how do we prove that a sum of two periods is still a period?optimizing Frobenius instance solutionsDoes...
how do we prove that a sum of two periods is still a period?
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Kontsevich and Zagier define periods as the values of absolutely convergent integrals $int_sigma f$ where $f$ is a rational function with rational coefficients and $sigma$ is a semi-algebraic subset of $mathbb{R}^n$. How do we prove that the sum of two such numbers is still of this form? I've tried a few things but they don't seem to work...
ag.algebraic-geometry nt.number-theory
asked Apr 2 at 14:27
periodsperiods
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Kontsevich and Zagier define periods as the values of absolutely convergent integrals $int_sigma f$ where $f$ is a rational function with rational coefficients and $sigma$ is a semi-algebraic subset of $mathbb{R}^n$. How do we prove that the sum of two such numbers is still of this form? I've tried a few things but they don't seem to work...
ag.algebraic-geometry nt.number-theory
asked Apr 2 at 14:27
periodsperiods
492
New contributor
periods is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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Kontsevich and Zagier define periods as the values of absolutely convergent integrals $int_sigma f$ where $f$ is a rational function with rational coefficients and $sigma$ is a semi-algebraic subset of $mathbb{R}^n$. How do we prove that the sum of two such numbers is still of this form? I've tried a few things but they don't seem to work...
ag.algebraic-geometry nt.number-theory
asked Apr 2 at 14:27
periodsperiods
492
New contributor
periods is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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Kontsevich and Zagier define periods as the values of absolutely convergent integrals $int_sigma f$ where $f$ is a rational function with rational coefficients and $sigma$ is a semi-algebraic subset of $mathbb{R}^n$. How do we prove that the sum of two such numbers is still of this form? I've tried a few things but they don't seem to work...
ag.algebraic-geometry nt.number-theory
ag.algebraic-geometry nt.number-theory
asked Apr 2 at 14:27
periodsperiods
492
New contributor
periods is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Apr 2 at 14:27
periodsperiods
492
New contributor
periods is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Apr 2 at 14:27
periodsperiods
492
New contributor
periods is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Apr 2 at 14:27
periodsperiods
492
asked Apr 2 at 14:27
periodsperiods
492
492
New contributor
periods is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
periods is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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periods is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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1 Answer
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Let $alpha$ and $beta$ be two periods corresponding respectively to two absolutely convergent integrals $int_sigma f(x)dx$ and $int_tau g(y)dy$, where $f$ (resp. $g$) is a rational function on $Bbb Q$ with $r$ (resp. $s$) variables and $sigma$ (resp. $tau$) is a semi-algebraic subset of $Bbb R^r$ (resp. $Bbb R^s$).
Setting $omega:=sigmatimesleftlbrace0rightrbracetimes(0,1)^scoprod(0,1)^rtimesleftlbrace1rightrbracetimestau$, one immediately gets that $$alpha+beta=int_omega left[(1-t)f(x)+tg(y)right]dxdydt$$which is again an absolutely convergent integral, so that $alpha+beta$ is a period.
answered Apr 2 at 15:15
GaussianGaussian
35318
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@periods: if you're not satisfied by the answer, please tell me how to improve it.
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– Gaussian
Apr 2 at 18:55
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Let $alpha$ and $beta$ be two periods corresponding respectively to two absolutely convergent integrals $int_sigma f(x)dx$ and $int_tau g(y)dy$, where $f$ (resp. $g$) is a rational function on $Bbb Q$ with $r$ (resp. $s$) variables and $sigma$ (resp. $tau$) is a semi-algebraic subset of $Bbb R^r$ (resp. $Bbb R^s$).
Setting $omega:=sigmatimesleftlbrace0rightrbracetimes(0,1)^scoprod(0,1)^rtimesleftlbrace1rightrbracetimestau$, one immediately gets that $$alpha+beta=int_omega left[(1-t)f(x)+tg(y)right]dxdydt$$which is again an absolutely convergent integral, so that $alpha+beta$ is a period.
answered Apr 2 at 15:15
GaussianGaussian
35318
$endgroup$
$begingroup$
@periods: if you're not satisfied by the answer, please tell me how to improve it.
$endgroup$
– Gaussian
Apr 2 at 18:55
add a comment |
$begingroup$
Let $alpha$ and $beta$ be two periods corresponding respectively to two absolutely convergent integrals $int_sigma f(x)dx$ and $int_tau g(y)dy$, where $f$ (resp. $g$) is a rational function on $Bbb Q$ with $r$ (resp. $s$) variables and $sigma$ (resp. $tau$) is a semi-algebraic subset of $Bbb R^r$ (resp. $Bbb R^s$).
Setting $omega:=sigmatimesleftlbrace0rightrbracetimes(0,1)^scoprod(0,1)^rtimesleftlbrace1rightrbracetimestau$, one immediately gets that $$alpha+beta=int_omega left[(1-t)f(x)+tg(y)right]dxdydt$$which is again an absolutely convergent integral, so that $alpha+beta$ is a period.
answered Apr 2 at 15:15
GaussianGaussian
35318
$endgroup$
$begingroup$
@periods: if you're not satisfied by the answer, please tell me how to improve it.
$endgroup$
– Gaussian
Apr 2 at 18:55
add a comment |
$begingroup$
Let $alpha$ and $beta$ be two periods corresponding respectively to two absolutely convergent integrals $int_sigma f(x)dx$ and $int_tau g(y)dy$, where $f$ (resp. $g$) is a rational function on $Bbb Q$ with $r$ (resp. $s$) variables and $sigma$ (resp. $tau$) is a semi-algebraic subset of $Bbb R^r$ (resp. $Bbb R^s$).
Setting $omega:=sigmatimesleftlbrace0rightrbracetimes(0,1)^scoprod(0,1)^rtimesleftlbrace1rightrbracetimestau$, one immediately gets that $$alpha+beta=int_omega left[(1-t)f(x)+tg(y)right]dxdydt$$which is again an absolutely convergent integral, so that $alpha+beta$ is a period.
answered Apr 2 at 15:15
GaussianGaussian
35318
$endgroup$
Let $alpha$ and $beta$ be two periods corresponding respectively to two absolutely convergent integrals $int_sigma f(x)dx$ and $int_tau g(y)dy$, where $f$ (resp. $g$) is a rational function on $Bbb Q$ with $r$ (resp. $s$) variables and $sigma$ (resp. $tau$) is a semi-algebraic subset of $Bbb R^r$ (resp. $Bbb R^s$).
Setting $omega:=sigmatimesleftlbrace0rightrbracetimes(0,1)^scoprod(0,1)^rtimesleftlbrace1rightrbracetimestau$, one immediately gets that $$alpha+beta=int_omega left[(1-t)f(x)+tg(y)right]dxdydt$$which is again an absolutely convergent integral, so that $alpha+beta$ is a period.
answered Apr 2 at 15:15
GaussianGaussian
35318
answered Apr 2 at 15:15
GaussianGaussian
35318
answered Apr 2 at 15:15
GaussianGaussian
35318
answered Apr 2 at 15:15
GaussianGaussian
35318
35318
$begingroup$
@periods: if you're not satisfied by the answer, please tell me how to improve it.
$endgroup$
– Gaussian
Apr 2 at 18:55
add a comment |
$begingroup$
@periods: if you're not satisfied by the answer, please tell me how to improve it.
$endgroup$
– Gaussian
Apr 2 at 18:55
$begingroup$
@periods: if you're not satisfied by the answer, please tell me how to improve it.
$endgroup$
– Gaussian
Apr 2 at 18:55
$begingroup$
@periods: if you're not satisfied by the answer, please tell me how to improve it.
$endgroup$
– Gaussian
Apr 2 at 18:55
add a comment |
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