Which of the following is the Taylor series of ln(1−2x) about x=0 up to and including the terms of order three?
To find the Taylor series of ln(1-2x) about x=0 up to the terms of order three, we can use the formula for the Taylor series expansion. The formula for the Taylor series expansion of a function f(x) about a point a is:
f(x) = f(a) + f'(a)(x-a) + f''(a)(x-a)^2/2! + f'''(a)(x-a)^3/3! + ...
To find the Taylor series expansion of ln(1-2x) about x=0 up to the terms of order three, we need to find the function value and the derivatives of ln(1-2x) at x=0.
1. Function value at x=0:
Substituting x=0 into ln(1-2x), we get ln(1) = 0. Therefore, f(0) = 0.
2. First derivative at x=0:
Differentiating ln(1-2x) with respect to x, we get d/dx[ln(1-2x)] = -2/(1-2x).
Substituting x=0 into this derivative, we get f'(0) = -2.
3. Second derivative at x=0:
Differentiating the first derivative with respect to x, we get d^2/dx^2[ln(1-2x)] = -4/(1-2x)^2.
Substituting x=0 into this derivative, we get f''(0) = -4.
4. Third derivative at x=0:
Differentiating the second derivative with respect to x, we get d^3/dx^3[ln(1-2x)] = 8/(1-2x)^3.
Substituting x=0 into this derivative, we get f'''(0) = 8.
Now we can substitute these values into the Taylor series expansion formula:
ln(1-2x) = f(0) + f'(0)(x-0) + f''(0)(x-0)^2/2! + f'''(0)(x-0)^3/3! + ...
= 0 + (-2)(x) + (-4)(x^2)/2! + 8(x^3)/3!
Simplifying this expression:
ln(1-2x) = -2x - 2(x^2) + 8(x^3)/3
Therefore, the Taylor series of ln(1-2x) about x=0 up to and including the terms of order three is -2x - 2(x^2) + 8(x^3)/3.