Are There Any Special Properties for Different Number of “Terms” in Equal Summations












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I found that solving for $e^{ix}$ gave a formula that produces two n "terms" of the starting summation



and defining terms as two n products like 1+ix being two terms of the summation



$e^{ix} $ =$sum_{n=0}^infty$ $frac{(ix)^n}{n!}$



$=1+ix-frac{x^2}{2!}-frac{ix^3}{3!}+...$



So solving for this using Euler's Identity $e^{i x}=cos(x)+isin(x)$,



$e^{ix}=sum_{n=0}^infty frac{(-1)^nx^{2n}}{(2n)!} + sum_{n=0}^infty frac{(-1)^nx^{2n+1}}{(2n+1)!}i $



$=sum_{n=0}^infty frac{(-1)^nx^{2n}}{(2n)!}+frac{(-1)^nx^{2n+1}}{(2n+1)(2n)!}i$



$=sum_{n=0}^infty frac{(2n+1)(-1)^nx^{2n}}{(2n)!(2n+1)}+frac{(-1)^nx^{2n+1}}{(2n+1)(2n)!}i$, so



$=sum_{n=0}^infty frac{(-1)^nx^{2n}(2n+ix+1)}{(2n+1)!}=e^{ix}$ ,



but unlike the first $e^{ix}$ series this one produces 2 parts of the first series with n=0 on the solved summation =1+ix, which were n=0 and n=1 for the original



from this I decided to make a summation that produced 3 terms per each n by grouping the original $e^{ix}$ summation by threes and got



$sum_{n=0}^infty frac{(-1)^{n+1}x^{3n}i^{n}(x^2-(3n+2)ix-(3n+1)(3n+2))}{(3n+2)!}$ ,



so n=0 produces $1+ix-frac{x^2}{2!}$, and n=1 producing $frac{-ix^3}{3!}+frac{x^4}{4!}+frac{ix^5}{5!}$, a continuation of the original infinite series. At the same time I haven't been able to prove this is equal to $e^{ix}$ through any other means



I also did this for 4 terms which created,



$sum_{n=0}^infty frac{-ix^{4n} (x^3-(4n+3)ix^2-(4n+2)(4n+3)x+(4n+1)(4n+2)(4n+3)i}{(4n+3)!}$



Which again n=0 produces $1+ix-frac{x^2}{2!}-frac{ix^3}{3!}$



and from these I tried to create a general formula to produce the infinite sum that produces a certain number of the original terms per n, as there seems to be a definite pattern, but I haven't quite found one yet.




So do all these have the same properties? Like would there be any theoretical way to produce a formula that has infinite terms, or truncate to the negatives, or zero? Or anything special that one summation is capable of over another?






sequences-and-series summation







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    I found that solving for $e^{ix}$ gave a formula that produces two n "terms" of the starting summation



    and defining terms as two n products like 1+ix being two terms of the summation



    $e^{ix} $ =$sum_{n=0}^infty$ $frac{(ix)^n}{n!}$



    $=1+ix-frac{x^2}{2!}-frac{ix^3}{3!}+...$



    So solving for this using Euler's Identity $e^{i x}=cos(x)+isin(x)$,



    $e^{ix}=sum_{n=0}^infty frac{(-1)^nx^{2n}}{(2n)!} + sum_{n=0}^infty frac{(-1)^nx^{2n+1}}{(2n+1)!}i $



    $=sum_{n=0}^infty frac{(-1)^nx^{2n}}{(2n)!}+frac{(-1)^nx^{2n+1}}{(2n+1)(2n)!}i$



    $=sum_{n=0}^infty frac{(2n+1)(-1)^nx^{2n}}{(2n)!(2n+1)}+frac{(-1)^nx^{2n+1}}{(2n+1)(2n)!}i$, so



    $=sum_{n=0}^infty frac{(-1)^nx^{2n}(2n+ix+1)}{(2n+1)!}=e^{ix}$ ,



    but unlike the first $e^{ix}$ series this one produces 2 parts of the first series with n=0 on the solved summation =1+ix, which were n=0 and n=1 for the original



    from this I decided to make a summation that produced 3 terms per each n by grouping the original $e^{ix}$ summation by threes and got



    $sum_{n=0}^infty frac{(-1)^{n+1}x^{3n}i^{n}(x^2-(3n+2)ix-(3n+1)(3n+2))}{(3n+2)!}$ ,



    so n=0 produces $1+ix-frac{x^2}{2!}$, and n=1 producing $frac{-ix^3}{3!}+frac{x^4}{4!}+frac{ix^5}{5!}$, a continuation of the original infinite series. At the same time I haven't been able to prove this is equal to $e^{ix}$ through any other means



    I also did this for 4 terms which created,



    $sum_{n=0}^infty frac{-ix^{4n} (x^3-(4n+3)ix^2-(4n+2)(4n+3)x+(4n+1)(4n+2)(4n+3)i}{(4n+3)!}$



    Which again n=0 produces $1+ix-frac{x^2}{2!}-frac{ix^3}{3!}$



    and from these I tried to create a general formula to produce the infinite sum that produces a certain number of the original terms per n, as there seems to be a definite pattern, but I haven't quite found one yet.




    So do all these have the same properties? Like would there be any theoretical way to produce a formula that has infinite terms, or truncate to the negatives, or zero? Or anything special that one summation is capable of over another?






    sequences-and-series summation







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      I found that solving for $e^{ix}$ gave a formula that produces two n "terms" of the starting summation



      and defining terms as two n products like 1+ix being two terms of the summation



      $e^{ix} $ =$sum_{n=0}^infty$ $frac{(ix)^n}{n!}$



      $=1+ix-frac{x^2}{2!}-frac{ix^3}{3!}+...$



      So solving for this using Euler's Identity $e^{i x}=cos(x)+isin(x)$,



      $e^{ix}=sum_{n=0}^infty frac{(-1)^nx^{2n}}{(2n)!} + sum_{n=0}^infty frac{(-1)^nx^{2n+1}}{(2n+1)!}i $



      $=sum_{n=0}^infty frac{(-1)^nx^{2n}}{(2n)!}+frac{(-1)^nx^{2n+1}}{(2n+1)(2n)!}i$



      $=sum_{n=0}^infty frac{(2n+1)(-1)^nx^{2n}}{(2n)!(2n+1)}+frac{(-1)^nx^{2n+1}}{(2n+1)(2n)!}i$, so



      $=sum_{n=0}^infty frac{(-1)^nx^{2n}(2n+ix+1)}{(2n+1)!}=e^{ix}$ ,



      but unlike the first $e^{ix}$ series this one produces 2 parts of the first series with n=0 on the solved summation =1+ix, which were n=0 and n=1 for the original



      from this I decided to make a summation that produced 3 terms per each n by grouping the original $e^{ix}$ summation by threes and got



      $sum_{n=0}^infty frac{(-1)^{n+1}x^{3n}i^{n}(x^2-(3n+2)ix-(3n+1)(3n+2))}{(3n+2)!}$ ,



      so n=0 produces $1+ix-frac{x^2}{2!}$, and n=1 producing $frac{-ix^3}{3!}+frac{x^4}{4!}+frac{ix^5}{5!}$, a continuation of the original infinite series. At the same time I haven't been able to prove this is equal to $e^{ix}$ through any other means



      I also did this for 4 terms which created,



      $sum_{n=0}^infty frac{-ix^{4n} (x^3-(4n+3)ix^2-(4n+2)(4n+3)x+(4n+1)(4n+2)(4n+3)i}{(4n+3)!}$



      Which again n=0 produces $1+ix-frac{x^2}{2!}-frac{ix^3}{3!}$



      and from these I tried to create a general formula to produce the infinite sum that produces a certain number of the original terms per n, as there seems to be a definite pattern, but I haven't quite found one yet.




      So do all these have the same properties? Like would there be any theoretical way to produce a formula that has infinite terms, or truncate to the negatives, or zero? Or anything special that one summation is capable of over another?






      sequences-and-series summation







      share|cite|improve this question









      $endgroup$




      I found that solving for $e^{ix}$ gave a formula that produces two n "terms" of the starting summation



      and defining terms as two n products like 1+ix being two terms of the summation



      $e^{ix} $ =$sum_{n=0}^infty$ $frac{(ix)^n}{n!}$



      $=1+ix-frac{x^2}{2!}-frac{ix^3}{3!}+...$



      So solving for this using Euler's Identity $e^{i x}=cos(x)+isin(x)$,



      $e^{ix}=sum_{n=0}^infty frac{(-1)^nx^{2n}}{(2n)!} + sum_{n=0}^infty frac{(-1)^nx^{2n+1}}{(2n+1)!}i $



      $=sum_{n=0}^infty frac{(-1)^nx^{2n}}{(2n)!}+frac{(-1)^nx^{2n+1}}{(2n+1)(2n)!}i$



      $=sum_{n=0}^infty frac{(2n+1)(-1)^nx^{2n}}{(2n)!(2n+1)}+frac{(-1)^nx^{2n+1}}{(2n+1)(2n)!}i$, so



      $=sum_{n=0}^infty frac{(-1)^nx^{2n}(2n+ix+1)}{(2n+1)!}=e^{ix}$ ,



      but unlike the first $e^{ix}$ series this one produces 2 parts of the first series with n=0 on the solved summation =1+ix, which were n=0 and n=1 for the original



      from this I decided to make a summation that produced 3 terms per each n by grouping the original $e^{ix}$ summation by threes and got



      $sum_{n=0}^infty frac{(-1)^{n+1}x^{3n}i^{n}(x^2-(3n+2)ix-(3n+1)(3n+2))}{(3n+2)!}$ ,



      so n=0 produces $1+ix-frac{x^2}{2!}$, and n=1 producing $frac{-ix^3}{3!}+frac{x^4}{4!}+frac{ix^5}{5!}$, a continuation of the original infinite series. At the same time I haven't been able to prove this is equal to $e^{ix}$ through any other means



      I also did this for 4 terms which created,



      $sum_{n=0}^infty frac{-ix^{4n} (x^3-(4n+3)ix^2-(4n+2)(4n+3)x+(4n+1)(4n+2)(4n+3)i}{(4n+3)!}$



      Which again n=0 produces $1+ix-frac{x^2}{2!}-frac{ix^3}{3!}$



      and from these I tried to create a general formula to produce the infinite sum that produces a certain number of the original terms per n, as there seems to be a definite pattern, but I haven't quite found one yet.




      So do all these have the same properties? Like would there be any theoretical way to produce a formula that has infinite terms, or truncate to the negatives, or zero? Or anything special that one summation is capable of over another?






      sequences-and-series summation




      sequences-and-series summation






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 4 '18 at 6:49









      2AM Aesthetic2AM Aesthetic

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