Question

How do you find the taylor series series at x=0 of `f(x) = 1/(1-2x)`?

Answer 1

Rather than differentiating, just write out a power series that when multiplied by `1-2x` gives `1`.

Explanation:

`f(x) = sum_(n=0)^oo 2^nx^n = 1 + 2x + 4x^2 + 8x^3 + ...`

Then we find:

`(sum_(n=0)^oo 2^nx^n)(1-2x)`

`= (sum_(n=0)^oo 2^nx^n) - 2x(sum_(n=0)^oo 2^nx^n)`

`= (sum_(n=0)^oo 2^nx^n) - (sum_(n=1)^oo 2^nx^n)`

`=2^0x^0 = 1`

Of course this only works when the sums converge - which works for `abs(x) < 1/2` at least.