Calculating $sum_{substack{k|r \ k leq n}} mu left({ {k}}right)$?
Background & Question
I recently thought of a combinatoric method to get an interesting result:
$$ sum_{r=n+1}^{n!} sum_{substack{k|r \ k leq n}} mu left({ {k}}right) = n! O(frac{1}{sqrt n}) -1 $$
I was wondering how other methods compared to mine in calculating the quantity:
$$sum_{substack{k|r \ k leq n}} mu left({ {k}}right)$$
and then summing over to see how it compared to my result? Also is there any method of calculating it without using the Riemann Hypothesis?
Proof of my method
Consider the following sums:
$$ 1+1 + 1+ 1+1 +underbracedots_{n! text{ times}} = n! $$
$$ 0 + 1+ 0+1 +0 +underbracedots_{n! text{ times}} = frac{n!}{2} $$
$$ 0 + 0+ 1+0 +0 +underbracedots_{n! text{ times}} = frac{n!}{3} $$
We proceed to do so $n$ times:
$$ 0+ 0 +underbracedots_{n text{ times}} +1+underbracedots_{n! text{ times}} = frac{n!}{n} $$
Multiplying the $r$th row with $a_r$ which is an arbitrary variable:
$$ a_1+ a_1 + a_1+ a_1+a_1 +underbracedots_{n! text{ times}} = n! a_1 $$
$$ 0 + a_2+ 0+a_2 +0 +underbracedots_{n! text{ times}} =n! frac{a_2}{2} $$
$$ 0 + 0+ a_3+0 +0 +underbracedots_{n! text{ times}} = n! frac{a_3}{3} $$
We proceed to do so $n$ times:
$$ 0+ 0 +underbracedots_{n-1 text{ times}} +a_n+underbracedots_{n! text{ times}} = n!frac{a_n}{n} $$
Adding all the above equations vertically (as it is a finite sum):
$$ underbrace{b_1}_{a_1} + underbrace{b_2}_{a_1+a_2}+ underbrace{b_3}_{a_1+a_3} + dots+ b_n +sum_{r=n+1}^{n!} tilde b_r = n!sum_{r=1}^n frac{a_r}{r}$$
Writing the above properly:
$$ sum_{r=1}^n b_r + sum_{r=n+1}^{n!} tilde b_r = n! sum_{r=1}^n frac{a_r}{r} $$
where $b_r = sum_{r|k} a_k$ and $tilde b_r = sum_{substack{k|r \ kleq n}} a_k $. Using the mobius function we know $ a_{n}=sum_{dmid n}mu left({frac {n}{d}}right)b_{d}$. Re-expressing the expression the above in terms of $b_r$:
$$ sum_{r=1}^n b_r + sum_{r=n+1}^{n!} sum_{substack{k|r \ kleq n}} sum_{dmid k}mu left({frac {k}{d}}right)b_{d} = n! sum_{r=1}^n frac{sum _{dmid r}mu left({frac {r}{d}}right)b_{d}}{r} $$
Applying $frac{partial}{partial b_1} $ both sides:
$$ 1 + sum_{r=n+1}^{n!} sum_{substack{k|r \ k leq n}} mu left({ {k}}right) = n! sum_{r=1}^n frac{mu left({{r}}right)}{r} $$
Using RH:
$$ 1 + sum_{r=n+1}^{n!} sum_{substack{k|r \ k leq n}} mu left({ {k}}right) = n! O(frac{1}{sqrt n}) $$
combinatorics number-theory mobius-function
asked Nov 25 '18 at 23:07
More Anonymous
34419
add a comment |
Background & Question
I recently thought of a combinatoric method to get an interesting result:
$$ sum_{r=n+1}^{n!} sum_{substack{k|r \ k leq n}} mu left({ {k}}right) = n! O(frac{1}{sqrt n}) -1 $$
I was wondering how other methods compared to mine in calculating the quantity:
$$sum_{substack{k|r \ k leq n}} mu left({ {k}}right)$$
and then summing over to see how it compared to my result? Also is there any method of calculating it without using the Riemann Hypothesis?
Proof of my method
Consider the following sums:
$$ 1+1 + 1+ 1+1 +underbracedots_{n! text{ times}} = n! $$
$$ 0 + 1+ 0+1 +0 +underbracedots_{n! text{ times}} = frac{n!}{2} $$
$$ 0 + 0+ 1+0 +0 +underbracedots_{n! text{ times}} = frac{n!}{3} $$
We proceed to do so $n$ times:
$$ 0+ 0 +underbracedots_{n text{ times}} +1+underbracedots_{n! text{ times}} = frac{n!}{n} $$
Multiplying the $r$th row with $a_r$ which is an arbitrary variable:
$$ a_1+ a_1 + a_1+ a_1+a_1 +underbracedots_{n! text{ times}} = n! a_1 $$
$$ 0 + a_2+ 0+a_2 +0 +underbracedots_{n! text{ times}} =n! frac{a_2}{2} $$
$$ 0 + 0+ a_3+0 +0 +underbracedots_{n! text{ times}} = n! frac{a_3}{3} $$
We proceed to do so $n$ times:
$$ 0+ 0 +underbracedots_{n-1 text{ times}} +a_n+underbracedots_{n! text{ times}} = n!frac{a_n}{n} $$
Adding all the above equations vertically (as it is a finite sum):
$$ underbrace{b_1}_{a_1} + underbrace{b_2}_{a_1+a_2}+ underbrace{b_3}_{a_1+a_3} + dots+ b_n +sum_{r=n+1}^{n!} tilde b_r = n!sum_{r=1}^n frac{a_r}{r}$$
Writing the above properly:
$$ sum_{r=1}^n b_r + sum_{r=n+1}^{n!} tilde b_r = n! sum_{r=1}^n frac{a_r}{r} $$
where $b_r = sum_{r|k} a_k$ and $tilde b_r = sum_{substack{k|r \ kleq n}} a_k $. Using the mobius function we know $ a_{n}=sum_{dmid n}mu left({frac {n}{d}}right)b_{d}$. Re-expressing the expression the above in terms of $b_r$:
$$ sum_{r=1}^n b_r + sum_{r=n+1}^{n!} sum_{substack{k|r \ kleq n}} sum_{dmid k}mu left({frac {k}{d}}right)b_{d} = n! sum_{r=1}^n frac{sum _{dmid r}mu left({frac {r}{d}}right)b_{d}}{r} $$
Applying $frac{partial}{partial b_1} $ both sides:
$$ 1 + sum_{r=n+1}^{n!} sum_{substack{k|r \ k leq n}} mu left({ {k}}right) = n! sum_{r=1}^n frac{mu left({{r}}right)}{r} $$
Using RH:
$$ 1 + sum_{r=n+1}^{n!} sum_{substack{k|r \ k leq n}} mu left({ {k}}right) = n! O(frac{1}{sqrt n}) $$
combinatorics number-theory mobius-function
asked Nov 25 '18 at 23:07
More Anonymous
34419
You’ve put $rmid k$ but it should be $kmid r$ in all instances.
– Jacob
Nov 26 '18 at 1:39
Also, if we go straight from our formula for $n!frac{a_k}{k}$ to $sum_{k=1}^{n!}sum_{dmid k \ dleq n}a_d=n!sum_{k=1}^nfrac{a_k}{k}$, and then use $a_k=mu(k)$ we have $n!sum_{k=1}^nfrac{mu(k)}{k}=sum_{k=1}^{n!}sum_{dmid k \ dleq n}mu(d)=left(sum_{k=1}^nsum_{dmid k}mu(d)right)+left(sum_{k=n+1}^{n!}sum_{dmid k \ dleq n}mu(d)right)$. Since $$sum_{dmid k}mu(d)=begin{cases} 1 & n=1 \ 0 & text{otherwise},end{cases}$$ we end up with $n!sum_{k=1}^nfrac{mu(k)}{k}=1+sum_{k=n+1}^{n!}sum_{dmid k \ dleq n}mu(d)$.
– Jacob
Nov 26 '18 at 2:03
Made it $k|r$ hopefully everywhere
– More Anonymous
Nov 26 '18 at 11:05
The title still has $rmid k$.
– Gerry Myerson
Nov 27 '18 at 2:49
My bad ... fixed now
– More Anonymous
Nov 27 '18 at 2:53
add a comment |
Background & Question
I recently thought of a combinatoric method to get an interesting result:
$$ sum_{r=n+1}^{n!} sum_{substack{k|r \ k leq n}} mu left({ {k}}right) = n! O(frac{1}{sqrt n}) -1 $$
I was wondering how other methods compared to mine in calculating the quantity:
$$sum_{substack{k|r \ k leq n}} mu left({ {k}}right)$$
and then summing over to see how it compared to my result? Also is there any method of calculating it without using the Riemann Hypothesis?
Proof of my method
Consider the following sums:
$$ 1+1 + 1+ 1+1 +underbracedots_{n! text{ times}} = n! $$
$$ 0 + 1+ 0+1 +0 +underbracedots_{n! text{ times}} = frac{n!}{2} $$
$$ 0 + 0+ 1+0 +0 +underbracedots_{n! text{ times}} = frac{n!}{3} $$
We proceed to do so $n$ times:
$$ 0+ 0 +underbracedots_{n text{ times}} +1+underbracedots_{n! text{ times}} = frac{n!}{n} $$
Multiplying the $r$th row with $a_r$ which is an arbitrary variable:
$$ a_1+ a_1 + a_1+ a_1+a_1 +underbracedots_{n! text{ times}} = n! a_1 $$
$$ 0 + a_2+ 0+a_2 +0 +underbracedots_{n! text{ times}} =n! frac{a_2}{2} $$
$$ 0 + 0+ a_3+0 +0 +underbracedots_{n! text{ times}} = n! frac{a_3}{3} $$
We proceed to do so $n$ times:
$$ 0+ 0 +underbracedots_{n-1 text{ times}} +a_n+underbracedots_{n! text{ times}} = n!frac{a_n}{n} $$
Adding all the above equations vertically (as it is a finite sum):
$$ underbrace{b_1}_{a_1} + underbrace{b_2}_{a_1+a_2}+ underbrace{b_3}_{a_1+a_3} + dots+ b_n +sum_{r=n+1}^{n!} tilde b_r = n!sum_{r=1}^n frac{a_r}{r}$$
Writing the above properly:
$$ sum_{r=1}^n b_r + sum_{r=n+1}^{n!} tilde b_r = n! sum_{r=1}^n frac{a_r}{r} $$
where $b_r = sum_{r|k} a_k$ and $tilde b_r = sum_{substack{k|r \ kleq n}} a_k $. Using the mobius function we know $ a_{n}=sum_{dmid n}mu left({frac {n}{d}}right)b_{d}$. Re-expressing the expression the above in terms of $b_r$:
$$ sum_{r=1}^n b_r + sum_{r=n+1}^{n!} sum_{substack{k|r \ kleq n}} sum_{dmid k}mu left({frac {k}{d}}right)b_{d} = n! sum_{r=1}^n frac{sum _{dmid r}mu left({frac {r}{d}}right)b_{d}}{r} $$
Applying $frac{partial}{partial b_1} $ both sides:
$$ 1 + sum_{r=n+1}^{n!} sum_{substack{k|r \ k leq n}} mu left({ {k}}right) = n! sum_{r=1}^n frac{mu left({{r}}right)}{r} $$
Using RH:
$$ 1 + sum_{r=n+1}^{n!} sum_{substack{k|r \ k leq n}} mu left({ {k}}right) = n! O(frac{1}{sqrt n}) $$
combinatorics number-theory mobius-function
asked Nov 25 '18 at 23:07
More Anonymous
34419
Background & Question
I recently thought of a combinatoric method to get an interesting result:
$$ sum_{r=n+1}^{n!} sum_{substack{k|r \ k leq n}} mu left({ {k}}right) = n! O(frac{1}{sqrt n}) -1 $$
I was wondering how other methods compared to mine in calculating the quantity:
$$sum_{substack{k|r \ k leq n}} mu left({ {k}}right)$$
and then summing over to see how it compared to my result? Also is there any method of calculating it without using the Riemann Hypothesis?
Proof of my method
Consider the following sums:
$$ 1+1 + 1+ 1+1 +underbracedots_{n! text{ times}} = n! $$
$$ 0 + 1+ 0+1 +0 +underbracedots_{n! text{ times}} = frac{n!}{2} $$
$$ 0 + 0+ 1+0 +0 +underbracedots_{n! text{ times}} = frac{n!}{3} $$
We proceed to do so $n$ times:
$$ 0+ 0 +underbracedots_{n text{ times}} +1+underbracedots_{n! text{ times}} = frac{n!}{n} $$
Multiplying the $r$th row with $a_r$ which is an arbitrary variable:
$$ a_1+ a_1 + a_1+ a_1+a_1 +underbracedots_{n! text{ times}} = n! a_1 $$
$$ 0 + a_2+ 0+a_2 +0 +underbracedots_{n! text{ times}} =n! frac{a_2}{2} $$
$$ 0 + 0+ a_3+0 +0 +underbracedots_{n! text{ times}} = n! frac{a_3}{3} $$
We proceed to do so $n$ times:
$$ 0+ 0 +underbracedots_{n-1 text{ times}} +a_n+underbracedots_{n! text{ times}} = n!frac{a_n}{n} $$
Adding all the above equations vertically (as it is a finite sum):
$$ underbrace{b_1}_{a_1} + underbrace{b_2}_{a_1+a_2}+ underbrace{b_3}_{a_1+a_3} + dots+ b_n +sum_{r=n+1}^{n!} tilde b_r = n!sum_{r=1}^n frac{a_r}{r}$$
Writing the above properly:
$$ sum_{r=1}^n b_r + sum_{r=n+1}^{n!} tilde b_r = n! sum_{r=1}^n frac{a_r}{r} $$
where $b_r = sum_{r|k} a_k$ and $tilde b_r = sum_{substack{k|r \ kleq n}} a_k $. Using the mobius function we know $ a_{n}=sum_{dmid n}mu left({frac {n}{d}}right)b_{d}$. Re-expressing the expression the above in terms of $b_r$:
$$ sum_{r=1}^n b_r + sum_{r=n+1}^{n!} sum_{substack{k|r \ kleq n}} sum_{dmid k}mu left({frac {k}{d}}right)b_{d} = n! sum_{r=1}^n frac{sum _{dmid r}mu left({frac {r}{d}}right)b_{d}}{r} $$
Applying $frac{partial}{partial b_1} $ both sides:
$$ 1 + sum_{r=n+1}^{n!} sum_{substack{k|r \ k leq n}} mu left({ {k}}right) = n! sum_{r=1}^n frac{mu left({{r}}right)}{r} $$
Using RH:
$$ 1 + sum_{r=n+1}^{n!} sum_{substack{k|r \ k leq n}} mu left({ {k}}right) = n! O(frac{1}{sqrt n}) $$
combinatorics number-theory mobius-function
combinatorics number-theory mobius-function
asked Nov 25 '18 at 23:07
More Anonymous
34419
asked Nov 25 '18 at 23:07
More Anonymous
34419
edited Nov 27 '18 at 2:51
asked Nov 25 '18 at 23:07
More Anonymous
34419
asked Nov 25 '18 at 23:07
More Anonymous
34419
asked Nov 25 '18 at 23:07
More Anonymous
34419
34419
You’ve put $rmid k$ but it should be $kmid r$ in all instances.
– Jacob
Nov 26 '18 at 1:39
Also, if we go straight from our formula for $n!frac{a_k}{k}$ to $sum_{k=1}^{n!}sum_{dmid k \ dleq n}a_d=n!sum_{k=1}^nfrac{a_k}{k}$, and then use $a_k=mu(k)$ we have $n!sum_{k=1}^nfrac{mu(k)}{k}=sum_{k=1}^{n!}sum_{dmid k \ dleq n}mu(d)=left(sum_{k=1}^nsum_{dmid k}mu(d)right)+left(sum_{k=n+1}^{n!}sum_{dmid k \ dleq n}mu(d)right)$. Since $$sum_{dmid k}mu(d)=begin{cases} 1 & n=1 \ 0 & text{otherwise},end{cases}$$ we end up with $n!sum_{k=1}^nfrac{mu(k)}{k}=1+sum_{k=n+1}^{n!}sum_{dmid k \ dleq n}mu(d)$.
– Jacob
Nov 26 '18 at 2:03
Made it $k|r$ hopefully everywhere
– More Anonymous
Nov 26 '18 at 11:05
The title still has $rmid k$.
– Gerry Myerson
Nov 27 '18 at 2:49
My bad ... fixed now
– More Anonymous
Nov 27 '18 at 2:53
add a comment |
You’ve put $rmid k$ but it should be $kmid r$ in all instances.
– Jacob
Nov 26 '18 at 1:39
Also, if we go straight from our formula for $n!frac{a_k}{k}$ to $sum_{k=1}^{n!}sum_{dmid k \ dleq n}a_d=n!sum_{k=1}^nfrac{a_k}{k}$, and then use $a_k=mu(k)$ we have $n!sum_{k=1}^nfrac{mu(k)}{k}=sum_{k=1}^{n!}sum_{dmid k \ dleq n}mu(d)=left(sum_{k=1}^nsum_{dmid k}mu(d)right)+left(sum_{k=n+1}^{n!}sum_{dmid k \ dleq n}mu(d)right)$. Since $$sum_{dmid k}mu(d)=begin{cases} 1 & n=1 \ 0 & text{otherwise},end{cases}$$ we end up with $n!sum_{k=1}^nfrac{mu(k)}{k}=1+sum_{k=n+1}^{n!}sum_{dmid k \ dleq n}mu(d)$.
– Jacob
Nov 26 '18 at 2:03
Made it $k|r$ hopefully everywhere
– More Anonymous
Nov 26 '18 at 11:05
The title still has $rmid k$.
– Gerry Myerson
Nov 27 '18 at 2:49
My bad ... fixed now
– More Anonymous
Nov 27 '18 at 2:53
You’ve put $rmid k$ but it should be $kmid r$ in all instances.
– Jacob
Nov 26 '18 at 1:39
You’ve put $rmid k$ but it should be $kmid r$ in all instances.
– Jacob
Nov 26 '18 at 1:39
Also, if we go straight from our formula for $n!frac{a_k}{k}$ to $sum_{k=1}^{n!}sum_{dmid k \ dleq n}a_d=n!sum_{k=1}^nfrac{a_k}{k}$, and then use $a_k=mu(k)$ we have $n!sum_{k=1}^nfrac{mu(k)}{k}=sum_{k=1}^{n!}sum_{dmid k \ dleq n}mu(d)=left(sum_{k=1}^nsum_{dmid k}mu(d)right)+left(sum_{k=n+1}^{n!}sum_{dmid k \ dleq n}mu(d)right)$. Since $$sum_{dmid k}mu(d)=begin{cases} 1 & n=1 \ 0 & text{otherwise},end{cases}$$ we end up with $n!sum_{k=1}^nfrac{mu(k)}{k}=1+sum_{k=n+1}^{n!}sum_{dmid k \ dleq n}mu(d)$.
– Jacob
Nov 26 '18 at 2:03
Also, if we go straight from our formula for $n!frac{a_k}{k}$ to $sum_{k=1}^{n!}sum_{dmid k \ dleq n}a_d=n!sum_{k=1}^nfrac{a_k}{k}$, and then use $a_k=mu(k)$ we have $n!sum_{k=1}^nfrac{mu(k)}{k}=sum_{k=1}^{n!}sum_{dmid k \ dleq n}mu(d)=left(sum_{k=1}^nsum_{dmid k}mu(d)right)+left(sum_{k=n+1}^{n!}sum_{dmid k \ dleq n}mu(d)right)$. Since $$sum_{dmid k}mu(d)=begin{cases} 1 & n=1 \ 0 & text{otherwise},end{cases}$$ we end up with $n!sum_{k=1}^nfrac{mu(k)}{k}=1+sum_{k=n+1}^{n!}sum_{dmid k \ dleq n}mu(d)$.
– Jacob
Nov 26 '18 at 2:03
Made it $k|r$ hopefully everywhere
– More Anonymous
Nov 26 '18 at 11:05
Made it $k|r$ hopefully everywhere
– More Anonymous
Nov 26 '18 at 11:05
The title still has $rmid k$.
– Gerry Myerson
Nov 27 '18 at 2:49
The title still has $rmid k$.
– Gerry Myerson
Nov 27 '18 at 2:49
My bad ... fixed now
– More Anonymous
Nov 27 '18 at 2:53
My bad ... fixed now
– More Anonymous
Nov 27 '18 at 2:53
add a comment |
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You’ve put $rmid k$ but it should be $kmid r$ in all instances.
– Jacob
Nov 26 '18 at 1:39
Also, if we go straight from our formula for $n!frac{a_k}{k}$ to $sum_{k=1}^{n!}sum_{dmid k \ dleq n}a_d=n!sum_{k=1}^nfrac{a_k}{k}$, and then use $a_k=mu(k)$ we have $n!sum_{k=1}^nfrac{mu(k)}{k}=sum_{k=1}^{n!}sum_{dmid k \ dleq n}mu(d)=left(sum_{k=1}^nsum_{dmid k}mu(d)right)+left(sum_{k=n+1}^{n!}sum_{dmid k \ dleq n}mu(d)right)$. Since $$sum_{dmid k}mu(d)=begin{cases} 1 & n=1 \ 0 & text{otherwise},end{cases}$$ we end up with $n!sum_{k=1}^nfrac{mu(k)}{k}=1+sum_{k=n+1}^{n!}sum_{dmid k \ dleq n}mu(d)$.
– Jacob
Nov 26 '18 at 2:03
Made it $k|r$ hopefully everywhere
– More Anonymous
Nov 26 '18 at 11:05
The title still has $rmid k$.
– Gerry Myerson
Nov 27 '18 at 2:49
My bad ... fixed now
– More Anonymous
Nov 27 '18 at 2:53