Riemann sums over dense countable sets












0












$begingroup$


Let $f$ and $g$ be positive, smooth and integrable functions in $mathbb{R}$, whose derivatives are also integrable.
Assume as well that the expression
$$
frac{sum_{qin mathbb{Q}} f(q)}{sum_{qin mathbb{Q}} g(q)} := lim_{n to infty} frac{sum_{i =1}^n f(q_i)}{sum_{i=1}^n g(q_i)}
$$
is well defined, where ${q_i}_i^infty = mathbb{Q}$.
I would like to evaluate
$$
left| frac{sum_{qin mathbb{Q}} f(q)}{sum_{qin mathbb{Q}} g(q)} - frac{int f(x) dx}{int g(x) dx} right|.
$$



When the sum is over a discrete set, I can use the hypothesis to bound the error of Riemman sums by the diameter of the partition generated by the set. What about in such case where the "diameter" is zero?






real-analysis integration analysis summation riemann-sum







share|cite|improve this question











$endgroup$












  • $begingroup$
    Related ... en.wikipedia.org/wiki/Equidistributed_sequence ... take functions on $[0,1]$ and $g(x) = 1$. Some enumerations of the rationals are equidistributed, others are not.
    $endgroup$
    – GEdgar
    Dec 6 '18 at 12:27










  • $begingroup$
    That is perfect; I think this will be really useful. Do you know if there is an error bound for those approximations of the integral?
    $endgroup$
    – Kernel
    Dec 6 '18 at 15:02
















0












$begingroup$


Let $f$ and $g$ be positive, smooth and integrable functions in $mathbb{R}$, whose derivatives are also integrable.
Assume as well that the expression
$$
frac{sum_{qin mathbb{Q}} f(q)}{sum_{qin mathbb{Q}} g(q)} := lim_{n to infty} frac{sum_{i =1}^n f(q_i)}{sum_{i=1}^n g(q_i)}
$$
is well defined, where ${q_i}_i^infty = mathbb{Q}$.
I would like to evaluate
$$
left| frac{sum_{qin mathbb{Q}} f(q)}{sum_{qin mathbb{Q}} g(q)} - frac{int f(x) dx}{int g(x) dx} right|.
$$



When the sum is over a discrete set, I can use the hypothesis to bound the error of Riemman sums by the diameter of the partition generated by the set. What about in such case where the "diameter" is zero?






real-analysis integration analysis summation riemann-sum







share|cite|improve this question











$endgroup$












  • $begingroup$
    Related ... en.wikipedia.org/wiki/Equidistributed_sequence ... take functions on $[0,1]$ and $g(x) = 1$. Some enumerations of the rationals are equidistributed, others are not.
    $endgroup$
    – GEdgar
    Dec 6 '18 at 12:27










  • $begingroup$
    That is perfect; I think this will be really useful. Do you know if there is an error bound for those approximations of the integral?
    $endgroup$
    – Kernel
    Dec 6 '18 at 15:02














0












0








0





$begingroup$


Let $f$ and $g$ be positive, smooth and integrable functions in $mathbb{R}$, whose derivatives are also integrable.
Assume as well that the expression
$$
frac{sum_{qin mathbb{Q}} f(q)}{sum_{qin mathbb{Q}} g(q)} := lim_{n to infty} frac{sum_{i =1}^n f(q_i)}{sum_{i=1}^n g(q_i)}
$$
is well defined, where ${q_i}_i^infty = mathbb{Q}$.
I would like to evaluate
$$
left| frac{sum_{qin mathbb{Q}} f(q)}{sum_{qin mathbb{Q}} g(q)} - frac{int f(x) dx}{int g(x) dx} right|.
$$



When the sum is over a discrete set, I can use the hypothesis to bound the error of Riemman sums by the diameter of the partition generated by the set. What about in such case where the "diameter" is zero?






real-analysis integration analysis summation riemann-sum







share|cite|improve this question











$endgroup$




Let $f$ and $g$ be positive, smooth and integrable functions in $mathbb{R}$, whose derivatives are also integrable.
Assume as well that the expression
$$
frac{sum_{qin mathbb{Q}} f(q)}{sum_{qin mathbb{Q}} g(q)} := lim_{n to infty} frac{sum_{i =1}^n f(q_i)}{sum_{i=1}^n g(q_i)}
$$
is well defined, where ${q_i}_i^infty = mathbb{Q}$.
I would like to evaluate
$$
left| frac{sum_{qin mathbb{Q}} f(q)}{sum_{qin mathbb{Q}} g(q)} - frac{int f(x) dx}{int g(x) dx} right|.
$$



When the sum is over a discrete set, I can use the hypothesis to bound the error of Riemman sums by the diameter of the partition generated by the set. What about in such case where the "diameter" is zero?






real-analysis integration analysis summation riemann-sum




real-analysis integration analysis summation riemann-sum






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 6 '18 at 10:52









A.Γ.

22.6k32656




22.6k32656










asked Dec 6 '18 at 8:09









KernelKernel

772521




772521












  • $begingroup$
    Related ... en.wikipedia.org/wiki/Equidistributed_sequence ... take functions on $[0,1]$ and $g(x) = 1$. Some enumerations of the rationals are equidistributed, others are not.
    $endgroup$
    – GEdgar
    Dec 6 '18 at 12:27










  • $begingroup$
    That is perfect; I think this will be really useful. Do you know if there is an error bound for those approximations of the integral?
    $endgroup$
    – Kernel
    Dec 6 '18 at 15:02


















  • $begingroup$
    Related ... en.wikipedia.org/wiki/Equidistributed_sequence ... take functions on $[0,1]$ and $g(x) = 1$. Some enumerations of the rationals are equidistributed, others are not.
    $endgroup$
    – GEdgar
    Dec 6 '18 at 12:27










  • $begingroup$
    That is perfect; I think this will be really useful. Do you know if there is an error bound for those approximations of the integral?
    $endgroup$
    – Kernel
    Dec 6 '18 at 15:02
















$begingroup$
Related ... en.wikipedia.org/wiki/Equidistributed_sequence ... take functions on $[0,1]$ and $g(x) = 1$. Some enumerations of the rationals are equidistributed, others are not.
$endgroup$
– GEdgar
Dec 6 '18 at 12:27




$begingroup$
Related ... en.wikipedia.org/wiki/Equidistributed_sequence ... take functions on $[0,1]$ and $g(x) = 1$. Some enumerations of the rationals are equidistributed, others are not.
$endgroup$
– GEdgar
Dec 6 '18 at 12:27












$begingroup$
That is perfect; I think this will be really useful. Do you know if there is an error bound for those approximations of the integral?
$endgroup$
– Kernel
Dec 6 '18 at 15:02




$begingroup$
That is perfect; I think this will be really useful. Do you know if there is an error bound for those approximations of the integral?
$endgroup$
– Kernel
Dec 6 '18 at 15:02










1 Answer
1






active

oldest

votes


















0












$begingroup$

If $f$ is not identically zero, there is some $ainBbb R$ such that $f(a)>0$. Since $f$ is continuous, there exists $delta>0$ such that
$$
|x-a|<deltaimplies f(x)>frac{f(a)}{2}>0.
$$
Since there are infinitely many rationals $q$ such that $|q-a|<delta$, we see that $sum_{qinBbb Q}f(q)=infty$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Yes, but notice I am taking the quotient of two such sums, so I am wondering if I could get a cancellation.
    $endgroup$
    – Kernel
    Dec 6 '18 at 9:27










  • $begingroup$
    The sums are either $infty$ or $0$; the possible quotients are $infty/infty$, $0/0$, $infty/0$ and $07infty$. I do not see any possible cancelation.
    $endgroup$
    – Julián Aguirre
    Dec 6 '18 at 9:34












  • $begingroup$
    Perhaps the new edit will make the cancellation idea more precise.
    $endgroup$
    – Kernel
    Dec 6 '18 at 9:54








  • 1




    $begingroup$
    This is a different question. And in any case, the solution will depend on the enumeration of $Bbb Q$.
    $endgroup$
    – Julián Aguirre
    Dec 6 '18 at 9:57











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1 Answer
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1 Answer
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active

oldest

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active

oldest

votes









0












$begingroup$

If $f$ is not identically zero, there is some $ainBbb R$ such that $f(a)>0$. Since $f$ is continuous, there exists $delta>0$ such that
$$
|x-a|<deltaimplies f(x)>frac{f(a)}{2}>0.
$$
Since there are infinitely many rationals $q$ such that $|q-a|<delta$, we see that $sum_{qinBbb Q}f(q)=infty$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Yes, but notice I am taking the quotient of two such sums, so I am wondering if I could get a cancellation.
    $endgroup$
    – Kernel
    Dec 6 '18 at 9:27










  • $begingroup$
    The sums are either $infty$ or $0$; the possible quotients are $infty/infty$, $0/0$, $infty/0$ and $07infty$. I do not see any possible cancelation.
    $endgroup$
    – Julián Aguirre
    Dec 6 '18 at 9:34












  • $begingroup$
    Perhaps the new edit will make the cancellation idea more precise.
    $endgroup$
    – Kernel
    Dec 6 '18 at 9:54








  • 1




    $begingroup$
    This is a different question. And in any case, the solution will depend on the enumeration of $Bbb Q$.
    $endgroup$
    – Julián Aguirre
    Dec 6 '18 at 9:57
















0












$begingroup$

If $f$ is not identically zero, there is some $ainBbb R$ such that $f(a)>0$. Since $f$ is continuous, there exists $delta>0$ such that
$$
|x-a|<deltaimplies f(x)>frac{f(a)}{2}>0.
$$
Since there are infinitely many rationals $q$ such that $|q-a|<delta$, we see that $sum_{qinBbb Q}f(q)=infty$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Yes, but notice I am taking the quotient of two such sums, so I am wondering if I could get a cancellation.
    $endgroup$
    – Kernel
    Dec 6 '18 at 9:27










  • $begingroup$
    The sums are either $infty$ or $0$; the possible quotients are $infty/infty$, $0/0$, $infty/0$ and $07infty$. I do not see any possible cancelation.
    $endgroup$
    – Julián Aguirre
    Dec 6 '18 at 9:34












  • $begingroup$
    Perhaps the new edit will make the cancellation idea more precise.
    $endgroup$
    – Kernel
    Dec 6 '18 at 9:54








  • 1




    $begingroup$
    This is a different question. And in any case, the solution will depend on the enumeration of $Bbb Q$.
    $endgroup$
    – Julián Aguirre
    Dec 6 '18 at 9:57














0












0








0





$begingroup$

If $f$ is not identically zero, there is some $ainBbb R$ such that $f(a)>0$. Since $f$ is continuous, there exists $delta>0$ such that
$$
|x-a|<deltaimplies f(x)>frac{f(a)}{2}>0.
$$
Since there are infinitely many rationals $q$ such that $|q-a|<delta$, we see that $sum_{qinBbb Q}f(q)=infty$.






share|cite|improve this answer









$endgroup$



If $f$ is not identically zero, there is some $ainBbb R$ such that $f(a)>0$. Since $f$ is continuous, there exists $delta>0$ such that
$$
|x-a|<deltaimplies f(x)>frac{f(a)}{2}>0.
$$
Since there are infinitely many rationals $q$ such that $|q-a|<delta$, we see that $sum_{qinBbb Q}f(q)=infty$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 6 '18 at 8:31









Julián AguirreJulián Aguirre

68k24094




68k24094












  • $begingroup$
    Yes, but notice I am taking the quotient of two such sums, so I am wondering if I could get a cancellation.
    $endgroup$
    – Kernel
    Dec 6 '18 at 9:27










  • $begingroup$
    The sums are either $infty$ or $0$; the possible quotients are $infty/infty$, $0/0$, $infty/0$ and $07infty$. I do not see any possible cancelation.
    $endgroup$
    – Julián Aguirre
    Dec 6 '18 at 9:34












  • $begingroup$
    Perhaps the new edit will make the cancellation idea more precise.
    $endgroup$
    – Kernel
    Dec 6 '18 at 9:54








  • 1




    $begingroup$
    This is a different question. And in any case, the solution will depend on the enumeration of $Bbb Q$.
    $endgroup$
    – Julián Aguirre
    Dec 6 '18 at 9:57


















  • $begingroup$
    Yes, but notice I am taking the quotient of two such sums, so I am wondering if I could get a cancellation.
    $endgroup$
    – Kernel
    Dec 6 '18 at 9:27










  • $begingroup$
    The sums are either $infty$ or $0$; the possible quotients are $infty/infty$, $0/0$, $infty/0$ and $07infty$. I do not see any possible cancelation.
    $endgroup$
    – Julián Aguirre
    Dec 6 '18 at 9:34












  • $begingroup$
    Perhaps the new edit will make the cancellation idea more precise.
    $endgroup$
    – Kernel
    Dec 6 '18 at 9:54








  • 1




    $begingroup$
    This is a different question. And in any case, the solution will depend on the enumeration of $Bbb Q$.
    $endgroup$
    – Julián Aguirre
    Dec 6 '18 at 9:57
















$begingroup$
Yes, but notice I am taking the quotient of two such sums, so I am wondering if I could get a cancellation.
$endgroup$
– Kernel
Dec 6 '18 at 9:27




$begingroup$
Yes, but notice I am taking the quotient of two such sums, so I am wondering if I could get a cancellation.
$endgroup$
– Kernel
Dec 6 '18 at 9:27












$begingroup$
The sums are either $infty$ or $0$; the possible quotients are $infty/infty$, $0/0$, $infty/0$ and $07infty$. I do not see any possible cancelation.
$endgroup$
– Julián Aguirre
Dec 6 '18 at 9:34






$begingroup$
The sums are either $infty$ or $0$; the possible quotients are $infty/infty$, $0/0$, $infty/0$ and $07infty$. I do not see any possible cancelation.
$endgroup$
– Julián Aguirre
Dec 6 '18 at 9:34














$begingroup$
Perhaps the new edit will make the cancellation idea more precise.
$endgroup$
– Kernel
Dec 6 '18 at 9:54






$begingroup$
Perhaps the new edit will make the cancellation idea more precise.
$endgroup$
– Kernel
Dec 6 '18 at 9:54






1




1




$begingroup$
This is a different question. And in any case, the solution will depend on the enumeration of $Bbb Q$.
$endgroup$
– Julián Aguirre
Dec 6 '18 at 9:57




$begingroup$
This is a different question. And in any case, the solution will depend on the enumeration of $Bbb Q$.
$endgroup$
– Julián Aguirre
Dec 6 '18 at 9:57


















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