To find the Taylor series of the function $f(x) = \ln(x)$ centered at $c = 1$, we use the Taylor series definition:

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

Step 1: Calculate the derivatives

1. First derivative: $f'(x) = \frac{1}{x}$ Evaluating at $c = 1$: $f'(1) = 1$

2. Second derivative: $f''(x) = -\frac{1}{x^2}$ Evaluating at $c = 1$: $f''(1) = -1$

3. Third derivative: $f'''(x) = \frac{2}{x^3}$ Evaluating at $c = 1$: $f'''(1) = 2$

4. Fourth derivative: $f^{(4)}(x) = -\frac{6}{x^4}$ Evaluating at $c = 1$: $f^{(4)}(1) = -6$

5. Fifth derivative: $f^{(5)}(x) = \frac{24}{x^5}$ Evaluating at $c = 1$: $f^{(5)}(1) = 24$

Step 2: Construct the Taylor series

Now, we can substitute these values into the Taylor series formula:

$f(x) = \ln(1) + 1(x - 1) - \frac{1}{2}(x - 1)^2 + \frac{2}{6}(x - 1)^3 - \frac{6}{24}(x - 1)^4 + \frac{24}{120}(x - 1)^5 + \cdots$

Since $\ln(1) = 0$, we have:

$f(x) = (x - 1) - \frac{1}{2}(x - 1)^2 + \frac{1}{3}(x - 1)^3 - \frac{1}{4}(x - 1)^4 + \frac{1}{5}(x - 1)^5 + \cdots$

Final Result

The Taylor series for $f(x) = \ln(x)$ centered at $c = 1$ is:

$\ln(x) = (x - 1) - \frac{1}{2}(x - 1)^2 + \frac{1}{3}(x - 1)^3 - \frac{1}{4}(x - 1)^4 + \frac{1}{5}(x - 1)^5 + \cdots$

Summary

1. Taylor Series: The series converges for $x > 0$.
2. Radius of Convergence: The series converges in the interval $(0, \infty)$.

Do you want details on any specific part of the calculation or have any questions? Here are some follow-up questions:

1. What is the significance of the Taylor series in calculus?
2. How would you derive the Taylor series for other functions, like $e^x$?
3. Can you explain the convergence criteria for Taylor series?
4. What is the difference between a Taylor series and a Maclaurin series?
5. How can you use the Taylor series for approximating values of $\ln(x)$?

Tip: When working with Taylor series, always check the interval of convergence for accurate results.