What is the best estimation of the factorial function?











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












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

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
    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





















      Your Answer





      StackExchange.ifUsing("editor", function () {
      return StackExchange.using("mathjaxEditing", function () {
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      });
      });
      }, "mathjax-editing");

      StackExchange.ready(function() {
      var channelOptions = {
      tags: "".split(" "),
      id: "69"
      };
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function() {
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled) {
      StackExchange.using("snippets", function() {
      createEditor();
      });
      }
      else {
      createEditor();
      }
      });

      function createEditor() {
      StackExchange.prepareEditor({
      heartbeatType: 'answer',
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });














      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3009040%2fwhat-is-the-best-estimation-of-the-factorial-function%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      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






























                      draft saved

                      draft discarded




















































                      Thanks for contributing an answer to Mathematics Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid …



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.





                      Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                      Please pay close attention to the following guidance:


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid …



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3009040%2fwhat-is-the-best-estimation-of-the-factorial-function%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      -

                      Popular posts from this blog

                      Ellipse (mathématiques)

                      Image
                      Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
                      Read more

                      Quarter-circle Tiles

                      Image
                      up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
                      Read more

                      Mont Emei

                      Image
                      Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
                      Read more