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 \(\ln(1 + x)\), we first write out the series expansion and then determine the number of terms needed to achieve the desired accuracy.

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

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

\[
\ln(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}x^n
\]

### Step 2: Substitute \( x = 0.4 \) into the series

To estimate \(\ln(1.4)\), we 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 number of terms needed

We need to determine the number of terms required such that the error (i.e., the magnitude of the next term in the series) is less than 0.000001.

The error after \(n\) terms in the alternating series can be estimated by the magnitude of the next term:

\[
\text{Error} \approx \left|\frac{(-1)^{n+1}}{n}x^n\right| = \frac{0.4^{n+1}}{n+1}
\]

We want:

\[
\frac{0.4^{n+1}}{n+1} < 0.000001
\]

### Step 4: Calculate the minimum \(n\)

Now we compute the minimum \(n\) such that this inequality is satisfied:

\[
\frac{0.4^{n+1}}{n+1} < 0.000001
\]

Let's test successive values of \(n\) until the inequality is satisfied.

1. For \(n = 1\):
   \[
   \frac{0.4^2}{2} = \frac{0.16}{2} = 0.08
   \]
   (Error is too large)
   
2. For \(n = 2\):
   \[
   \frac{0.4^3}{3} = \frac{0.064}{3} \approx 0.0213
   \]
   (Error is too large)

3. For \(n = 3\):
   \[
   \frac{0.4^4}{4} = \frac{0.0256}{4} = 0.0064
   \]
   (Error is still too large)

4. For \(n = 4\):
   \[
   \frac{0.4^5}{5} = \frac{0.01024}{5} \approx 0.002048
   \]
   (Error is still too large)

5. For \(n = 5\):
   \[
   \frac{0.4^6}{6} = \frac{0.004096}{6} \approx 0.0006827
   \]
   (Error is still too large)

6. For \(n = 6\):
   \[
   \frac{0.4^7}{7} = \frac{0.0016384}{7} \approx 0.0002341
   \]
   (Error is still too large)

7. For \(n = 7\):
   \[
   \frac{0.4^8}{8} = \frac{0.00065536}{8} \approx 0.00008192
   \]
   (Error is still too large)

8. For \(n = 8\):
   \[
   \frac{0.4^9}{9} = \frac{0.000262144}{9} \approx 0.00002912
   \]
   (Error is still too large)

9. For \(n = 9\):
   \[
   \frac{0.4^{10}}{10} = \frac{0.0001048576}{10} \approx 0.00001048576
   \]
   (Error is still too large)

10. For \(n = 10\):
    \[
    \frac{0.4^{11}}{11} = \frac{0.00004194304}{11} \approx 0.000003813
    \]
    (Error is sufficiently small)

### Conclusion:
You need **10 nonzero terms** of the Maclaurin series to estimate \(\ln(1.4)\) to within 0.000001.

Would you like more details or have any questions? Here are some related questions to explore:

1. How does the error bound for an alternating series help in such estimations?
2. What is the significance of the radius of convergence in Maclaurin series expansions?
3. Can the same method be applied to estimate \(\ln(2)\)?
4. How does the value of \( x \) affect the convergence of the series?
5. What if we used a higher precision, such as 0.0000001, how many terms would be required?
6. How does the convergence rate compare for \(\ln(1+x)\) versus other logarithmic expansions?
7. How would the series change if we approximated \(\ln(1.5)\) instead?
8. How do we generalize this method to other functions like \(\sin(x)\) or \(e^x\)?

**Tip:** Always check the error term when using partial sums of a series to ensure the approximation meets the desired accuracy.

Learn how to estimate ln(1.4) to within 0.000001 using Maclaurin series for ln(1 + x). This detailed solution covers the steps to determine the number of terms required and discusses error estimation in series expansions.

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