The Correct Answer and Explanation is:
The problem asks to find the power series expansion of y=ln(x)y = \ln(x)y=ln(x) around x=1x = 1x=1 in terms of a series involving (x−1)(x – 1)(x−1), i.e., in the form:ln(x)=∑n=0∞an(x−1)n\ln(x) = \sum_{n=0}^{\infty} a_n (x – 1)^nln(x)=n=0∑∞an(x−1)n
We will derive the coefficients ana_nan by differentiating y=ln(x)y = \ln(x)y=ln(x) and evaluating each derivative at x=1x = 1x=1. Here are the steps to find these coefficients:
Step 1: Define the function and its derivatives.
The function is f(x)=ln(x)f(x) = \ln(x)f(x)=ln(x).
- First derivative: f′(x)=1x=x−1f'(x) = \frac{1}{x} = x^{-1}f′(x)=x1=x−1.
- Second derivative: f′′(x)=−x−2f”(x) = -x^{-2}f′′(x)=−x−2.
- Third derivative: f(3)(x)=2x−3f^{(3)}(x) = 2x^{-3}f(3)(x)=2x−3.
- Fourth derivative: f(4)(x)=−6x−4f^{(4)}(x) = -6x^{-4}f(4)(x)=−6x−4.
- Continue this process for higher derivatives.
Step 2: Evaluate the derivatives at x=1x = 1x=1.
For each derivative, we will evaluate it at x=1x = 1x=1 to find an=f(n)(1)n!a_n = \frac{f^{(n)}(1)}{n!}an=n!f(n)(1):
- f(1)=ln(1)=0f(1) = \ln(1) = 0f(1)=ln(1)=0.
- f′(1)=1f'(1) = 1f′(1)=1.
- f′′(1)=−1f”(1) = -1f′′(1)=−1.
- f(3)(1)=2f^{(3)}(1) = 2f(3)(1)=2.
- f(4)(1)=−6f^{(4)}(1) = -6f(4)(1)=−6.
- f(5)(1)=24f^{(5)}(1) = 24f(5)(1)=24, and so on.
Step 3: Write the power series.
Now we can write the power series expansion. We will use the formula for the coefficients ana_nan:an=f(n)(1)n!a_n = \frac{f^{(n)}(1)}{n!}an=n!f(n)(1)
Substitute the values of the derivatives evaluated at x=1x = 1x=1:
- a0=0a_0 = 0a0=0
- a1=1a_1 = 1a1=1
- a2=−12a_2 = -\frac{1}{2}a2=−21
- a3=13a_3 = \frac{1}{3}a3=31
- a4=−14a_4 = -\frac{1}{4}a4=−41
- a5=15a_5 = \frac{1}{5}a5=51
- And so on.
Thus, the power series expansion of ln(x)\ln(x)ln(x) at x=1x = 1x=1 is:ln(x)=(x−1)−12(x−1)2+13(x−1)3−14(x−1)4+⋯\ln(x) = (x – 1) – \frac{1}{2}(x – 1)^2 + \frac{1}{3}(x – 1)^3 – \frac{1}{4}(x – 1)^4 + \cdotsln(x)=(x−1)−21(x−1)2+31(x−1)3−41(x−1)4+⋯
This is the desired power series expansion of y=ln(x)y = \ln(x)y=ln(x) at x=1x = 1x=1.
