The Taylor series expansion of $\ln(1+x)$ about $x = 0$ is derived by taking the derivatives of $\ln(1+x)$ and evaluating them at $x = 0$. The series can be written as:

$\ln(1+x) = \sum_{n=1}^\infty (-1)^{n+1} \frac{x^n}{n}, \quad \text{for } |x| < 1.$

Step-by-step Derivation:

1. Function: $f(x) = \ln(1+x)$.
2. First derivative: $f'(x) = \frac{1}{1+x}$.
3. Higher derivatives: $f^{(n)}(x) = (-1)^{n-1} \frac{(n-1)!}{(1+x)^n}.$
4. Evaluate derivatives at $x = 0$: $f^{(n)}(0) = (-1)^{n-1} (n-1)!.$
5. Taylor series formula: $f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!} x^n.$
6. Substituting $f^{(n)}(0)$: $\ln(1+x) = \sum_{n=1}^\infty (-1)^{n+1} \frac{x^n}{n}.$

The First Few Terms:

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

Radius of Convergence:

The series converges for $|x| < 1$.

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Related Questions:

1. How is the radius of convergence determined for this series?
2. Can you derive this series using integration techniques?
3. How does the Taylor series change if expanded about $x = a$ instead of $x = 0$?
4. What are practical applications of the $\ln(1+x)$ expansion?
5. How does this series compare to the expansion of $e^x$?

Tip:

To approximate $\ln(1+x)$ numerically, use a few terms of the series. Increasing terms improves accuracy for $|x| < 1$.