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


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?


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