Upper bound like $O(log T)$ for the following sum of square root.
1
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I'm wondering if an upper bound in the order of $O(log T)$ can be gotten for this sum:
$sum_{t=1}^{T}sqrt{sum_{j=1}^{d}frac{a_j(t+1)-a_j(t)}{a_j^2(t)}}$,
where $d>0$ is a fixed integer, $a_j(t+1)ge a_j(t)>1$, $sum_j a_j(t+1)-a_j(t)le 1$ and $sum_j a_j(t)le t+d$.
real-analysis sequences-and-series inequality summation
asked Jan 14 at 17:13
Jimmy KangJimmy Kang
818
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add a comment |
1
$begingroup$
I'm wondering if an upper bound in the order of $O(log T)$ can be gotten for this sum:
$sum_{t=1}^{T}sqrt{sum_{j=1}^{d}frac{a_j(t+1)-a_j(t)}{a_j^2(t)}}$,
where $d>0$ is a fixed integer, $a_j(t+1)ge a_j(t)>1$, $sum_j a_j(t+1)-a_j(t)le 1$ and $sum_j a_j(t)le t+d$.
real-analysis sequences-and-series inequality summation
asked Jan 14 at 17:13
Jimmy KangJimmy Kang
818
$endgroup$
$begingroup$
Random comments: Various constructions all seem to obey the $O(log T)$ bound. WLOG we can take $a_j(1)=1$. The case $d=1$ already seems to contain the essential complexity. Setting $delta(t) = a_1(t+1)-a_1(t) le 1$, we see we want an upper bound for $$frac{sqrt{delta_1}}1 + frac{sqrt{delta_2}}{1+delta_1} + frac{sqrt{delta_3}}{1+delta_1+delta_2} + cdots + frac{sqrt{delta_T}}{1+delta_1+cdots+delta_{T-1}}.$$ Some examination suggests that this function is increasing in each $delta_t$, which would mean that the maximum is given by $a_1(t) = t$ which satisfies $O(log T)$.
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– Greg Martin
Jan 15 at 8:56
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I find for $d = 1$, if $a(t+1)-a(t)=1/T$, $sum_{t=1}^{T}frac{sqrt{1/T}}{a(1)+t/T} ge frac{sqrt{T}}{a(1)+1}$. So essentially this sum is not upper bounded by $O(log T)$.
$endgroup$
– Jimmy Kang
Jan 15 at 9:41
add a comment |
1
1
1
$begingroup$
I'm wondering if an upper bound in the order of $O(log T)$ can be gotten for this sum:
$sum_{t=1}^{T}sqrt{sum_{j=1}^{d}frac{a_j(t+1)-a_j(t)}{a_j^2(t)}}$,
where $d>0$ is a fixed integer, $a_j(t+1)ge a_j(t)>1$, $sum_j a_j(t+1)-a_j(t)le 1$ and $sum_j a_j(t)le t+d$.
real-analysis sequences-and-series inequality summation
asked Jan 14 at 17:13
Jimmy KangJimmy Kang
818
$endgroup$
I'm wondering if an upper bound in the order of $O(log T)$ can be gotten for this sum:
$sum_{t=1}^{T}sqrt{sum_{j=1}^{d}frac{a_j(t+1)-a_j(t)}{a_j^2(t)}}$,
where $d>0$ is a fixed integer, $a_j(t+1)ge a_j(t)>1$, $sum_j a_j(t+1)-a_j(t)le 1$ and $sum_j a_j(t)le t+d$.
real-analysis sequences-and-series inequality summation
real-analysis sequences-and-series inequality summation
asked Jan 14 at 17:13
Jimmy KangJimmy Kang
818
asked Jan 14 at 17:13
Jimmy KangJimmy Kang
818
edited Jan 15 at 8:11
Jimmy Kang
asked Jan 14 at 17:13
Jimmy KangJimmy Kang
818
asked Jan 14 at 17:13
Jimmy KangJimmy Kang
818
asked Jan 14 at 17:13
Jimmy KangJimmy Kang
818
818
$begingroup$
Random comments: Various constructions all seem to obey the $O(log T)$ bound. WLOG we can take $a_j(1)=1$. The case $d=1$ already seems to contain the essential complexity. Setting $delta(t) = a_1(t+1)-a_1(t) le 1$, we see we want an upper bound for $$frac{sqrt{delta_1}}1 + frac{sqrt{delta_2}}{1+delta_1} + frac{sqrt{delta_3}}{1+delta_1+delta_2} + cdots + frac{sqrt{delta_T}}{1+delta_1+cdots+delta_{T-1}}.$$ Some examination suggests that this function is increasing in each $delta_t$, which would mean that the maximum is given by $a_1(t) = t$ which satisfies $O(log T)$.
$endgroup$
– Greg Martin
Jan 15 at 8:56
$begingroup$
I find for $d = 1$, if $a(t+1)-a(t)=1/T$, $sum_{t=1}^{T}frac{sqrt{1/T}}{a(1)+t/T} ge frac{sqrt{T}}{a(1)+1}$. So essentially this sum is not upper bounded by $O(log T)$.
$endgroup$
– Jimmy Kang
Jan 15 at 9:41
add a comment |
$begingroup$
Random comments: Various constructions all seem to obey the $O(log T)$ bound. WLOG we can take $a_j(1)=1$. The case $d=1$ already seems to contain the essential complexity. Setting $delta(t) = a_1(t+1)-a_1(t) le 1$, we see we want an upper bound for $$frac{sqrt{delta_1}}1 + frac{sqrt{delta_2}}{1+delta_1} + frac{sqrt{delta_3}}{1+delta_1+delta_2} + cdots + frac{sqrt{delta_T}}{1+delta_1+cdots+delta_{T-1}}.$$ Some examination suggests that this function is increasing in each $delta_t$, which would mean that the maximum is given by $a_1(t) = t$ which satisfies $O(log T)$.
$endgroup$
– Greg Martin
Jan 15 at 8:56
$begingroup$
I find for $d = 1$, if $a(t+1)-a(t)=1/T$, $sum_{t=1}^{T}frac{sqrt{1/T}}{a(1)+t/T} ge frac{sqrt{T}}{a(1)+1}$. So essentially this sum is not upper bounded by $O(log T)$.
$endgroup$
– Jimmy Kang
Jan 15 at 9:41
$begingroup$
Random comments: Various constructions all seem to obey the $O(log T)$ bound. WLOG we can take $a_j(1)=1$. The case $d=1$ already seems to contain the essential complexity. Setting $delta(t) = a_1(t+1)-a_1(t) le 1$, we see we want an upper bound for $$frac{sqrt{delta_1}}1 + frac{sqrt{delta_2}}{1+delta_1} + frac{sqrt{delta_3}}{1+delta_1+delta_2} + cdots + frac{sqrt{delta_T}}{1+delta_1+cdots+delta_{T-1}}.$$ Some examination suggests that this function is increasing in each $delta_t$, which would mean that the maximum is given by $a_1(t) = t$ which satisfies $O(log T)$.
$endgroup$
– Greg Martin
Jan 15 at 8:56
$begingroup$
Random comments: Various constructions all seem to obey the $O(log T)$ bound. WLOG we can take $a_j(1)=1$. The case $d=1$ already seems to contain the essential complexity. Setting $delta(t) = a_1(t+1)-a_1(t) le 1$, we see we want an upper bound for $$frac{sqrt{delta_1}}1 + frac{sqrt{delta_2}}{1+delta_1} + frac{sqrt{delta_3}}{1+delta_1+delta_2} + cdots + frac{sqrt{delta_T}}{1+delta_1+cdots+delta_{T-1}}.$$ Some examination suggests that this function is increasing in each $delta_t$, which would mean that the maximum is given by $a_1(t) = t$ which satisfies $O(log T)$.
$endgroup$
– Greg Martin
Jan 15 at 8:56
$begingroup$
I find for $d = 1$, if $a(t+1)-a(t)=1/T$, $sum_{t=1}^{T}frac{sqrt{1/T}}{a(1)+t/T} ge frac{sqrt{T}}{a(1)+1}$. So essentially this sum is not upper bounded by $O(log T)$.
$endgroup$
– Jimmy Kang
Jan 15 at 9:41
$begingroup$
I find for $d = 1$, if $a(t+1)-a(t)=1/T$, $sum_{t=1}^{T}frac{sqrt{1/T}}{a(1)+t/T} ge frac{sqrt{T}}{a(1)+1}$. So essentially this sum is not upper bounded by $O(log T)$.
$endgroup$
– Jimmy Kang
Jan 15 at 9:41
add a comment |
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$begingroup$
Random comments: Various constructions all seem to obey the $O(log T)$ bound. WLOG we can take $a_j(1)=1$. The case $d=1$ already seems to contain the essential complexity. Setting $delta(t) = a_1(t+1)-a_1(t) le 1$, we see we want an upper bound for $$frac{sqrt{delta_1}}1 + frac{sqrt{delta_2}}{1+delta_1} + frac{sqrt{delta_3}}{1+delta_1+delta_2} + cdots + frac{sqrt{delta_T}}{1+delta_1+cdots+delta_{T-1}}.$$ Some examination suggests that this function is increasing in each $delta_t$, which would mean that the maximum is given by $a_1(t) = t$ which satisfies $O(log T)$.
$endgroup$
– Greg Martin
Jan 15 at 8:56
$begingroup$
I find for $d = 1$, if $a(t+1)-a(t)=1/T$, $sum_{t=1}^{T}frac{sqrt{1/T}}{a(1)+t/T} ge frac{sqrt{T}}{a(1)+1}$. So essentially this sum is not upper bounded by $O(log T)$.
$endgroup$
– Jimmy Kang
Jan 15 at 9:41