Let's solve the problem step by step.

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Part (a): Taylor Series for $f(x) = \frac{1}{x}$ centered at $a = 3$.

1. Taylor series formula:

The Taylor series expansion of a function $f(x)$ about $a = 3$ is: $f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(3)}{n!}(x-3)^n$

2. Compute the derivatives of $f(x) = \frac{1}{x}$:

• First derivative: $f'(x) = -\frac{1}{x^2}$
• Second derivative: $f''(x) = \frac{2}{x^3}$
• Third derivative: $f'''(x) = -\frac{6}{x^4}$
• $n$-th derivative: $f^{(n)}(x) = (-1)^n \cdot \frac{n!}{x^{n+1}}$

3. Evaluate $f^{(n)}(3)$:

$f^{(n)}(3) = (-1)^n \cdot \frac{n!}{3^{n+1}}$

4. Substitute into the Taylor series formula:

$f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(3)}{n!}(x-3)^n$ $f(x) = \sum_{n=0}^\infty \frac{(-1)^n \cdot \frac{n!}{3^{n+1}}}{n!}(x-3)^n$ $f(x) = \sum_{n=0}^\infty \frac{(-1)^n}{3^{n+1}}(x-3)^n$

Thus, the Taylor series is: $f(x) = \sum_{n=0}^\infty \frac{(-1)^n}{3^{n+1}}(x-3)^n$

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Part (b): Interval of Convergence

The interval of convergence is determined by the radius of convergence $R$, which can be found using the ratio test. For the series: $\sum_{n=0}^\infty c_n (x-3)^n \quad \text{where} \quad c_n = \frac{(-1)^n}{3^{n+1}}$

1. Apply the ratio test:

$\lim_{n \to \infty} \left| \frac{c_{n+1}(x-3)^{n+1}}{c_n (x-3)^n} \right| < 1$ $\lim_{n \to \infty} \left| \frac{\frac{(-1)^{n+1}}{3^{n+2}}}{\frac{(-1)^n}{3^{n+1}}} \cdot (x-3) \right| < 1$ $\lim_{n \to \infty} \left| \frac{1}{3} \cdot (x-3) \right| < 1$ $\frac{|x-3|}{3} < 1$

2. Solve for $|x-3|$:

$|x-3| < 3$

3. Interval of convergence:

The series converges for $x$ in the interval: $(3-3, 3+3) = (0, 6)$

Now, check the endpoints $x = 0$ and $x = 6$:

• At $x = 0$: The series becomes: $\sum_{n=0}^\infty \frac{(-1)^n}{3^{n+1}}(-3)^n$ This diverges because the terms do not approach 0.

• At $x = 6$: The series becomes: $\sum_{n=0}^\infty \frac{(-1)^n}{3^{n+1}}(3)^n$ This diverges as well for similar reasons.

Thus, the interval of convergence is: $(0, 6)$

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Final Answers:

(a) The Taylor series is: $f(x) = \sum_{n=0}^\infty \frac{(-1)^n}{3^{n+1}}(x-3)^n$

(b) The interval of convergence is: $(0, 6)$

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Would you like a detailed explanation for any step?

Here are some related questions for practice:

1. How would the Taylor series change if the center were $a = 2$?
2. What happens to the radius of convergence if $f(x)$ were $\frac{1}{x^2}$?
3. How do you test convergence at the endpoints of an interval?
4. What is the general formula for a Taylor series of $\ln(x)$?
5. How do the derivatives of a rational function like $\frac{1}{x}$ behave?

Tip: Always use the ratio test to check the radius and interval of convergence for Taylor series.