Find a power series for f(x)=lnx, centered at 1
The Correct Answer and Explanation is:
To find the power series expansion of f(x)=ln(x)f(x) = \ln(x)f(x)=ln(x) centered at x=1x = 1x=1, we can use the Taylor series formula, which is given by:f(x)=f(a)+f′(a)(x−a)+f′′(a)2!(x−a)2+f(3)(a)3!(x−a)3+⋯f(x) = f(a) + f'(a)(x – a) + \frac{f”(a)}{2!}(x – a)^2 + \frac{f^{(3)}(a)}{3!}(x – a)^3 + \cdotsf(x)=f(a)+f′(a)(x−a)+2!f′′(a)(x−a)2+3!f(3)(a)(x−a)3+⋯
For the function f(x)=ln(x)f(x) = \ln(x)f(x)=ln(x), we want the series centered at a=1a = 1a=1. Let’s first compute the necessary derivatives of f(x)f(x)f(x):
- Zeroth derivative: f(x)=ln(x)f(x) = \ln(x)f(x)=ln(x) Evaluating at x=1x = 1x=1: f(1)=ln(1)=0f(1) = \ln(1) = 0f(1)=ln(1)=0
- First derivative: f′(x)=1xf'(x) = \frac{1}{x}f′(x)=x1 Evaluating at x=1x = 1x=1: f′(1)=11=1f'(1) = \frac{1}{1} = 1f′(1)=11=1
- Second derivative: f′′(x)=−1x2f”(x) = -\frac{1}{x^2}f′′(x)=−x21 Evaluating at x=1x = 1x=1: f′′(1)=−1f”(1) = -1f′′(1)=−1
- Third derivative: f(3)(x)=2x3f^{(3)}(x) = \frac{2}{x^3}f(3)(x)=x32 Evaluating at x=1x = 1x=1: f(3)(1)=2f^{(3)}(1) = 2f(3)(1)=2
Now, we can plug these values into the Taylor series expansion:ln(x)=0+1(x−1)+−12!(x−1)2+23!(x−1)3+⋯\ln(x) = 0 + 1(x – 1) + \frac{-1}{2!}(x – 1)^2 + \frac{2}{3!}(x – 1)^3 + \cdotsln(x)=0+1(x−1)+2!−1(x−1)2+3!2(x−1)3+⋯
This simplifies to: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+⋯
Thus, the power series for ln(x)\ln(x)ln(x) centered at x=1x = 1x=1 is:ln(x)=∑n=1∞(−1)n+1n(x−1)n\ln(x) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}(x – 1)^nln(x)=n=1∑∞n(−1)n+1(x−1)n
This is a valid expansion for ln(x)\ln(x)ln(x) for ∣x−1∣<1|x – 1| < 1∣x−1∣<1, i.e., xxx must lie in the interval (0,2)(0, 2)(0,2). The series converges to ln(x)\ln(x)ln(x) within this interval.
