How many nonzero terms of the Maclaurin series for

ln(1 + x)

do you need to use to estimate

ln(1.4)

to within 0.000001?

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.

Learn how to estimate ln(1.4) using the Maclaurin series for ln(1 + x) with a detailed step-by-step solution. This problem involves determining the number of nonzero terms required for an accurate estimation within 0.000001. Suitable for advanced undergraduate students and anyone interested in series approximation.

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