Problem 11
Question
In Exercises 11-14, prove that the Maclaurin series for the function converges to the function for all \(x\). $$ f(x)=\cos x $$
Step-by-Step Solution
Verified Answer
The Maclaurin series for \(f(x) = \cos x\) is: \[ f(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \ldots \] It converges for all \(x\) due to the Alternating Series Test.
1Step 1: Write down the Maclaurin Series
The Maclaurin series for a function \(f(x)\) is given by: \[ f(x) = f(0) + f'(0)x + f''(0)\frac{x^2}{2!} + f'''(0)\frac{x^3}{3!} + \ldots \]
2Step 2: Calculate the derivatives
For the function \(f(x) = \cos x\), calculate the first few derivatives at \(0\): \(f(0) = \cos 0 = 1\), \(f'(0) = -\sin 0 = 0\), \(f''(0) = -\cos 0 = -1\), \(f'''(0) = \sin 0 = 0\). Note that the pattern repeats every four terms.
3Step 3: Construct the Maclaurin series
Substitute these values into the Maclaurin series. This yields: \[ f(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \ldots \] This is the Maclaurin series for \(f(x) = \cos x\)
4Step 4: Prove convergence
The terms of the series become smaller and smaller as \(x\) increases, and alternate in sign. Therefore, this series converges for all \(x\), according to the Alternating Series Test
Key Concepts
ConvergenceCosine FunctionAlternating Series Test
Convergence
Understanding convergence is crucial when working with series, particularly in relation to the Maclaurin series. Convergence refers to the behavior of an infinite series as more terms are added. A series converges if the sum of its terms approaches a finite number as more terms are included.
One way to determine if a series converges is to examine the size of its terms and how they behave as the series progresses. If the terms get smaller and approach zero, there's a good chance the series will converge.
This is particularly important for functions like the cosine function, whose Maclaurin series creates an infinite sum.
One way to determine if a series converges is to examine the size of its terms and how they behave as the series progresses. If the terms get smaller and approach zero, there's a good chance the series will converge.
This is particularly important for functions like the cosine function, whose Maclaurin series creates an infinite sum.
- The Maclaurin series for cosine is represented by an alternating series, where terms switch between positive and negative.
- Each term gets smaller because it is divided by a factorial, like in \(\frac{x^2}{2!}\) and \(\frac{x^4}{4!}\). Factorials grow very quickly, making these terms shrink rapidly as well.
- The "Alternating Series Test" helps determine convergence for series with alternating signs.
Cosine Function
The cosine function, denoted as \(\cos x\), is a fundamental trigonometric function that appears frequently in mathematics. It describes the cosine of an angle in a right-angled triangle, representing the adjacent side over the hypotenuse.
More broadly, it reflects the horizontal component of a point moving around a circle in the unit circle model. This trigonometric function is periodic with a period of \(2\pi\), which means it repeats its values every \(2\pi\) units.
This is particularly useful for cosine, whose derivatives lead directly to a simple repeating series pattern, helping us approximate \(\cos x\) effectively, particularly near \(x = 0\).
More broadly, it reflects the horizontal component of a point moving around a circle in the unit circle model. This trigonometric function is periodic with a period of \(2\pi\), which means it repeats its values every \(2\pi\) units.
- One key property of the cosine function is its symmetry, making it an even function: \(\cos(-x) = \cos(x)\).
- It oscillates between -1 and 1, without ever exceeding these boundaries.
This is particularly useful for cosine, whose derivatives lead directly to a simple repeating series pattern, helping us approximate \(\cos x\) effectively, particularly near \(x = 0\).
Alternating Series Test
The Alternating Series Test is a method used to determine if an alternating series converges. An alternating series is one where the signs of the terms switch between positive and negative.
For example, the Maclaurin series for \(\cos x\), \(1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \ldots\), alternates in sign from term to term.For the Alternating Series Test to confirm that a series converges, two main conditions must be met:
By recognizing both these patterns and the decreasing nature of the terms, the Alternating Series Test confirms that the Maclaurin series for \(\cos x\) converges. This convergence efficiently approximates the actual \(\cos x\) value, as more terms create greater accuracy.
For example, the Maclaurin series for \(\cos x\), \(1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \ldots\), alternates in sign from term to term.For the Alternating Series Test to confirm that a series converges, two main conditions must be met:
- The absolute value of the terms must decrease steadily (each term is smaller than the one before it).
- The terms must approach zero as the series progresses. This means that as more terms are added, they contribute less and less to the total sum.
By recognizing both these patterns and the decreasing nature of the terms, the Alternating Series Test confirms that the Maclaurin series for \(\cos x\) converges. This convergence efficiently approximates the actual \(\cos x\) value, as more terms create greater accuracy.
Other exercises in this chapter
Problem 11
In Exercises \(7-28,\) find the interval of convergence of the power series. (Be sure to include a check for convergence at the endpoints of the interval.) $$ \
View solution Problem 11
In Exercises \(7-18\), find the Maclaurin polynomial of degree \(n\) for the function. $$ f(x)=\sin x, \quad n=5 $$
View solution Problem 11
Determine the convergence or divergence of the series. $$ \sum_{n=0}^{\infty} \frac{(-1)^{n}}{n !} $$
View solution Problem 11
Find a power series for the function, centered at \(c,\) and determine the interval of convergence. $$ g(x)=\frac{3 x}{x^{2}+x-2}, \quad c=0 $$
View solution