Question

How do you find the Maclaurin Series for `f(x)= 1/ (1-x)`?

Answer 1

`1/(1-x) = sum_(k=0)^oo x^k`

Explanation:

Given:

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

It seems to me that the easiest way to find the Maclaurin Series is basically to start to write down the multiplier for `(1-x)` that results in a value of `1`...

Write down:

`1 = (1-x)(...`

The first term of the multiplier will be `1`, in order to get `1` when multiplied, so add that to the right hand side:

`1 = (1-x)(1...`

When `-x` is multiplied by `1` it will give us `-x` to cancel out. So the next term on the right hand side is `x`...

`1 = (1-x)(1+x...`

When `-x` is multiplied by `x` it will give us `-x^2` to cancel out. So the next term on the right hand side is `x^2`...

`1 = (1-x)(1+x+x^2...`

Continuing in this way, we get:

`1 = (1-x)(1+x+x^2+x^3+x^4+...)`

So:

`1/(1-x) = 1+x+x^2+x^3+x^4+... = sum_(k=0)^oo x^k`

Note that if `abs(x) < 1` then the remainder gets smaller each time we add a term on the right hand side. Hence the Maclaurin series converges for `abs(x) < 1`.