Problem 18
Question
In Exercises \(7-18\), find the Maclaurin polynomial of degree \(n\) for the function. $$ f(x)=\tan x, \quad n=3 $$
Step-by-Step Solution
Verified Answer
The Maclaurin polynomial of degree 3 for the function \( f(x) = \tan x \) is \( x + \frac{x^3}{3} \).
1Step 1: Compute the function value and derivative at \( x = 0 \)
Compute the function value at \( x = 0 \): \( f(0) = \tan 0 = 0 \). Calculate the first derivative of \( f(x) = \tan x \): \( f'(x) = \sec^2 x \), and then evaluate at \( x = 0 \): \( f'(0) = \sec^2 0 = 1 \).
2Step 2: Compute the second derivative
The second derivative of \( f(x) \) is given by the derivative of \( f'(x) \). Thus, \( f''(x) = 2\tan x \cdot \sec^2 x \) and when evaluated at \( x = 0 \), we get \( f''(0) = 0 \).
3Step 3: Compute the third derivative
The third derivative of \( f(x) \) is then \( f'''(x) = 2\sec^2 x + 4\tan^2 x \sec^2 x \). Evaluate this at \( x = 0 \) to get \( f'''(0) = 2 \).
4Step 4: Construct the Maclaurin Polynomial
Combine these results to form the Maclaurin polynomial of degree 3: \( P_3(x) = f(0) + f'(0)x + \frac{f''(0)x^2}{2!} + \frac{f'''(0)x^3}{3!} = 0 + 1x + 0x^2 + \frac{2x^3}{6} = x + \frac{x^3}{3} \).
Key Concepts
CalculusDerivativeTaylor Series Expansion
Calculus
Calculus is a branch of mathematics that studies continuous change. It provides a framework to model systems that change and helps in calculating various important quantities such as areas, volumes, and critical points. In the context of Maclaurin polynomials, calculus allows us to approximate functions using a series expansion.
A Maclaurin polynomial is a type of Taylor polynomial centered at zero. It is specifically useful when evaluating a function or its derivatives at zero. The procedure involves taking successive derivatives of a function, evaluating them at zero, and then constructing a polynomial based on these values.
In this exercise, we are constructing a Maclaurin polynomial for the function \( f(x) = \tan x \) up to degree 3. Calculus provides the tools to find the necessary derivatives, which are crucial as they determine the coefficients of the polynomial terms.
A Maclaurin polynomial is a type of Taylor polynomial centered at zero. It is specifically useful when evaluating a function or its derivatives at zero. The procedure involves taking successive derivatives of a function, evaluating them at zero, and then constructing a polynomial based on these values.
In this exercise, we are constructing a Maclaurin polynomial for the function \( f(x) = \tan x \) up to degree 3. Calculus provides the tools to find the necessary derivatives, which are crucial as they determine the coefficients of the polynomial terms.
Derivative
In calculus, a derivative represents the rate at which a function is changing at any given point. It is a fundamental concept for understanding the behavior of functions. Calculating derivatives allows us to derive formulas that describe how functions change and predict values.
To construct a Maclaurin polynomial, it is essential to calculate derivatives systematically. For example, for \( f(x) = \tan x \), we use the fact that the first derivative is \( f'(x) = \sec^2 x \), which simplifies to 1 when \( x = 0 \). Second and third derivatives are computed by taking higher-order derivatives and evaluating them at zero to find \( f''(0) \) and \( f'''(0) \), respectively.
Each derivative provides the next term in the Taylor series expansion. Understanding how to calculate derivatives and their roles in constructing the polynomial is crucial for solving such problems efficiently.
To construct a Maclaurin polynomial, it is essential to calculate derivatives systematically. For example, for \( f(x) = \tan x \), we use the fact that the first derivative is \( f'(x) = \sec^2 x \), which simplifies to 1 when \( x = 0 \). Second and third derivatives are computed by taking higher-order derivatives and evaluating them at zero to find \( f''(0) \) and \( f'''(0) \), respectively.
Each derivative provides the next term in the Taylor series expansion. Understanding how to calculate derivatives and their roles in constructing the polynomial is crucial for solving such problems efficiently.
Taylor Series Expansion
The Taylor Series Expansion is a way of representing a function as an infinite sum of its derivatives at a particular point. The Maclaurin series is a special case of the Taylor Series, where the expansion is taken around zero. This type of series allows us to approximate complex functions using simpler polynomial forms.
The expansion provides a simplified way to handle functions using only a few terms, which is especially helpful in computational applications or when handling functions that are otherwise complicated to compute directly. By understanding how to form a Maclaurin series, you gain insight into the behavior of functions and the power of polynomial approximations.
- Start with the function and evaluate it at the expansion point.
- Take the first derivative, evaluate, and use it to find the linear term.
- Continue with higher derivatives for more terms.
The expansion provides a simplified way to handle functions using only a few terms, which is especially helpful in computational applications or when handling functions that are otherwise complicated to compute directly. By understanding how to form a Maclaurin series, you gain insight into the behavior of functions and the power of polynomial approximations.
Other exercises in this chapter
Problem 18
Write the next two apparent terms of the sequence. Describe the pattern you used to find these terms. \(1,-\frac{1}{2}, \frac{1}{4},-\frac{1}{8}, \ldots\)
View solution Problem 18
Explain why the Integral Test does not apply to the series. $$ \sum_{n=1}^{\infty}\left(\frac{\sin n}{n}\right)^{2} $$
View solution Problem 18
Use the binomial series to find the Maclaurin series for the function. $$ f(x)=\sqrt[4]{1+x} $$
View solution Problem 18
Approximate the sum of the series by using the first six terms. (See Example 4.) $$ \sum_{n=1}^{\infty} \frac{(-1)^{n+1} 4}{\ln (n+1)} $$
View solution