Find the first four non-zero terms and the general term for the Taylor Series centered at c = 1 for the integral of ln(x).
since ln(x) = -SUM[k>=1] (-1)^k * (x-1)^k/k
the integral is -SUM[k>=1] (-1)^k * (x-1)^(k-1)
To find the Taylor Series for the integral of ln(x) centered at c = 1, we need to compute the derivatives of ln(x) at x = 1.
The integral of ln(x) can be expressed as the antiderivative or the indefinite integral of ln(x). Since we are only interested in finding the Taylor Series centered at c = 1, we can ignore the constant of integration for now.
Let's start by finding the derivatives of ln(x) at x = 1. The general derivative of ln(x) is given by:
d/dx ln(x) = 1/x
To find the derivatives at x = 1, we substitute x = 1 into the derivative expression:
d/dx ln(x)|x=1 = 1/1 = 1
Next, we find the second derivative of ln(x) and evaluate it at x = 1:
d^2/dx^2 ln(x) = d/dx (1/x) = -1/x^2
d^2/dx^2 ln(x)|x=1 = -1/1^2 = -1
Similarly, we find the third derivative:
d^3/dx^3 ln(x) = d/dx (-1/x^2) = 2/x^3
d^3/dx^3 ln(x)|x=1 = 2/1^3 = 2
Continuing, we find the fourth derivative:
d^4/dx^4 ln(x) = d/dx (2/x^3) = -6/x^4
d^4/dx^4 ln(x)|x=1 = -6/1^4 = -6
Now we have the first four derivatives of ln(x) evaluated at x = 1:
f'(1) = 1
f''(1) = -1
f'''(1) = 2
f''''(1) = -6
To find the first four non-zero terms of the Taylor Series, we can use the formula:
f(x) ≈ f(c) + f'(c)(x - c)/1! + f''(c)(x - c)^2/2! + f'''(c)(x - c)^3/3! + ...
For our case, since c = 1:
f(x) ≈ f(1) + f'(1)(x - 1)/1! + f''(1)(x - 1)^2/2! + f'''(1)(x - 1)^3/3!
Substituting the evaluated derivatives, we have:
f(x) ≈ 0 + 1(x - 1)/1! + (-1)(x - 1)^2/2! + 2(x - 1)^3/3!
Simplifying, we get:
f(x) ≈ (x - 1) - (x - 1)^2/2 + 2(x - 1)^3/6
This is the general form of the Taylor Series for the integral of ln(x) centered at c = 1.