Let's solve the problem step by step:

Part 1: Taylor Series Expansion of $f(x)$ at $x = 2$

The given function is: $f(x) = \frac{2}{1-x} + \frac{1}{1+x}.$

Step 1: Expand each term about $x = 2$.

1. For $\frac{2}{1-x}$: The Taylor series expansion for $\frac{1}{1-x}$ around $x = 2$ can be written as: $\frac{1}{1-x} = \frac{1}{1-2} \cdot \frac{1}{1 - \frac{x-2}{1-2}} = -\frac{1}{1 - (x-2)} = -\sum_{n=0}^\infty (x-2)^n,$ where this series converges for $|x-2| < 1$.

   Hence: $\frac{2}{1-x} = -2\sum_{n=0}^\infty (x-2)^n = -2 - 2(x-2) - 2(x-2)^2 - 2(x-2)^3 + \dots$

2. For $\frac{1}{1+x}$: The Taylor series expansion for $\frac{1}{1+x}$ about $x = 2$ is given by: $\frac{1}{1+x} = \frac{1}{1+2} \cdot \frac{1}{1 + \frac{x-2}{1+2}} = \frac{1}{3} \cdot \frac{1}{1 + \frac{x-2}{3}} = \frac{1}{3} \sum_{n=0}^\infty (-1)^n \left(\frac{x-2}{3}\right)^n.$ Expanding the first few terms, we have: $\frac{1}{1+x} = \frac{1}{3} - \frac{x-2}{9} + \frac{(x-2)^2}{27} - \frac{(x-2)^3}{81} + \dots$

Step 2: Combine the results.

Add the two expansions to get the Taylor series for $f(x)$ about $x = 2$: $f(x) = -2 - 2(x-2) - 2(x-2)^2 - 2(x-2)^3 + \dots + \frac{1}{3} - \frac{x-2}{9} + \frac{(x-2)^2}{27} - \frac{(x-2)^3}{81} + \dots$

Combine like terms for the first four terms: $f(x) = -\frac{5}{3} - \left(2 + \frac{1}{9}\right)(x-2) - \left(2 - \frac{1}{27}\right)(x-2)^2 - \left(2 + \frac{1}{81}\right)(x-2)^3 + \dots$

Part 2: Taylor Series for $g(x) = \ln \left(\frac{1+x}{(1-x)^2}\right)$

Step 1: Separate the terms.

We know that: $g(x) = \ln(1+x) - 2\ln(1-x).$

Expand each term as follows:

1. For $\ln(1+x)$: The Taylor series for $\ln(1+x)$ around $x = 0$ is: $\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots$

2. For $\ln(1-x)$: The Taylor series for $-\ln(1-x)$ around $x = 0$ is: $-\ln(1-x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + \dots$

Combine these: $g(x) = (x - \frac{x^2}{2} + \frac{x^3}{3} - \dots) - 2(x + \frac{x^2}{2} + \frac{x^3}{3} + \dots).$

Simplify: $g(x) = -x - 2x^2 - \frac{5x^3}{3} + \dots$

Part 3: Approximation of $\ln\left(\frac{14}{9}\right)$

Set $x = \frac{5}{9}$ in $g(x)$ and use the first four terms: $g\left(\frac{5}{9}\right) \approx -\frac{5}{9} - 2\left(\frac{5}{9}\right)^2 - \frac{5}{3}\left(\frac{5}{9}\right)^3.$

Evaluate this to approximate $\ln\left(\frac{14}{9}\right)$.

Would you like me to compute the numerical value or clarify any step?

Related Questions:

1. What are the general conditions for the convergence of a Taylor series?
2. How can you compute higher-order derivatives for Taylor expansions systematically?
3. Can we approximate the error in truncating a Taylor series after a certain number of terms?
4. How is the radius of convergence determined for a Taylor series?
5. What alternative methods exist for approximating functions (e.g., Padé approximations)?

Tip:

When approximating a function with its Taylor series, always check the radius of convergence to ensure your approximation is valid within the desired interval.