Deriving a geometric series

1. Oct 23, 2006

Elec68

I could use some help with this question:

Derive the geometric series representation of 1/(1-x) by finding a0, a1,
a2,... such that
(1-x)(a0+a1x+a2x^2+a3X^3+...)=1

Thank you.

2. Oct 23, 2006

AKG

$$(1 - x)(a_0 + a_1x + a_2x^2 + a_3x^3 + \dots) = ? = \dots = ? = a_0 + (a_1-a_0)x + (a_2-a_1)x^2 + (a_3-a_2)x^3 + \dots$$

3. Oct 23, 2006

Office_Shredder

Staff Emeritus
I recommend multiplying through by 1, then multiplying through by -x, then rearranging the terms

4. Oct 23, 2006

mathman

How about (1-x)/(1-x)=1. Expand the denominator into a power series (binomial expansion of 1/(1-x)).

5. Oct 27, 2006

fopc

If |x| <1, (1-x)*(1+x+x^2+x^3+x^4+...) = 1.
So your coefficients a_i = 1, for all i.

6. Oct 27, 2006

Hurkyl

Staff Emeritus
(1) You (presumably) know how to multiply.
(2) You (presumably) know how to tell when two power series are equal.

There's no "trick" to this one -- you just do exactly what the equation suggests.