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 = ?

To find the Taylor series expansion of the function \( f(x) = \ln(x) \) centered at \( a = 2 \), we follow these steps:

### Step 1: Calculate the derivatives of \( f(x) = \ln(x) \)
First, we calculate the derivatives of \( f(x) = \ln(x) \):

\[
f(x) = \ln(x)
\]
\[
f'(x) = \frac{1}{x}
\]
\[
f''(x) = -\frac{1}{x^2}
\]
\[
f^{(3)}(x) = \frac{2}{x^3}
\]
\[
f^{(4)}(x) = -\frac{6}{x^4}
\]
and so on.

### Step 2: Evaluate the derivatives at \( x = 2 \)
Now, evaluate each derivative at \( x = 2 \):

\[
f(2) = \ln(2)
\]
\[
f'(2) = \frac{1}{2}
\]
\[
f''(2) = -\frac{1}{4}
\]
\[
f^{(3)}(2) = \frac{2}{8} = \frac{1}{4}
\]
\[
f^{(4)}(2) = -\frac{6}{16} = -\frac{3}{8}
\]
and so on.

### Step 3: Form the Taylor series
The Taylor series for \( f(x) \) centered at \( a = 2 \) is given by:

\[
f(x) = f(2) + f'(2)(x - 2) + \frac{f''(2)}{2!}(x - 2)^2 + \frac{f^{(3)}(2)}{3!}(x - 2)^3 + \cdots
\]

Substituting the evaluated derivatives:

\[
f(x) = \ln(2) + \frac{1}{2}(x - 2) - \frac{1}{4 \cdot 2}(x - 2)^2 + \frac{1}{4 \cdot 6}(x - 2)^3 - \cdots
\]

This simplifies to:

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

### Step 4: Determine the radius of convergence
The radius of convergence \( R \) of the Taylor series can be found using the formula:

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

Here, the general term is:

\[
a_n = \frac{(-1)^{n+1}}{n \cdot 2^n}
\]

So, 

\[
\left| \frac{a_n}{a_{n+1}} \right| = \lim_{n \to \infty} \left| \frac{\frac{(-1)^{n+1}}{n \cdot 2^n}}{\frac{(-1)^{(n+2)}}{(n+1) \cdot 2^{n+1}}} \right| = \lim_{n \to \infty} \frac{2(n+1)}{n} = 2
\]

Thus, the radius of convergence \( R \) is:

\[
R = 2
\]

### Final Answer:
- The Taylor series for \( f(x) = \ln(x) \) centered at \( a = 2 \) is:

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

- The radius of convergence \( R \) is:

\[
R = 2
\]

Would you like more details or have any questions? Here are some related questions to explore:

1. How would you calculate the Taylor series expansion for other logarithmic functions?
2. What happens to the Taylor series if we choose a different center \( a \)?
3. Can we approximate \( \ln(x) \) for values close to 2 using the first few terms of the series?
4. How is the radius of convergence determined for a general power series?
5. How would the series change if the function was \( \ln(2x) \) instead of \( \ln(x) \)?
6. How does the Taylor series behave for points outside the radius of convergence?
7. What is the general formula for the nth derivative of \( \ln(x) \)?
8. Can the Taylor series be used to approximate \( \ln(x) \) at negative values of \( x \)?

**Tip:** When working with Taylor series, always check the radius of convergence to know the interval where the series accurately represents the function.

Learn how to find the Taylor series expansion for ln(x) centered at a = 2 with a step-by-step solution. Understand the concept of Taylor series, power series, and radius of convergence. Suitable for advanced undergraduate students studying calculus.

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