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Maximum Value of $|sum_{k=1}^n (-1)^{lfloor ak/brfloor}|$
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Let $a,binmathbb{N}$ such that $(a,b)=1$. If $2nmid a$, then one can show that
$$rho_{a/b}(n)=sum_{k=1}^n (-1)^{lfloor ak/brfloor}$$
is periodic. Is there any explicit way to give the maximum of $|rho_{a/b}(n)|$ in terms of $a$ and $b$? As a first attempt we can show that
$$rho_{a/b}(n+b)=sum_{k=1}^{n+b}(-1)^{lfloor ak/brfloor}=rho_{a/b}(b)+sum_{k=1}^n(-1)^{lfloor a(k+b)/brfloor}=rho_{a/b}(b)+(-1)^arho_{a/b}(n).$$
Since we assume that $2nmid a$ we have that $rho_{a/b}(n+2b)=rho_{a/b}(n)$ meaning the period is $le 2b$. Since the function starts at $0$ and $rho_{a/b}(2b)=0$, combined with the fact that it moves in discrete steps, gives
$$max|rho_{a/b}(n)|le b.$$
However, I've noticed that the maximum doesn't seem to depend too reliably on just $b$. If you fix $b$ you can achieve a whole spectrum of bounds by adjusting $a$. Hence, I've wondering if we can improve this bound, or even better explicitly find the max.
Edit
Since $2b$ is a valid period, if $T_0$ is a fundamental period we know that $T_0mid 2b$. Moreover, we know that any period must be even due to $rho_{a/b}(T_0)=0$ so that there can be an equal number of negative terms as even terms. Thus $T_0=2d$ for some $dmid b$. We know that for whatever $T_0$ we find that
$$max |rho_{a/b}(n)|le T_0/2$$
however, the function rarely achieves this maximum of $T_0/2$.
Edit 2 Let $M(a,b)=max |rho_{a/b}(n)|$. As noted we have that $M(a_1,b)=M(a_2,b)$ if $a_1equiv a_2mod 2b$. Hence for fixed $b$, it suffices to analyze $ain [-b,b]$. Notice that we have
$$leftlfloorfrac{-ak}{b}rightrfloor=begin{cases}
-leftlfloorfrac{ak}{b}rightrfloor - 1 & bnmid k \
-leftlfloorfrac{ak}{b}rightrfloor & bmid k
end{cases}.$$
Hence,
$$rho_{a/b}(n)+rho_{-a/b}(n)=2sum_{substack{k=1 \ bmid k}}^n(-1)^{leftlfloorfrac{ak}{b}rightrfloor}=2sum_{k=1}^{lfloor n/brfloor}(-1)^k=begin{cases}
-2 & 2nmid lfloor n/brfloor \
0 & 2mid lfloor n/brfloor
end{cases}.$$
Thus
$$|M(a,b)-M(-a,b)|le 2.$$
Thus to properly analyze bounds on $M(a,b)$ it suffices to analyze $M(a,b)$ for $ain [1,b]$.
real-analysis number-theory elementary-number-theory
asked Nov 18 at 22:03
Will Fisher
3,547729
add a comment |
up vote
3
down vote
favorite
Let $a,binmathbb{N}$ such that $(a,b)=1$. If $2nmid a$, then one can show that
$$rho_{a/b}(n)=sum_{k=1}^n (-1)^{lfloor ak/brfloor}$$
is periodic. Is there any explicit way to give the maximum of $|rho_{a/b}(n)|$ in terms of $a$ and $b$? As a first attempt we can show that
$$rho_{a/b}(n+b)=sum_{k=1}^{n+b}(-1)^{lfloor ak/brfloor}=rho_{a/b}(b)+sum_{k=1}^n(-1)^{lfloor a(k+b)/brfloor}=rho_{a/b}(b)+(-1)^arho_{a/b}(n).$$
Since we assume that $2nmid a$ we have that $rho_{a/b}(n+2b)=rho_{a/b}(n)$ meaning the period is $le 2b$. Since the function starts at $0$ and $rho_{a/b}(2b)=0$, combined with the fact that it moves in discrete steps, gives
$$max|rho_{a/b}(n)|le b.$$
However, I've noticed that the maximum doesn't seem to depend too reliably on just $b$. If you fix $b$ you can achieve a whole spectrum of bounds by adjusting $a$. Hence, I've wondering if we can improve this bound, or even better explicitly find the max.
Edit
Since $2b$ is a valid period, if $T_0$ is a fundamental period we know that $T_0mid 2b$. Moreover, we know that any period must be even due to $rho_{a/b}(T_0)=0$ so that there can be an equal number of negative terms as even terms. Thus $T_0=2d$ for some $dmid b$. We know that for whatever $T_0$ we find that
$$max |rho_{a/b}(n)|le T_0/2$$
however, the function rarely achieves this maximum of $T_0/2$.
Edit 2 Let $M(a,b)=max |rho_{a/b}(n)|$. As noted we have that $M(a_1,b)=M(a_2,b)$ if $a_1equiv a_2mod 2b$. Hence for fixed $b$, it suffices to analyze $ain [-b,b]$. Notice that we have
$$leftlfloorfrac{-ak}{b}rightrfloor=begin{cases}
-leftlfloorfrac{ak}{b}rightrfloor - 1 & bnmid k \
-leftlfloorfrac{ak}{b}rightrfloor & bmid k
end{cases}.$$
Hence,
$$rho_{a/b}(n)+rho_{-a/b}(n)=2sum_{substack{k=1 \ bmid k}}^n(-1)^{leftlfloorfrac{ak}{b}rightrfloor}=2sum_{k=1}^{lfloor n/brfloor}(-1)^k=begin{cases}
-2 & 2nmid lfloor n/brfloor \
0 & 2mid lfloor n/brfloor
end{cases}.$$
Thus
$$|M(a,b)-M(-a,b)|le 2.$$
Thus to properly analyze bounds on $M(a,b)$ it suffices to analyze $M(a,b)$ for $ain [1,b]$.
real-analysis number-theory elementary-number-theory
asked Nov 18 at 22:03
Will Fisher
3,547729
It’s interesting to note that $rho_{(a+2b)/b}(x)=rho_{a/b}(x)$ since $(-1)^{lfloor (a+2b)k/b rfloor}=(-1)^{lfloor ak/b + 2krfloor}=(-1)^{lfloor ak/b rfloor}$, so we have another period of $2b$. Also, by comparing terms, we have $rho_{a/b}(x)+rho_{a/b+1}(x)=rho_{2a/b}(lfloor x/2 rfloor)$, which may be of use as well.
– Jacob
2 days ago
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Let $a,binmathbb{N}$ such that $(a,b)=1$. If $2nmid a$, then one can show that
$$rho_{a/b}(n)=sum_{k=1}^n (-1)^{lfloor ak/brfloor}$$
is periodic. Is there any explicit way to give the maximum of $|rho_{a/b}(n)|$ in terms of $a$ and $b$? As a first attempt we can show that
$$rho_{a/b}(n+b)=sum_{k=1}^{n+b}(-1)^{lfloor ak/brfloor}=rho_{a/b}(b)+sum_{k=1}^n(-1)^{lfloor a(k+b)/brfloor}=rho_{a/b}(b)+(-1)^arho_{a/b}(n).$$
Since we assume that $2nmid a$ we have that $rho_{a/b}(n+2b)=rho_{a/b}(n)$ meaning the period is $le 2b$. Since the function starts at $0$ and $rho_{a/b}(2b)=0$, combined with the fact that it moves in discrete steps, gives
$$max|rho_{a/b}(n)|le b.$$
However, I've noticed that the maximum doesn't seem to depend too reliably on just $b$. If you fix $b$ you can achieve a whole spectrum of bounds by adjusting $a$. Hence, I've wondering if we can improve this bound, or even better explicitly find the max.
Edit
Since $2b$ is a valid period, if $T_0$ is a fundamental period we know that $T_0mid 2b$. Moreover, we know that any period must be even due to $rho_{a/b}(T_0)=0$ so that there can be an equal number of negative terms as even terms. Thus $T_0=2d$ for some $dmid b$. We know that for whatever $T_0$ we find that
$$max |rho_{a/b}(n)|le T_0/2$$
however, the function rarely achieves this maximum of $T_0/2$.
Edit 2 Let $M(a,b)=max |rho_{a/b}(n)|$. As noted we have that $M(a_1,b)=M(a_2,b)$ if $a_1equiv a_2mod 2b$. Hence for fixed $b$, it suffices to analyze $ain [-b,b]$. Notice that we have
$$leftlfloorfrac{-ak}{b}rightrfloor=begin{cases}
-leftlfloorfrac{ak}{b}rightrfloor - 1 & bnmid k \
-leftlfloorfrac{ak}{b}rightrfloor & bmid k
end{cases}.$$
Hence,
$$rho_{a/b}(n)+rho_{-a/b}(n)=2sum_{substack{k=1 \ bmid k}}^n(-1)^{leftlfloorfrac{ak}{b}rightrfloor}=2sum_{k=1}^{lfloor n/brfloor}(-1)^k=begin{cases}
-2 & 2nmid lfloor n/brfloor \
0 & 2mid lfloor n/brfloor
end{cases}.$$
Thus
$$|M(a,b)-M(-a,b)|le 2.$$
Thus to properly analyze bounds on $M(a,b)$ it suffices to analyze $M(a,b)$ for $ain [1,b]$.
real-analysis number-theory elementary-number-theory
asked Nov 18 at 22:03
Will Fisher
3,547729
Let $a,binmathbb{N}$ such that $(a,b)=1$. If $2nmid a$, then one can show that
$$rho_{a/b}(n)=sum_{k=1}^n (-1)^{lfloor ak/brfloor}$$
is periodic. Is there any explicit way to give the maximum of $|rho_{a/b}(n)|$ in terms of $a$ and $b$? As a first attempt we can show that
$$rho_{a/b}(n+b)=sum_{k=1}^{n+b}(-1)^{lfloor ak/brfloor}=rho_{a/b}(b)+sum_{k=1}^n(-1)^{lfloor a(k+b)/brfloor}=rho_{a/b}(b)+(-1)^arho_{a/b}(n).$$
Since we assume that $2nmid a$ we have that $rho_{a/b}(n+2b)=rho_{a/b}(n)$ meaning the period is $le 2b$. Since the function starts at $0$ and $rho_{a/b}(2b)=0$, combined with the fact that it moves in discrete steps, gives
$$max|rho_{a/b}(n)|le b.$$
However, I've noticed that the maximum doesn't seem to depend too reliably on just $b$. If you fix $b$ you can achieve a whole spectrum of bounds by adjusting $a$. Hence, I've wondering if we can improve this bound, or even better explicitly find the max.
Edit
Since $2b$ is a valid period, if $T_0$ is a fundamental period we know that $T_0mid 2b$. Moreover, we know that any period must be even due to $rho_{a/b}(T_0)=0$ so that there can be an equal number of negative terms as even terms. Thus $T_0=2d$ for some $dmid b$. We know that for whatever $T_0$ we find that
$$max |rho_{a/b}(n)|le T_0/2$$
however, the function rarely achieves this maximum of $T_0/2$.
Edit 2 Let $M(a,b)=max |rho_{a/b}(n)|$. As noted we have that $M(a_1,b)=M(a_2,b)$ if $a_1equiv a_2mod 2b$. Hence for fixed $b$, it suffices to analyze $ain [-b,b]$. Notice that we have
$$leftlfloorfrac{-ak}{b}rightrfloor=begin{cases}
-leftlfloorfrac{ak}{b}rightrfloor - 1 & bnmid k \
-leftlfloorfrac{ak}{b}rightrfloor & bmid k
end{cases}.$$
Hence,
$$rho_{a/b}(n)+rho_{-a/b}(n)=2sum_{substack{k=1 \ bmid k}}^n(-1)^{leftlfloorfrac{ak}{b}rightrfloor}=2sum_{k=1}^{lfloor n/brfloor}(-1)^k=begin{cases}
-2 & 2nmid lfloor n/brfloor \
0 & 2mid lfloor n/brfloor
end{cases}.$$
Thus
$$|M(a,b)-M(-a,b)|le 2.$$
Thus to properly analyze bounds on $M(a,b)$ it suffices to analyze $M(a,b)$ for $ain [1,b]$.
real-analysis number-theory elementary-number-theory
real-analysis number-theory elementary-number-theory
asked Nov 18 at 22:03
Will Fisher
3,547729
asked Nov 18 at 22:03
Will Fisher
3,547729
edited 14 hours ago
asked Nov 18 at 22:03
Will Fisher
3,547729
asked Nov 18 at 22:03
Will Fisher
3,547729
asked Nov 18 at 22:03
Will Fisher
3,547729
3,547729
It’s interesting to note that $rho_{(a+2b)/b}(x)=rho_{a/b}(x)$ since $(-1)^{lfloor (a+2b)k/b rfloor}=(-1)^{lfloor ak/b + 2krfloor}=(-1)^{lfloor ak/b rfloor}$, so we have another period of $2b$. Also, by comparing terms, we have $rho_{a/b}(x)+rho_{a/b+1}(x)=rho_{2a/b}(lfloor x/2 rfloor)$, which may be of use as well.
– Jacob
2 days ago
add a comment |
It’s interesting to note that $rho_{(a+2b)/b}(x)=rho_{a/b}(x)$ since $(-1)^{lfloor (a+2b)k/b rfloor}=(-1)^{lfloor ak/b + 2krfloor}=(-1)^{lfloor ak/b rfloor}$, so we have another period of $2b$. Also, by comparing terms, we have $rho_{a/b}(x)+rho_{a/b+1}(x)=rho_{2a/b}(lfloor x/2 rfloor)$, which may be of use as well.
– Jacob
2 days ago
It’s interesting to note that $rho_{(a+2b)/b}(x)=rho_{a/b}(x)$ since $(-1)^{lfloor (a+2b)k/b rfloor}=(-1)^{lfloor ak/b + 2krfloor}=(-1)^{lfloor ak/b rfloor}$, so we have another period of $2b$. Also, by comparing terms, we have $rho_{a/b}(x)+rho_{a/b+1}(x)=rho_{2a/b}(lfloor x/2 rfloor)$, which may be of use as well.
– Jacob
2 days ago
It’s interesting to note that $rho_{(a+2b)/b}(x)=rho_{a/b}(x)$ since $(-1)^{lfloor (a+2b)k/b rfloor}=(-1)^{lfloor ak/b + 2krfloor}=(-1)^{lfloor ak/b rfloor}$, so we have another period of $2b$. Also, by comparing terms, we have $rho_{a/b}(x)+rho_{a/b+1}(x)=rho_{2a/b}(lfloor x/2 rfloor)$, which may be of use as well.
– Jacob
2 days ago
add a comment |
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It’s interesting to note that $rho_{(a+2b)/b}(x)=rho_{a/b}(x)$ since $(-1)^{lfloor (a+2b)k/b rfloor}=(-1)^{lfloor ak/b + 2krfloor}=(-1)^{lfloor ak/b rfloor}$, so we have another period of $2b$. Also, by comparing terms, we have $rho_{a/b}(x)+rho_{a/b+1}(x)=rho_{2a/b}(lfloor x/2 rfloor)$, which may be of use as well.
– Jacob
2 days ago