Question

How do you find taylor polynomials of degree n approximating `5/(2-2x)` for x near 0?

Answer 1

`5/(2-2x) = sum_(n=0)^oo 5/2x^n`

Explanation:

If `abs(x) < 1` then:

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

Hence:

`5/(2-2x) = 5/(2(1-x)) = 5/2*1/(1-x) = 5/2 sum_(n=0)^oo x^n = sum_(n=0)^oo 5/2x^n`