Numerical integration with arbitrary precision

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I have to perform a numerical integral on $[0,1]$ with a complicated integrand (possibly with integrable singularities at $0$ and/or $1$) and I am looking for a platform (sagemath, maple, python, mathematica, pari, matlab,...) that allows to do this in a satisfactory way. Ideally, though, I would like to go on working with sagemath.



The (real) integrand I am considering is a linear combination of functions of the form



$$mathrmImleft(logleft(frac1-zeta^a(1-t)^1/Nzeta^bt^1/Nright) cdotfraczeta^bt^1/N-1-zeta^a(1-t)^1/N-1zeta^bt^1/N+zeta^a(1-t)^1/N right),$$



where $t$ is the variable, $N,a,b in mathbbN$ are fixed and $zeta=e^2pi i /N$.



Concretely, I would like to be able to:



  1. Perform the integral with arbitrary precision, i.e., I will set the number of digits for the result and the program guarantees that the result is exact up to this digit. Probably this is just hopeless: one can always have pathological functions and since all numerical integrals are (to my knowledge) based on pointwise evaluation on some grid, one can never obtain a maximal bound on the precision, am I right?

  2. Alternatively and more humbly, I would also appreciate a way to set the number of digits of the result (even if they are not guaranteed to be all exact) and to get an error estimation (even if it is not a maximal bound).

Sagemath's "numerical_integral" deals with integrable singularities (can perform this integral) and provides an estimation of the error, although it is too big in some cases: that is why I need a way to reduce it. In principle it can also adapt the number of digits, but apparently this last feature does not work with the numerical integration: even changing the number of digits, the result is always the same. Also, one can in principle set the tolerance for the absolute and/or relative error of the numerical integration, but also this does not seem to change the result.



Any help/hint is appreciated, in particular on how to implement (part of) my goal with sagemath, but also for other platforms or for theoretical results.




numerical-methods numerical-calculus




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  • 1




    In boost.math, you can do complex-valued integrals with endpoint singularities to arbitrary precision, with error estimates and L1-norm evaluation. Use tanh-sinh quadrature. See boost.org/users/download
    – user14717
    Aug 3 at 9:32










  • Thanks! I will have a look
    – 57Jimmy
    Aug 4 at 12:28














up vote
0
down vote

favorite












I have to perform a numerical integral on $[0,1]$ with a complicated integrand (possibly with integrable singularities at $0$ and/or $1$) and I am looking for a platform (sagemath, maple, python, mathematica, pari, matlab,...) that allows to do this in a satisfactory way. Ideally, though, I would like to go on working with sagemath.



The (real) integrand I am considering is a linear combination of functions of the form



$$mathrmImleft(logleft(frac1-zeta^a(1-t)^1/Nzeta^bt^1/Nright) cdotfraczeta^bt^1/N-1-zeta^a(1-t)^1/N-1zeta^bt^1/N+zeta^a(1-t)^1/N right),$$



where $t$ is the variable, $N,a,b in mathbbN$ are fixed and $zeta=e^2pi i /N$.



Concretely, I would like to be able to:



  1. Perform the integral with arbitrary precision, i.e., I will set the number of digits for the result and the program guarantees that the result is exact up to this digit. Probably this is just hopeless: one can always have pathological functions and since all numerical integrals are (to my knowledge) based on pointwise evaluation on some grid, one can never obtain a maximal bound on the precision, am I right?

  2. Alternatively and more humbly, I would also appreciate a way to set the number of digits of the result (even if they are not guaranteed to be all exact) and to get an error estimation (even if it is not a maximal bound).

Sagemath's "numerical_integral" deals with integrable singularities (can perform this integral) and provides an estimation of the error, although it is too big in some cases: that is why I need a way to reduce it. In principle it can also adapt the number of digits, but apparently this last feature does not work with the numerical integration: even changing the number of digits, the result is always the same. Also, one can in principle set the tolerance for the absolute and/or relative error of the numerical integration, but also this does not seem to change the result.



Any help/hint is appreciated, in particular on how to implement (part of) my goal with sagemath, but also for other platforms or for theoretical results.




numerical-methods numerical-calculus




share|cite|improve this question















  • 1




    In boost.math, you can do complex-valued integrals with endpoint singularities to arbitrary precision, with error estimates and L1-norm evaluation. Use tanh-sinh quadrature. See boost.org/users/download
    – user14717
    Aug 3 at 9:32










  • Thanks! I will have a look
    – 57Jimmy
    Aug 4 at 12:28












up vote
0
down vote

favorite









up vote
0
down vote

favorite











I have to perform a numerical integral on $[0,1]$ with a complicated integrand (possibly with integrable singularities at $0$ and/or $1$) and I am looking for a platform (sagemath, maple, python, mathematica, pari, matlab,...) that allows to do this in a satisfactory way. Ideally, though, I would like to go on working with sagemath.



The (real) integrand I am considering is a linear combination of functions of the form



$$mathrmImleft(logleft(frac1-zeta^a(1-t)^1/Nzeta^bt^1/Nright) cdotfraczeta^bt^1/N-1-zeta^a(1-t)^1/N-1zeta^bt^1/N+zeta^a(1-t)^1/N right),$$



where $t$ is the variable, $N,a,b in mathbbN$ are fixed and $zeta=e^2pi i /N$.



Concretely, I would like to be able to:



  1. Perform the integral with arbitrary precision, i.e., I will set the number of digits for the result and the program guarantees that the result is exact up to this digit. Probably this is just hopeless: one can always have pathological functions and since all numerical integrals are (to my knowledge) based on pointwise evaluation on some grid, one can never obtain a maximal bound on the precision, am I right?

  2. Alternatively and more humbly, I would also appreciate a way to set the number of digits of the result (even if they are not guaranteed to be all exact) and to get an error estimation (even if it is not a maximal bound).

Sagemath's "numerical_integral" deals with integrable singularities (can perform this integral) and provides an estimation of the error, although it is too big in some cases: that is why I need a way to reduce it. In principle it can also adapt the number of digits, but apparently this last feature does not work with the numerical integration: even changing the number of digits, the result is always the same. Also, one can in principle set the tolerance for the absolute and/or relative error of the numerical integration, but also this does not seem to change the result.



Any help/hint is appreciated, in particular on how to implement (part of) my goal with sagemath, but also for other platforms or for theoretical results.




numerical-methods numerical-calculus




share|cite|improve this question











I have to perform a numerical integral on $[0,1]$ with a complicated integrand (possibly with integrable singularities at $0$ and/or $1$) and I am looking for a platform (sagemath, maple, python, mathematica, pari, matlab,...) that allows to do this in a satisfactory way. Ideally, though, I would like to go on working with sagemath.



The (real) integrand I am considering is a linear combination of functions of the form



$$mathrmImleft(logleft(frac1-zeta^a(1-t)^1/Nzeta^bt^1/Nright) cdotfraczeta^bt^1/N-1-zeta^a(1-t)^1/N-1zeta^bt^1/N+zeta^a(1-t)^1/N right),$$



where $t$ is the variable, $N,a,b in mathbbN$ are fixed and $zeta=e^2pi i /N$.



Concretely, I would like to be able to:



  1. Perform the integral with arbitrary precision, i.e., I will set the number of digits for the result and the program guarantees that the result is exact up to this digit. Probably this is just hopeless: one can always have pathological functions and since all numerical integrals are (to my knowledge) based on pointwise evaluation on some grid, one can never obtain a maximal bound on the precision, am I right?

  2. Alternatively and more humbly, I would also appreciate a way to set the number of digits of the result (even if they are not guaranteed to be all exact) and to get an error estimation (even if it is not a maximal bound).

Sagemath's "numerical_integral" deals with integrable singularities (can perform this integral) and provides an estimation of the error, although it is too big in some cases: that is why I need a way to reduce it. In principle it can also adapt the number of digits, but apparently this last feature does not work with the numerical integration: even changing the number of digits, the result is always the same. Also, one can in principle set the tolerance for the absolute and/or relative error of the numerical integration, but also this does not seem to change the result.



Any help/hint is appreciated, in particular on how to implement (part of) my goal with sagemath, but also for other platforms or for theoretical results.





numerical-methods numerical-calculus





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asked Aug 3 at 9:09









57Jimmy

3,110421




3,110421







  • 1




    In boost.math, you can do complex-valued integrals with endpoint singularities to arbitrary precision, with error estimates and L1-norm evaluation. Use tanh-sinh quadrature. See boost.org/users/download
    – user14717
    Aug 3 at 9:32










  • Thanks! I will have a look
    – 57Jimmy
    Aug 4 at 12:28












  • 1




    In boost.math, you can do complex-valued integrals with endpoint singularities to arbitrary precision, with error estimates and L1-norm evaluation. Use tanh-sinh quadrature. See boost.org/users/download
    – user14717
    Aug 3 at 9:32










  • Thanks! I will have a look
    – 57Jimmy
    Aug 4 at 12:28







1




1




In boost.math, you can do complex-valued integrals with endpoint singularities to arbitrary precision, with error estimates and L1-norm evaluation. Use tanh-sinh quadrature. See boost.org/users/download
– user14717
Aug 3 at 9:32




In boost.math, you can do complex-valued integrals with endpoint singularities to arbitrary precision, with error estimates and L1-norm evaluation. Use tanh-sinh quadrature. See boost.org/users/download
– user14717
Aug 3 at 9:32












Thanks! I will have a look
– 57Jimmy
Aug 4 at 12:28




Thanks! I will have a look
– 57Jimmy
Aug 4 at 12:28










1 Answer
1






active

oldest

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up vote
2
down vote













mpmath (an arbitrary precision mathematical library for Python 2.5 or higher, including Python 3.x) sounds exactly like what you'd be interested in. The numerical integration documentation is here. I used it to good effect in this MSE answer - not quite the same application, but I was computing an oscillatory integral on the interval $[0,infty)$. The result was accurate to within $sim10^-50$ after $sim33$ seconds of calculation.



I haven't personally used Sagemath, however, at least part of mpmath is documented on Sagemath.org, see this link.



Edit: Yes, if you pass the option "error=True" to the quad function, then you will receive an estimate of the integral & an estimate of the error. Below I've posted an example, from this MSE question, computed in Python 2.7.12. The only change I've made is to format the long decimal numbers in a more readable format. Unfortunately I don't know the details of the error estimation routine. Also, you can see in this instance that the estimate is not very precise.



>>> from mpmath import *
>>> f = lambda x: log(1 + pow(x, 2 + sqrt(3)))/(1 + x)
>>> mp.dps = 240
>>> I = pi*pi*(1 - sqrt(3))/12 + log(2)*log(1 + sqrt(3))
>>> mp.pretty = True
>>> print I
0.094 561 677 526 995 723 016 194 274 812 869 488 153 163 
 479 510 430 611 733 992 647 539 552 027 774 635 759 786 
 503 446 184 802 723 491 332 330 132 858 978 766 871 567 
 356 331 135 943 122 717 636 925 391 051 154 737 292 763 
 310 732 514 933 830 590 857 033 702 491 725 001 769 229 
 295 367 639 103 717 378 474 427 104 071 6
>>> mp.dps = 15
>>> Iapprox = quad(f, [0,1], error=True)
>>> print Iapprox[0]
0.094 561 677 526 995 7 
>>> print Iapprox[1]
1.0e-28
>>> mp.dps = 30
>>> Iapprox = quad(f, [0,1], error=True)
>>> print Iapprox[0]
0.094 561 677 526 995 723 016 194 274 812 9 
>>> print Iapprox[1]
11.0e-62
>>> mp.dps = 60
>>> Iapprox = quad(f, [0,1], error=True)
>>> print Iapprox[0]
0.094 561 677 526 995 723 016 194 274 812 869 488 153 163 
 479 510 430 611 733 992 8 
>>> print Iapprox[1]
1.0e-62





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  • 1




    Thank you for your answer. Is there any easy way to obtain an estimate of the error committed in the numerical integration?
    – 57Jimmy
    Aug 3 at 19:27










  • Yes it is possible @57Jimmy, I've edited my answer to include an example.
    – Kyle
    yesterday










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1 Answer
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active

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votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
2
down vote













mpmath (an arbitrary precision mathematical library for Python 2.5 or higher, including Python 3.x) sounds exactly like what you'd be interested in. The numerical integration documentation is here. I used it to good effect in this MSE answer - not quite the same application, but I was computing an oscillatory integral on the interval $[0,infty)$. The result was accurate to within $sim10^-50$ after $sim33$ seconds of calculation.



I haven't personally used Sagemath, however, at least part of mpmath is documented on Sagemath.org, see this link.



Edit: Yes, if you pass the option "error=True" to the quad function, then you will receive an estimate of the integral & an estimate of the error. Below I've posted an example, from this MSE question, computed in Python 2.7.12. The only change I've made is to format the long decimal numbers in a more readable format. Unfortunately I don't know the details of the error estimation routine. Also, you can see in this instance that the estimate is not very precise.



>>> from mpmath import *
>>> f = lambda x: log(1 + pow(x, 2 + sqrt(3)))/(1 + x)
>>> mp.dps = 240
>>> I = pi*pi*(1 - sqrt(3))/12 + log(2)*log(1 + sqrt(3))
>>> mp.pretty = True
>>> print I
0.094 561 677 526 995 723 016 194 274 812 869 488 153 163 
 479 510 430 611 733 992 647 539 552 027 774 635 759 786 
 503 446 184 802 723 491 332 330 132 858 978 766 871 567 
 356 331 135 943 122 717 636 925 391 051 154 737 292 763 
 310 732 514 933 830 590 857 033 702 491 725 001 769 229 
 295 367 639 103 717 378 474 427 104 071 6
>>> mp.dps = 15
>>> Iapprox = quad(f, [0,1], error=True)
>>> print Iapprox[0]
0.094 561 677 526 995 7 
>>> print Iapprox[1]
1.0e-28
>>> mp.dps = 30
>>> Iapprox = quad(f, [0,1], error=True)
>>> print Iapprox[0]
0.094 561 677 526 995 723 016 194 274 812 9 
>>> print Iapprox[1]
11.0e-62
>>> mp.dps = 60
>>> Iapprox = quad(f, [0,1], error=True)
>>> print Iapprox[0]
0.094 561 677 526 995 723 016 194 274 812 869 488 153 163 
 479 510 430 611 733 992 8 
>>> print Iapprox[1]
1.0e-62





share|cite|improve this answer



















  • 1




    Thank you for your answer. Is there any easy way to obtain an estimate of the error committed in the numerical integration?
    – 57Jimmy
    Aug 3 at 19:27










  • Yes it is possible @57Jimmy, I've edited my answer to include an example.
    – Kyle
    yesterday














up vote
2
down vote













mpmath (an arbitrary precision mathematical library for Python 2.5 or higher, including Python 3.x) sounds exactly like what you'd be interested in. The numerical integration documentation is here. I used it to good effect in this MSE answer - not quite the same application, but I was computing an oscillatory integral on the interval $[0,infty)$. The result was accurate to within $sim10^-50$ after $sim33$ seconds of calculation.



I haven't personally used Sagemath, however, at least part of mpmath is documented on Sagemath.org, see this link.



Edit: Yes, if you pass the option "error=True" to the quad function, then you will receive an estimate of the integral & an estimate of the error. Below I've posted an example, from this MSE question, computed in Python 2.7.12. The only change I've made is to format the long decimal numbers in a more readable format. Unfortunately I don't know the details of the error estimation routine. Also, you can see in this instance that the estimate is not very precise.



>>> from mpmath import *
>>> f = lambda x: log(1 + pow(x, 2 + sqrt(3)))/(1 + x)
>>> mp.dps = 240
>>> I = pi*pi*(1 - sqrt(3))/12 + log(2)*log(1 + sqrt(3))
>>> mp.pretty = True
>>> print I
0.094 561 677 526 995 723 016 194 274 812 869 488 153 163 
 479 510 430 611 733 992 647 539 552 027 774 635 759 786 
 503 446 184 802 723 491 332 330 132 858 978 766 871 567 
 356 331 135 943 122 717 636 925 391 051 154 737 292 763 
 310 732 514 933 830 590 857 033 702 491 725 001 769 229 
 295 367 639 103 717 378 474 427 104 071 6
>>> mp.dps = 15
>>> Iapprox = quad(f, [0,1], error=True)
>>> print Iapprox[0]
0.094 561 677 526 995 7 
>>> print Iapprox[1]
1.0e-28
>>> mp.dps = 30
>>> Iapprox = quad(f, [0,1], error=True)
>>> print Iapprox[0]
0.094 561 677 526 995 723 016 194 274 812 9 
>>> print Iapprox[1]
11.0e-62
>>> mp.dps = 60
>>> Iapprox = quad(f, [0,1], error=True)
>>> print Iapprox[0]
0.094 561 677 526 995 723 016 194 274 812 869 488 153 163 
 479 510 430 611 733 992 8 
>>> print Iapprox[1]
1.0e-62





share|cite|improve this answer



















  • 1




    Thank you for your answer. Is there any easy way to obtain an estimate of the error committed in the numerical integration?
    – 57Jimmy
    Aug 3 at 19:27










  • Yes it is possible @57Jimmy, I've edited my answer to include an example.
    – Kyle
    yesterday












up vote
2
down vote










up vote
2
down vote









mpmath (an arbitrary precision mathematical library for Python 2.5 or higher, including Python 3.x) sounds exactly like what you'd be interested in. The numerical integration documentation is here. I used it to good effect in this MSE answer - not quite the same application, but I was computing an oscillatory integral on the interval $[0,infty)$. The result was accurate to within $sim10^-50$ after $sim33$ seconds of calculation.



I haven't personally used Sagemath, however, at least part of mpmath is documented on Sagemath.org, see this link.



Edit: Yes, if you pass the option "error=True" to the quad function, then you will receive an estimate of the integral & an estimate of the error. Below I've posted an example, from this MSE question, computed in Python 2.7.12. The only change I've made is to format the long decimal numbers in a more readable format. Unfortunately I don't know the details of the error estimation routine. Also, you can see in this instance that the estimate is not very precise.



>>> from mpmath import *
>>> f = lambda x: log(1 + pow(x, 2 + sqrt(3)))/(1 + x)
>>> mp.dps = 240
>>> I = pi*pi*(1 - sqrt(3))/12 + log(2)*log(1 + sqrt(3))
>>> mp.pretty = True
>>> print I
0.094 561 677 526 995 723 016 194 274 812 869 488 153 163 
 479 510 430 611 733 992 647 539 552 027 774 635 759 786 
 503 446 184 802 723 491 332 330 132 858 978 766 871 567 
 356 331 135 943 122 717 636 925 391 051 154 737 292 763 
 310 732 514 933 830 590 857 033 702 491 725 001 769 229 
 295 367 639 103 717 378 474 427 104 071 6
>>> mp.dps = 15
>>> Iapprox = quad(f, [0,1], error=True)
>>> print Iapprox[0]
0.094 561 677 526 995 7 
>>> print Iapprox[1]
1.0e-28
>>> mp.dps = 30
>>> Iapprox = quad(f, [0,1], error=True)
>>> print Iapprox[0]
0.094 561 677 526 995 723 016 194 274 812 9 
>>> print Iapprox[1]
11.0e-62
>>> mp.dps = 60
>>> Iapprox = quad(f, [0,1], error=True)
>>> print Iapprox[0]
0.094 561 677 526 995 723 016 194 274 812 869 488 153 163 
 479 510 430 611 733 992 8 
>>> print Iapprox[1]
1.0e-62





share|cite|improve this answer















mpmath (an arbitrary precision mathematical library for Python 2.5 or higher, including Python 3.x) sounds exactly like what you'd be interested in. The numerical integration documentation is here. I used it to good effect in this MSE answer - not quite the same application, but I was computing an oscillatory integral on the interval $[0,infty)$. The result was accurate to within $sim10^-50$ after $sim33$ seconds of calculation.



I haven't personally used Sagemath, however, at least part of mpmath is documented on Sagemath.org, see this link.



Edit: Yes, if you pass the option "error=True" to the quad function, then you will receive an estimate of the integral & an estimate of the error. Below I've posted an example, from this MSE question, computed in Python 2.7.12. The only change I've made is to format the long decimal numbers in a more readable format. Unfortunately I don't know the details of the error estimation routine. Also, you can see in this instance that the estimate is not very precise.



>>> from mpmath import *
>>> f = lambda x: log(1 + pow(x, 2 + sqrt(3)))/(1 + x)
>>> mp.dps = 240
>>> I = pi*pi*(1 - sqrt(3))/12 + log(2)*log(1 + sqrt(3))
>>> mp.pretty = True
>>> print I
0.094 561 677 526 995 723 016 194 274 812 869 488 153 163 
 479 510 430 611 733 992 647 539 552 027 774 635 759 786 
 503 446 184 802 723 491 332 330 132 858 978 766 871 567 
 356 331 135 943 122 717 636 925 391 051 154 737 292 763 
 310 732 514 933 830 590 857 033 702 491 725 001 769 229 
 295 367 639 103 717 378 474 427 104 071 6
>>> mp.dps = 15
>>> Iapprox = quad(f, [0,1], error=True)
>>> print Iapprox[0]
0.094 561 677 526 995 7 
>>> print Iapprox[1]
1.0e-28
>>> mp.dps = 30
>>> Iapprox = quad(f, [0,1], error=True)
>>> print Iapprox[0]
0.094 561 677 526 995 723 016 194 274 812 9 
>>> print Iapprox[1]
11.0e-62
>>> mp.dps = 60
>>> Iapprox = quad(f, [0,1], error=True)
>>> print Iapprox[0]
0.094 561 677 526 995 723 016 194 274 812 869 488 153 163 
 479 510 430 611 733 992 8 
>>> print Iapprox[1]
1.0e-62






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share|cite|improve this answer



share|cite|improve this answer








edited yesterday


























answered Aug 3 at 12:57









Kyle

1,079617




1,079617







  • 1




    Thank you for your answer. Is there any easy way to obtain an estimate of the error committed in the numerical integration?
    – 57Jimmy
    Aug 3 at 19:27










  • Yes it is possible @57Jimmy, I've edited my answer to include an example.
    – Kyle
    yesterday












  • 1




    Thank you for your answer. Is there any easy way to obtain an estimate of the error committed in the numerical integration?
    – 57Jimmy
    Aug 3 at 19:27










  • Yes it is possible @57Jimmy, I've edited my answer to include an example.
    – Kyle
    yesterday







1




1




Thank you for your answer. Is there any easy way to obtain an estimate of the error committed in the numerical integration?
– 57Jimmy
Aug 3 at 19:27




Thank you for your answer. Is there any easy way to obtain an estimate of the error committed in the numerical integration?
– 57Jimmy
Aug 3 at 19:27












Yes it is possible @57Jimmy, I've edited my answer to include an example.
– Kyle
yesterday




Yes it is possible @57Jimmy, I've edited my answer to include an example.
– Kyle
yesterday












 

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Color the edges and diagonals of a regular polygon

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Clash Royale CLAN TAG #URR8PPP up vote 5 down vote favorite 1 Here is the problem: For what $n$ is it possible to color the edges and diagonals of an $n$-side regular polygon with $dfracbinomn23$ colors, such that you use every color exactly three times and for every color the three segments (edges or diagonals) with that color form a triangle? Trivially $nequiv 0 text or 1 pmod3$. I can also prove that if the statement is true for $k$ than it is true for $3k$ as well. How to finish? Please help! Thanks combinatorial-geometry share | cite | improve this question edited Aug 6 at 21:57 alcana 118 4 asked Aug 6 at 21:15 Leo Gardner 356 11 Even assuming "polyhpn" in the title is a typo for polygon , it is unclear what you mean by "the edges and sides". Aren't the side of a regular polygon the same as the edges? A regular $n$-sided polygon has $n$ edges. – hardmath Aug 6 at 21:20 3 $n$ ne...
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Relationship between determinant of matrix and determinant of adjoint?

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Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite We are studying adjoints in class, and I was curious if there is a relationship between the determinant of matrix A, and the determinant of the adjoint of matrix A? I assume there would be a relationship because finding the adjoint requires creating a cofactor matrix and then transposing it. linear-algebra determinant adjoint-operators share | cite | improve this question asked Nov 17 '17 at 17:32 CluelessCoder 32 5 For a square matrix $A$ of order $n$, we have $A(operatornameadj A)=(operatornameadj A)A=|A|I_n$ where $I_n$ is the identity matrix of order $n$. This should tell you about the determinant of the adjoint in terms of that of $A$ (use multiplicative property and other elementary properties of determinants). – Prasun Biswas Nov 17 '17 at 17:35 1 @CluelessCoder: see here: math.stackexchange.com/questions/516127/…...
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