4

I'm trying to find a closed form of $ S(n)= \sum_{i=0}^n a^{q^i} $ for positive constant params $a, q, n$. I searched it in google but I couldn't find any. I searched it in wolframalpha, but I didn't get any result.I thought there is some set like geometric progression but no results in google.

I tried finding more fundamental properties for the set $a_1,a_2...,a_n$ as $a_{k+1} = a_k^q$, but with not much of a success.

The one thing I managed to find is the sum $A(n) = \sum_{i=0}^n log(a^{q^i})$ which was trivial but still couldn't get any results with this sum in wolframalpha for some reason...

So my question is : Is it possible to find closed form of $S(n)$ and if so what method should I use to find it?

EDIT:

A theory of mine: If there is closed form function: $f(\sum_{i=0}^n x_i^q) = \sum_{i=0}^n x_i^{q^2} $ then $S(n+1) = f(S(n))+a$ however there might be set $Y$ where: $S(n) =\sum_{i=0}^n y_i^{q^n}$ then $S(n+1) = f(S(n)) + a$ is different from the old $S(n+1)$. So in order for $f(x)$ to exist $X$ must be a set of $a_1,a_2...,a_n$ elements and as such $f(x)$ can't be closed form and $S(n)$ respectively.

Community
  • 1
user2377766
  • 41
  • Welcome to Math SE! Take a look here for information about formatting your question with MathJax to make it more readable – DMcMor Sep 07 '17 at 14:18
  • @DMcMor I tried it but for some reason $\sum_{i=0}^n a^(q^i) $ didn't look good, so I thought it would be more clear using plain text. If you have any Ideas about the mathjax I should use please write it to me so I can fix my question's appearance – user2377766 Sep 07 '17 at 14:22
  • @user2377766 Try a^{q^i} instead it formats like this $a^{q^i}$ – kingW3 Sep 07 '17 at 14:24
  • 1
    I guess you might add some conditions like $a>0$ etc – MAN-MADE Sep 07 '17 at 14:35
  • @MANMAID changed them all to positive – user2377766 Sep 07 '17 at 14:42
  • Try multiplying $S(n)$ by $a^q$ and then rewrite what you get in terms of S(n). You should get something like $a^q S(n) = S(n) + x - y$ then solve for S(n). This is how you do it for the regular geometric series – FullofDill Sep 07 '17 at 14:43
  • @FullofDill thats the first thing I tried but since $a^q*a^{q^k}= a^{q^k+q}$ it didn't quite work for me in that direction... – user2377766 Sep 07 '17 at 14:45
  • oi, total brain fart. sorry – FullofDill Sep 07 '17 at 14:47
  • @FullofDill Hahah. I get confused too. Not used to that level of powers ... :D – user2377766 Sep 07 '17 at 14:48
  • It's $\Delta S(n)=a+(a^q-1) S(n)$ with $\Delta S(n):=S(n+1)-S(n)$ . You can search how such problems are handled in the literature of Concrete Mathematics . Perhaps you will find out that there is no closed form or that it's necessary to use Analysis (see: Euler-Maclaurin formula) . – user90369 Sep 07 '17 at 15:15
  • @ user90369 will do – user2377766 Sep 07 '17 at 15:17

0 Answers0