Question

What is the Maclaurin series for `(1-x)ln(1-x)`?

Answer 1

`-x + 1/2x^2 + 1/6x^3 + 1/12x^4`

Explanation:

Start with the known Maclaurin series for `ln(1-x)`, which is:

`ln(1-x) = -x-1/2x^2-1/3x^3-1/4x^4 -1/5x^5- ...`

Then we can just use algebra to multiply this series by `(1-x)`, hence:

`f(x) = (1-x)ln(1-x)`
`" " = (1-x){-x-1/2x^2-1/3x^3-1/4x^4 - 1/5x^5... }`

`" " = (1){-x-1/2x^2-1/3x^3-1/4x^4 - 1/5x^5 ... }`
`" " - x{-x-1/2x^2-1/3x^3-1/4x^4 - 1/5x^5... }`

`" " = -x-1/2x^2-1/3x^3-1/4x^4 - 1/5x^5... +`
`" " x^2+1/2x^3+1/3x^4+ 1/4x^5...`

`" " = -x + 1/2x^2 + 1/6x^3 + 1/12x^4 + 1/20x^5...`

So the required polynomial of degree `4` is

`-x + 1/2x^2 + 1/6x^3 + 1/12x^4`