Creating evenly spaced numbers on a log scale (a geometric progression) can easily be done for a given base and number of elements if the starting and final values of the sequence are known, e.g., with numpy.logspace and numpy.geomspace. Now assume I want to define the geometric progression the other way around, i.e., based on the properties of the resulting geometric series. If I know the sum of the series as well as the first and last element of the progression, can I compute the quotient and number of elements?
For instance, assume the first and last elements of the progression are and
and the sum of the series should be equal to
. This works out for
and
as I know from trial and error, but how could these values be computed?
CodePudding user response：
You have enough information to solve it:
Sum of series = a a*r a*(r^2) ... a*(r^(n-1))
= a*((r^n)-1)/(r-1)
= a*((last element * r) - 1)/(r-1)
Given the sum of series, a, and the last element, you can use the above equation to find the value of r.
Plugging in values for the given example:
50 = 1 * ((15*r)-1) / (r-1)
50r - 50 = 15r - 1
35r = 49
r = 1.4
Then, using sum of series = a*((r^n)-1)/(r-1):
50 = 1*((1.4^n)-1)(1.4-1)
21 = 1.4^n
n = log(21)/log(1.4) = 9.04
You can approximate n and recalculate r if n isn't an integer.
CodePudding user response：
We have to reconstruct geometric progesssion, i.e. obtain a, q, m (here ^ means raise into power):
a, a * q, a * q^2, ..., a * q^(m - 1)
if we know first, last, total:
first = a # first item
last = a * q^(m - 1) # last item
total = a * (q^m - 1) / (q - 1) # sum
Solving these equation we can find
a = first
q = (total - first) / (total - last)
m = log(last / a) / log(q)
if you want to get number of items n, note that n == m 1
Code:
import math
...
def Solve(first, last, total):
a = first
q = (total - first) / (total - last)
n = math.log(last / a) / math.log(q) 1
return (a, q, n);