Use the Alternating Series Estimation Theorem or Taylor's Inequality to estimate the range of values of $x$ for which the given approximation is accurate to within the stated error. Check your answer graphically. (Enter your answer using interval notation. Round your answers to three decimal places.) $\sin(x) \approx x - \frac{x^3}{6} \quad (\text{error} < 0.01)$

Neetesh Kumar | December 22, 2024

Calculus Homework Help

This is the solution to Math 1c
Assignment: 11.11 Question Number 7
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Step-by-step solution:

Step 1: Recall the Taylor series for $\sin(x)$:

The Taylor series expansion for $\sin(x)$ is:

$\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots.$

The approximation given is:

$\sin(x) \approx x - \frac{x^3}{6}.$

This truncates the series after the second term, so the next term in the series (the remainder term) is:

$R_3(x) = \frac{x^5}{5!} = \frac{x^5}{120}.$

Step 2: Use Taylor's Inequality to bound the error:

Taylor's Inequality states that the error of the approximation is bounded by the magnitude of the first omitted term. Thus, the error is:

$|R_3(x)| = \left|\frac{x^5}{120}\right|.$

We need this error to satisfy:

$\left|\frac{x^5}{120}\right| < 0.01.$

Step 3: Solve for $|x|$:

Rearrange the inequality:

$\frac{|x|^5}{120} < 0.01.$

Multiply through by $120$:

$|x|^5 < 1.2.$

Take the fifth root of both sides:

$|x| < \sqrt[5]{1.2}.$

Using a calculator:

$|x| \approx 1.037.$

Thus, the range of $x$ is:

$-1.037 < x < 1.037.$

Step 4: Express the range in interval notation:

The range of values of $x$ for which the error is less than $0.01$ is:

$(-1.037, 1.037).$

Final Answer:

The range of $x$ is:

$\boxed{(-1.037, 1.037)}$

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