Geometric series: calculate quotient and number of elements from sum and first & last element-CodePudding

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 $a_0=1$ and $a_n=15$ and the sum of the series should be equal to $\sum_{k=0}^{n} a \cdot r^k = 50$ . I know from trial and error that it works out for n=9 and r≈1.404, 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);

Fiddle

If you put your data (1, 15, 50) you'll get the solution

a = 1 
q = 1.4 
n = 9.04836151801382 # not integer

since n is not an integer you, probably want to adjust; let last == 15 be exact, when total can vary. In this case q = (last / first) ^ (1 / (n - 1)) and total = first * (q ^ n - 1) / (q - 1)

a = 1
q = 1.402850552006674
n = 9

total = 49.752 # now n is integer, but total <> 50

CodePudding user response：

You have to solve the following two equations for r and n:

a:= An / Ao = r^(n - 1)

and

s:= Sn / Ao = (r^n - 1) / (r - 1)

You can eliminate n by

s = (r a - 1) / (r - 1)

and solve for r. Then n follows by log(a) / log(r) 1.

In your case, r = 7 / 5 = 1.4 and n = 9.048...

It makes sense to round n to 9, but then r^8 = 15 (r ~ 1.40285) and r = 1.4 are not quite compatible.