#math - Wed 14 Mar 2007 between 23:44 and 23:50

PuppiesOnAcidOr would that even be possible
Because as I understand it, the first term of your geometric series has to be just your "a" value
keltPuppiesOnAcid: there is a rule (geometric Ithink) like for all p > 2 (1/p)^x converges?
PuppiesOnAcidThat's not rly helping me here though
KasadkadPuppiesOnAcid: sum(x^n, n=0..infinity) = 1/(1-x) if |x| < 1, that's the geometric series formula
PuppiesOnAcidKasadkad: I have a geometric series as being sum(ar^(n-1), n=1..oo)
KasadkadThat's the same as a*sum(r^n, n=0..infinity)
Just shift the indices
You can pull the a out of the sum too; it doesn't affect convergence at all
PuppiesOnAcidKasadkad: The summation I'm working with is sum(1/3^n, n=1..oo)
Is that in geometric form?
If so, what is the a and the r value?
Kasadkadr = 1/e
r = 1/3
a = 1
keltwell if you do sqrt(b-4ac) = sqrt(1+4) = sqrt(5) so x^2+x-1 shouldn't have a complex number but x^2+x+1 should I guess
PuppiesOnAcidRight, but the first term of your geometric series has to be your a value
GoplatKasadkad: actually since you start from n=1, a isn't 1
pengratekelt: Yeah, you can find the factors by using the quadratic formula
KasadkadEhm, yeah
PuppiesOnAcidGoplat: Exactly, that is my hang up
Kasadkada = 1/3 then
But it doesn't matter
PuppiesOnAcidif a = 1/3, is your r value still 1/3?
KasadkadYou don't need the exact form anyway
keltpengrate: you solve for x?
then do (x+a1)(x-a2) or whatever?
PuppiesOnAcidKasadkad: I need the a and r values so I can find the sum
pengratekelt: Well, (x-a1)(x-a2), but yeah
KasadkadThe geometric series formula tells you that sum(1/3^n, n=0..infinity) = 1/(1-1/3) = 3/2
But what's sum(1/3^n, n=1..infinity) in terms of that?
if i say m E integers
do i say
m is either even or odd
Echelonm is either even, off, or 0

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