Numbers in Geometric Progression

Asked 19/6, 2012 at 5:23 Answered 31/10, 2014 at 19:34

How can i generate a sequence of numbers which are in Geometric Progression in R? for example i need to generate the sequence : 1, 2,4,8,16,32 and so on....till say a finite value?

Epilate answered 19/6, 2012 at 5:23 Comment(1)

2^seq(0, 5, by=1) – Hematology 19/6, 2012 at 5:32

Here's what I'd do:

geomSeries <- function(base, max) {
    base^(0:floor(log(max, base)))
}

geomSeries(base=2, max=2000)
# [1]    1    2    4    8   16   32   64  128  256  512 1024

geomSeries(3, 100)
# [1]  1  3  9 27 81

Crampton answered 19/6, 2012 at 5:37 Comment(4)

thank you, can u please see my comment above and tell me if i can write this code in that fashion? i am a newbie and hence have a lot of doubts :( – Epilate 19/6, 2012 at 5:39

@Epilate - Josh's answer is what you want I reckon. geomSeries(2,32) as outlined in his response will give you a base 2 progression up until the max value of 32. – Expressly 19/6, 2012 at 5:46

thank you everybody! yes, @josh's answer was indeed helpful! cheers guys! – Epilate 19/6, 2012 at 5:49

(Do be aware that the function above would need a much more checking and adjusting for edge and corner cases to be really ready for prime time. Try for instance geomSeries(.1, 100), geomSeries(1, 1.001), and geomSeries(2, 0.8) to see what I mean.) – Lodicule 19/6, 2012 at 5:54

Why not just enter 2^(0:n)? E.g. 2^(0:5) gets you from 1 to 32 and so on. Capture the vector by assigning to a variable like so: x <- 2^(0:5)

Navarro answered 19/6, 2012 at 5:32 Comment(3)

when i want a sequence from say 1 to 100, incrementing by 10, i write: seq(1, 100, by=10). so now i want a sequence from 1 to say 1000 that increments geometrically like 1, then 2, then 4 and so on. – Epilate 19/6, 2012 at 5:35

Something like this to get a geometric progression that always ends less than the number specified (the 1000 in this case):2^(1:floor(log(1000,2))) – Expressly 19/6, 2012 at 5:39

Suggest you have a look at seq(), as per baptiste's comment above. Do ? and the function name to get help on an R function, so ?seq. In this case you want something like seq(0, 10, by = 10). Note that goes from 0 to 100, not 1 to 100, which is not a regularly spaced series. – Navarro 19/6, 2012 at 5:39

You can find any term in a geometric sequence with this mathematical function:

term = start * ratio ** (n-1)

Where:

term = the term in the sequence that you want
start = the first term in the sequence
ratio = the common ratio (i.e. the multiple that defines the sequence)
n = the number of the term in the sequence that you want

Using this information, write up a function in R that provides any subset of a geometric sequence for any start and ratio:

#begin = beginning of subset
#end = end of subset

geomSeq <- function(start,ratio,begin,end){
  begin=begin-1
  end=end-1
  start*ratio**(begin:end)
}

geomSeq(1, 2, 1, 10)
# [1]   1   2   4   8  16  32  64 128 256 512

geomSeq(10,3,1,8)
# [1]    10    30    90   270   810  2430  7290 21870

geomSeq(10,3,4,8)
# [1]   270   810  2430  7290 21870

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