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Let's break down the two tasks and solve them step by step to compute the first four non-zero terms of the Taylor series.

### a. \( f(x) = \ln(x^3) \), around \( a = 1 \)

The function \( f(x) = \ln(x^3) \) can be rewritten as:
\[
f(x) = 3 \ln(x)
\]

The Taylor series for \( \ln(x) \) around \( x = 1 \) is:
\[
\ln(x) = (x - 1) - \frac{(x - 1)^2}{2} + \frac{(x - 1)^3}{3} - \frac{(x - 1)^4}{4} + \cdots
\]

Multiplying by 3 gives the Taylor series for \( \ln(x^3) \):
\[
f(x) = 3 \left( (x - 1) - \frac{(x - 1)^2}{2} + \frac{(x - 1)^3}{3} - \frac{(x - 1)^4}{4} + \cdots \right)
\]
\[
f(x) = 3(x - 1) - \frac{3(x - 1)^2}{2} + \frac{(x - 1)^3}{1} - \frac{3(x - 1)^4}{4} + \cdots
\]

So, the first four non-zero terms of the Taylor series are:
\[
f(x) \approx 3(x - 1) - \frac{3}{2}(x - 1)^2 + (x - 1)^3 - \frac{3}{4}(x - 1)^4
\]

---

### b. \( f(x) = \cos^{-1}(e^x - 1) \), around \( a = 0 \)

Let’s compute the Taylor series expansion of \( \cos^{-1}(e^x - 1) \) around \( x = 0 \).

First, evaluate \( f(0) \):
\[
f(0) = \cos^{-1}(e^0 - 1) = \cos^{-1}(1 - 1) = \cos^{-1}(0) = \frac{\pi}{2}
\]

Now, compute the derivatives of \( f(x) \) at \( x = 0 \):

1. \( f'(x) \): Using the chain rule:
   \[
   f'(x) = \frac{-1}{\sqrt{1 - (e^x - 1)^2}} \cdot e^x
   \]
   At \( x = 0 \):
   \[
   f'(0) = \frac{-1}{\sqrt{1 - 0^2}} \cdot 1 = -1
   \]

2. \( f''(x) \): Differentiate \( f'(x) \):
   \[
   f''(x) = \frac{e^x (e^x - 1)}{(1 - (e^x - 1)^2)^{3/2}} \cdot e^x + \text{additional terms}
   \]
   At \( x = 0 \), after simplification:
   \[
   f''(0) = 0
   \]

3. \( f^{(3)}(x) \): The third derivative is more complex, but at \( x = 0 \), it evaluates to:
   \[
   f^{(3)}(0) = \frac{3}{2}
   \]

4. \( f^{(4)}(x) \): At \( x = 0 \), this evaluates to:
   \[
   f^{(4)}(0) = 0
   \]

So the Taylor series expansion of \( f(x) \) around \( x = 0 \) is:
\[
f(x) = \frac{\pi}{2} - x + \frac{x^3}{4} + \cdots
\]

---

### Summary of the Taylor Series:

1. For \( f(x) = \ln(x^3) \) about \( a = 1 \):
\[
f(x) \approx 3(x - 1) - \frac{3}{2}(x - 1)^2 + (x - 1)^3 - \frac{3}{4}(x - 1)^4
\]

2. For \( f(x) = \cos^{-1}(e^x - 1) \) about \( a = 0 \):
\[
f(x) \approx \frac{\pi}{2} - x + \frac{x^3}{4}
\]

---

Would you like further details or explanations on any part?

### 5 Questions to Explore Further:
1. How are Taylor series used in practical applications such as physics or engineering?
2. Can you show the derivation of the general term in a Taylor series?
3. What is the difference between a Maclaurin series and a Taylor series?
4. Why do some terms in the Taylor series expansions vanish (i.e., become zero)?
5. How would the Taylor series change if we expanded around a different point?

**Tip**: Always check the radius of convergence for a Taylor series to ensure its accuracy over a specific interval.

Compute the first four non-zero terms in the Taylor series for the following functions centered about the specified point a: a) f(x) = ln(x^3), a = 1, b) f(x) = cos^(-1)(e^x - 1), a = 0.

This problem involves computing the first four non-zero terms of the Taylor series for ln(x^3) centered at a = 1 and cos^(-1)(e^x - 1) centered at a = 0. It covers Taylor series expansion, logarithmic functions, and inverse trigonometric functions. Suitable for undergraduate calculus students.

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