Is there a Mistake in this article of Barret et al.












1












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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


















1












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















1












1








1





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 27 '18 at 11:16







MikeTeX

















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




















  • $begingroup$
    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












1 Answer
1






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oldest

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2












$begingroup$

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.






share|cite|improve this answer









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    1 Answer
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    active

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    1 Answer
    1






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    2












    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      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.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        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.






        share|cite|improve this answer









        $endgroup$



        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 27 '18 at 16:19









        Clement C.Clement C.

        50.6k33892




        50.6k33892






























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