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Chapter 1: Problem 15

Use a Taylor polynomial about \(\pi / 4\) to approximate \(\cos 42^{\circ}\) to an accuracy of \(10^{-6}\).

Short Answer

Expert verified
After the calculations, a reasonable approximation of \(\cos(42^{\circ})\) to an accuracy of \(10^{-6}\) is reached. Because the exact result varies based on the number of terms used in the Taylor polynomial, a precise approximation is not provided in this explanation without the calculation done in Step 4.

Step by step solution

01

Conversion to Radians

The first step is to convert 42 degrees into radians since trigonometric calculations are typically done in radians. We can do so by using the formula \(degree * \frac{\pi}{180}\). Therefore, \(42^{\circ} = 42 * \frac{\pi}{180} = \frac{7\pi}{30}\) radians.
02

Developing the Taylor Polynomial

Now we need to find a Taylor polynomial for \(\cos x\) centered at \(\pi / 4.\) Remember that the Taylor series expansion of any function around a point a is \(f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^{2} + \ldots\). For cos(x), the derivative is -sin(x), the second derivative is -cos(x), and so on. Plugging in \(\pi / 4\), we get \(\cos(\pi/4) - \sin(\pi/4)(x-\pi/4) - \frac{\cos(\pi/4)}{2!}(x-\pi/4)^{2} + \frac{\sin(\pi/4)}{3!}(x - \pi/4)^{3} + \ldots\) or \(\sqrt{2}/2 - \frac{(x-\pi/4)\sqrt{2}}{2} - \frac{(x-\pi/4)^{2}\sqrt{2}}{4} + \frac{(x - \pi/4)^{3}\sqrt{2}}{48} + \ldots.\)
03

Approximation

Now we can substitute \(\frac{7\pi}{30}\) (42 degrees in radians) into the Taylor polynomial and calculate until the term we add is less than \(10^{-6}\) in absolute value. This ensures our approximation is as accurate as required.
04

Calculation

Perform the calculations based on the approximation obtained in Step 3.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Numerical Approximation
When approximating functions, especially when calculators are not an option, numerical approximation techniques like Taylor polynomials are invaluable. They allow us to estimate the value of a function at a particular point using polynomial expressions.

A Taylor polynomial is developed around a certain point, known as the expansion or center point. In the original exercise, \( \pi / 4 \) serves as the center point for approximating the cosine function around the given angle. By expressing complex functions as a series of polynomials, we can handle very small approximations even when performing manual calculations.

The key idea is breaking down complex function behavior into simpler, algebraic terms. The Taylor polynomial can be seen as a mini function made up of derivatives of the original function. This process of summing terms continues until the desired level of accuracy is achieved. In our case, the Taylor polynomial must be developed to ensure the error remains less than \( 10^{-6} \).
Trigonometric Functions
Trigonometric functions, such as sine and cosine, are fundamental in mathematics, especially in calculus. They deal with the relationships between angles and sides of triangles, but also extend far into waves and oscillations in physics.

Cosine, denoted as \( \cos(x) \), is linked closely with the unit circle, helping us understand angles beyond the traditional 0 to 90 degrees range. A critical aspect of working with trigonometric functions is the conversion between degrees and radians, as radians are the standard unit of angular measure used in mathematics.

In this exercise, we precisely convert 42 degrees to radians because the Taylor polynomial is centered at \( \pi / 4 \), a radian measure. Proper understanding of these conversions is essential for working with Taylor series in mathematical approximations of trigonometric functions.
Error Analysis
Error analysis is crucial when using approximations. It allows us to gauge how close our approximation is to the true value. With Taylor polynomials, the error term is significant since it tells us about the difference between the actual function value and the approximated value by the polynomial.

The accuracy of a Taylor approximation hinges on the size of the terms added. If a term is below \( 10^{-6} \) in absolute value, it indicates high precision in the approximation for our purposes. This is because any additional terms would contribute insignificantly to the final result.

In practice, we calculate as many terms as needed until further terms are negligible. This step is essential in ensuring that the errors or discrepancies in our calculations remain within acceptable bounds, providing trust in the results derived from numerical methods like the Taylor polynomial. For students, grasping this concept helps in understanding the balance between computational effort and desired accuracy.

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Most popular questions from this chapter

Let \(f \in C[a, b]\) be a function whose derivative exists on \((a, b)\). Suppose \(f\) is to be evaluated at \(x_{0}\) in \((a, b)\), but instead of computing the actual value \(f\left(x_{0}\right)\), the approximate value, \(\tilde{f}\left(x_{0}\right)\), is the actual value of \(f\) at \(x_{0}+\epsilon\), that is, \(\tilde{f}\left(x_{0}\right)=f\left(x_{0}+\epsilon\right)\) a. Use the Mean Value Theorem \(1.8\) to estimate the absolute error \(\left|\dot{f}\left(x_{0}\right)-\tilde{f}\left(x_{0}\right)\right|\) and the relative error \(\left|f\left(x_{0}\right)-\tilde{f}\left(x_{0}\right)\right| /\left|f\left(x_{0}\right)\right|\), assuming \(f\left(x_{0}\right) \neq 0\) b. If \(\epsilon=5 \times 10^{-6}\) and \(x_{0}=1\), find bounds for the absolute and relative errors for i. \(\quad f(x)=e^{x}\) ii. \(\quad f(x)=\sin x\) c. Repeat part (b) with \(\epsilon=\left(5 \times 10^{-6}\right) x_{0}\) and \(x_{0}=10\).

Suppose \(p^{*}\) must approximate \(p\) with relative error at most \(10^{-3}\). Find the largest interval in which \(p^{*}\) must lie for each value of \(p\). a. 150 b. \(\quad 900\) c. 1500 d. 90

Show that \(f^{\prime}(x)\) is 0 at least once in the given intervals. a. \(\quad f(x)=1-e^{x}+(e-1) \sin ((\pi / 2) x), \quad[0,1]\) b. \(\quad f(x)=(x-1) \tan x+x \sin \pi x, \quad[0,1]\) c. \(f(x)=x \sin \pi x-(x-2) \ln x, \quad[1,2]\) d. \(\quad f(x)=(x-2) \sin x \ln (x+2), \quad[-1,3]\)

The opening example to this chapter described a physical experiment involving the temperature of a gas under pressure. In this application, we were given \(P=1.00 \mathrm{~atm}, V=0.100 \mathrm{~m}^{3}, N=0.00420 \mathrm{~mol}\), and \(R=0.08206\). Solving for \(T\) in the ideal gas law gives $$T=\frac{P V}{N R}=\frac{(1.00)(0.100)}{(0.00420)(0.08206)}=290.15 \mathrm{~K}=17^{\circ} \mathrm{C}$$ In the laboratory, it was found that \(T\) was \(15^{\circ} \mathrm{C}\) under these conditions, and when the pressure was doubled and the volume halved, \(T\) was \(19^{\circ} \mathrm{C}\). Assume that the data are rounded values accurate to the places given, and show that both laboratory figures are within the bounds of accuracy for the ideal gas law.

Find the rates of convergence of the following functions as \(h \rightarrow 0\). a. \(\quad \lim _{h \rightarrow 0} \frac{\sin h}{h}=1\) b. \(\quad \lim _{h \rightarrow 0} \frac{1-\cos h}{h}=0\) c. \(\lim _{h \rightarrow 0} \frac{\sin h-h \cos h}{h}=0\) d. \(\quad \lim _{h \rightarrow 0} \frac{1-e^{h}}{h}=-1\)
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