To analyze this approximation of $\sin x$ using the Taylor series expansion, let's break down the question and solve each part:

Question Breakdown

1. The expression given for approximating $\sin x$ is: $\sin x \approx x - \frac{x^3}{3!} = x - \frac{x^3}{6}$
2. The range for $x$ is given by $|x| < 10^{-2}$.
3. We need to determine how close this approximation is to the actual $\sin x$.
4. Additionally, we are asked to find values of $x$ for which the approximation $x - \frac{x^3}{6}$ is greater than $\sin x$.

Step 1: Analyzing the Error in Approximation

The Taylor series expansion of $\sin x$ around $x = 0$ is: $\sin x = x - \frac{x^3}{6} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots$ The approximation provided neglects higher-order terms starting from $\frac{x^5}{120}$. For small values of $x$ (where $|x| < 10^{-2}$), the error can be approximated by the next term in the series, $\frac{x^5}{120}$.

Therefore, the error $E$ in the approximation $x - \frac{x^3}{6}$ compared to $\sin x$ can be roughly estimated as: $E \approx \left| \frac{x^5}{120} \right|$

Step 2: Calculate the Error for $|x| < 10^{-2}$

If $|x| = 10^{-2}$, then the error would be: $E \approx \frac{(10^{-2})^5}{120} = \frac{10^{-10}}{120} = 8.33 \times 10^{-13}$ This shows that the approximation is very close to $\sin x$ within this range, with an error in the order of $10^{-13}$.

Step 3: Finding Values of $x$ Where $x - \frac{x^3}{6} > \sin x$

Since $\sin x = x - \frac{x^3}{6} + \frac{x^5}{120} - \dots$, the term $\frac{x^5}{120}$ is positive for small values of $x$. This implies that: $x - \frac{x^3}{6} > \sin x$ because we are neglecting a positive term, $\frac{x^5}{120}$, in the approximation.

Thus, for values of $x$ within $|x| < 10^{-2}$, $x - \frac{x^3}{6}$ will be slightly greater than $\sin x$.

Summary

1. The approximation $x - \frac{x^3}{6}$ is extremely close to $\sin x$ within $|x| < 10^{-2}$, with an error of approximately $8.33 \times 10^{-13}$.
2. For all values of $x$ within this range, $x - \frac{x^3}{6} > \sin x$ because the neglected terms in the series are positive.

Would you like more details on this approximation, or have any specific questions about Taylor series or error estimation?

Follow-up Questions:

1. What is the Taylor series expansion of $\cos x$ and how is it derived?
2. How does increasing the range of $x$ affect the accuracy of this approximation?
3. What would be the next term in the Taylor series for $\sin x$, and how does it influence the approximation?
4. How would the error change if $|x|$ were allowed to be larger than $10^{-2}$?
5. Can you explain how to derive the error bound for higher-order Taylor approximations?

Tip:

When using Taylor series approximations, always consider the range of $x$ since the error grows with higher values of $x$.