Math Problem Statement

Find the Taylor series for f(x) centered at the given value of a. [Assume that f has a power series expansion. Do not show that Rn(x) → 0.]

f(x) = ln(x), a = 2

∞ f(x) = ln(2) + ∑ ( ? ) n = 1

Find the associated radius of convergence, R. R = ?

Solution

To find the Taylor series for the function f(x)=ln(x)f(x) = \ln(x) centered at a=2a = 2, we need to compute the derivatives of f(x)f(x) at x=2x = 2 and then express the series in the form:

f(x)=ln(2)+n=1f(n)(2)n!(x2)nf(x) = \ln(2) + \sum_{n=1}^{\infty} \frac{f^{(n)}(2)}{n!} (x - 2)^n

Step 1: Compute the derivatives of f(x)=ln(x)f(x) = \ln(x)

  1. First derivative: f(x)=1xf'(x) = \frac{1}{x} At x=2x = 2: f(2)=12f'(2) = \frac{1}{2}

  2. Second derivative: f(x)=1x2f''(x) = -\frac{1}{x^2} At x=2x = 2: f(2)=14f''(2) = -\frac{1}{4}

  3. Third derivative: f(3)(x)=2x3f^{(3)}(x) = \frac{2}{x^3} At x=2x = 2: f(3)(2)=28=14f^{(3)}(2) = \frac{2}{8} = \frac{1}{4}

  4. Fourth derivative: f(4)(x)=6x4f^{(4)}(x) = -\frac{6}{x^4} At x=2x = 2: f(4)(2)=616=38f^{(4)}(2) = -\frac{6}{16} = -\frac{3}{8}

Step 2: Write the Taylor series

Now, substitute these derivatives into the Taylor series formula:

f(x)=ln(2)+n=1f(n)(2)n!(x2)nf(x) = \ln(2) + \sum_{n=1}^{\infty} \frac{f^{(n)}(2)}{n!} (x - 2)^n

Substituting the computed derivatives:

f(x)=ln(2)+(12(x2))(142(x2)2)+(143!(x2)3)(384!(x2)4)+f(x) = \ln(2) + \left(\frac{1}{2}(x - 2)\right) - \left(\frac{1}{4 \cdot 2}(x - 2)^2\right) + \left(\frac{1}{4 \cdot 3!}(x - 2)^3\right) - \left(\frac{3}{8 \cdot 4!}(x - 2)^4\right) + \cdots

Simplifying, we get:

f(x)=ln(2)+n=1(1)n+1(x2)nn2nf(x) = \ln(2) + \sum_{n=1}^{\infty} (-1)^{n+1} \frac{(x - 2)^n}{n \cdot 2^n}

Step 3: Radius of Convergence RR

The radius of convergence RR of the Taylor series can be found using the formula:

1R=limnan+1an\frac{1}{R} = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|

Here, the series is:

n=1(1)n+1(x2)nn2n\sum_{n=1}^{\infty} (-1)^{n+1} \frac{(x - 2)^n}{n \cdot 2^n}

Thus, the radius of convergence RR is:

R=2R = 2

Final Answers:

  1. Taylor Series: f(x)=ln(2)+n=1(1)n+1(x2)nn2nf(x) = \ln(2) + \sum_{n=1}^{\infty} (-1)^{n+1} \frac{(x - 2)^n}{n \cdot 2^n}

  2. Radius of Convergence: R=2R = 2

Would you like any additional details or have any questions?

Here are some related questions you might find interesting:

  1. How do you derive the formula for the radius of convergence using the ratio test?
  2. What happens to the Taylor series if we expand around a=1a = 1 instead?
  3. Can you calculate the first few terms of the Taylor series expansion for a different function, like exe^x?
  4. How does the error term Rn(x)R_n(x) behave in Taylor series approximations?
  5. What are some practical applications of Taylor series in physics or engineering?
  6. How can you use Taylor series to approximate ln(2.1)\ln(2.1)?
  7. How does the Taylor series relate to the concept of analytic functions?
  8. What are the conditions under which a function has a Taylor series expansion?

Tip: The Taylor series is especially useful for approximating functions near a point where the function is difficult to evaluate directly.

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Math Problem Analysis

Mathematical Concepts

Calculus
Taylor series
Power series
Radius of convergence

Formulas

Taylor series expansion
Radius of convergence formula

Theorems

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Suitable Grade Level

College