What is the best estimation of the factorial function?











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I have to calculate factorials of an arbitrarily large integers mod another arbitrarily large integer. i have considered Stirling's Approximation and Ramanujan’s factorial approximation. is it possible to get better estimates?






factorial







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  • 3




    If you are using modular arithmetic, analytic estimates might not be the most useful thing for you, depending on what exactly your problem is. Instead things like en.wikipedia.org/wiki/Wilson%27s_theorem might help.
    – Alex J Best
    Nov 22 at 11:53






  • 1




    There's the gamma function but I don't think it will help you much in contexts of modular arithmetic...
    – Prasun Biswas
    Nov 22 at 11:57






  • 1




    Have a look at math.stackexchange.com/questions/676952/…. For $n=20$, what I wrote gives $2432902008176665737$ to be compared to the exact $2432902008176640000$
    – Claude Leibovici
    Nov 22 at 12:32






  • 1




    I don't know if this could be of interest since the coefficients start to be huge. Using the same kind of expression with degree $2$ in numerator and degree $3$ in denominator, for $n=20$, I get $2432902008176639896$. If you want, I could post the numbers in an answer.
    – Claude Leibovici
    Nov 22 at 13:05






  • 1




    Comparing to high order expansion of Stirling formula.
    – Claude Leibovici
    Nov 22 at 13:41















up vote
0
down vote

favorite
1












I have to calculate factorials of an arbitrarily large integers mod another arbitrarily large integer. i have considered Stirling's Approximation and Ramanujan’s factorial approximation. is it possible to get better estimates?






factorial







share|cite|improve this question


















  • 3




    If you are using modular arithmetic, analytic estimates might not be the most useful thing for you, depending on what exactly your problem is. Instead things like en.wikipedia.org/wiki/Wilson%27s_theorem might help.
    – Alex J Best
    Nov 22 at 11:53






  • 1




    There's the gamma function but I don't think it will help you much in contexts of modular arithmetic...
    – Prasun Biswas
    Nov 22 at 11:57






  • 1




    Have a look at math.stackexchange.com/questions/676952/…. For $n=20$, what I wrote gives $2432902008176665737$ to be compared to the exact $2432902008176640000$
    – Claude Leibovici
    Nov 22 at 12:32






  • 1




    I don't know if this could be of interest since the coefficients start to be huge. Using the same kind of expression with degree $2$ in numerator and degree $3$ in denominator, for $n=20$, I get $2432902008176639896$. If you want, I could post the numbers in an answer.
    – Claude Leibovici
    Nov 22 at 13:05






  • 1




    Comparing to high order expansion of Stirling formula.
    – Claude Leibovici
    Nov 22 at 13:41













up vote
0
down vote

favorite
1









up vote
0
down vote

favorite
1






1





I have to calculate factorials of an arbitrarily large integers mod another arbitrarily large integer. i have considered Stirling's Approximation and Ramanujan’s factorial approximation. is it possible to get better estimates?






factorial







share|cite|improve this question













I have to calculate factorials of an arbitrarily large integers mod another arbitrarily large integer. i have considered Stirling's Approximation and Ramanujan’s factorial approximation. is it possible to get better estimates?






factorial




factorial






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 22 at 11:47









Malik Hamza Murtaza

247




247








  • 3




    If you are using modular arithmetic, analytic estimates might not be the most useful thing for you, depending on what exactly your problem is. Instead things like en.wikipedia.org/wiki/Wilson%27s_theorem might help.
    – Alex J Best
    Nov 22 at 11:53






  • 1




    There's the gamma function but I don't think it will help you much in contexts of modular arithmetic...
    – Prasun Biswas
    Nov 22 at 11:57






  • 1




    Have a look at math.stackexchange.com/questions/676952/…. For $n=20$, what I wrote gives $2432902008176665737$ to be compared to the exact $2432902008176640000$
    – Claude Leibovici
    Nov 22 at 12:32






  • 1




    I don't know if this could be of interest since the coefficients start to be huge. Using the same kind of expression with degree $2$ in numerator and degree $3$ in denominator, for $n=20$, I get $2432902008176639896$. If you want, I could post the numbers in an answer.
    – Claude Leibovici
    Nov 22 at 13:05






  • 1




    Comparing to high order expansion of Stirling formula.
    – Claude Leibovici
    Nov 22 at 13:41














  • 3




    If you are using modular arithmetic, analytic estimates might not be the most useful thing for you, depending on what exactly your problem is. Instead things like en.wikipedia.org/wiki/Wilson%27s_theorem might help.
    – Alex J Best
    Nov 22 at 11:53






  • 1




    There's the gamma function but I don't think it will help you much in contexts of modular arithmetic...
    – Prasun Biswas
    Nov 22 at 11:57






  • 1




    Have a look at math.stackexchange.com/questions/676952/…. For $n=20$, what I wrote gives $2432902008176665737$ to be compared to the exact $2432902008176640000$
    – Claude Leibovici
    Nov 22 at 12:32






  • 1




    I don't know if this could be of interest since the coefficients start to be huge. Using the same kind of expression with degree $2$ in numerator and degree $3$ in denominator, for $n=20$, I get $2432902008176639896$. If you want, I could post the numbers in an answer.
    – Claude Leibovici
    Nov 22 at 13:05






  • 1




    Comparing to high order expansion of Stirling formula.
    – Claude Leibovici
    Nov 22 at 13:41








3




3




If you are using modular arithmetic, analytic estimates might not be the most useful thing for you, depending on what exactly your problem is. Instead things like en.wikipedia.org/wiki/Wilson%27s_theorem might help.
– Alex J Best
Nov 22 at 11:53




If you are using modular arithmetic, analytic estimates might not be the most useful thing for you, depending on what exactly your problem is. Instead things like en.wikipedia.org/wiki/Wilson%27s_theorem might help.
– Alex J Best
Nov 22 at 11:53




1




1




There's the gamma function but I don't think it will help you much in contexts of modular arithmetic...
– Prasun Biswas
Nov 22 at 11:57




There's the gamma function but I don't think it will help you much in contexts of modular arithmetic...
– Prasun Biswas
Nov 22 at 11:57




1




1




Have a look at math.stackexchange.com/questions/676952/…. For $n=20$, what I wrote gives $2432902008176665737$ to be compared to the exact $2432902008176640000$
– Claude Leibovici
Nov 22 at 12:32




Have a look at math.stackexchange.com/questions/676952/…. For $n=20$, what I wrote gives $2432902008176665737$ to be compared to the exact $2432902008176640000$
– Claude Leibovici
Nov 22 at 12:32




1




1




I don't know if this could be of interest since the coefficients start to be huge. Using the same kind of expression with degree $2$ in numerator and degree $3$ in denominator, for $n=20$, I get $2432902008176639896$. If you want, I could post the numbers in an answer.
– Claude Leibovici
Nov 22 at 13:05




I don't know if this could be of interest since the coefficients start to be huge. Using the same kind of expression with degree $2$ in numerator and degree $3$ in denominator, for $n=20$, I get $2432902008176639896$. If you want, I could post the numbers in an answer.
– Claude Leibovici
Nov 22 at 13:05




1




1




Comparing to high order expansion of Stirling formula.
– Claude Leibovici
Nov 22 at 13:41




Comparing to high order expansion of Stirling formula.
– Claude Leibovici
Nov 22 at 13:41










2 Answers
2






active

oldest

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up vote
1
down vote



accepted










Writing
$$n!approxsqrt{pi}left(frac{n}{e}right)^nrootLARGE{6}of{8n^3+4n^2+n+frac 1 {30}color{red}{-}x(n)}$$ with
$$x(n)=frac{a_0+a_1n+a_2n^2}{b_0+b_1n+b_2n^2+b_3n^3}$$ the coefficients are
$$left(
begin{array}{cc}
a_0 & 12521740840824081\
a_1 & 132077016740516320 \
a_2 & 261892615461486240 \
b_0 &3339455095907419720 \
b_1 & 7902477164268212400 \
b_2 & 5812898776788230400
end{array}
right)$$
A few values
$$left(
begin{array}{ccc}
n & text{approximation} & text{exact} \
10 & 3628800 & 3628800 \
15 & 1307674368000 & 1307674368000 \
20 & 2432902008176639896 & 2432902008176640000 \
25 & 15511210043330985910414618 & 15511210043330985984000000 \
30 & 265252859812191058429178640362769 & 265252859812191058636308480000000
end{array}
right)$$



Edit



Using ratios of polyomials (as done in this answer and the previous one) leads to incredibly huge coefficients. Thinking more about it, I thought that is would be better to just write
$$x(n)=sum_{k=1}^m frac{a_k}{n^k}$$ and, when required, transform this expansion to the desired $[p,p+1]$ Padé approximant. The coefficients so obtained are listed below
$$left(
begin{array}{cc}
k & a_k \
1 & frac{11}{240} \
2 & -frac{79}{3360} \
3 & -frac{3539}{201600} \
4 & frac{9511}{403200} \
5 & frac{10051}{716800} \
6 & -frac{233934691}{6386688000} \
7 & -frac{3595113569}{178827264000} \
8 & frac{403527851669}{4649508864000} \
9 & frac{25622861661869}{557941063680000} \
10 & -frac{30016604936501}{101443829760000} \
11 & -frac{685661227463561}{4463528509440000} \
12 & frac{109896661164737049961}{79673983893504000000}
end{array}
right)$$ For $n=25$, this would lead to $color{blue}{1551121004333098598400000}5$.



For $n=30$ , this would lead to $color{blue}{26525285981219105863630848}5359781$.



This seems to be significantly better.






share|cite|improve this answer






























    up vote
    0
    down vote













    Stirlings approximation is good ... especially if in the expression for $ln(n!)$ you include the series in odd powers of 1/n. Are you aware of that series - Stirlings aporoximation is all-too-often given without it. The coefficients of it are not too compliated - Bernoulli numbers multiplied by simple factors. Ah yes! maybe it is troublesome for very large n then because of that - the way the Bernoulli numbers 'turn round'.



    This might answer your question though.






    share|cite|improve this answer





















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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      1
      down vote



      accepted










      Writing
      $$n!approxsqrt{pi}left(frac{n}{e}right)^nrootLARGE{6}of{8n^3+4n^2+n+frac 1 {30}color{red}{-}x(n)}$$ with
      $$x(n)=frac{a_0+a_1n+a_2n^2}{b_0+b_1n+b_2n^2+b_3n^3}$$ the coefficients are
      $$left(
      begin{array}{cc}
      a_0 & 12521740840824081\
      a_1 & 132077016740516320 \
      a_2 & 261892615461486240 \
      b_0 &3339455095907419720 \
      b_1 & 7902477164268212400 \
      b_2 & 5812898776788230400
      end{array}
      right)$$
      A few values
      $$left(
      begin{array}{ccc}
      n & text{approximation} & text{exact} \
      10 & 3628800 & 3628800 \
      15 & 1307674368000 & 1307674368000 \
      20 & 2432902008176639896 & 2432902008176640000 \
      25 & 15511210043330985910414618 & 15511210043330985984000000 \
      30 & 265252859812191058429178640362769 & 265252859812191058636308480000000
      end{array}
      right)$$



      Edit



      Using ratios of polyomials (as done in this answer and the previous one) leads to incredibly huge coefficients. Thinking more about it, I thought that is would be better to just write
      $$x(n)=sum_{k=1}^m frac{a_k}{n^k}$$ and, when required, transform this expansion to the desired $[p,p+1]$ Padé approximant. The coefficients so obtained are listed below
      $$left(
      begin{array}{cc}
      k & a_k \
      1 & frac{11}{240} \
      2 & -frac{79}{3360} \
      3 & -frac{3539}{201600} \
      4 & frac{9511}{403200} \
      5 & frac{10051}{716800} \
      6 & -frac{233934691}{6386688000} \
      7 & -frac{3595113569}{178827264000} \
      8 & frac{403527851669}{4649508864000} \
      9 & frac{25622861661869}{557941063680000} \
      10 & -frac{30016604936501}{101443829760000} \
      11 & -frac{685661227463561}{4463528509440000} \
      12 & frac{109896661164737049961}{79673983893504000000}
      end{array}
      right)$$       For $n=25$, this would lead to $color{blue}{1551121004333098598400000}5$.



      For $n=30$ , this would lead to $color{blue}{26525285981219105863630848}5359781$.



      This seems to be significantly better.






      share|cite|improve this answer



























        up vote
        1
        down vote



        accepted










        Writing
        $$n!approxsqrt{pi}left(frac{n}{e}right)^nrootLARGE{6}of{8n^3+4n^2+n+frac 1 {30}color{red}{-}x(n)}$$ with
        $$x(n)=frac{a_0+a_1n+a_2n^2}{b_0+b_1n+b_2n^2+b_3n^3}$$ the coefficients are
        $$left(
        begin{array}{cc}
        a_0 & 12521740840824081\
        a_1 & 132077016740516320 \
        a_2 & 261892615461486240 \
        b_0 &3339455095907419720 \
        b_1 & 7902477164268212400 \
        b_2 & 5812898776788230400
        end{array}
        right)$$
        A few values
        $$left(
        begin{array}{ccc}
        n & text{approximation} & text{exact} \
        10 & 3628800 & 3628800 \
        15 & 1307674368000 & 1307674368000 \
        20 & 2432902008176639896 & 2432902008176640000 \
        25 & 15511210043330985910414618 & 15511210043330985984000000 \
        30 & 265252859812191058429178640362769 & 265252859812191058636308480000000
        end{array}
        right)$$



        Edit



        Using ratios of polyomials (as done in this answer and the previous one) leads to incredibly huge coefficients. Thinking more about it, I thought that is would be better to just write
        $$x(n)=sum_{k=1}^m frac{a_k}{n^k}$$ and, when required, transform this expansion to the desired $[p,p+1]$ Padé approximant. The coefficients so obtained are listed below
        $$left(
        begin{array}{cc}
        k & a_k \
        1 & frac{11}{240} \
        2 & -frac{79}{3360} \
        3 & -frac{3539}{201600} \
        4 & frac{9511}{403200} \
        5 & frac{10051}{716800} \
        6 & -frac{233934691}{6386688000} \
        7 & -frac{3595113569}{178827264000} \
        8 & frac{403527851669}{4649508864000} \
        9 & frac{25622861661869}{557941063680000} \
        10 & -frac{30016604936501}{101443829760000} \
        11 & -frac{685661227463561}{4463528509440000} \
        12 & frac{109896661164737049961}{79673983893504000000}
        end{array}
        right)$$         For $n=25$, this would lead to $color{blue}{1551121004333098598400000}5$.



        For $n=30$ , this would lead to $color{blue}{26525285981219105863630848}5359781$.



        This seems to be significantly better.






        share|cite|improve this answer

























          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          Writing
          $$n!approxsqrt{pi}left(frac{n}{e}right)^nrootLARGE{6}of{8n^3+4n^2+n+frac 1 {30}color{red}{-}x(n)}$$ with
          $$x(n)=frac{a_0+a_1n+a_2n^2}{b_0+b_1n+b_2n^2+b_3n^3}$$ the coefficients are
          $$left(
          begin{array}{cc}
          a_0 & 12521740840824081\
          a_1 & 132077016740516320 \
          a_2 & 261892615461486240 \
          b_0 &3339455095907419720 \
          b_1 & 7902477164268212400 \
          b_2 & 5812898776788230400
          end{array}
          right)$$
          A few values
          $$left(
          begin{array}{ccc}
          n & text{approximation} & text{exact} \
          10 & 3628800 & 3628800 \
          15 & 1307674368000 & 1307674368000 \
          20 & 2432902008176639896 & 2432902008176640000 \
          25 & 15511210043330985910414618 & 15511210043330985984000000 \
          30 & 265252859812191058429178640362769 & 265252859812191058636308480000000
          end{array}
          right)$$



          Edit



          Using ratios of polyomials (as done in this answer and the previous one) leads to incredibly huge coefficients. Thinking more about it, I thought that is would be better to just write
          $$x(n)=sum_{k=1}^m frac{a_k}{n^k}$$ and, when required, transform this expansion to the desired $[p,p+1]$ Padé approximant. The coefficients so obtained are listed below
          $$left(
          begin{array}{cc}
          k & a_k \
          1 & frac{11}{240} \
          2 & -frac{79}{3360} \
          3 & -frac{3539}{201600} \
          4 & frac{9511}{403200} \
          5 & frac{10051}{716800} \
          6 & -frac{233934691}{6386688000} \
          7 & -frac{3595113569}{178827264000} \
          8 & frac{403527851669}{4649508864000} \
          9 & frac{25622861661869}{557941063680000} \
          10 & -frac{30016604936501}{101443829760000} \
          11 & -frac{685661227463561}{4463528509440000} \
          12 & frac{109896661164737049961}{79673983893504000000}
          end{array}
          right)$$           For $n=25$, this would lead to $color{blue}{1551121004333098598400000}5$.



          For $n=30$ , this would lead to $color{blue}{26525285981219105863630848}5359781$.



          This seems to be significantly better.






          share|cite|improve this answer














          Writing
          $$n!approxsqrt{pi}left(frac{n}{e}right)^nrootLARGE{6}of{8n^3+4n^2+n+frac 1 {30}color{red}{-}x(n)}$$ with
          $$x(n)=frac{a_0+a_1n+a_2n^2}{b_0+b_1n+b_2n^2+b_3n^3}$$ the coefficients are
          $$left(
          begin{array}{cc}
          a_0 & 12521740840824081\
          a_1 & 132077016740516320 \
          a_2 & 261892615461486240 \
          b_0 &3339455095907419720 \
          b_1 & 7902477164268212400 \
          b_2 & 5812898776788230400
          end{array}
          right)$$
          A few values
          $$left(
          begin{array}{ccc}
          n & text{approximation} & text{exact} \
          10 & 3628800 & 3628800 \
          15 & 1307674368000 & 1307674368000 \
          20 & 2432902008176639896 & 2432902008176640000 \
          25 & 15511210043330985910414618 & 15511210043330985984000000 \
          30 & 265252859812191058429178640362769 & 265252859812191058636308480000000
          end{array}
          right)$$



          Edit



          Using ratios of polyomials (as done in this answer and the previous one) leads to incredibly huge coefficients. Thinking more about it, I thought that is would be better to just write
          $$x(n)=sum_{k=1}^m frac{a_k}{n^k}$$ and, when required, transform this expansion to the desired $[p,p+1]$ Padé approximant. The coefficients so obtained are listed below
          $$left(
          begin{array}{cc}
          k & a_k \
          1 & frac{11}{240} \
          2 & -frac{79}{3360} \
          3 & -frac{3539}{201600} \
          4 & frac{9511}{403200} \
          5 & frac{10051}{716800} \
          6 & -frac{233934691}{6386688000} \
          7 & -frac{3595113569}{178827264000} \
          8 & frac{403527851669}{4649508864000} \
          9 & frac{25622861661869}{557941063680000} \
          10 & -frac{30016604936501}{101443829760000} \
          11 & -frac{685661227463561}{4463528509440000} \
          12 & frac{109896661164737049961}{79673983893504000000}
          end{array}
          right)$$           For $n=25$, this would lead to $color{blue}{1551121004333098598400000}5$.



          For $n=30$ , this would lead to $color{blue}{26525285981219105863630848}5359781$.



          This seems to be significantly better.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 23 at 5:40

























          answered Nov 22 at 13:39









          Claude Leibovici

          118k1156131




          118k1156131






















              up vote
              0
              down vote













              Stirlings approximation is good ... especially if in the expression for $ln(n!)$ you include the series in odd powers of 1/n. Are you aware of that series - Stirlings aporoximation is all-too-often given without it. The coefficients of it are not too compliated - Bernoulli numbers multiplied by simple factors. Ah yes! maybe it is troublesome for very large n then because of that - the way the Bernoulli numbers 'turn round'.



              This might answer your question though.






              share|cite|improve this answer

























                up vote
                0
                down vote













                Stirlings approximation is good ... especially if in the expression for $ln(n!)$ you include the series in odd powers of 1/n. Are you aware of that series - Stirlings aporoximation is all-too-often given without it. The coefficients of it are not too compliated - Bernoulli numbers multiplied by simple factors. Ah yes! maybe it is troublesome for very large n then because of that - the way the Bernoulli numbers 'turn round'.



                This might answer your question though.






                share|cite|improve this answer























                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  Stirlings approximation is good ... especially if in the expression for $ln(n!)$ you include the series in odd powers of 1/n. Are you aware of that series - Stirlings aporoximation is all-too-often given without it. The coefficients of it are not too compliated - Bernoulli numbers multiplied by simple factors. Ah yes! maybe it is troublesome for very large n then because of that - the way the Bernoulli numbers 'turn round'.



                  This might answer your question though.






                  share|cite|improve this answer












                  Stirlings approximation is good ... especially if in the expression for $ln(n!)$ you include the series in odd powers of 1/n. Are you aware of that series - Stirlings aporoximation is all-too-often given without it. The coefficients of it are not too compliated - Bernoulli numbers multiplied by simple factors. Ah yes! maybe it is troublesome for very large n then because of that - the way the Bernoulli numbers 'turn round'.



                  This might answer your question though.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 23 at 7:06









                  AmbretteOrrisey

                  51110




                  51110






























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