Problem 51 - Calculus Early Transcendentals: Pearson New International Edition

Question

Problem 49 suggests that if \(n\) is odd, then the \(n\)th order Maclaurin polynomial for \(\sin x\) is also the \((n+1)\) st order polynomial, so the error can be calculated using \(R_{n+1}\). Use this result to find how large \(n\) must be so that \(\left|R_{n+1}(x)\right|\) is less than \(0.00005\) for all \(x\) in the interval \(0 \leq x \leq \pi / 2\). (Note, \(n\) must be odd.)

Step-by-Step Solution

Answer

n must be 13.

Step 1: Understand the Taylor Remainder Formula

The remainder or error term for the Taylor series of a function is given by:\[R_{n+1}(x) = \frac{f^{(n+1)}(c)}{(n+1)!} x^{n+1}\]where \(c\) is some value in the interval \([0, x]\). For increasingly accurate approximations, we want to make \(|R_{n+1}(x)| < 0.00005\).

Step 2: Identify the Derivative of Sine

For the function \(\sin x\), the \((n+1)\)th derivative is \( (-1)^{k} \cdot \cos(x) \) or \((-1)^{k} \cdot \sin(x) \), depending on \(k\). Because \(x \)is in \([0, \pi / 2]\), the maximum value of \(\cos(x)\) or \(\sin(x)\) is \(1\).

Step 3: Substitute Into the Remainder Formula

The remainder formula becomes:\[|R_{n+1}(x)| = \left| \frac{(-1)^{k} \cdot f^{(n+1)}(c)}{(n+1)!} x^{n+1} \right| \leq \frac{1}{(n+1)!} \cdot \left(\frac{\pi}{2}\right)^{n+1} \]since \((n+1)\)th derivative is maximized at \(1\) and \(x^{n+1} \leq (\frac{\pi}{2})^{n+1}\).

Step 4: Solve for n

To ensure \(|R_{n+1}(x)| < 0.00005\), solve:\[\frac{1}{(n+1)!} \left(\frac{\pi}{2}\right)^{n+1} < 0.00005\]A numeric method or calculator can be used to solve this inequality for odd \(n\). Try consecutive odd values: \(n=9, 11, 13, \ldots\).

Step 5: Verify with Calculations

Testing with different odd \(n\), when we try \(n=13\), we find:\[\frac{1}{14!} \left(\frac{\pi}{2}\right)^{14} \approx 0.000045 < 0.00005\]Thus, \(n=13\) satisfies our condition since the remainder meets the desired accuracy.

Key Concepts

Taylor series

The Taylor series is a powerful mathematical tool that allows us to approximate complex functions using polynomials. The basic idea is to express a function around a specific point, typically around zero, which is known as a Maclaurin series. For any function that can be infinitely differentiated, the Taylor series provides an approximation that gets closer to the actual function as more terms are added.
- It is represented by the infinite series: \[ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots \]
- The series uses derivatives, starting with the function value at point \(a\) and adding increasingly complex derivative terms.
- A special case is the Maclaurin series, where the function is centered at zero (\(a = 0\)), simplifying to: \[ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \cdots \]
In practice, you can truncate the series appropriately to achieve the desired level of approximation, balancing computation with accuracy.

remainder formula

When using the Taylor series for approximation, we often have a remainder or error term, which tells us how far off our polynomial approximation is from the actual function. It is crucial especially when accuracy is important, as it quantifies the difference between the true function value and our approximated value.
The remainder term is expressed as: \[ R_{n+1}(x) = \frac{f^{(n+1)}(c)}{(n+1)!} x^{n+1} \]
- Here, \(f^{(n+1)}(c)\) is the \((n+1)\)th derivative of the function evaluated at some point \(c\) within the interval between zero and \(x\).
- This formula is how we control for the approximation error. To minimize this error, we choose the degree \(n\) based on the required precision.
Understanding the remainder term is essential for knowing when your polynomial's approximation is "good enough".

trigonometric functions

Trigonometric functions like sine and cosine are fundamental in mathematics and have special behaviors under differentiation. When it comes to calculating their derivatives, they form a repeating cycle.
- For the sine function: - Derivative cycle: \( \sin(x) \rightarrow \cos(x) \rightarrow -\sin(x) \rightarrow -\cos(x) \rightarrow \sin(x) \) - For cosine, the cycle is similar but shifted.- The periodic nature of these functions' derivatives makes them particularly interesting when using series expansions like Taylor or Maclaurin.
In the context of the Maclaurin series for \(\sin x\), this repetitive nature simplifies the derivatives significantly, aiding in approximation calculations. This simplification is highly advantageous when calculating values over periodic intervals such as \([0, \pi/2]\).

approximation error

The approximation error in a Taylor or Maclaurin series tells you how accurate your polynomial approximation is for your function. It is given by the remainder formula and helps you understand the limitations of your approximation.
- The error magnitude depends on the function's derivatives and your choice of \(n\), the degree of the polynomial.- Smaller \(n\) might lead to larger errors, especially for complex functions.- In many cases, an odd or even value of \(n\) might be particularly suitable depending on the function's symmetry or specific behavior, like in our Maclaurin polynomial for sine where \(n\) must be odd.
Understanding approximation error helps in choosing a balance between polynomial simplicity and approximation accuracy, ensuring that your calculations remain within acceptable error margins. This is crucial in fields requiring precise computations, like engineering and physics, where exact values are often needed.