How to find first 32 digits of $displaystyle prod_{n=1}^{infty} (1-gamma^n)$?












1












$begingroup$


I need to find first 32 digits of $displaystyle prod_{n=1}^{infty} (1-gamma^n)$ but wolframalpha's brain is too narrow to contain the result, and I don't know any software and programming to find the result. $gamma$ is the Euler's constant. Or does it converge to a known number? Please help! Thanks



Added for Bounty. Does the mentioned infinite product have any closed form in terms of known mathematical constants?






infinite-product q-series







share|cite|improve this question











$endgroup$












  • $begingroup$
    Using Euler's pentagonal number formula gives a very rapidly convergent series. Of course, you need the value of $gamma$ quite accurately.
    $endgroup$
    – Lord Shark the Unknown
    Jan 11 at 5:23










  • $begingroup$
    Pentagonal number theorem.
    $endgroup$
    – Alex Ravsky
    Jan 15 at 6:40
















1












$begingroup$


I need to find first 32 digits of $displaystyle prod_{n=1}^{infty} (1-gamma^n)$ but wolframalpha's brain is too narrow to contain the result, and I don't know any software and programming to find the result. $gamma$ is the Euler's constant. Or does it converge to a known number? Please help! Thanks



Added for Bounty. Does the mentioned infinite product have any closed form in terms of known mathematical constants?






infinite-product q-series







share|cite|improve this question











$endgroup$












  • $begingroup$
    Using Euler's pentagonal number formula gives a very rapidly convergent series. Of course, you need the value of $gamma$ quite accurately.
    $endgroup$
    – Lord Shark the Unknown
    Jan 11 at 5:23










  • $begingroup$
    Pentagonal number theorem.
    $endgroup$
    – Alex Ravsky
    Jan 15 at 6:40














1












1








1


1



$begingroup$


I need to find first 32 digits of $displaystyle prod_{n=1}^{infty} (1-gamma^n)$ but wolframalpha's brain is too narrow to contain the result, and I don't know any software and programming to find the result. $gamma$ is the Euler's constant. Or does it converge to a known number? Please help! Thanks



Added for Bounty. Does the mentioned infinite product have any closed form in terms of known mathematical constants?






infinite-product q-series







share|cite|improve this question











$endgroup$




I need to find first 32 digits of $displaystyle prod_{n=1}^{infty} (1-gamma^n)$ but wolframalpha's brain is too narrow to contain the result, and I don't know any software and programming to find the result. $gamma$ is the Euler's constant. Or does it converge to a known number? Please help! Thanks



Added for Bounty. Does the mentioned infinite product have any closed form in terms of known mathematical constants?






infinite-product q-series




infinite-product q-series






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 13 at 9:39







72D

















asked Jan 11 at 5:07









72D72D

299117




299117












  • $begingroup$
    Using Euler's pentagonal number formula gives a very rapidly convergent series. Of course, you need the value of $gamma$ quite accurately.
    $endgroup$
    – Lord Shark the Unknown
    Jan 11 at 5:23










  • $begingroup$
    Pentagonal number theorem.
    $endgroup$
    – Alex Ravsky
    Jan 15 at 6:40


















  • $begingroup$
    Using Euler's pentagonal number formula gives a very rapidly convergent series. Of course, you need the value of $gamma$ quite accurately.
    $endgroup$
    – Lord Shark the Unknown
    Jan 11 at 5:23










  • $begingroup$
    Pentagonal number theorem.
    $endgroup$
    – Alex Ravsky
    Jan 15 at 6:40
















$begingroup$
Using Euler's pentagonal number formula gives a very rapidly convergent series. Of course, you need the value of $gamma$ quite accurately.
$endgroup$
– Lord Shark the Unknown
Jan 11 at 5:23




$begingroup$
Using Euler's pentagonal number formula gives a very rapidly convergent series. Of course, you need the value of $gamma$ quite accurately.
$endgroup$
– Lord Shark the Unknown
Jan 11 at 5:23












$begingroup$
Pentagonal number theorem.
$endgroup$
– Alex Ravsky
Jan 15 at 6:40




$begingroup$
Pentagonal number theorem.
$endgroup$
– Alex Ravsky
Jan 15 at 6:40










2 Answers
2






active

oldest

votes


















5












$begingroup$

Using Pochhammer symbols $$displaystyle prod_{n=1}^{infty} (1-gamma^n)=(gamma ;gamma )_{infty }$$ A very fast way to compute it (have a look here) is
$$(gamma ;gamma )_{infty }=sum_{k=-infty}^infty (-1)^k ,gamma ^{frac{3k^2-k}{2} }$$
Consider partial sums
$$S_p=sum_{k=-p}^p (-1)^k ,gamma ^{frac{3k^2-k}{2} }$$ and compute for $50$ decimal places
$$left(
begin{array}{cc}
5 & 0.17340542156115797462125386932229340638693672477775 \
6 & 0.17340542156185623295922558757637084613978106011987 \
7 & 0.17340542156185621287562583984349502239524993758969 \
8 & 0.17340542156185621287573763551045706783782602427578 \
9 & 0.17340542156185621287573763539033111094663184209447 \
10 & 0.17340542156185621287573763539033113582404282775260 \
11 & 0.17340542156185621287573763539033113582404183569900 \
12 & 0.17340542156185621287573763539033113582404183569901 \
13 & 0.17340542156185621287573763539033113582404183569901 \
14 & 0.17340542156185621287573763539033113582404183569900 \
15 & 0.17340542156185621287573763539033113582404183569900
end{array}
right)$$ For $32$ decimal places $S_9$ would be fine.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    I asked Alpha and got
    $0.1734054215618562128757376353903311358240418\
    356990083565526180089819971434621977210396477\
    9552159698610211689009709...$



    with one click on More Digits. I have found that sometimes when a calculation fails, close the tab, open a new one, and try again sometimes works.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      So I wasn't asking it probably! Would you let me know if there is a 'closed form' for it, I mean the product as a function of other known numbers?
      $endgroup$
      – 72D
      Jan 11 at 5:19












    • $begingroup$
      I would guess not, but don't really know.
      $endgroup$
      – Ross Millikan
      Jan 11 at 5:19












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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    5












    $begingroup$

    Using Pochhammer symbols $$displaystyle prod_{n=1}^{infty} (1-gamma^n)=(gamma ;gamma )_{infty }$$ A very fast way to compute it (have a look here) is
    $$(gamma ;gamma )_{infty }=sum_{k=-infty}^infty (-1)^k ,gamma ^{frac{3k^2-k}{2} }$$
    Consider partial sums
    $$S_p=sum_{k=-p}^p (-1)^k ,gamma ^{frac{3k^2-k}{2} }$$ and compute for $50$ decimal places
    $$left(
    begin{array}{cc}
    5 & 0.17340542156115797462125386932229340638693672477775 \
    6 & 0.17340542156185623295922558757637084613978106011987 \
    7 & 0.17340542156185621287562583984349502239524993758969 \
    8 & 0.17340542156185621287573763551045706783782602427578 \
    9 & 0.17340542156185621287573763539033111094663184209447 \
    10 & 0.17340542156185621287573763539033113582404282775260 \
    11 & 0.17340542156185621287573763539033113582404183569900 \
    12 & 0.17340542156185621287573763539033113582404183569901 \
    13 & 0.17340542156185621287573763539033113582404183569901 \
    14 & 0.17340542156185621287573763539033113582404183569900 \
    15 & 0.17340542156185621287573763539033113582404183569900
    end{array}
    right)$$     For $32$ decimal places $S_9$ would be fine.






    share|cite|improve this answer









    $endgroup$


















      5












      $begingroup$

      Using Pochhammer symbols $$displaystyle prod_{n=1}^{infty} (1-gamma^n)=(gamma ;gamma )_{infty }$$ A very fast way to compute it (have a look here) is
      $$(gamma ;gamma )_{infty }=sum_{k=-infty}^infty (-1)^k ,gamma ^{frac{3k^2-k}{2} }$$
      Consider partial sums
      $$S_p=sum_{k=-p}^p (-1)^k ,gamma ^{frac{3k^2-k}{2} }$$ and compute for $50$ decimal places
      $$left(
      begin{array}{cc}
      5 & 0.17340542156115797462125386932229340638693672477775 \
      6 & 0.17340542156185623295922558757637084613978106011987 \
      7 & 0.17340542156185621287562583984349502239524993758969 \
      8 & 0.17340542156185621287573763551045706783782602427578 \
      9 & 0.17340542156185621287573763539033111094663184209447 \
      10 & 0.17340542156185621287573763539033113582404282775260 \
      11 & 0.17340542156185621287573763539033113582404183569900 \
      12 & 0.17340542156185621287573763539033113582404183569901 \
      13 & 0.17340542156185621287573763539033113582404183569901 \
      14 & 0.17340542156185621287573763539033113582404183569900 \
      15 & 0.17340542156185621287573763539033113582404183569900
      end{array}
      right)$$       For $32$ decimal places $S_9$ would be fine.






      share|cite|improve this answer









      $endgroup$
















        5












        5








        5





        $begingroup$

        Using Pochhammer symbols $$displaystyle prod_{n=1}^{infty} (1-gamma^n)=(gamma ;gamma )_{infty }$$ A very fast way to compute it (have a look here) is
        $$(gamma ;gamma )_{infty }=sum_{k=-infty}^infty (-1)^k ,gamma ^{frac{3k^2-k}{2} }$$
        Consider partial sums
        $$S_p=sum_{k=-p}^p (-1)^k ,gamma ^{frac{3k^2-k}{2} }$$ and compute for $50$ decimal places
        $$left(
        begin{array}{cc}
        5 & 0.17340542156115797462125386932229340638693672477775 \
        6 & 0.17340542156185623295922558757637084613978106011987 \
        7 & 0.17340542156185621287562583984349502239524993758969 \
        8 & 0.17340542156185621287573763551045706783782602427578 \
        9 & 0.17340542156185621287573763539033111094663184209447 \
        10 & 0.17340542156185621287573763539033113582404282775260 \
        11 & 0.17340542156185621287573763539033113582404183569900 \
        12 & 0.17340542156185621287573763539033113582404183569901 \
        13 & 0.17340542156185621287573763539033113582404183569901 \
        14 & 0.17340542156185621287573763539033113582404183569900 \
        15 & 0.17340542156185621287573763539033113582404183569900
        end{array}
        right)$$         For $32$ decimal places $S_9$ would be fine.






        share|cite|improve this answer









        $endgroup$



        Using Pochhammer symbols $$displaystyle prod_{n=1}^{infty} (1-gamma^n)=(gamma ;gamma )_{infty }$$ A very fast way to compute it (have a look here) is
        $$(gamma ;gamma )_{infty }=sum_{k=-infty}^infty (-1)^k ,gamma ^{frac{3k^2-k}{2} }$$
        Consider partial sums
        $$S_p=sum_{k=-p}^p (-1)^k ,gamma ^{frac{3k^2-k}{2} }$$ and compute for $50$ decimal places
        $$left(
        begin{array}{cc}
        5 & 0.17340542156115797462125386932229340638693672477775 \
        6 & 0.17340542156185623295922558757637084613978106011987 \
        7 & 0.17340542156185621287562583984349502239524993758969 \
        8 & 0.17340542156185621287573763551045706783782602427578 \
        9 & 0.17340542156185621287573763539033111094663184209447 \
        10 & 0.17340542156185621287573763539033113582404282775260 \
        11 & 0.17340542156185621287573763539033113582404183569900 \
        12 & 0.17340542156185621287573763539033113582404183569901 \
        13 & 0.17340542156185621287573763539033113582404183569901 \
        14 & 0.17340542156185621287573763539033113582404183569900 \
        15 & 0.17340542156185621287573763539033113582404183569900
        end{array}
        right)$$         For $32$ decimal places $S_9$ would be fine.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 11 at 6:10









        Claude LeiboviciClaude Leibovici

        125k1158135




        125k1158135























            1












            $begingroup$

            I asked Alpha and got
            $0.1734054215618562128757376353903311358240418\
            356990083565526180089819971434621977210396477\
            9552159698610211689009709...$



            with one click on More Digits. I have found that sometimes when a calculation fails, close the tab, open a new one, and try again sometimes works.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              So I wasn't asking it probably! Would you let me know if there is a 'closed form' for it, I mean the product as a function of other known numbers?
              $endgroup$
              – 72D
              Jan 11 at 5:19












            • $begingroup$
              I would guess not, but don't really know.
              $endgroup$
              – Ross Millikan
              Jan 11 at 5:19
















            1












            $begingroup$

            I asked Alpha and got
            $0.1734054215618562128757376353903311358240418\
            356990083565526180089819971434621977210396477\
            9552159698610211689009709...$



            with one click on More Digits. I have found that sometimes when a calculation fails, close the tab, open a new one, and try again sometimes works.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              So I wasn't asking it probably! Would you let me know if there is a 'closed form' for it, I mean the product as a function of other known numbers?
              $endgroup$
              – 72D
              Jan 11 at 5:19












            • $begingroup$
              I would guess not, but don't really know.
              $endgroup$
              – Ross Millikan
              Jan 11 at 5:19














            1












            1








            1





            $begingroup$

            I asked Alpha and got
            $0.1734054215618562128757376353903311358240418\
            356990083565526180089819971434621977210396477\
            9552159698610211689009709...$



            with one click on More Digits. I have found that sometimes when a calculation fails, close the tab, open a new one, and try again sometimes works.






            share|cite|improve this answer









            $endgroup$



            I asked Alpha and got
            $0.1734054215618562128757376353903311358240418\
            356990083565526180089819971434621977210396477\
            9552159698610211689009709...$



            with one click on More Digits. I have found that sometimes when a calculation fails, close the tab, open a new one, and try again sometimes works.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 11 at 5:16









            Ross MillikanRoss Millikan

            301k24200375




            301k24200375












            • $begingroup$
              So I wasn't asking it probably! Would you let me know if there is a 'closed form' for it, I mean the product as a function of other known numbers?
              $endgroup$
              – 72D
              Jan 11 at 5:19












            • $begingroup$
              I would guess not, but don't really know.
              $endgroup$
              – Ross Millikan
              Jan 11 at 5:19


















            • $begingroup$
              So I wasn't asking it probably! Would you let me know if there is a 'closed form' for it, I mean the product as a function of other known numbers?
              $endgroup$
              – 72D
              Jan 11 at 5:19












            • $begingroup$
              I would guess not, but don't really know.
              $endgroup$
              – Ross Millikan
              Jan 11 at 5:19
















            $begingroup$
            So I wasn't asking it probably! Would you let me know if there is a 'closed form' for it, I mean the product as a function of other known numbers?
            $endgroup$
            – 72D
            Jan 11 at 5:19






            $begingroup$
            So I wasn't asking it probably! Would you let me know if there is a 'closed form' for it, I mean the product as a function of other known numbers?
            $endgroup$
            – 72D
            Jan 11 at 5:19














            $begingroup$
            I would guess not, but don't really know.
            $endgroup$
            – Ross Millikan
            Jan 11 at 5:19




            $begingroup$
            I would guess not, but don't really know.
            $endgroup$
            – Ross Millikan
            Jan 11 at 5:19


















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