find the third taylor polynomial at x=0 of f(x)=ln(1-x)?

Question

x=0 of f(x)=ln(1-x)

Answer 1

find the third taylor polynomial at x=0 of f(x)=ln(1-x)?

x=0 of f(x)=ln(1-x)

you need the derivatives of f
f'(x) = -1/(1-x) =-(1-x)^{-1}
f''(x) = (1-x)^{-2}
f'''(x) = -2(1-x)^{=3}
f''''(x) = 2(3) (1-x)^{-4}

f(0)=0
f'(0) = -1/(1-0) =-1
f''(0) = (1-0)^{-2} =1
f'''(0) = -2(1-0)^{=3}=-2
f''''(0) = 2(3) (1-0)^{-4}=6

the taylor polynomial is:
-1x/1! + 1 x^2 /2! -2x^3/3! = -x + x^2/2 -x^3/3 .

Answer 2

if f(x) = ln(1-x)

Then f'(x) = 1/(1-x)*-1 = -1/(1-x)
f''(x) = -1*-1/(1-x)^2 * -1 = -1/(1-x)^2
f'''(x) = -1*-2*1/(1-x)^3*-1 = -2/(1-x)^3

So, at x = 0,
f'(0) = -1
f''(0) = -1
f''(0) = -2
Therefore, for x close to zero, ln(1-x) = ln(1) -x - x^2/2! - x^3/3!
but since ln(1) = 0, we can drop this term:
ln(1-x) is approximately -x-x^2/2! - x^3/3!