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Chapter 12: Problem 32

Use Taylor polynomials to estimate the following within 0.01. $$\cos 6^{\circ}$$

Short Answer

Expert verified
We can estimate \(\cos(6^{\circ})\) within 0.01 using Taylor polynomials by first converting the angle to radians, \(6^{\circ} = \frac{\pi}{30}\) radians, and then approximating the value using the Taylor series expansion for the cosine function up to the fourth-degree term: \(\cos(\frac{\pi}{30}) \approx 1 - \frac{(\frac{\pi}{30})^2}{2!} + \frac{(\frac{\pi}{30})^4}{4!}\) which yields an estimate of approximately 0.9945.

Step by step solution

01

Convert Angle to Radians

We need to convert 6 degrees to radians. Recall that \(1^{\circ} = \frac{\pi}{180} \text{ radians}\), so: $$ 6^{\circ} = 6 \cdot \frac{\pi}{180} = \frac{\pi}{30} \text{ radians} $$
02

Recall the Taylor Series Expansion for Cosine Function

The Taylor series expansion for the cosine function around 0 is given by: $$ \cos(x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!} x^{2n} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \dots $$
03

Determine the Number of Terms Needed for the Given Tolerance

We are asked to approximate \(\cos(6^{\circ}) = \cos(\frac{\pi}{30})\) within 0.01. To do this, we will find the degree of the Taylor polynomial needed to achieve the given tolerance. We know that the error in the Taylor polynomial approximation is given by the next term in the series. To achieve an error less than 0.01, we'll find the smallest n such that: $$ \frac{(\frac{\pi}{30})^{2n+2}}{(2n+2)!} < 0.01 $$ By trying different n values, we find that n = 2 satisfies the inequality: $$ \frac{(\frac{\pi}{30})^6}{6!} \approx 0.000821 < 0.01 $$
04

Approximate the Value of the Cosine Function

Using the Taylor series expansion with n = 2, we can approximate \(\cos(\frac{\pi}{30})\) as: $$ \cos(\frac{\pi}{30}) \approx 1 - \frac{(\frac{\pi}{30})^2}{2!} + \frac{(\frac{\pi}{30})^4}{4!} $$ Plug in the values: $$ \cos(\frac{\pi}{30}) \approx 1 - \frac{(\frac{\pi}{30})^2}{2!} + \frac{(\frac{\pi}{30})^4}{4!} \approx 0.9945 $$ Therefore, \(\cos(6^{\circ})\) can be estimated to be approximately 0.9945 within an error of 0.01.

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Key Concepts

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

Cosine Function
The cosine function, often simply called "cosine," is a fundamental trigonometric function, represented as \(\cos(x)\). It is primarily used to relate the angles and sides of right-angled triangles. The function is periodic and even, which means it repeats every \(2\pi\) radians, and its graph is symmetric about the y-axis.
  • Properties: Defined for all real numbers; ranges from -1 to 1.
  • Applications: Used in signal processing, physics for wave equations, and engineering for alternating current circuits.
  • Key Identity: \(\cos^2(x) + \sin^2(x) = 1\).
The cosine function is particularly useful for approximations through its Taylor series expansion, enabling us to calculate the value of cosine for angles that aren鈥檛 standard, such as 6 degrees.
Angle Conversion to Radians
Converting angles from degrees to radians is crucial in mathematics, especially when dealing with trigonometric functions and calculus. Radians provide a natural way of expressing angles, as they relate directly to the arc length of circles.
  • Conversion Formula: To convert degrees to radians, use the formula: \[\text{radians} = \text{degrees} \times \frac{\pi}{180}\]
  • Why Radians?: In calculus, radians are often preferred because they simplify derivative and integral calculations. For example, the derivative of \(\sin(x)\) in radians is \(\cos(x)\), a far simpler result than other units might provide.
For example, converting 6 degrees into radians using the formula results in \(\frac{\pi}{30}\), which enables easier calculations when approximating the cosine of this angle.
Taylor Series Expansion
Taylor series expansion is a mathematical method that allows us to express a function as an infinite sum of terms calculated from the values of its derivatives at a single point. This is especially useful for approximating functions that are complex to calculate directly.
  • General Formula: For a function \(f(x)\) expanded about \(x = 0\) (Maclaurin series),\[f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n\]
  • Cosine Series: For the cosine function, the expansion is:\[\cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \ldots\]This allows us to calculate \(\cos(6^{\circ})\) by substituting \(x = \frac{\pi}{30}\).
Taylor series provide a powerful tool to approximate functions with desired accuracy by retaining a limited number of terms.
Approximation Error
When using Taylor polynomials to estimate the value of a function, an important consideration is the approximation error, which represents the difference between the actual value and the approximation.
  • Definition: The approximation error at a given point can be understood as the magnitude of the next term in the Taylor series that wasn't included in the sum.
  • Error Bound: For our example of approximating \(\cos(6^{\circ})\), the error is found by ensuring the next term of the Taylor polynomial is less than 0.01, calculated as:\[\frac{(\frac{\pi}{30})^6}{6!} \approx 0.000821\]
  • Managing Error: Choosing sufficient terms until the error term is smaller than the desired tolerance helps keep the approximation precise.
Thus, achieving the correct approximation depends on understanding and applying the concept of approximation error effectively.

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Most popular questions from this chapter

(a) Prove that if \(\sum_{k=0}^{\infty} a_{k}\) is a convergent series with all terms nonzero, then \(\sum_{x=0}^{\infty}\left(1 / a_{k}\right)\) diverges. (b) Suppose that \(a_{k}>0\) for all \(k\) and \(\sum_{k=1}^{\infty} a_{k}\) diverges. Show by example that \(\sum_{k=0}^{\infty}\left(1 / a_{k}\right)\) may converge and it may diverge.

Find a series expansion for the expression. $$\frac{x}{1+4 x^{2}} \text { for }|x|<\frac{1}{2}$$

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Derive a series expansion in \(x\) for the function and specify the numbers \(x\) for which the expansion is valid. Take \(a>0\). $$f(x)=e^{2 x}$$
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