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Chapter 8: Problem 4

Find the first four terms of the Taylor series for the function \(\cos (x)\) about the point \(a=\) \(-\pi / 4\). (Your answers should include the variable \(\mathrm{x}\) when appropriate.) \(\cos (x)=\) \(+\ldots\)

Short Answer

Expert verified
The first four terms of the Taylor series for \(\cos(x)\) about \(a = -\pi/4\) are \[T(x) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}(x + \frac{\pi}{4}) - \frac{\sqrt{2}}{4}(x + \frac{\pi}{4})^2 - \frac{\sqrt{2}}{12}(x + \frac{\pi}{4})^3\]

Step by step solution

01

Identify the function and the point of expansion

The given function is \(\cos(x)\) and the point of expansion is \(a = -\pi / 4\). So \(f(x) = \cos(x)\) and \(a = -\pi / 4\).
02

Find the function's derivatives

The first few derivatives of \(\cos(x)\) are:\[f'(x) = -\sin(x)\]\[f''(x) = -\cos(x)\]\[f'''(x) = \sin(x)\]\[f''''(x) = \cos(x)\]
03

Evaluate the derivatives at the point \(a = -\pi / 4\)

Next, evaluate each derivative at \(x = -\pi / 4\):\[f(-\pi / 4) = \cos(-\pi / 4) = \frac{\sqrt{2}}{2}\]\[f'(-\pi / 4) = -\sin(-\pi / 4) = \frac{\sqrt{2}}{2}\]\[f''(-\pi / 4) = -\cos(-\pi / 4) = -\frac{\sqrt{2}}{2}\]\[f'''(-\pi / 4) = \sin(-\pi / 4) = -\frac{\sqrt{2}}{2}\]
04

Construct the Taylor series

Using the Taylor series formula \[T(x) = f(a) + \frac{f'(a)}{1!} (x - a) + \frac{f''(a)}{2!} (x - a)^2 + \frac{f'''(a)}{3!} (x - a)^3 + \ldots\]substitute \(-\pi / 4\) for \(a\) and the evaluated derivatives to get:\[T(x) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}(x + \frac{\pi}{4}) - \frac{\sqrt{2}}{4}(x + \frac{\pi}{4})^2 - \frac{\sqrt{2}}{12}(x + \frac{\pi}{4})^3\]

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Key Concepts

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

Trigonometric Functions
Trigonometric functions are a cornerstone in mathematics, especially in the study of periodic phenomena such as waves and circular motion. Functions like sine, cosine, and tangent relate the angles of a triangle to the lengths of its sides. The cosine function, denoted as \(\text{cos}(x)\), is particularly vital. It tells us the horizontal position of a point on the unit circle as it moves around. For a given angle \(x\), \(\text{cos}(x)\) provides the x-coordinate of the corresponding point on the circle. Beyond geometry, trigonometric functions have applications in physics, engineering, and computer science.
Understanding the behavior of these functions is crucial for solving many real-world problems.
Derivatives
Derivatives measure how a function changes as its input changes. In simpler terms, they tell us the slope of the function at any given point. For the cosine function, its derivatives reveal a cyclic pattern:
  • The first derivative of \(\text{cos}(x)\) is \(-\text{sin}(x)\).
  • The second derivative brings us back to \(-\text{cos}(x)\).
  • As we calculate higher-order derivatives, we alternate between sine and cosine functions. This recursive nature allows us to predict subsequent derivatives easily.
  • The third derivative is \( \text{sin}(x) \).
  • The fourth derivative circles back to \( \text{cos}(x) \).
    Using derivatives is essential in creating Taylor series and understanding how functions behave near a point.
Series Expansion
Series expansion lets us express complex functions as infinite sums of simpler terms. With Taylor series, we can represent functions as power series around a particular point, called the expansion point. For the cosine function, using Taylor series helps us approximate its values near any point we choose. The Taylor series for \(\text{cos}(x)\) around a point \(a\) is given by:
\[\text{T}(x) = f(a) + \frac{f'(a)}{1!} (x - a) + \frac{f''(a)}{2!} (x - a)^2 + \frac{f'''(a)}{3!} (x - a)^3 + \text{...} \].

This formula uses the function's derivatives at the expansion point. Calculating the first few terms of this series gives a good approximation of the function near that point. The more terms we include, the more accurate our approximation becomes. This is incredibly useful in physics and engineering, where exact solutions can be complex or impossible to find.
Cosine Function
The cosine function, denoted as \(\text{cos}(x)\), is a fundamental trigonometric function. It is periodic with a period of \ (2\text{蟺}) \, meaning \(\text{cos}(x + 2\text{蟺}) = \text{cos}(x)\).
The Taylor series for \(\text{cos}(x)\) at a point \(a\) uses its derivatives evaluated at that point. In our example, we expanded around \(a = -\text{蟺}/4\).
Calculating the derivatives of \(\text{cos}(x)\) at \(a = -\text{蟺}/4\):
  • \(f(-\text{蟺}/4) = \text{cos}(-\text{蟺}/4) = \frac{\text{鈭殅2}{2}\)
  • \(f'(-\text{蟺}/4) = -\text{sin}(-\text{蟺}/4) = \frac{\text{鈭殅2}{2}\)
  • \(f''(-\text{蟺}/4) = -\text{cos}(-\text{蟺}/4) = -\frac{\text{鈭殅2}{2}\)
  • \(f'''(-\text{蟺}/4) = \text{sin}(-\text{蟺}/4) = -\frac{\text{鈭殅2}{2}\).
Substituting these values into the Taylor series formula gives us an excellent approximation of the cosine function near \(a = -\text{蟺}/4\).

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Most popular questions from this chapter

Given: \(A_{n}=\frac{80}{8^{n}}\) Determine: (a) whether \(\sum_{n=1}^{\infty}\left(A_{n}\right)\) is convergent. (b) whether \(\left\\{A_{n}\right\\}\) is convergent. If convergent, enter the limit of convergence. If not, enter DIV.

In the Limit Comparison Test we compared the behavior of a series to one whose behavior we know. In that test we use the limit of the ratio of corresponding terms of the series to determine if the comparison is valid. In this exercise we see how we can compare two series directly, term by term, without using a limit of sequence. First we consider an example. a. Consider the series \(\sum \frac{1}{k^{2}}\) and \(\sum \frac{1}{k^{2}+k}\) We know that the series \(\sum \frac{1}{k^{2}}\) is a \(p\) -series with \(p=2>1\) and so \(\sum \frac{1}{k^{2}}\) converges. In this part of the exercise we will see how to use information about \(\sum \frac{1}{k^{2}}\) to determine information about \(\sum \frac{1}{k^{2}+k} .\) Let \(a_{k}=\frac{1}{k^{2}}\) and \(b_{k}=\frac{1}{k^{2}+k} .\) i) Let \(S_{n}\) be the \(n\) th partial sum of \(\sum \frac{1}{k^{2}}\) and \(T_{n}\) the \(n\) th partial sum of \(\sum \frac{1}{k^{2}+k}\). Which is larger, \(S_{1}\) or \(T_{1}\) ? Why? ii) Recall that $$ S_{2}=S_{1}+a_{2} \text { and } T_{2}=T_{1}+b_{2} $$ Which is larger, \(a_{2}\) or \(b_{2}\) ? Based on that answer, which is larger, \(S_{2}\) or \(T_{2}\) ? iii) Recall that $$ S_{3}=S_{2}+a_{3} \text { and } T_{3}=T_{2}+b_{3} $$ Which is larger, \(a_{3}\) or \(b_{3}\) ? Based on that answer, which is larger, \(S_{3}\) or \(T_{3}\) ? iv) Which is larger, \(a_{n}\) or \(b_{n}\) ? Explain. Based on that answer, which is larger, \(S_{n}\) or \(T_{n}\) ? v) Based on your response to the previous part of this exercise, what relationship do you expect there to be between \(\sum \frac{1}{k^{2}}\) and \(\sum \frac{1}{k^{2}+k} ?\) Do you expect \(\sum \frac{1}{k^{2}+k}\) to converge or diverge? Why? b. The example in the previous part of this exercise illustrates a more general result. Explain why the Direct Comparison Test, stated here, works.

In this exercise, we examine one of the conditions of the Alternating Series Test. Consider the alternating series $$ 1-1+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{9}+\frac{1}{4}-\frac{1}{16}+\cdots $$ where the terms are selected alternately from the sequences \(\left\\{\frac{1}{n}\right\\}\) and \(\left\\{-\frac{1}{n^{2}}\right\\}\). a. Explain why the \(n\) th term of the given series converges to 0 as \(n\) goes to infinity. b. Rewrite the given series by grouping terms in the following manner: $$ (1-1)+\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{3}-\frac{1}{9}\right)+\left(\frac{1}{4}-\frac{1}{16}\right)+\cdots $$ Use this regrouping to determine if the series converges or diverges. c. Explain why the condition that the sequence \(\left\\{a_{n}\right\\}\) decreases to a limit of 0 is included in the Alternating Series Test.

Determine the sum of the following series. $$ \sum_{n=1}^{\infty}\left(\frac{3^{n}+5^{n}}{9^{n}}\right) $$

Suppose you play a game with a friend that involves rolling a standard six- sided die. Before a player can participate in the game, he or she must roll a six with the die. Assume that you roll first and that you and your friend take alternate rolls. In this exercise we will determine the probability that you roll the first six. a. Explain why the probability of rolling a six on any single roll (including your first turn) is \(\frac{1}{6}\) b. If you don't roll a six on your first turn, then in order for you to roll the first six on your second turn, both you and your friend had to fail to roll a six on your first turns, and then you had to succeed in rolling a six on your second turn. Explain why the probability of this event is $$ \left(\frac{5}{6}\right)\left(\frac{5}{6}\right)\left(\frac{1}{6}\right)=\left(\frac{5}{6}\right)^{2}\left(\frac{1}{6}\right) $$ c. Now suppose you fail to roll the first six on your second turn. Explain why the probability is $$ \left(\frac{5}{6}\right)\left(\frac{5}{6}\right)\left(\frac{5}{6}\right)\left(\frac{5}{6}\right)\left(\frac{1}{6}\right)=\left(\frac{5}{6}\right)^{4}\left(\frac{1}{6}\right) $$ that you to roll the first six on your third turn. d. The probability of you rolling the first six is the probability that you roll the first six on your first turn plus the probability that you roll the first six on your second turn plus the probability that your roll the first six on your third turn, and so on. Explain why this probability is $$ \frac{1}{6}+\left(\frac{5}{6}\right)^{2}\left(\frac{1}{6}\right)+\left(\frac{5}{6}\right)^{4}\left(\frac{1}{6}\right)+\cdots $$ Find the sum of this series and determine the probability that you roll the first six.
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