Geometric progression

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In mathematics, a geometric progression (also known as a geometric sequence, and, inaccurately, as a geometric series; see below) is a sequence of numbers such that the quotient of any two successive members of the sequence is constant. This ratio is called the common ratio of the sequence.

Thus without loss of generality a geometric sequence can be written as

where r ≠ 0 is the common ratio and a is a scale factor. Thus the common ratio gives a family of geometric sequences whose starting value is determined by the scale factor. Pedantically speaking, the case r = 0 ought to be excluded, since the common ratio is not even defined; but the sequence that is always 0 is included, by convention.

Euclid's books VIII and IX analyze a geometric progression. A geometric progression gains its geometric character from the very important mathematical theorem that two similar plane figures or numbers are in duplicate ratio to their corresponding sides. Further two similar solid figures and numbers are in triplicate ratio of their corresponding sides.

For example the square numbers 4 and 9 are in the ratio 4 to 6 to 9, which is the duplicate ratio of their sides, 2 and 3. The cube numbers 8 and 125 are in the ratio of 8 to 20 to 50 to 125, which is the triplicate ratio of their sides. In the geometric progression 4,6,9 the number 6 is the mean proportional between 4 and 9, and in the geometric progression 8,20,50,125 the numbers 20 and 50 are the mean proportionals between 8 and 125. All squares have one mean proportional between them all cubes have two mean proportionals between and so on to higher dimensions.

1 Formula
2 Examples
3 Geometric series
- 3.1 Infinite geometric series
- 3.2 Complex numbers
4 Product
5 Book IX, Proposition 36
6 See also

[edit] Formula

Progressions allow the use of a few simple formulas to find each term. The nth term can be defined as

<math>a_n = a_1\,r^{n-1}</math></dd>	where n is an integer such that <math>n \ge 1</math> </dd>
The common ratio is then
<math>r=\left(\frac{a_n}{a}\right)^\frac{1}{n-1}</math> or <math>r=\sqrt[n-1]{\frac{a_n}{a}}</math></dd>	where n is an integer such that <math>n \ge 1</math> </dd>

and the scale factor is

[edit] Examples

A sequence with a common ratio of 2 and a scale factor of 1 is

1, 2, 4, 8, 16, 32, ....

A sequence with a common ratio of 2/3 and a scale factor of 729 is

729 (1, 2/3, 4/9, 8/27, 16/81, 32/243, 64/729, ....) = 729, 486, 324, 216, 144, 96, 64, ....

A sequence with a common ratio of −1 and a scale factor of 3 is

3 (1, −1, 1, −1, 1, −1, 1, −1, 1, −1, ....) = 3, −3, 3, −3, 3, −3, 3, −3, 3, −3, ....

This sequence's behaviour depends on the value of the common ratio.

If the common ratio is:

Positive, the terms will all be the same sign as the initial term.
Negative, the terms will alternate between positive and negative.
0, the results will remain at zero.
Greater than 1, there will be exponential growth towards infinity (positive).
1, the progression is a constant sequence.
Between 1 and −1 but not zero, there will be exponential decay towards zero.
−1, the progression is an alternating sequence (see alternating series)
Less than −1, there will be exponential growth towards infinity (positive and negative).

A geometric progression with common ratio <math>\notin \{0,\pm1\}</math> shows exponential growth or exponential decay, as opposed to the linear growth (or decline) of an arithmetic progression such as 4, 15, 26, 37, 48, ....
This result was taken by T.R. Malthus as the mathematical foundation of his Principle of Population.

Note that the two kinds of progression are related: taking the logarithm of each term in a geometric progression yields an arithmetic one.

[edit] Geometric series

A geometric series is the sum of the numbers in a geometric progression:

<math>\sum_{k=0}^{n} ar^k = ar^0+ar^1+ar^2+ar^3+\cdots+ar^n \,</math>

We can find a simpler formula for this sum by multiplying both sides of the above equation by <math>(1-r)</math>, and we'll see that

since all the other terms cancel. Rearranging (for <math>r\ne1</math>) gives the convenient formula for a geometric series:

Note: If one were to begin the sum not from 0, but from a higher term, say m, then

Differentiating the sum with respect to r allows us to arrive at formulae for sums of the form

For example:

<math>\frac{d}{dr}\sum_{k=0}^nr^k = \sum_{k=0}^nkr^{k-1}=

\frac{1-r^{n+1}}{(1-r)^2}-\frac{(n+1)r^n}{1-r}</math>

[edit] Infinite geometric series

An infinite geometric series is an infinite series whose successive terms have a common ratio. Such a series converges if and only if the absolute value of the common ratio is less than one ( | r | < 1 ). Its value can then be computed from the finite sum formulae

<math>\sum_{k=0}^\infty ar^k = \lim_{n\to\infty}{\sum_{k=0}^{n} ar^k} = \lim_{n\to\infty}\frac{a(1-r^{n+1})}{1-r} = \lim_{n\to\infty}\frac{a}{1-r} - \lim_{n\to\infty}{\frac{ar^{n+1}}{1-r}} = \frac{a}{1-r}</math>

For example,

<math>\sum_{k=0}^\infty (191) \left(\frac{6}{7}\right)^k = \frac{191}{1-\frac{6}{7}} = 1337</math>

In cases where the sum does not start at k = 0,

<math>\sum_{k=m}^\infty ar^k=\frac{ar^m}{1-r}</math>

Both formulae are valid only for | r | < 1. The latter formula is actually valid in every Banach algebra, as long as the norm of r is less than one, and also in the field of p-adic numbers if | r |_p < 1. As in the case for a finite sum, we can differentiate to calculate formulae for related sums. For example,

<math>\frac{d}{dr}\sum_{k=0}^\infty r^k = \sum_{k=0}^\infty kr^{k-1}=

\frac{1}{(1-r)^2}</math>

This formula only works for | r | < 1 as well. From this, it follows that, for | r | < 1,

<math>\sum_{k=0}^{\infty} k r^k = \frac{r}{\left(1-r\right)^2} \,;\, \sum_{k=0}^{\infty} k^2 r^k = \frac{r \left( 1+r \right)}{\left(1-r\right)^3} \, ; \, \sum_{k=0}^{\infty} k^3 r^k = \frac{r \left( 1+4 r + r^2\right)}{\left( 1-r\right)^4}</math>

[edit] Complex numbers

The summation formula for geometric series remains valid even when the common ratio is a complex number. This fact can be used, along with Euler's formula, to calculate some sums of non-obvious geometric series, such as:

<math> \sum_{k=0}^{\infty} \frac{\sin(kx)}{r^k} = \frac{1}{2 i} \left[ \sum_{k=0}^{\infty} \left( \frac{e^{ix}}{r} \right)^k - \sum_{k=0}^{\infty} \left(\frac{e^{-ix}}{r}\right)^k\right]</math>.

It is clear that this is just the difference of two geometric series. From here, it is straighforward formula application to calculate that

<math> \sum_{k=0}^{\infty} \frac{\sin(kx)}{r^k} = \frac{r \sin(x)}{1 + r^2 - 3 r \cos(x)} </math>

[edit] Product

The product of a geometric progression is the product of all terms, and can be mathematically defined as

<math>\prod_{i=0}^{n} ar^i = (a_1 \cdot a_{n+1})^{\frac{n+1}{2}}</math>.

Proof:

Let the product be represented by P:

<math>P=a \cdot ar \cdot ar^2 \cdots ar^{n-1} \cdot ar^{n}</math>.

Now, carrying out the multiplications, we conclude that

<math>P=a^{n+1} r^{1+2+3+ \cdots +(n-1)+n}</math>.

Applying the sum of arithmetic series, the expression will yield

<math>P=a^{n+1} r^{\frac{n(n+1)}{2}}</math>.

<math>P=(ar^{\frac{n}{2}})^{n+1}</math>.

We raise both sides to the second power:

<math>P^2=(a^2 r^{n})^{n+1}=(a\cdot ar^n)^{n+1}</math>.

Consequently

<math>P^2=(a_1 \cdot a_{n+1})^{n+1}</math> and

<math>P=(a_1 \cdot a_{n+1})^{\frac{n+1}{2}}</math>,

which concludes the proof.

[edit] Book IX, Proposition 36

The geometric progression 1,2,4,8,16... or in binary 1,10,100,1000,10000... are very interesting and important in number theory. In Euclid's book IX, proposition 36 it is proved that if the sum of the first n terms of this progression is prime, then this sum times the nth term is a perfect number.

In order to facilitate an understanding of this theoerm lets prove that 496 is perfect number using Euclid's method. The series 1,2,4,8,16 sums to 31 which is a prime number and this number multiplied by 16 equals 496. The geometric progression 31,62,124,248,496 is a progression with same ratio as 1,2,4,8,16. In Book IX, proposition 35 Euclid proves that in a geometric series if the first term is subtracted from the second and last term in the sequence then as the excess of the second is to the first, so will the excess of the last be to all of those before it. Therefore 496 minus 31 is to the sum of 31,62,124,248 as 62 minus 31 is to 31. therefore the numbers 1,2,4,8,16,31,62,124,248 add up to 496 and further these are all the numbers which measure 496. For suppose that P measures 496 and it is not amongst these numbers. Let P measure it according to Q and therefore P*Q equals 16*31, or 31 is to Q as P is to 16. Now P cannot measure 16 or it would be amongst the numbers 1,2,4,8,16. Therefore 31 cannot measure Q. And since 31 does not measure Q and Q measures 496, the fundumental theory of arithmatic implies that Q must divide 16 and be amongst the numbers 1,2,4,8,16. Let Q be 4, then P must be 124, which is impossible since by hypothesis P is not amongst the numbers 1,2,4,8,16,31,62,124,248.