Chapter 6: Problem 145
Find the Taylor series of the given function centered at the indicated point. $$ \cos x \text { at } x=\frac{\pi}{2} $$
Short Answer
Expert verified
The Taylor series for \( \cos x \) at \( x = \frac{\pi}{2} \) is \(- (x-\frac{\pi}{2}) + \frac{(x-\frac{\pi}{2})^3}{3!} - \frac{(x-\frac{\pi}{2})^5}{5!} + \cdots\)
Step by step solution
01
Identify the Function and Center
We need to find the Taylor series for the function \( f(x) = \cos x \) centered at \( x = \frac{\pi}{2} \). This means we need the derivatives of \( \cos x \) evaluated at \( x = \frac{\pi}{2} \).
02
Compute the First Few Derivatives
The derivatives of \( \cos x \) are:- \( f(x) = \cos x \)- \( f'(x) = -\sin x \)- \( f''(x) = -\cos x \)- \( f'''(x) = \sin x \)- \( f^{(4)}(x) = \cos x \)These derivatives repeat every four terms.
03
Evaluate the Derivatives at the Center
Calculate the function and its derivatives at \( x = \frac{\pi}{2} \):- \( f\left(\frac{\pi}{2}\right) = \cos\left(\frac{\pi}{2}\right) = 0 \)- \( f'\left(\frac{\pi}{2}\right) = -\sin\left(\frac{\pi}{2}\right) = -1 \)- \( f''\left(\frac{\pi}{2}\right) = -\cos\left(\frac{\pi}{2}\right) = 0 \)- \( f'''\left(\frac{\pi}{2}\right) = \sin\left(\frac{\pi}{2}\right) = 1 \)- \( f^{(4)}(\frac{\pi}{2}) = \cos\left(\frac{\pi}{2}\right) = 0 \)Repeat the derivatives as necessary.
04
Construct the Taylor Series
The Taylor series formula is:\[T(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x-a)^n\]For \( f(x) = \cos x \) at \( a = \frac{\pi}{2} \):\[T(x) = 0 \cdot (x-\frac{\pi}{2})^0 + (-1) \cdot \frac{(x-\frac{\pi}{2})^1}{1!} + 0 \cdot \frac{(x-\frac{\pi}{2})^2}{2!} + 1 \cdot \frac{(x-\frac{\pi}{2})^3}{3!} + \cdots\]Establishing the pattern, we get:\[T(x) = - (x-\frac{\pi}{2}) + \frac{(x-\frac{\pi}{2})^3}{3!} - \frac{(x-\frac{\pi}{2})^5}{5!} + \cdots\]The terms where the derivatives are zero do not appear in the series.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
The Cosine Function
The cosine function, represented as \( \cos x \), is one of the fundamental trigonometric functions. It describes the horizontal coordinate of a point on the unit circle as you move around it. More intuitively, it tells us how far "right" or "left" a point lies on the circle as an angle \( x \) changes. The cosine function has the key property of being periodic with a period of \( 2\pi \), meaning every \( 2\pi \), its values repeat. Some important characteristics of \( \cos x \):
- It is an even function: \( \cos(-x) = \cos(x) \).
- It has a range of \(-1\) to \(1\), and the graph oscillates within this range.
- It is continuous and differentiable everywhere on the real line.
Derivatives of Cosine Function
When expanding functions such as \( \cos x \) using a Taylor series, understanding derivatives is crucial. The derivative of a function reflects its rate of change. For \( \cos x \), the sequence of derivatives forms a repeating pattern every four calculations:
- \( f(x) = \cos x \)
- \( f'(x) = -\sin x \)
- \( f''(x) = -\cos x \)
- \( f'''(x) = \sin x \)
- \( f^{(4)}(x) = \cos x \)
Series Expansion
Series expansion allows us to express functions as sums of simpler terms. The Taylor series is a powerful form of series expansion that expresses any function as an infinite sum based on its derivatives at a single point. The general formula is:\[T(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x-a)^n\]For \( f(x) = \cos x \) centered at \( a = \frac{\pi}{2} \), it can be written as:\[T(x) = 0 \cdot (x-\frac{\pi}{2})^0 + (-1) \cdot \frac{(x-\frac{\pi}{2})^1}{1!} + 0 \cdot \frac{(x-\frac{\pi}{2})^2}{2!} + 1 \cdot \frac{(x-\frac{\pi}{2})^3}{3!} + \cdots\]From the pattern of derivatives, some terms vanish because their coefficients are zero. This leaves us with a pattern:\[T(x) = - (x-\frac{\pi}{2}) + \frac{(x-\frac{\pi}{2})^3}{3!} - \frac{(x-\frac{\pi}{2})^5}{5!} + \cdots\]This pattern highlights how only odd powers contribute to the expansion when centered at \( \frac{\pi}{2} \).
Centered at a Point
A Taylor series is centered at a specific point \( x = a \). This point is where the function and its derivatives are evaluated. By centering the series at this point, we create a polynomial that closely approximates the original function near \( a \). In this exercise, the Taylor series is centered at \( x = \frac{\pi}{2} \).Evaluating derivatives at this point:
- \( \cos\left(\frac{\pi}{2}\right) = 0 \)
- \( -\sin\left(\frac{\pi}{2}\right) = -1 \)
- \( -\cos\left(\frac{\pi}{2}\right) = 0 \)
- \( \sin\left(\frac{\pi}{2}\right) = 1 \)
- \( \cos\left(\frac{\pi}{2}\right) = 0 \)