To estimate $\ln(1.4)$ using the Maclaurin series for $\ln(1 + x)$ to within $0.000001$, we need to determine how many nonzero terms of the series are required to achieve this level of accuracy.

Step 1: Write the Maclaurin series for $\ln(1 + x)$

The Maclaurin series for $\ln(1 + x)$ is:

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

Step 2: Evaluate the series at $x = 0.4$

To estimate $\ln(1.4)$, substitute $x = 0.4$ into the series:

$\ln(1.4) = 0.4 - \frac{0.4^2}{2} + \frac{0.4^3}{3} - \frac{0.4^4}{4} + \cdots$

Step 3: Determine the remainder (error) term $R_n$

The error in the Taylor series approximation is given by the remainder term $R_n(x)$:

$R_n(x) = \frac{(-1)^{n+1} x^{n+1}}{(n+1)}$

We need to find the smallest $n$ such that the absolute value of the remainder term is less than or equal to $0.000001$:

$\left|\frac{(-1)^{n+1} (0.4)^{n+1}}{n+1}\right| \leq 0.000001$

Simplifying the inequality:

$\frac{(0.4)^{n+1}}{n+1} \leq 0.000001$

Step 4: Find the required number of terms

We can solve this inequality by trying different values of $n$:

1. For $n = 1$: $\frac{(0.4)^2}{2} = \frac{0.16}{2} = 0.08 \quad (\text{Not small enough})$

2. For $n = 2$: $\frac{(0.4)^3}{3} = \frac{0.064}{3} \approx 0.02133 \quad (\text{Not small enough})$

3. For $n = 3$: $\frac{(0.4)^4}{4} = \frac{0.0256}{4} = 0.0064 \quad (\text{Not small enough})$

4. For $n = 4$: $\frac{(0.4)^5}{5} = \frac{0.01024}{5} \approx 0.002048 \quad (\text{Not small enough})$

5. For $n = 5$: $\frac{(0.4)^6}{6} = \frac{0.004096}{6} \approx 0.00068267 \quad (\text{Not small enough})$

6. For $n = 6$: $\frac{(0.4)^7}{7} = \frac{0.0016384}{7} \approx 0.00023406 \quad (\text{Not small enough})$

7. For $n = 7$: $\frac{(0.4)^8}{8} = \frac{0.00065536}{8} \approx 0.00008192 \quad (\text{Not small enough})$

8. For $n = 8$: $\frac{(0.4)^9}{9} = \frac{0.000262144}{9} \approx 0.00002913 \quad (\text{Not small enough})$

9. For $n = 9$: $\frac{(0.4)^{10}}{10} = \frac{0.0001048576}{10} \approx 0.00001049 \quad (\text{Not small enough})$

10. For $n = 10$: $\frac{(0.4)^{11}}{11} = \frac{0.00004194304}{11} \approx 0.00000381 \quad (\text{Small enough!})$

Conclusion:

You need to use 11 nonzero terms of the Maclaurin series for $\ln(1 + x)$ to estimate $\ln(1.4)$ to within $0.000001$.

Would you like any further details or explanations?

Here are some related questions you might find interesting:

1. What is the general formula for the remainder term in a Taylor series?
2. How does the convergence of the Maclaurin series differ for different values of $x$?
3. Can you derive the Maclaurin series for other logarithmic functions, like $\ln(x)$?
4. What are some real-world applications of using Maclaurin series for approximations?
5. How does the error in a Taylor series expansion depend on the order of the term?
6. What is the interval of convergence for the Maclaurin series of $\ln(1 + x)$?
7. How would the series change if you wanted to estimate $\ln(1.5)$ instead?
8. What is the role of alternating series in ensuring the accuracy of the approximation?

Tip: The more rapidly the terms of a series decrease in magnitude, the fewer terms are needed for an accurate approximation.