, so I reopen the question because I wish to have detailed delucidation on the passages on the first row of his good proof (summatory index manipulations and t outside the first sum):
We have $$sum_{dmid n}sigma(d)=sum_{dmid n}sum_{tmid d}t=sum_{tmid n}sum_{kmid frac{n}{t}}t=sum_{tmid n}tsum_{kmid frac{n}t}1,$$
where $k=frac{d}{t}$. So
$$sum_{dmid n}sigma(d)=sum_{tmid n}ttauleft(frac{n}tright)=sum_{delta mid n}frac{n}{delta}tau(delta)=nsum_{deltamid n}frac{tau(delta)}{delta},$$
where $delta =frac{n}{t}$.
elementary-number-theory summation divisor-sum arithmetic-functions divisor-counting-function
elementary-number-theory summation divisor-sum arithmetic-functions divisor-counting-function
asked Nov 27 '18 at 14:46
AlessarAlessar
313115
asked Nov 27 '18 at 14:46
AlessarAlessar
313115
edited Jan 15 at 13:32
Alessar
asked Nov 27 '18 at 14:46
AlessarAlessar
313115
asked Nov 27 '18 at 14:46
AlessarAlessar
313115
asked Nov 27 '18 at 14:46
AlessarAlessar
313115
313115
$begingroup$
Use math.stackexchange.com/questions/1948422/…
$endgroup$
– lab bhattacharjee
Nov 27 '18 at 14:51
1
$begingroup$
The sums express two different ways to count the sum of how many divisors $cmid n$ each divisor $dmid n$ divides.
$endgroup$
– Servaes
Nov 27 '18 at 14:55
add a comment |
$begingroup$
Use math.stackexchange.com/questions/1948422/…
$endgroup$
– lab bhattacharjee
Nov 27 '18 at 14:51
1
$begingroup$
The sums express two different ways to count the sum of how many divisors $cmid n$ each divisor $dmid n$ divides.
$endgroup$
– Servaes
Nov 27 '18 at 14:55
$begingroup$
Use math.stackexchange.com/questions/1948422/…
$endgroup$
– lab bhattacharjee
Nov 27 '18 at 14:51
$begingroup$
Use math.stackexchange.com/questions/1948422/…
$endgroup$
– lab bhattacharjee
Nov 27 '18 at 14:51
1
1
$begingroup$
The sums express two different ways to count the sum of how many divisors $cmid n$ each divisor $dmid n$ divides.
$endgroup$
– Servaes
Nov 27 '18 at 14:55
$begingroup$
The sums express two different ways to count the sum of how many divisors $cmid n$ each divisor $dmid n$ divides.
$endgroup$
– Servaes
Nov 27 '18 at 14:55
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Here is a derivation with some intermediate steps which might be helpful.
We obtain
begin{align*}
color{blue}{sum_{dmid n}sigma(d)}&=sum_{dmid n}sum_{tmid d}ttag{1}\
&=sum_{{tmid dmid n}atop{t,dgeq 1}}ttag{2}\
&=sum_{{t mid tk mid n}atop{t,kgeq 1}}ttag{3}\
&=sum_{{t mid n,k mid frac{n}{t}}atop{t,kgeq 1}}ttag{4}\
&,,color{blue}{=sum_{t mid n}tsum_{k mid frac{n}{t}}1}tag{5}
end{align*}
Comment:
In (1) we use the definition of the divisor function $sigma(d)=sum_{t| d}t$.
In (2) we write the index region somewhat more compactly. This does not change the sum, as it is only a rearrangement of the summands.
In (3) we use $t|d Longleftrightarrow exists kgeq 1 : tk=d$, assuming $t,d,k$ are positive integers.
In (4) we use the transitivity of the $|$ operator: $t| d| nLongrightarrow t|n$ and we also use $t cdot k| nLongleftrightarrow t | frac{n}{k}$.
In (5) we rearrange the summands again by summing at first over $t| n$. We also factor out $t$ from the inner sum, since $t$ does not depend on the index variable $k$.
[Add-on 2019-01-17]: A derivation using Dirichlet convolution due to OPs comment.
Taking the unit-function $u(n)=1$ for all $n$ and the function $N(n)=n$ for all $n$ we have
begin{align*}
tau(n)&=sum_{d|n}1=(ustar u)(n)\
sigma(n)&=sum_{d|n}d=(ustar N)(n)
end{align*}
We obtain
begin{align*}
color{blue}{sum_{d|n}sigma(d)}&=(sigma star u)(n)\
&=((ustar N)star u)(n)\
&=(ustar(N star u))(n)\
&=(ustar (ustar N))(n)\
&=((ustar u) star N)(n)\
&=(tau star N)(n)\
&,,color{blue}{=sum_{d|n}tau(d)frac{n}{d}}
end{align*}
and the claim follows.
answered Jan 16 at 10:53
Markus ScheuerMarkus Scheuer
61.9k457148
$endgroup$
$begingroup$
Thank you for this precise explanation; one thing, the term $t$ can be carried out of the fist sum according to some kind of properties or constraint?
$endgroup$
– Alessar
Jan 16 at 11:44
1
$begingroup$
@Alessar: You're welcome and you're right, we can factor out $t$ since it does not depend on the index of the inner sum. I've updated the answer accordingly.
$endgroup$
– Markus Scheuer
Jan 16 at 12:14
$begingroup$
Sheuer, one curious question; this kind of derivation in this proof can be done only using Dirichlet convolutions?
$endgroup$
– Alessar
Jan 17 at 7:39
1
$begingroup$
@Alessar: Many thanks for accepting my answer and granting the bounty. I've added a derivation based upon Dirichlet convolution.
$endgroup$
– Markus Scheuer
Jan 17 at 9:55
1
$begingroup$
Wow! Thanks! Wish I could give 100 times that bounty!
$endgroup$
– Alessar
Jan 17 at 9:58
add a comment |
$begingroup$
We have $$sum_{dmid n}sigma(d)=sum_{dmid n}sum_{tmid d}t=sum_{tmid n}sum_{kmid frac{n}{t}}t=sum_{tmid n}tsum_{kmid frac{n}t}1,$$
where $k=frac{d}{t}$. So
$$sum_{dmid n}sigma(d)=sum_{tmid n}ttauleft(frac{n}tright)=sum_{delta mid n}frac{n}{delta}tau(delta)=nsum_{deltamid n}frac{tau(delta)}{delta},$$
where $delta =frac{n}{t}$.
$endgroup$
$begingroup$
In the final result $delta$ has the same exact meaning of $d$, identify the same divisors, right?
$endgroup$
– Alessar
Nov 27 '18 at 15:01
1
$begingroup$
Yep. I used a different variable because I wanted to define $delta=frac{n}{t}$, which could be confusing if I say $d=frac{n}{t}$, since $d$ is used in $k=frac{d}{t}$ already. But then, which ever variables you are using as the indices of summation don't really matter.
$endgroup$
– user593746
Nov 27 '18 at 15:03
$begingroup$
You're welcome, thanks for the very good answer!
$endgroup$
– Alessar
Jan 16 at 15:00
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here is a derivation with some intermediate steps which might be helpful.
We obtain
begin{align*}
color{blue}{sum_{dmid n}sigma(d)}&=sum_{dmid n}sum_{tmid d}ttag{1}\
&=sum_{{tmid dmid n}atop{t,dgeq 1}}ttag{2}\
&=sum_{{t mid tk mid n}atop{t,kgeq 1}}ttag{3}\
&=sum_{{t mid n,k mid frac{n}{t}}atop{t,kgeq 1}}ttag{4}\
&,,color{blue}{=sum_{t mid n}tsum_{k mid frac{n}{t}}1}tag{5}
end{align*}
Comment:
In (1) we use the definition of the divisor function $sigma(d)=sum_{t| d}t$.
In (2) we write the index region somewhat more compactly. This does not change the sum, as it is only a rearrangement of the summands.
In (3) we use $t|d Longleftrightarrow exists kgeq 1 : tk=d$, assuming $t,d,k$ are positive integers.
In (4) we use the transitivity of the $|$ operator: $t| d| nLongrightarrow t|n$ and we also use $t cdot k| nLongleftrightarrow t | frac{n}{k}$.
In (5) we rearrange the summands again by summing at first over $t| n$. We also factor out $t$ from the inner sum, since $t$ does not depend on the index variable $k$.
[Add-on 2019-01-17]: A derivation using Dirichlet convolution due to OPs comment.
Taking the unit-function $u(n)=1$ for all $n$ and the function $N(n)=n$ for all $n$ we have
begin{align*}
tau(n)&=sum_{d|n}1=(ustar u)(n)\
sigma(n)&=sum_{d|n}d=(ustar N)(n)
end{align*}
We obtain
begin{align*}
color{blue}{sum_{d|n}sigma(d)}&=(sigma star u)(n)\
&=((ustar N)star u)(n)\
&=(ustar(N star u))(n)\
&=(ustar (ustar N))(n)\
&=((ustar u) star N)(n)\
&=(tau star N)(n)\
&,,color{blue}{=sum_{d|n}tau(d)frac{n}{d}}
end{align*}
and the claim follows.
answered Jan 16 at 10:53
Markus ScheuerMarkus Scheuer
61.9k457148
$endgroup$
$begingroup$
Thank you for this precise explanation; one thing, the term $t$ can be carried out of the fist sum according to some kind of properties or constraint?
$endgroup$
– Alessar
Jan 16 at 11:44
1
$begingroup$
@Alessar: You're welcome and you're right, we can factor out $t$ since it does not depend on the index of the inner sum. I've updated the answer accordingly.
$endgroup$
– Markus Scheuer
Jan 16 at 12:14
$begingroup$
Sheuer, one curious question; this kind of derivation in this proof can be done only using Dirichlet convolutions?
$endgroup$
– Alessar
Jan 17 at 7:39
1
$begingroup$
@Alessar: Many thanks for accepting my answer and granting the bounty. I've added a derivation based upon Dirichlet convolution.
$endgroup$
– Markus Scheuer
Jan 17 at 9:55
1
$begingroup$
Wow! Thanks! Wish I could give 100 times that bounty!
$endgroup$
– Alessar
Jan 17 at 9:58
add a comment |
$begingroup$
Here is a derivation with some intermediate steps which might be helpful.
We obtain
begin{align*}
color{blue}{sum_{dmid n}sigma(d)}&=sum_{dmid n}sum_{tmid d}ttag{1}\
&=sum_{{tmid dmid n}atop{t,dgeq 1}}ttag{2}\
&=sum_{{t mid tk mid n}atop{t,kgeq 1}}ttag{3}\
&=sum_{{t mid n,k mid frac{n}{t}}atop{t,kgeq 1}}ttag{4}\
&,,color{blue}{=sum_{t mid n}tsum_{k mid frac{n}{t}}1}tag{5}
end{align*}
Comment:
In (1) we use the definition of the divisor function $sigma(d)=sum_{t| d}t$.
In (2) we write the index region somewhat more compactly. This does not change the sum, as it is only a rearrangement of the summands.
In (3) we use $t|d Longleftrightarrow exists kgeq 1 : tk=d$, assuming $t,d,k$ are positive integers.
In (4) we use the transitivity of the $|$ operator: $t| d| nLongrightarrow t|n$ and we also use $t cdot k| nLongleftrightarrow t | frac{n}{k}$.
In (5) we rearrange the summands again by summing at first over $t| n$. We also factor out $t$ from the inner sum, since $t$ does not depend on the index variable $k$.
[Add-on 2019-01-17]: A derivation using Dirichlet convolution due to OPs comment.
Taking the unit-function $u(n)=1$ for all $n$ and the function $N(n)=n$ for all $n$ we have
begin{align*}
tau(n)&=sum_{d|n}1=(ustar u)(n)\
sigma(n)&=sum_{d|n}d=(ustar N)(n)
end{align*}
We obtain
begin{align*}
color{blue}{sum_{d|n}sigma(d)}&=(sigma star u)(n)\
&=((ustar N)star u)(n)\
&=(ustar(N star u))(n)\
&=(ustar (ustar N))(n)\
&=((ustar u) star N)(n)\
&=(tau star N)(n)\
&,,color{blue}{=sum_{d|n}tau(d)frac{n}{d}}
end{align*}
and the claim follows.
answered Jan 16 at 10:53
Markus ScheuerMarkus Scheuer
61.9k457148
$endgroup$
$begingroup$
Thank you for this precise explanation; one thing, the term $t$ can be carried out of the fist sum according to some kind of properties or constraint?
$endgroup$
– Alessar
Jan 16 at 11:44
1
$begingroup$
@Alessar: You're welcome and you're right, we can factor out $t$ since it does not depend on the index of the inner sum. I've updated the answer accordingly.
$endgroup$
– Markus Scheuer
Jan 16 at 12:14
$begingroup$
Sheuer, one curious question; this kind of derivation in this proof can be done only using Dirichlet convolutions?
$endgroup$
– Alessar
Jan 17 at 7:39
1
$begingroup$
@Alessar: Many thanks for accepting my answer and granting the bounty. I've added a derivation based upon Dirichlet convolution.
$endgroup$
– Markus Scheuer
Jan 17 at 9:55
1
$begingroup$
Wow! Thanks! Wish I could give 100 times that bounty!
$endgroup$
– Alessar
Jan 17 at 9:58
add a comment |
$begingroup$
Here is a derivation with some intermediate steps which might be helpful.
We obtain
begin{align*}
color{blue}{sum_{dmid n}sigma(d)}&=sum_{dmid n}sum_{tmid d}ttag{1}\
&=sum_{{tmid dmid n}atop{t,dgeq 1}}ttag{2}\
&=sum_{{t mid tk mid n}atop{t,kgeq 1}}ttag{3}\
&=sum_{{t mid n,k mid frac{n}{t}}atop{t,kgeq 1}}ttag{4}\
&,,color{blue}{=sum_{t mid n}tsum_{k mid frac{n}{t}}1}tag{5}
end{align*}
Comment:
In (1) we use the definition of the divisor function $sigma(d)=sum_{t| d}t$.
In (2) we write the index region somewhat more compactly. This does not change the sum, as it is only a rearrangement of the summands.
In (3) we use $t|d Longleftrightarrow exists kgeq 1 : tk=d$, assuming $t,d,k$ are positive integers.
In (4) we use the transitivity of the $|$ operator: $t| d| nLongrightarrow t|n$ and we also use $t cdot k| nLongleftrightarrow t | frac{n}{k}$.
In (5) we rearrange the summands again by summing at first over $t| n$. We also factor out $t$ from the inner sum, since $t$ does not depend on the index variable $k$.
[Add-on 2019-01-17]: A derivation using Dirichlet convolution due to OPs comment.
Taking the unit-function $u(n)=1$ for all $n$ and the function $N(n)=n$ for all $n$ we have
begin{align*}
tau(n)&=sum_{d|n}1=(ustar u)(n)\
sigma(n)&=sum_{d|n}d=(ustar N)(n)
end{align*}
We obtain
begin{align*}
color{blue}{sum_{d|n}sigma(d)}&=(sigma star u)(n)\
&=((ustar N)star u)(n)\
&=(ustar(N star u))(n)\
&=(ustar (ustar N))(n)\
&=((ustar u) star N)(n)\
&=(tau star N)(n)\
&,,color{blue}{=sum_{d|n}tau(d)frac{n}{d}}
end{align*}
and the claim follows.
answered Jan 16 at 10:53
Markus ScheuerMarkus Scheuer
61.9k457148
$endgroup$
Here is a derivation with some intermediate steps which might be helpful.
We obtain
begin{align*}
color{blue}{sum_{dmid n}sigma(d)}&=sum_{dmid n}sum_{tmid d}ttag{1}\
&=sum_{{tmid dmid n}atop{t,dgeq 1}}ttag{2}\
&=sum_{{t mid tk mid n}atop{t,kgeq 1}}ttag{3}\
&=sum_{{t mid n,k mid frac{n}{t}}atop{t,kgeq 1}}ttag{4}\
&,,color{blue}{=sum_{t mid n}tsum_{k mid frac{n}{t}}1}tag{5}
end{align*}
Comment:
In (1) we use the definition of the divisor function $sigma(d)=sum_{t| d}t$.
In (2) we write the index region somewhat more compactly. This does not change the sum, as it is only a rearrangement of the summands.
In (3) we use $t|d Longleftrightarrow exists kgeq 1 : tk=d$, assuming $t,d,k$ are positive integers.
In (4) we use the transitivity of the $|$ operator: $t| d| nLongrightarrow t|n$ and we also use $t cdot k| nLongleftrightarrow t | frac{n}{k}$.
In (5) we rearrange the summands again by summing at first over $t| n$. We also factor out $t$ from the inner sum, since $t$ does not depend on the index variable $k$.
[Add-on 2019-01-17]: A derivation using Dirichlet convolution due to OPs comment.
Taking the unit-function $u(n)=1$ for all $n$ and the function $N(n)=n$ for all $n$ we have
begin{align*}
tau(n)&=sum_{d|n}1=(ustar u)(n)\
sigma(n)&=sum_{d|n}d=(ustar N)(n)
end{align*}
We obtain
begin{align*}
color{blue}{sum_{d|n}sigma(d)}&=(sigma star u)(n)\
&=((ustar N)star u)(n)\
&=(ustar(N star u))(n)\
&=(ustar (ustar N))(n)\
&=((ustar u) star N)(n)\
&=(tau star N)(n)\
&,,color{blue}{=sum_{d|n}tau(d)frac{n}{d}}
end{align*}
and the claim follows.
answered Jan 16 at 10:53
Markus ScheuerMarkus Scheuer
61.9k457148
edited Jan 17 at 9:59
answered Jan 16 at 10:53
Markus ScheuerMarkus Scheuer
61.9k457148
answered Jan 16 at 10:53
Markus ScheuerMarkus Scheuer
61.9k457148
answered Jan 16 at 10:53
Markus ScheuerMarkus Scheuer
61.9k457148
61.9k457148
$begingroup$
Thank you for this precise explanation; one thing, the term $t$ can be carried out of the fist sum according to some kind of properties or constraint?
$endgroup$
– Alessar
Jan 16 at 11:44
1
$begingroup$
@Alessar: You're welcome and you're right, we can factor out $t$ since it does not depend on the index of the inner sum. I've updated the answer accordingly.
$endgroup$
– Markus Scheuer
Jan 16 at 12:14
$begingroup$
Sheuer, one curious question; this kind of derivation in this proof can be done only using Dirichlet convolutions?
$endgroup$
– Alessar
Jan 17 at 7:39
1
$begingroup$
@Alessar: Many thanks for accepting my answer and granting the bounty. I've added a derivation based upon Dirichlet convolution.
$endgroup$
– Markus Scheuer
Jan 17 at 9:55
1
$begingroup$
Wow! Thanks! Wish I could give 100 times that bounty!
$endgroup$
– Alessar
Jan 17 at 9:58
add a comment |
$begingroup$
Thank you for this precise explanation; one thing, the term $t$ can be carried out of the fist sum according to some kind of properties or constraint?
$endgroup$
– Alessar
Jan 16 at 11:44
1
$begingroup$
@Alessar: You're welcome and you're right, we can factor out $t$ since it does not depend on the index of the inner sum. I've updated the answer accordingly.
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– Markus Scheuer
Jan 16 at 12:14
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Sheuer, one curious question; this kind of derivation in this proof can be done only using Dirichlet convolutions?
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– Alessar
Jan 17 at 7:39
1
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@Alessar: Many thanks for accepting my answer and granting the bounty. I've added a derivation based upon Dirichlet convolution.
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– Markus Scheuer
Jan 17 at 9:55
1
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Wow! Thanks! Wish I could give 100 times that bounty!
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– Alessar
Jan 17 at 9:58
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Thank you for this precise explanation; one thing, the term $t$ can be carried out of the fist sum according to some kind of properties or constraint?
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– Alessar
Jan 16 at 11:44
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Thank you for this precise explanation; one thing, the term $t$ can be carried out of the fist sum according to some kind of properties or constraint?
$endgroup$
– Alessar
Jan 16 at 11:44
1
1
$begingroup$
@Alessar: You're welcome and you're right, we can factor out $t$ since it does not depend on the index of the inner sum. I've updated the answer accordingly.
$endgroup$
– Markus Scheuer
Jan 16 at 12:14
$begingroup$
@Alessar: You're welcome and you're right, we can factor out $t$ since it does not depend on the index of the inner sum. I've updated the answer accordingly.
$endgroup$
– Markus Scheuer
Jan 16 at 12:14
$begingroup$
Sheuer, one curious question; this kind of derivation in this proof can be done only using Dirichlet convolutions?
$endgroup$
– Alessar
Jan 17 at 7:39
$begingroup$
Sheuer, one curious question; this kind of derivation in this proof can be done only using Dirichlet convolutions?
$endgroup$
– Alessar
Jan 17 at 7:39
1
1
$begingroup$
@Alessar: Many thanks for accepting my answer and granting the bounty. I've added a derivation based upon Dirichlet convolution.
$endgroup$
– Markus Scheuer
Jan 17 at 9:55
$begingroup$
@Alessar: Many thanks for accepting my answer and granting the bounty. I've added a derivation based upon Dirichlet convolution.
$endgroup$
– Markus Scheuer
Jan 17 at 9:55
1
1
$begingroup$
Wow! Thanks! Wish I could give 100 times that bounty!
$endgroup$
– Alessar
Jan 17 at 9:58
$begingroup$
Wow! Thanks! Wish I could give 100 times that bounty!
$endgroup$
– Alessar
Jan 17 at 9:58
add a comment |
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We have $$sum_{dmid n}sigma(d)=sum_{dmid n}sum_{tmid d}t=sum_{tmid n}sum_{kmid frac{n}{t}}t=sum_{tmid n}tsum_{kmid frac{n}t}1,$$
where $k=frac{d}{t}$. So
$$sum_{dmid n}sigma(d)=sum_{tmid n}ttauleft(frac{n}tright)=sum_{delta mid n}frac{n}{delta}tau(delta)=nsum_{deltamid n}frac{tau(delta)}{delta},$$
where $delta =frac{n}{t}$.
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$begingroup$
In the final result $delta$ has the same exact meaning of $d$, identify the same divisors, right?
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– Alessar
Nov 27 '18 at 15:01
1
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Yep. I used a different variable because I wanted to define $delta=frac{n}{t}$, which could be confusing if I say $d=frac{n}{t}$, since $d$ is used in $k=frac{d}{t}$ already. But then, which ever variables you are using as the indices of summation don't really matter.
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– user593746
Nov 27 '18 at 15:03
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You're welcome, thanks for the very good answer!
$endgroup$
– Alessar
Jan 16 at 15:00
add a comment |
$begingroup$
We have $$sum_{dmid n}sigma(d)=sum_{dmid n}sum_{tmid d}t=sum_{tmid n}sum_{kmid frac{n}{t}}t=sum_{tmid n}tsum_{kmid frac{n}t}1,$$
where $k=frac{d}{t}$. So
$$sum_{dmid n}sigma(d)=sum_{tmid n}ttauleft(frac{n}tright)=sum_{delta mid n}frac{n}{delta}tau(delta)=nsum_{deltamid n}frac{tau(delta)}{delta},$$
where $delta =frac{n}{t}$.
$endgroup$
$begingroup$
In the final result $delta$ has the same exact meaning of $d$, identify the same divisors, right?
$endgroup$
– Alessar
Nov 27 '18 at 15:01
1
$begingroup$
Yep. I used a different variable because I wanted to define $delta=frac{n}{t}$, which could be confusing if I say $d=frac{n}{t}$, since $d$ is used in $k=frac{d}{t}$ already. But then, which ever variables you are using as the indices of summation don't really matter.
$endgroup$
– user593746
Nov 27 '18 at 15:03
$begingroup$
You're welcome, thanks for the very good answer!
$endgroup$
– Alessar
Jan 16 at 15:00
add a comment |
$begingroup$
We have $$sum_{dmid n}sigma(d)=sum_{dmid n}sum_{tmid d}t=sum_{tmid n}sum_{kmid frac{n}{t}}t=sum_{tmid n}tsum_{kmid frac{n}t}1,$$
where $k=frac{d}{t}$. So
$$sum_{dmid n}sigma(d)=sum_{tmid n}ttauleft(frac{n}tright)=sum_{delta mid n}frac{n}{delta}tau(delta)=nsum_{deltamid n}frac{tau(delta)}{delta},$$
where $delta =frac{n}{t}$.
$endgroup$
We have $$sum_{dmid n}sigma(d)=sum_{dmid n}sum_{tmid d}t=sum_{tmid n}sum_{kmid frac{n}{t}}t=sum_{tmid n}tsum_{kmid frac{n}t}1,$$
where $k=frac{d}{t}$. So
$$sum_{dmid n}sigma(d)=sum_{tmid n}ttauleft(frac{n}tright)=sum_{delta mid n}frac{n}{delta}tau(delta)=nsum_{deltamid n}frac{tau(delta)}{delta},$$
where $delta =frac{n}{t}$.
answered Nov 27 '18 at 14:51
user593746
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In the final result $delta$ has the same exact meaning of $d$, identify the same divisors, right?
$endgroup$
– Alessar
Nov 27 '18 at 15:01
1
$begingroup$
Yep. I used a different variable because I wanted to define $delta=frac{n}{t}$, which could be confusing if I say $d=frac{n}{t}$, since $d$ is used in $k=frac{d}{t}$ already. But then, which ever variables you are using as the indices of summation don't really matter.
$endgroup$
– user593746
Nov 27 '18 at 15:03
$begingroup$
You're welcome, thanks for the very good answer!
$endgroup$
– Alessar
Jan 16 at 15:00
add a comment |
$begingroup$
In the final result $delta$ has the same exact meaning of $d$, identify the same divisors, right?
$endgroup$
– Alessar
Nov 27 '18 at 15:01
1
$begingroup$
Yep. I used a different variable because I wanted to define $delta=frac{n}{t}$, which could be confusing if I say $d=frac{n}{t}$, since $d$ is used in $k=frac{d}{t}$ already. But then, which ever variables you are using as the indices of summation don't really matter.
$endgroup$
– user593746
Nov 27 '18 at 15:03
$begingroup$
You're welcome, thanks for the very good answer!
$endgroup$
– Alessar
Jan 16 at 15:00
$begingroup$
In the final result $delta$ has the same exact meaning of $d$, identify the same divisors, right?
$endgroup$
– Alessar
Nov 27 '18 at 15:01
$begingroup$
In the final result $delta$ has the same exact meaning of $d$, identify the same divisors, right?
$endgroup$
– Alessar
Nov 27 '18 at 15:01
1
1
$begingroup$
Yep. I used a different variable because I wanted to define $delta=frac{n}{t}$, which could be confusing if I say $d=frac{n}{t}$, since $d$ is used in $k=frac{d}{t}$ already. But then, which ever variables you are using as the indices of summation don't really matter.
$endgroup$
– user593746
Nov 27 '18 at 15:03
$begingroup$
Yep. I used a different variable because I wanted to define $delta=frac{n}{t}$, which could be confusing if I say $d=frac{n}{t}$, since $d$ is used in $k=frac{d}{t}$ already. But then, which ever variables you are using as the indices of summation don't really matter.
$endgroup$
– user593746
Nov 27 '18 at 15:03
$begingroup$
You're welcome, thanks for the very good answer!
$endgroup$
– Alessar
Jan 16 at 15:00
$begingroup$
You're welcome, thanks for the very good answer!
$endgroup$
– Alessar
Jan 16 at 15:00
add a comment |
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Use math.stackexchange.com/questions/1948422/…
$endgroup$
– lab bhattacharjee
Nov 27 '18 at 14:51
1
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The sums express two different ways to count the sum of how many divisors $cmid n$ each divisor $dmid n$ divides.
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– Servaes
Nov 27 '18 at 14:55