I have found a way of computing Euler's number. Is there any possible intuition of how that might be the...












4












$begingroup$


So a few days ago I just kind of messed around with my calculator, when I had an idea about a new continued fraction. I inputted it, and I found that it converged really quickly, and, quite wondrously, converged to the base of the natural logarithm.
Then, the day later, I tried computing a similar continued fraction, this time starting with the number 2 instead of 1, without the increment of the numerator by 1.



My first computation
My second computation



Interestingly, the second computation converges to EXACTLY the same values, except it's faster in the sense that the first method is delayed by 1 value per time.
Does anybody possibly have an intuition / explanation as for why this might be the case? I've looked at the general ways of computing e: the sum of integer factorials and (1+1/n)^n, which I suspect might be more relevant to this.






sequences-and-series convergence exponential-function continued-fractions conjectures







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$endgroup$








  • 1




    $begingroup$
    I feel like the first one can be rewritten as $sum_{i=0}^infty frac{1}{i!}$ somehow. Then, the second sum is $2+sum_{i=2}^infty frac{1}{i!}$, which is equivalent to the first sum.
    $endgroup$
    – Noble Mushtak
    Dec 29 '18 at 17:09








  • 3




    $begingroup$
    Both are just forms of the factorial series. (Why images, not formulas?)
    $endgroup$
    – metamorphy
    Dec 29 '18 at 17:10






  • 2




    $begingroup$
    The first thing I notice is that the denominators get multiplied together and make factorials. I've not tried on paper, but it looks as though a bit of cancelling turns it straight into the usual series.
    $endgroup$
    – timtfj
    Jan 29 at 20:20
















4












$begingroup$


So a few days ago I just kind of messed around with my calculator, when I had an idea about a new continued fraction. I inputted it, and I found that it converged really quickly, and, quite wondrously, converged to the base of the natural logarithm.
Then, the day later, I tried computing a similar continued fraction, this time starting with the number 2 instead of 1, without the increment of the numerator by 1.



My first computation
My second computation



Interestingly, the second computation converges to EXACTLY the same values, except it's faster in the sense that the first method is delayed by 1 value per time.
Does anybody possibly have an intuition / explanation as for why this might be the case? I've looked at the general ways of computing e: the sum of integer factorials and (1+1/n)^n, which I suspect might be more relevant to this.






sequences-and-series convergence exponential-function continued-fractions conjectures







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I feel like the first one can be rewritten as $sum_{i=0}^infty frac{1}{i!}$ somehow. Then, the second sum is $2+sum_{i=2}^infty frac{1}{i!}$, which is equivalent to the first sum.
    $endgroup$
    – Noble Mushtak
    Dec 29 '18 at 17:09








  • 3




    $begingroup$
    Both are just forms of the factorial series. (Why images, not formulas?)
    $endgroup$
    – metamorphy
    Dec 29 '18 at 17:10






  • 2




    $begingroup$
    The first thing I notice is that the denominators get multiplied together and make factorials. I've not tried on paper, but it looks as though a bit of cancelling turns it straight into the usual series.
    $endgroup$
    – timtfj
    Jan 29 at 20:20














4












4








4





$begingroup$


So a few days ago I just kind of messed around with my calculator, when I had an idea about a new continued fraction. I inputted it, and I found that it converged really quickly, and, quite wondrously, converged to the base of the natural logarithm.
Then, the day later, I tried computing a similar continued fraction, this time starting with the number 2 instead of 1, without the increment of the numerator by 1.



My first computation
My second computation



Interestingly, the second computation converges to EXACTLY the same values, except it's faster in the sense that the first method is delayed by 1 value per time.
Does anybody possibly have an intuition / explanation as for why this might be the case? I've looked at the general ways of computing e: the sum of integer factorials and (1+1/n)^n, which I suspect might be more relevant to this.






sequences-and-series convergence exponential-function continued-fractions conjectures







share|cite|improve this question











$endgroup$




So a few days ago I just kind of messed around with my calculator, when I had an idea about a new continued fraction. I inputted it, and I found that it converged really quickly, and, quite wondrously, converged to the base of the natural logarithm.
Then, the day later, I tried computing a similar continued fraction, this time starting with the number 2 instead of 1, without the increment of the numerator by 1.



My first computation
My second computation



Interestingly, the second computation converges to EXACTLY the same values, except it's faster in the sense that the first method is delayed by 1 value per time.
Does anybody possibly have an intuition / explanation as for why this might be the case? I've looked at the general ways of computing e: the sum of integer factorials and (1+1/n)^n, which I suspect might be more relevant to this.






sequences-and-series convergence exponential-function continued-fractions conjectures




sequences-and-series convergence exponential-function continued-fractions conjectures






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 29 at 20:09







Avtarás Karîm Elymés̱er

















asked Dec 29 '18 at 17:02









Avtarás Karîm Elymés̱erAvtarás Karîm Elymés̱er

243




243








  • 1




    $begingroup$
    I feel like the first one can be rewritten as $sum_{i=0}^infty frac{1}{i!}$ somehow. Then, the second sum is $2+sum_{i=2}^infty frac{1}{i!}$, which is equivalent to the first sum.
    $endgroup$
    – Noble Mushtak
    Dec 29 '18 at 17:09








  • 3




    $begingroup$
    Both are just forms of the factorial series. (Why images, not formulas?)
    $endgroup$
    – metamorphy
    Dec 29 '18 at 17:10






  • 2




    $begingroup$
    The first thing I notice is that the denominators get multiplied together and make factorials. I've not tried on paper, but it looks as though a bit of cancelling turns it straight into the usual series.
    $endgroup$
    – timtfj
    Jan 29 at 20:20














  • 1




    $begingroup$
    I feel like the first one can be rewritten as $sum_{i=0}^infty frac{1}{i!}$ somehow. Then, the second sum is $2+sum_{i=2}^infty frac{1}{i!}$, which is equivalent to the first sum.
    $endgroup$
    – Noble Mushtak
    Dec 29 '18 at 17:09








  • 3




    $begingroup$
    Both are just forms of the factorial series. (Why images, not formulas?)
    $endgroup$
    – metamorphy
    Dec 29 '18 at 17:10






  • 2




    $begingroup$
    The first thing I notice is that the denominators get multiplied together and make factorials. I've not tried on paper, but it looks as though a bit of cancelling turns it straight into the usual series.
    $endgroup$
    – timtfj
    Jan 29 at 20:20








1




1




$begingroup$
I feel like the first one can be rewritten as $sum_{i=0}^infty frac{1}{i!}$ somehow. Then, the second sum is $2+sum_{i=2}^infty frac{1}{i!}$, which is equivalent to the first sum.
$endgroup$
– Noble Mushtak
Dec 29 '18 at 17:09






$begingroup$
I feel like the first one can be rewritten as $sum_{i=0}^infty frac{1}{i!}$ somehow. Then, the second sum is $2+sum_{i=2}^infty frac{1}{i!}$, which is equivalent to the first sum.
$endgroup$
– Noble Mushtak
Dec 29 '18 at 17:09






3




3




$begingroup$
Both are just forms of the factorial series. (Why images, not formulas?)
$endgroup$
– metamorphy
Dec 29 '18 at 17:10




$begingroup$
Both are just forms of the factorial series. (Why images, not formulas?)
$endgroup$
– metamorphy
Dec 29 '18 at 17:10




2




2




$begingroup$
The first thing I notice is that the denominators get multiplied together and make factorials. I've not tried on paper, but it looks as though a bit of cancelling turns it straight into the usual series.
$endgroup$
– timtfj
Jan 29 at 20:20




$begingroup$
The first thing I notice is that the denominators get multiplied together and make factorials. I've not tried on paper, but it looks as though a bit of cancelling turns it straight into the usual series.
$endgroup$
– timtfj
Jan 29 at 20:20










2 Answers
2






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10












$begingroup$

Your second computation is equal to your first one:
$$1 + frac{2+frac{3+frac{4+frac{5+ frac{cdots}{cdots}}{5}}{4}}{3}}{2} = 1 + frac{2}{2} + frac{frac{3}{3} + frac{frac{4}{4} + frac{frac{5}{5} + frac{frac{6}{6} + frac{cdots}{cdots}}{5}}{4}}{3}}{2} = 2 + frac{1+frac{1+frac{1+frac{1+ frac{cdots}{cdots}}{5}}{4}}{3}}{2} $$



I hope this is clear.






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$endgroup$





















    7












    $begingroup$

    Your first calculation can be written as



    $$1+frac12(2+frac13(3+frac14(4+dots)))$$



    which expands to



    $$1+frac{2}{2}+frac{3}{2cdot 3}+frac{4}{2cdot 3 cdot 4} + dots$$



    Cancelling the repeated number in each term turns it into



    $$1+frac11+frac1{1cdot2}+frac{1}{1cdot2cdot 3} + dots$$



    $$=frac{1}{0!}+frac{1}{1!}+frac{1}{2!}+frac{1}{3!}+dots$$
    $$=e$$



    You've effectively factorised the series and written the result as a big fraction.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thank you for the explanation! Yes, I find this answer more useful than most other answers here.
      $endgroup$
      – Avtarás Karîm Elymés̱er
      Jan 30 at 21:45












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    2 Answers
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    2 Answers
    2






    active

    oldest

    votes









    active

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    active

    oldest

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    10












    $begingroup$

    Your second computation is equal to your first one:
    $$1 + frac{2+frac{3+frac{4+frac{5+ frac{cdots}{cdots}}{5}}{4}}{3}}{2} = 1 + frac{2}{2} + frac{frac{3}{3} + frac{frac{4}{4} + frac{frac{5}{5} + frac{frac{6}{6} + frac{cdots}{cdots}}{5}}{4}}{3}}{2} = 2 + frac{1+frac{1+frac{1+frac{1+ frac{cdots}{cdots}}{5}}{4}}{3}}{2} $$



    I hope this is clear.






    share|cite|improve this answer









    $endgroup$


















      10












      $begingroup$

      Your second computation is equal to your first one:
      $$1 + frac{2+frac{3+frac{4+frac{5+ frac{cdots}{cdots}}{5}}{4}}{3}}{2} = 1 + frac{2}{2} + frac{frac{3}{3} + frac{frac{4}{4} + frac{frac{5}{5} + frac{frac{6}{6} + frac{cdots}{cdots}}{5}}{4}}{3}}{2} = 2 + frac{1+frac{1+frac{1+frac{1+ frac{cdots}{cdots}}{5}}{4}}{3}}{2} $$



      I hope this is clear.






      share|cite|improve this answer









      $endgroup$
















        10












        10








        10





        $begingroup$

        Your second computation is equal to your first one:
        $$1 + frac{2+frac{3+frac{4+frac{5+ frac{cdots}{cdots}}{5}}{4}}{3}}{2} = 1 + frac{2}{2} + frac{frac{3}{3} + frac{frac{4}{4} + frac{frac{5}{5} + frac{frac{6}{6} + frac{cdots}{cdots}}{5}}{4}}{3}}{2} = 2 + frac{1+frac{1+frac{1+frac{1+ frac{cdots}{cdots}}{5}}{4}}{3}}{2} $$



        I hope this is clear.






        share|cite|improve this answer









        $endgroup$



        Your second computation is equal to your first one:
        $$1 + frac{2+frac{3+frac{4+frac{5+ frac{cdots}{cdots}}{5}}{4}}{3}}{2} = 1 + frac{2}{2} + frac{frac{3}{3} + frac{frac{4}{4} + frac{frac{5}{5} + frac{frac{6}{6} + frac{cdots}{cdots}}{5}}{4}}{3}}{2} = 2 + frac{1+frac{1+frac{1+frac{1+ frac{cdots}{cdots}}{5}}{4}}{3}}{2} $$



        I hope this is clear.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 29 '18 at 17:16









        Jonas De SchouwerJonas De Schouwer

        4089




        4089























            7












            $begingroup$

            Your first calculation can be written as



            $$1+frac12(2+frac13(3+frac14(4+dots)))$$



            which expands to



            $$1+frac{2}{2}+frac{3}{2cdot 3}+frac{4}{2cdot 3 cdot 4} + dots$$



            Cancelling the repeated number in each term turns it into



            $$1+frac11+frac1{1cdot2}+frac{1}{1cdot2cdot 3} + dots$$



            $$=frac{1}{0!}+frac{1}{1!}+frac{1}{2!}+frac{1}{3!}+dots$$
            $$=e$$



            You've effectively factorised the series and written the result as a big fraction.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Thank you for the explanation! Yes, I find this answer more useful than most other answers here.
              $endgroup$
              – Avtarás Karîm Elymés̱er
              Jan 30 at 21:45
















            7












            $begingroup$

            Your first calculation can be written as



            $$1+frac12(2+frac13(3+frac14(4+dots)))$$



            which expands to



            $$1+frac{2}{2}+frac{3}{2cdot 3}+frac{4}{2cdot 3 cdot 4} + dots$$



            Cancelling the repeated number in each term turns it into



            $$1+frac11+frac1{1cdot2}+frac{1}{1cdot2cdot 3} + dots$$



            $$=frac{1}{0!}+frac{1}{1!}+frac{1}{2!}+frac{1}{3!}+dots$$
            $$=e$$



            You've effectively factorised the series and written the result as a big fraction.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Thank you for the explanation! Yes, I find this answer more useful than most other answers here.
              $endgroup$
              – Avtarás Karîm Elymés̱er
              Jan 30 at 21:45














            7












            7








            7





            $begingroup$

            Your first calculation can be written as



            $$1+frac12(2+frac13(3+frac14(4+dots)))$$



            which expands to



            $$1+frac{2}{2}+frac{3}{2cdot 3}+frac{4}{2cdot 3 cdot 4} + dots$$



            Cancelling the repeated number in each term turns it into



            $$1+frac11+frac1{1cdot2}+frac{1}{1cdot2cdot 3} + dots$$



            $$=frac{1}{0!}+frac{1}{1!}+frac{1}{2!}+frac{1}{3!}+dots$$
            $$=e$$



            You've effectively factorised the series and written the result as a big fraction.






            share|cite|improve this answer











            $endgroup$



            Your first calculation can be written as



            $$1+frac12(2+frac13(3+frac14(4+dots)))$$



            which expands to



            $$1+frac{2}{2}+frac{3}{2cdot 3}+frac{4}{2cdot 3 cdot 4} + dots$$



            Cancelling the repeated number in each term turns it into



            $$1+frac11+frac1{1cdot2}+frac{1}{1cdot2cdot 3} + dots$$



            $$=frac{1}{0!}+frac{1}{1!}+frac{1}{2!}+frac{1}{3!}+dots$$
            $$=e$$



            You've effectively factorised the series and written the result as a big fraction.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 29 at 21:01

























            answered Jan 29 at 20:37









            timtfjtimtfj

            2,533420




            2,533420












            • $begingroup$
              Thank you for the explanation! Yes, I find this answer more useful than most other answers here.
              $endgroup$
              – Avtarás Karîm Elymés̱er
              Jan 30 at 21:45


















            • $begingroup$
              Thank you for the explanation! Yes, I find this answer more useful than most other answers here.
              $endgroup$
              – Avtarás Karîm Elymés̱er
              Jan 30 at 21:45
















            $begingroup$
            Thank you for the explanation! Yes, I find this answer more useful than most other answers here.
            $endgroup$
            – Avtarás Karîm Elymés̱er
            Jan 30 at 21:45




            $begingroup$
            Thank you for the explanation! Yes, I find this answer more useful than most other answers here.
            $endgroup$
            – Avtarás Karîm Elymés̱er
            Jan 30 at 21:45


















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