# Math(s) question

Discussion in 'Spigot Plugin Development' started by novucs, Oct 5, 2016.

1. ### novucs

I know this isn't technically plugin development, though I came across this problem while developing a plugin and has been bugging me for a while now. How could you non-programatically resolve the common ratio of a geometric series when provided the sum, initial and number in sequence?

Relevant equation:
S=a((1-r^n)/(1-r))

S: Sum of series
a: Initial value
r: Common ratio
n: Number in sequence to count up to

In this case, we need to resolve the equation for "r".

2. ### DarkSeraphim

What variables are actually known?

3. ### novucs

S, a and n. I am aware this could be worked out programmatically, though I would really like to know if there is anything out there to resolve the common ratio in one short but sweet equation.

4. ### DarkSeraphim

Shouldn't you be able to rewrite it to "r = ..."?

5. ### _NotTheWither_

Good Luck With Your Math Expidition! I Hope You Do Spigot Proud!

• Funny x 1
6. ### novucs

That is what I have been attempting, and what I'm asking for in this thread. Apologies if my original post was hard to deconstruct. I would appreciate if you or anyone else could help try to work that out for me.

7. ### Briboy

I think he's trying to find what "r" would equal to, so you are solving for "r" and in order to do that it needs to equal to "s"

@novucs Don't we need to know what these variables are storing what values so we know what's what?

8. ### novucs

No, all I need is the equation above to be written to solve "r" instead of "S". It's an algebra question.

10. ### insou

You wrote the equation incorrectly.

You wrote
Code (Text):
s=a*((1/r^n)/(1/r)) for r
when it should be
Code (Text):
s=a*((1-r^n)/(1-r)) for r
When you try and put the correct equation in, it times out and says you need pro, which I don't have

11. ### 567legodude

From what I'm seeing, there is no possible way to solve the equation for just r. You can get r^n and r on one side of the equation, but since those are two different things, you can't factor out the r and they are left separate.

The smallest you can simplify it is:
-ar^n + sr = -a + s

12. ### Roree

The fact that you have the r as a divisor and dividend and both of them contain two terms (1 and negative r) on the right side of the equation is the issue.

13. ### novucs

Thanks! I've managed to take it a little further than yours and I think I might have solved it if I'm not missing anything.

Workings:
Code (Text):
s = a((1-r^n)/(1-r))
s/a = (1-r^n)/(1-r)
(s-sr)/a = 1 - r^n
s-sr = a - ar^n
s + ar^n  = a + sr
ar^n = a + sr - s
r^n = (a + sr -s)/a
n * log(r) = log((a + sr - s) / a)
log(r) = log((a + sr - s) / a) / n
r = 10^(log((a + sr - s) / a) / n)
Edit: Oh I'm dumb, I still have "r" on the other side...

14. ### DarkSeraphim

how did you even get this? There's no way you get -a on the right side.

@novucs don't you happen to have the series instead of the sum? That would probably help a lot.

15. ### novucs

No, I only have the sum, the first value and the size of the series.

16. ### DarkSeraphim

Well first and last might actually help a lot, lol.