Geometric progression

From Maths
Jump to: navigation, search
Note: a geometric series is just a series derived from the terms of a geometric progression
Note: geometric sequence redirects here.
Stub grade: A
This page is a stub
This page is a stub, so it contains little or minimal information and is on a to-do list for being expanded.The message provided is:
Needs some work....


A geometric progression is any sequence, [ilmath](c_n)_{n\in\mathbb{N}_{\ge 1} } [/ilmath][Note 1] where each term is of the form:

Explicitly the sequence goes:

  • [ilmath](a,\ ar,\ ar^2,\ ar^3,\ \ldots\ ,\underbrace{\ ar^{k-1},\ }_{k^\text{th}\text{ term} }\ldots)[/ilmath]

As such we can characterise any geometric progression as a pair of numbers:

  • [ilmath]G\eq (a,r)\in\mathbb{R}^2[/ilmath], which we identify with a function:
    • [ilmath]G:\mathbb{N}_{\ge 1}\rightarrow\mathbb{R} [/ilmath] by [ilmath]G:k\mapsto ar^{k-1} [/ilmath]

This is natural considering that a sequence is a function which maps each integer, [ilmath]k[/ilmath], to the [ilmath]k^\text{th} [/ilmath] term (as explained on the sequence page)

See geometric series (the series from a geometric progression) for information on the sum of a geometric sequence, or a sub-sequence of it.

Canonical Geometric Progression

The canonical geometric progression will refer to [ilmath](1,r)[/ilmath], although any [ilmath](a,r)[/ilmath] such that [ilmath]a\neq 0[/ilmath] would do. As we now demonstrate:

  • Let [ilmath](u_k)_{k\in\mathbb{N}_{\ge 1} } [/ilmath] be the sequence of the geometric progression [ilmath](1,r)[/ilmath], meaning:
    • [ilmath](u_k)[/ilmath] is the sequence: [ilmath]1,\ r,\ r^2,\ r^3,\ \ldots,\ \underbrace{r^{k-1} }_{k^\text{th}\text{ term} },\ \ldots[/ilmath]
    • Let [ilmath](a,r)[/ilmath] be any other geometric progression with the same ratio, [ilmath]r[/ilmath], denote its terms by the sequence [ilmath](v_k)_{k\in\mathbb{N}_{\ge 1} } [/ilmath]
      • Then [ilmath](v_k)[/ilmath] is the sequence [ilmath]a,\ ar,\ ar^2,\ ar^3,\ \ldots,\ \underbrace{ar^{k-1} }_{k^\text{th}\text{ term} },\ \ldots[/ilmath]
        • We see that [ilmath]v_k\eq a u_k[/ilmath] for each [ilmath]k\in\mathbb{N}_{\ge 1} [/ilmath]
          • We abuse notation by writing [ilmath](v_k)_k\eq a(u_k)_k[/ilmath] or even [ilmath](a,r)\eq a(1,r)[/ilmath]

Formally, this shows: [ilmath]\forall (a,r)\in\mathbb{R}^2\exists b\in\mathbb{R}\big[b(1,r)\eq(a,r)\big][/ilmath] - namely [ilmath]b\eq a[/ilmath] itself.

As mentioned, we need not use [ilmath](1,r)[/ilmath] as our canonical progression, any non-zero value in place of [ilmath]1[/ilmath] would do, however [ilmath]1[/ilmath] is the natural choice over [ilmath]\sqrt{2} [/ilmath], 5, or even [ilmath]-1[/ilmath]

See also


  1. Notice the sequence goes:
    • [ilmath]c_1,\ c_2,\ c_3,\ \ldots[/ilmath] (starting from [ilmath]k\eq 1[/ilmath]), not:
    • [ilmath]c_0,\ c_1,\ c_2,\ \ldots[/ilmath]
    Either way is fine, arguably starting from [ilmath]0[/ilmath] is cleaner, however we will start from [ilmath]1[/ilmath], as is our convention