Mixing
Proof of Mobius inversion formula - the other direction
$begingroup$
The Mobius inversion formula states that if we define $f$ as $$f(m)=sum_{d|m}g(d)$$ then $$g(m)=sum_{d|m}fleft(frac mdright)mu(d)$$
We know that the other direction is also true: if we define $g$ as above, then $f$ also satisfies the above.
I've searched for the proof, but failed to find any that does not rely on convolutions. My question is whether there is a proof that relies on manipulating sums that proves the second direction of the theorem.
elementary-number-theory mobius-inversion
asked Jan 27 at 10:39
windircursewindircurse
1,154820
$endgroup$
add a comment |
$begingroup$
The Mobius inversion formula states that if we define $f$ as $$f(m)=sum_{d|m}g(d)$$ then $$g(m)=sum_{d|m}fleft(frac mdright)mu(d)$$
We know that the other direction is also true: if we define $g$ as above, then $f$ also satisfies the above.
I've searched for the proof, but failed to find any that does not rely on convolutions. My question is whether there is a proof that relies on manipulating sums that proves the second direction of the theorem.
elementary-number-theory mobius-inversion
asked Jan 27 at 10:39
windircursewindircurse
1,154820
$endgroup$
1
$begingroup$
How about just doing it? Put the second into the first, and hack away....
$endgroup$
– Lord Shark the Unknown
Jan 27 at 10:41
$begingroup$
@LordSharktheUnknown not sure how I would do that, sorry
$endgroup$
– windircurse
Jan 27 at 10:54
2
$begingroup$
It's kind of hard to avoid using convolutions when there's one in the problem's statement.
$endgroup$
– jmerry
Jan 27 at 10:57
add a comment |
$begingroup$
The Mobius inversion formula states that if we define $f$ as $$f(m)=sum_{d|m}g(d)$$ then $$g(m)=sum_{d|m}fleft(frac mdright)mu(d)$$
We know that the other direction is also true: if we define $g$ as above, then $f$ also satisfies the above.
I've searched for the proof, but failed to find any that does not rely on convolutions. My question is whether there is a proof that relies on manipulating sums that proves the second direction of the theorem.
elementary-number-theory mobius-inversion
asked Jan 27 at 10:39
windircursewindircurse
1,154820
$endgroup$
The Mobius inversion formula states that if we define $f$ as $$f(m)=sum_{d|m}g(d)$$ then $$g(m)=sum_{d|m}fleft(frac mdright)mu(d)$$
We know that the other direction is also true: if we define $g$ as above, then $f$ also satisfies the above.
I've searched for the proof, but failed to find any that does not rely on convolutions. My question is whether there is a proof that relies on manipulating sums that proves the second direction of the theorem.
elementary-number-theory mobius-inversion
elementary-number-theory mobius-inversion
asked Jan 27 at 10:39
windircursewindircurse
1,154820
asked Jan 27 at 10:39
windircursewindircurse
1,154820
asked Jan 27 at 10:39
windircursewindircurse
1,154820
asked Jan 27 at 10:39
windircursewindircurse
1,154820
asked Jan 27 at 10:39
windircursewindircurse
1,154820
1,154820
1
$begingroup$
How about just doing it? Put the second into the first, and hack away....
$endgroup$
– Lord Shark the Unknown
Jan 27 at 10:41
$begingroup$
@LordSharktheUnknown not sure how I would do that, sorry
$endgroup$
– windircurse
Jan 27 at 10:54
2
$begingroup$
It's kind of hard to avoid using convolutions when there's one in the problem's statement.
$endgroup$
– jmerry
Jan 27 at 10:57
add a comment |
1
$begingroup$
How about just doing it? Put the second into the first, and hack away....
$endgroup$
– Lord Shark the Unknown
Jan 27 at 10:41
$begingroup$
@LordSharktheUnknown not sure how I would do that, sorry
$endgroup$
– windircurse
Jan 27 at 10:54
2
$begingroup$
It's kind of hard to avoid using convolutions when there's one in the problem's statement.
$endgroup$
– jmerry
Jan 27 at 10:57
1
1
$begingroup$
How about just doing it? Put the second into the first, and hack away....
$endgroup$
– Lord Shark the Unknown
Jan 27 at 10:41
$begingroup$
How about just doing it? Put the second into the first, and hack away....
$endgroup$
– Lord Shark the Unknown
Jan 27 at 10:41
$begingroup$
@LordSharktheUnknown not sure how I would do that, sorry
$endgroup$
– windircurse
Jan 27 at 10:54
$begingroup$
@LordSharktheUnknown not sure how I would do that, sorry
$endgroup$
– windircurse
Jan 27 at 10:54
2
2
$begingroup$
It's kind of hard to avoid using convolutions when there's one in the problem's statement.
$endgroup$
– jmerry
Jan 27 at 10:57
$begingroup$
It's kind of hard to avoid using convolutions when there's one in the problem's statement.
$endgroup$
– jmerry
Jan 27 at 10:57
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
We assume
begin{align*}
g(m)=sum_{d|m}fleft(frac mdright)mu(d)tag{1}
end{align*}
and show the validity of
begin{align*}
f(m)=sum_{d|m}g(d)tag{2}
end{align*}
It is convenient to use the unit-function $u$ defined as $u(n)=1, ngeq 1$.
We start with the right-hand side of (2) and obtain
begin{align*}
color{blue}{sum_{m|n}g(m)}&=sum_{m|n}left(sum_{d|m}fleft(frac {m}{d}right)mu(d)right)uleft(frac{n}{m}right)tag{3}\
&=sum_{acdot m=n}left(sum_{bcdot d=m}f(b)mu(d)right)u(a)\
&=sum_{acdot bcdot d=n}f(b)mu(d)u(a)\
&=sum_{bcdot m=n}f(b)left(sum_{acdot d=m}mu(d)u(a)right)\
&=sum_{m|n}fleft(frac{n}{m}right)left(sum_{d|m}mu(d)uleft(frac{m}{d}right)right)\
&=sum_{m|n}fleft(frac{n}{m}right)sum_{d|m}mu(d)\
&=sum_{m|n}fleft(frac{n}{m}right)leftlfloorfrac{1}{m}rightrfloortag{4}\
&,,color{blue}{=f(n)}tag{5}
end{align*}
and the claim (2) follows.
Comment:
In (3) we use the identity (1) and multiply for convenience only with $1=uleft(frac{n}{m}right)$.
In (4) we use the identity $sum_{d|m}mu(d)=leftlfloorfrac{1}{m}rightrfloor=begin{cases}1&m=1\0&m>1end{cases}$.
In (5) we obtain $f(n)$ since all other summands with $m>1$ in (4) give zero.
Note: We can omit the usage of $u$ in the derivation, but this might reduce readability.
answered Jan 27 at 23:17
Markus ScheuerMarkus Scheuer
63k460150
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We assume
begin{align*}
g(m)=sum_{d|m}fleft(frac mdright)mu(d)tag{1}
end{align*}
and show the validity of
begin{align*}
f(m)=sum_{d|m}g(d)tag{2}
end{align*}
It is convenient to use the unit-function $u$ defined as $u(n)=1, ngeq 1$.
We start with the right-hand side of (2) and obtain
begin{align*}
color{blue}{sum_{m|n}g(m)}&=sum_{m|n}left(sum_{d|m}fleft(frac {m}{d}right)mu(d)right)uleft(frac{n}{m}right)tag{3}\
&=sum_{acdot m=n}left(sum_{bcdot d=m}f(b)mu(d)right)u(a)\
&=sum_{acdot bcdot d=n}f(b)mu(d)u(a)\
&=sum_{bcdot m=n}f(b)left(sum_{acdot d=m}mu(d)u(a)right)\
&=sum_{m|n}fleft(frac{n}{m}right)left(sum_{d|m}mu(d)uleft(frac{m}{d}right)right)\
&=sum_{m|n}fleft(frac{n}{m}right)sum_{d|m}mu(d)\
&=sum_{m|n}fleft(frac{n}{m}right)leftlfloorfrac{1}{m}rightrfloortag{4}\
&,,color{blue}{=f(n)}tag{5}
end{align*}
and the claim (2) follows.
Comment:
In (3) we use the identity (1) and multiply for convenience only with $1=uleft(frac{n}{m}right)$.
In (4) we use the identity $sum_{d|m}mu(d)=leftlfloorfrac{1}{m}rightrfloor=begin{cases}1&m=1\0&m>1end{cases}$.
In (5) we obtain $f(n)$ since all other summands with $m>1$ in (4) give zero.
Note: We can omit the usage of $u$ in the derivation, but this might reduce readability.
answered Jan 27 at 23:17
Markus ScheuerMarkus Scheuer
63k460150
$endgroup$
add a comment |
$begingroup$
We assume
begin{align*}
g(m)=sum_{d|m}fleft(frac mdright)mu(d)tag{1}
end{align*}
and show the validity of
begin{align*}
f(m)=sum_{d|m}g(d)tag{2}
end{align*}
It is convenient to use the unit-function $u$ defined as $u(n)=1, ngeq 1$.
We start with the right-hand side of (2) and obtain
begin{align*}
color{blue}{sum_{m|n}g(m)}&=sum_{m|n}left(sum_{d|m}fleft(frac {m}{d}right)mu(d)right)uleft(frac{n}{m}right)tag{3}\
&=sum_{acdot m=n}left(sum_{bcdot d=m}f(b)mu(d)right)u(a)\
&=sum_{acdot bcdot d=n}f(b)mu(d)u(a)\
&=sum_{bcdot m=n}f(b)left(sum_{acdot d=m}mu(d)u(a)right)\
&=sum_{m|n}fleft(frac{n}{m}right)left(sum_{d|m}mu(d)uleft(frac{m}{d}right)right)\
&=sum_{m|n}fleft(frac{n}{m}right)sum_{d|m}mu(d)\
&=sum_{m|n}fleft(frac{n}{m}right)leftlfloorfrac{1}{m}rightrfloortag{4}\
&,,color{blue}{=f(n)}tag{5}
end{align*}
and the claim (2) follows.
Comment:
In (3) we use the identity (1) and multiply for convenience only with $1=uleft(frac{n}{m}right)$.
In (4) we use the identity $sum_{d|m}mu(d)=leftlfloorfrac{1}{m}rightrfloor=begin{cases}1&m=1\0&m>1end{cases}$.
In (5) we obtain $f(n)$ since all other summands with $m>1$ in (4) give zero.
Note: We can omit the usage of $u$ in the derivation, but this might reduce readability.
answered Jan 27 at 23:17
Markus ScheuerMarkus Scheuer
63k460150
$endgroup$
add a comment |
$begingroup$
We assume
begin{align*}
g(m)=sum_{d|m}fleft(frac mdright)mu(d)tag{1}
end{align*}
and show the validity of
begin{align*}
f(m)=sum_{d|m}g(d)tag{2}
end{align*}
It is convenient to use the unit-function $u$ defined as $u(n)=1, ngeq 1$.
We start with the right-hand side of (2) and obtain
begin{align*}
color{blue}{sum_{m|n}g(m)}&=sum_{m|n}left(sum_{d|m}fleft(frac {m}{d}right)mu(d)right)uleft(frac{n}{m}right)tag{3}\
&=sum_{acdot m=n}left(sum_{bcdot d=m}f(b)mu(d)right)u(a)\
&=sum_{acdot bcdot d=n}f(b)mu(d)u(a)\
&=sum_{bcdot m=n}f(b)left(sum_{acdot d=m}mu(d)u(a)right)\
&=sum_{m|n}fleft(frac{n}{m}right)left(sum_{d|m}mu(d)uleft(frac{m}{d}right)right)\
&=sum_{m|n}fleft(frac{n}{m}right)sum_{d|m}mu(d)\
&=sum_{m|n}fleft(frac{n}{m}right)leftlfloorfrac{1}{m}rightrfloortag{4}\
&,,color{blue}{=f(n)}tag{5}
end{align*}
and the claim (2) follows.
Comment:
In (3) we use the identity (1) and multiply for convenience only with $1=uleft(frac{n}{m}right)$.
In (4) we use the identity $sum_{d|m}mu(d)=leftlfloorfrac{1}{m}rightrfloor=begin{cases}1&m=1\0&m>1end{cases}$.
In (5) we obtain $f(n)$ since all other summands with $m>1$ in (4) give zero.
Note: We can omit the usage of $u$ in the derivation, but this might reduce readability.
answered Jan 27 at 23:17
Markus ScheuerMarkus Scheuer
63k460150
$endgroup$
We assume
begin{align*}
g(m)=sum_{d|m}fleft(frac mdright)mu(d)tag{1}
end{align*}
and show the validity of
begin{align*}
f(m)=sum_{d|m}g(d)tag{2}
end{align*}
It is convenient to use the unit-function $u$ defined as $u(n)=1, ngeq 1$.
We start with the right-hand side of (2) and obtain
begin{align*}
color{blue}{sum_{m|n}g(m)}&=sum_{m|n}left(sum_{d|m}fleft(frac {m}{d}right)mu(d)right)uleft(frac{n}{m}right)tag{3}\
&=sum_{acdot m=n}left(sum_{bcdot d=m}f(b)mu(d)right)u(a)\
&=sum_{acdot bcdot d=n}f(b)mu(d)u(a)\
&=sum_{bcdot m=n}f(b)left(sum_{acdot d=m}mu(d)u(a)right)\
&=sum_{m|n}fleft(frac{n}{m}right)left(sum_{d|m}mu(d)uleft(frac{m}{d}right)right)\
&=sum_{m|n}fleft(frac{n}{m}right)sum_{d|m}mu(d)\
&=sum_{m|n}fleft(frac{n}{m}right)leftlfloorfrac{1}{m}rightrfloortag{4}\
&,,color{blue}{=f(n)}tag{5}
end{align*}
and the claim (2) follows.
Comment:
In (3) we use the identity (1) and multiply for convenience only with $1=uleft(frac{n}{m}right)$.
In (4) we use the identity $sum_{d|m}mu(d)=leftlfloorfrac{1}{m}rightrfloor=begin{cases}1&m=1\0&m>1end{cases}$.
In (5) we obtain $f(n)$ since all other summands with $m>1$ in (4) give zero.
Note: We can omit the usage of $u$ in the derivation, but this might reduce readability.
answered Jan 27 at 23:17
Markus ScheuerMarkus Scheuer
63k460150
edited Jan 28 at 17:23
answered Jan 27 at 23:17
Markus ScheuerMarkus Scheuer
63k460150
answered Jan 27 at 23:17
Markus ScheuerMarkus Scheuer
63k460150
answered Jan 27 at 23:17
Markus ScheuerMarkus Scheuer
63k460150
63k460150
add a comment |
add a comment |
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$begingroup$
How about just doing it? Put the second into the first, and hack away....
$endgroup$
– Lord Shark the Unknown
Jan 27 at 10:41
$begingroup$
@LordSharktheUnknown not sure how I would do that, sorry
$endgroup$
– windircurse
Jan 27 at 10:54
2
$begingroup$
It's kind of hard to avoid using convolutions when there's one in the problem's statement.
$endgroup$
– jmerry
Jan 27 at 10:57