Is there a Mistake in this article of Barret et al.
$begingroup$
In their article "On the spectral radius of a {0,1) Matrix Related to Mertens' Function", Barret et al. assert an inequality just at the end of p. 156.
Apparently, this inequality comes from the Abel summation formula applied to
$C(n, k+1) = sum_w ( C(w, k) - C(w, k-1)) ([n/w] - 1)$,
which yields in this case to
$C(n,k+1) = sum_w C(w,k) ([n/w] - [n/(w+1)])$.
And apparently, they allow themselves to replace the term $[n/w]-[n/(w+1)]$ by $n/w - n/(w+1)$, upon transforming the equality into an inequality.
That is, they consider that
$[n/w]-[n/(w+1)] leq n/w - n/(w+1)$.
But this is false, as can be seen for example with n = 10 and w = 3 or n = 15 and w = 3. Am I missing something?
real-analysis analysis summation summation-method
asked Dec 27 '18 at 10:23
MikeTeXMikeTeX
1,278412
$endgroup$
|
show 6 more comments
$begingroup$
In their article "On the spectral radius of a {0,1) Matrix Related to Mertens' Function", Barret et al. assert an inequality just at the end of p. 156.
Apparently, this inequality comes from the Abel summation formula applied to
$C(n, k+1) = sum_w ( C(w, k) - C(w, k-1)) ([n/w] - 1)$,
which yields in this case to
$C(n,k+1) = sum_w C(w,k) ([n/w] - [n/(w+1)])$.
And apparently, they allow themselves to replace the term $[n/w]-[n/(w+1)]$ by $n/w - n/(w+1)$, upon transforming the equality into an inequality.
That is, they consider that
$[n/w]-[n/(w+1)] leq n/w - n/(w+1)$.
But this is false, as can be seen for example with n = 10 and w = 3 or n = 15 and w = 3. Am I missing something?
real-analysis analysis summation summation-method
asked Dec 27 '18 at 10:23
MikeTeXMikeTeX
1,278412
$endgroup$
$begingroup$
I can't see what you say happens. What I see at the bottom of page 156 is that they have on the left hand $$left[c(w,k)-c(w,k-1)right]left(leftlfloor frac nwrightrfloor-1right)$$ and apparently they evaluate this as less than or equal $$c(w,k)left(frac nw-frac n{w+1}right)$$ This doesn't look like what you wrote...
$endgroup$
– DonAntonio
Dec 27 '18 at 10:36
$begingroup$
I am trying to explain this inequality: In my opinion, they first apply an Abel transform to the first expression you wrote, and then mistakenly remove the "brackets". See my question for more details.
$endgroup$
– MikeTeX
Dec 27 '18 at 11:08
$begingroup$
I did read your question. I just can't see why you think that "they consider" that inequality.
$endgroup$
– DonAntonio
Dec 27 '18 at 11:10
$begingroup$
I have edited this question to make it more understandable.
$endgroup$
– MikeTeX
Dec 27 '18 at 11:14
$begingroup$
Are you sure it yields what you say it does? Because if you are then they obviously have a mistake, and a rather big one, with that inequality, as you show with the values of $;n,,w;$ that you give as example.
$endgroup$
– DonAntonio
Dec 27 '18 at 11:16
|
show 6 more comments
$begingroup$
In their article "On the spectral radius of a {0,1) Matrix Related to Mertens' Function", Barret et al. assert an inequality just at the end of p. 156.
Apparently, this inequality comes from the Abel summation formula applied to
$C(n, k+1) = sum_w ( C(w, k) - C(w, k-1)) ([n/w] - 1)$,
which yields in this case to
$C(n,k+1) = sum_w C(w,k) ([n/w] - [n/(w+1)])$.
And apparently, they allow themselves to replace the term $[n/w]-[n/(w+1)]$ by $n/w - n/(w+1)$, upon transforming the equality into an inequality.
That is, they consider that
$[n/w]-[n/(w+1)] leq n/w - n/(w+1)$.
But this is false, as can be seen for example with n = 10 and w = 3 or n = 15 and w = 3. Am I missing something?
real-analysis analysis summation summation-method
asked Dec 27 '18 at 10:23
MikeTeXMikeTeX
1,278412
$endgroup$
In their article "On the spectral radius of a {0,1) Matrix Related to Mertens' Function", Barret et al. assert an inequality just at the end of p. 156.
Apparently, this inequality comes from the Abel summation formula applied to
$C(n, k+1) = sum_w ( C(w, k) - C(w, k-1)) ([n/w] - 1)$,
which yields in this case to
$C(n,k+1) = sum_w C(w,k) ([n/w] - [n/(w+1)])$.
And apparently, they allow themselves to replace the term $[n/w]-[n/(w+1)]$ by $n/w - n/(w+1)$, upon transforming the equality into an inequality.
That is, they consider that
$[n/w]-[n/(w+1)] leq n/w - n/(w+1)$.
But this is false, as can be seen for example with n = 10 and w = 3 or n = 15 and w = 3. Am I missing something?
real-analysis analysis summation summation-method
real-analysis analysis summation summation-method
asked Dec 27 '18 at 10:23
MikeTeXMikeTeX
1,278412
asked Dec 27 '18 at 10:23
MikeTeXMikeTeX
1,278412
edited Dec 27 '18 at 11:16
MikeTeX
asked Dec 27 '18 at 10:23
MikeTeXMikeTeX
1,278412
asked Dec 27 '18 at 10:23
MikeTeXMikeTeX
1,278412
asked Dec 27 '18 at 10:23
MikeTeXMikeTeX
1,278412
1,278412
$begingroup$
I can't see what you say happens. What I see at the bottom of page 156 is that they have on the left hand $$left[c(w,k)-c(w,k-1)right]left(leftlfloor frac nwrightrfloor-1right)$$ and apparently they evaluate this as less than or equal $$c(w,k)left(frac nw-frac n{w+1}right)$$ This doesn't look like what you wrote...
$endgroup$
– DonAntonio
Dec 27 '18 at 10:36
$begingroup$
I am trying to explain this inequality: In my opinion, they first apply an Abel transform to the first expression you wrote, and then mistakenly remove the "brackets". See my question for more details.
$endgroup$
– MikeTeX
Dec 27 '18 at 11:08
$begingroup$
I did read your question. I just can't see why you think that "they consider" that inequality.
$endgroup$
– DonAntonio
Dec 27 '18 at 11:10
$begingroup$
I have edited this question to make it more understandable.
$endgroup$
– MikeTeX
Dec 27 '18 at 11:14
$begingroup$
Are you sure it yields what you say it does? Because if you are then they obviously have a mistake, and a rather big one, with that inequality, as you show with the values of $;n,,w;$ that you give as example.
$endgroup$
– DonAntonio
Dec 27 '18 at 11:16
|
show 6 more comments
$begingroup$
I can't see what you say happens. What I see at the bottom of page 156 is that they have on the left hand $$left[c(w,k)-c(w,k-1)right]left(leftlfloor frac nwrightrfloor-1right)$$ and apparently they evaluate this as less than or equal $$c(w,k)left(frac nw-frac n{w+1}right)$$ This doesn't look like what you wrote...
$endgroup$
– DonAntonio
Dec 27 '18 at 10:36
$begingroup$
I am trying to explain this inequality: In my opinion, they first apply an Abel transform to the first expression you wrote, and then mistakenly remove the "brackets". See my question for more details.
$endgroup$
– MikeTeX
Dec 27 '18 at 11:08
$begingroup$
I did read your question. I just can't see why you think that "they consider" that inequality.
$endgroup$
– DonAntonio
Dec 27 '18 at 11:10
$begingroup$
I have edited this question to make it more understandable.
$endgroup$
– MikeTeX
Dec 27 '18 at 11:14
$begingroup$
Are you sure it yields what you say it does? Because if you are then they obviously have a mistake, and a rather big one, with that inequality, as you show with the values of $;n,,w;$ that you give as example.
$endgroup$
– DonAntonio
Dec 27 '18 at 11:16
$begingroup$
I can't see what you say happens. What I see at the bottom of page 156 is that they have on the left hand $$left[c(w,k)-c(w,k-1)right]left(leftlfloor frac nwrightrfloor-1right)$$ and apparently they evaluate this as less than or equal $$c(w,k)left(frac nw-frac n{w+1}right)$$ This doesn't look like what you wrote...
$endgroup$
– DonAntonio
Dec 27 '18 at 10:36
$begingroup$
I can't see what you say happens. What I see at the bottom of page 156 is that they have on the left hand $$left[c(w,k)-c(w,k-1)right]left(leftlfloor frac nwrightrfloor-1right)$$ and apparently they evaluate this as less than or equal $$c(w,k)left(frac nw-frac n{w+1}right)$$ This doesn't look like what you wrote...
$endgroup$
– DonAntonio
Dec 27 '18 at 10:36
$begingroup$
I am trying to explain this inequality: In my opinion, they first apply an Abel transform to the first expression you wrote, and then mistakenly remove the "brackets". See my question for more details.
$endgroup$
– MikeTeX
Dec 27 '18 at 11:08
$begingroup$
I am trying to explain this inequality: In my opinion, they first apply an Abel transform to the first expression you wrote, and then mistakenly remove the "brackets". See my question for more details.
$endgroup$
– MikeTeX
Dec 27 '18 at 11:08
$begingroup$
I did read your question. I just can't see why you think that "they consider" that inequality.
$endgroup$
– DonAntonio
Dec 27 '18 at 11:10
$begingroup$
I did read your question. I just can't see why you think that "they consider" that inequality.
$endgroup$
– DonAntonio
Dec 27 '18 at 11:10
$begingroup$
I have edited this question to make it more understandable.
$endgroup$
– MikeTeX
Dec 27 '18 at 11:14
$begingroup$
I have edited this question to make it more understandable.
$endgroup$
– MikeTeX
Dec 27 '18 at 11:14
$begingroup$
Are you sure it yields what you say it does? Because if you are then they obviously have a mistake, and a rather big one, with that inequality, as you show with the values of $;n,,w;$ that you give as example.
$endgroup$
– DonAntonio
Dec 27 '18 at 11:16
$begingroup$
Are you sure it yields what you say it does? Because if you are then they obviously have a mistake, and a rather big one, with that inequality, as you show with the values of $;n,,w;$ that you give as example.
$endgroup$
– DonAntonio
Dec 27 '18 at 11:16
|
show 6 more comments
1 Answer
1
active
oldest
votes
$begingroup$
The key here is to perform the two steps (Abel transform, and dealing with the floors) in the other order.
Namely, first, handle the floors:
$$
sum_{w} c(w,k) left( leftlfloor frac{n}{w} rightrfloor - 1right)
leq sum_{w} c(w,k) cdot frac{n}{w}
$$
and then perform the Abel transform. This way, you avoid the troublesome (and, indeed, false) inequality at the end.
answered Dec 27 '18 at 16:19
Clement C.Clement C.
50.6k33892
$endgroup$
add a comment |
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1 Answer
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$begingroup$
The key here is to perform the two steps (Abel transform, and dealing with the floors) in the other order.
Namely, first, handle the floors:
$$
sum_{w} c(w,k) left( leftlfloor frac{n}{w} rightrfloor - 1right)
leq sum_{w} c(w,k) cdot frac{n}{w}
$$
and then perform the Abel transform. This way, you avoid the troublesome (and, indeed, false) inequality at the end.
answered Dec 27 '18 at 16:19
Clement C.Clement C.
50.6k33892
$endgroup$
add a comment |
$begingroup$
The key here is to perform the two steps (Abel transform, and dealing with the floors) in the other order.
Namely, first, handle the floors:
$$
sum_{w} c(w,k) left( leftlfloor frac{n}{w} rightrfloor - 1right)
leq sum_{w} c(w,k) cdot frac{n}{w}
$$
and then perform the Abel transform. This way, you avoid the troublesome (and, indeed, false) inequality at the end.
answered Dec 27 '18 at 16:19
Clement C.Clement C.
50.6k33892
$endgroup$
add a comment |
$begingroup$
The key here is to perform the two steps (Abel transform, and dealing with the floors) in the other order.
Namely, first, handle the floors:
$$
sum_{w} c(w,k) left( leftlfloor frac{n}{w} rightrfloor - 1right)
leq sum_{w} c(w,k) cdot frac{n}{w}
$$
and then perform the Abel transform. This way, you avoid the troublesome (and, indeed, false) inequality at the end.
answered Dec 27 '18 at 16:19
Clement C.Clement C.
50.6k33892
$endgroup$
The key here is to perform the two steps (Abel transform, and dealing with the floors) in the other order.
Namely, first, handle the floors:
$$
sum_{w} c(w,k) left( leftlfloor frac{n}{w} rightrfloor - 1right)
leq sum_{w} c(w,k) cdot frac{n}{w}
$$
and then perform the Abel transform. This way, you avoid the troublesome (and, indeed, false) inequality at the end.
answered Dec 27 '18 at 16:19
Clement C.Clement C.
50.6k33892
answered Dec 27 '18 at 16:19
Clement C.Clement C.
50.6k33892
answered Dec 27 '18 at 16:19
Clement C.Clement C.
50.6k33892
answered Dec 27 '18 at 16:19
Clement C.Clement C.
50.6k33892
50.6k33892
add a comment |
add a comment |
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$begingroup$
I can't see what you say happens. What I see at the bottom of page 156 is that they have on the left hand $$left[c(w,k)-c(w,k-1)right]left(leftlfloor frac nwrightrfloor-1right)$$ and apparently they evaluate this as less than or equal $$c(w,k)left(frac nw-frac n{w+1}right)$$ This doesn't look like what you wrote...
$endgroup$
– DonAntonio
Dec 27 '18 at 10:36
$begingroup$
I am trying to explain this inequality: In my opinion, they first apply an Abel transform to the first expression you wrote, and then mistakenly remove the "brackets". See my question for more details.
$endgroup$
– MikeTeX
Dec 27 '18 at 11:08
$begingroup$
I did read your question. I just can't see why you think that "they consider" that inequality.
$endgroup$
– DonAntonio
Dec 27 '18 at 11:10
$begingroup$
I have edited this question to make it more understandable.
$endgroup$
– MikeTeX
Dec 27 '18 at 11:14
$begingroup$
Are you sure it yields what you say it does? Because if you are then they obviously have a mistake, and a rather big one, with that inequality, as you show with the values of $;n,,w;$ that you give as example.
$endgroup$
– DonAntonio
Dec 27 '18 at 11:16