# Homework Help: How do I produce a function from a series?

1. Nov 11, 2007

### giant016

A simple example is 1+x+x^2=x^3+...=1/(1-x)

I know that is equals 1/(1-x), but how do you arrive at that? Thanks.

2. Nov 11, 2007

### ZioX

There are lots of ways. The analysis route would be to show that 1+x+...+x^n=(1-x^{n+1})/(1-x). If |x|<1, then x^{n+1} -> 0 so the limit becomes just 1/(1-x).

3. Nov 11, 2007

### rock.freak667

well 1+x+x^2=x^3+... is like a+ar+ar^2+...
so that a=1 and r=x
and well if they stated that |x|<1 so this would imply that |r|<1

so then the sum to infinity(which is the sum of the infinite series1+x+x^2+x^3+...)

would be $$S_\infty=\frac{a}{1-r}$$

which is $$\frac{1}{1-x}$$

Last edited: Nov 11, 2007
4. Nov 12, 2007

### antiemptyv

sort of a generating function approach.

define a function f by the series:

f(x) = 1 + x + x^2 + x^3 + ...

now take a look at x*f(x):

xf(x) = x + x^2 + x^3 + ...

f(x) - xf(x) = (1 + x + x^2 + x^3 + ... ) - (x + x^2 + x^3 + ... ) = 1.
notice the terms cancel out. so f(x) - xf(x) = 1.

factor out f(x) from the terms in the left-hand-side:

f(x)(1-x) = 1.

dividing both sides by (1-x) yields the result:

f(x) = 1/(1-x).

Last edited: Nov 12, 2007
5. Nov 12, 2007

### Gib Z

the simplest way is do use polynomial division =]

6. Nov 12, 2007

### ZioX

This is doesn't tell you anything about the radius of convergence, which may or may not be relevant.

7. Nov 12, 2007

### antiemptyv

true. i just think it's a nice example of being able to play with a series to find an explicit formula, though this isn't the the most telling of its nature.