To analyze $f(x) = \tan^{-1}(4x)$, we will address each part step by step.

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Part (a): Find the first three nonzero terms of the Maclaurin series.

The Maclaurin series for $\tan^{-1}(x)$ is:

$\tan^{-1}(x) = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots$

For $\tan^{-1}(4x)$, substitute $4x$ in place of $x$:

$\tan^{-1}(4x) = 4x - \frac{(4x)^3}{3} + \frac{(4x)^5}{5} - \cdots$

Simplify the terms:

1. The first term: $4x$
2. The second term: $-\frac{(4x)^3}{3} = -\frac{64x^3}{3}$
3. The third term: $\frac{(4x)^5}{5} = \frac{1024x^5}{5}$

Thus, the first three nonzero terms of the Maclaurin series are:

$f(x) \approx 4x - \frac{64x^3}{3} + \frac{1024x^5}{5}$

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Part (b): Write the power series using summation notation.

The general term of the Maclaurin series for $\tan^{-1}(4x)$ is derived from the Maclaurin series for $\tan^{-1}(x)$:

$\tan^{-1}(x) = \sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{2n+1}$

Substitute $4x$ for $x$:

$\tan^{-1}(4x) = \sum_{n=0}^\infty (-1)^n \frac{(4x)^{2n+1}}{2n+1}$

Simplify:

$\tan^{-1}(4x) = \sum_{n=0}^\infty (-1)^n \frac{4^{2n+1}x^{2n+1}}{2n+1}$

This is the power series in summation notation.

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Part (c): Determine the interval of convergence.

For the series $\sum_{n=0}^\infty (-1)^n \frac{(4x)^{2n+1}}{2n+1}$, the convergence is based on the original series $\tan^{-1}(x)$, which converges for $|x| \leq 1$.

In this case, $|4x| \leq 1$, which simplifies to:

$|x| \leq \frac{1}{4}$

At the endpoints $x = \pm \frac{1}{4}$, the series becomes conditionally convergent because $\tan^{-1}(x)$ is continuous and alternates.

Thus, the interval of convergence is:

$\boxed{-\frac{1}{4} \leq x \leq \frac{1}{4}}$

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Summary of Results:

1. First three nonzero terms: $4x - \frac{64x^3}{3} + \frac{1024x^5}{5}$
2. Power series in summation notation: $\sum_{n=0}^\infty (-1)^n \frac{4^{2n+1}x^{2n+1}}{2n+1}$
3. Interval of convergence: $-\frac{1}{4} \leq x \leq \frac{1}{4}$

Would you like further clarifications or additional examples?

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5 Related Questions:

1. How is the Maclaurin series for $\tan^{-1}(x)$ derived?
2. What is the radius of convergence, and how is it determined for power series?
3. How can you approximate $\tan^{-1}(4x)$ using only the first few terms of the series?
4. What happens to the interval of convergence if the argument of $\tan^{-1}(x)$ changes to $ax$?
5. How do alternating series test conditions affect the convergence of series?

1 Tip:

For functions like $\tan^{-1}(x)$, always check the domain of the original function before determining convergence.