Let's break this problem down step by step:

Step 1: Finding the series for $\ln \left( \frac{1+x}{1-x} \right)$

We can start by recognizing that $\ln \left( \frac{1+x}{1-x} \right)$ can be rewritten using the properties of logarithms:

$\ln \left( \frac{1+x}{1-x} \right) = \ln(1+x) - \ln(1-x)$

Now, we'll use the Taylor series expansions for $\ln(1+x)$ and $\ln(1-x)$.

• The Taylor series for $\ln(1+x)$ is: $\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots$ valid for $|x| < 1$.

• The Taylor series for $\ln(1-x)$ is: $\ln(1-x) = -x - \frac{x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4} - \cdots$ also valid for $|x| < 1$.

Now subtract the series for $\ln(1-x)$ from $\ln(1+x)$:

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

Simplifying the subtraction:

$\ln \left( \frac{1+x}{1-x} \right) = 2x + \frac{2x^3}{3} + \frac{2x^5}{5} + \frac{2x^7}{7} + \cdots$

Thus, the Taylor series for $\ln \left( \frac{1+x}{1-x} \right)$ is:

$\ln \left( \frac{1+x}{1-x} \right) = 2 \sum_{n=0}^{\infty} \frac{x^{2n+1}}{2n+1}$

Step 2: Using the series to approximate $\ln(3)$

We want to approximate $\ln(3)$ using the first four terms of this series. Note that:

$\ln(3) = \ln \left( \frac{1 + \frac{1}{2}}{1 - \frac{1}{2}} \right) = \ln \left( \frac{3}{1} \right) = \ln 3$

This corresponds to using $x = \frac{1}{2}$ in the series.

Now substitute $x = \frac{1}{2}$ into the series:

$\ln 3 = 2 \left( \frac{1}{2} + \frac{1}{3} \left( \frac{1}{2} \right)^3 + \frac{1}{5} \left( \frac{1}{2} \right)^5 + \frac{1}{7} \left( \frac{1}{2} \right)^7 \right)$

Simplifying each term:

• First term: $\frac{1}{2}$
• Second term: $\frac{1}{3} \times \frac{1}{8} = \frac{1}{24}$
• Third term: $\frac{1}{5} \times \frac{1}{32} = \frac{1}{160}$
• Fourth term: $\frac{1}{7} \times \frac{1}{128} = \frac{1}{896}$

Adding these up:

$\ln 3 \approx 2 \left( \frac{1}{2} + \frac{1}{24} + \frac{1}{160} + \frac{1}{896} \right)$

Now calculate the sum inside the parentheses:

$\frac{1}{2} + \frac{1}{24} + \frac{1}{160} + \frac{1}{896} = 0.5 + 0.04167 + 0.00625 + 0.001116 = 0.548936$

Finally, multiplying by 2:

$\ln 3 \approx 2 \times 0.548936 = 1.097872$

So, using the first four terms of the series, we approximate $\ln(3)$ as $\approx 1.097872$.

Questions:

1. What would the approximation be if we included more terms in the series?
2. How accurate is this approximation compared to the exact value of $\ln(3)$?
3. How does the convergence of this series change as $x$ increases or decreases?
4. Can you derive similar series expansions for other logarithmic functions, such as $\ln(1+x)$ directly?
5. What are the limitations of using Taylor series expansions for approximations like this?

Tip:

When approximating values with series, always check the domain of convergence and try using as many terms as practical for increased accuracy.