Chapter 9: Problem 44
Use a CAS to find the first four nonzero terms in the Maclaurin series for each of the following. Check Problems \(43-48\) to see that you get the same answers using the methods of Section \(9.7 .\) \(\exp \left(x^{2}\right)\)
Short Answer
Expert verified
The first four nonzero terms are 1, \(x^2\), and \(\frac{1}{2}x^4\).
Step by step solution
01
Understand the Maclaurin Series
The Maclaurin series for a function \( f(x) \) is given by \( f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots \). This series is an expansion of the function about 0.
02
Define the Function and Compute Derivatives
Our function is \( f(x) = e^{x^2} \). We need to find the first four derivatives and evaluate them at 0:- \( f(x) = e^{x^2} \)- \( f'(x) = 2xe^{x^2} \)- \( f''(x) = (4x^2 + 2)e^{x^2} \)- \( f'''(x) = (8x^3 + 12x)e^{x^2} \)- \( f^{(4)}(x) = (16x^4 + 48x^2 + 12)e^{x^2} \).
03
Evaluate Derivatives at x = 0
Substitute \( x = 0 \) into each derivative to find:- \( f(0) = e^{0^2} = 1 \)- \( f'(0) = 2(0)e^{0^2} = 0 \)- \( f''(0) = (4(0)^2 + 2)e^{0^2} = 2 \)- \( f'''(0) = (8(0)^3 + 12(0))e^{0^2} = 0 \)- \( f^{(4)}(0) = (16(0)^4 + 48(0)^2 + 12)e^{0^2} = 12 \).
04
Construct the Maclaurin Series
Using the derivatives, build the series:- The zeroth term is \( 1 \).- The first term is \( 0 \).- The second term is \( \frac{2}{2!}x^2 = x^2 \).- The third term is \( 0 \).- The fourth term is \( \frac{12}{4!}x^4 = \frac{1}{2}x^4 \).- Therefore, the Maclaurin series is \( 1 + x^2 + \frac{1}{2}x^4 + \cdots \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Taylor Series
The Taylor series is a foundational idea in calculus that allows us to represent complex functions with simpler polynomial forms. It expands a function at a point, providing an infinite sum of terms particularly useful in approximations. The general formula for the Taylor series of a function \( f(x) \) at a point \( a \) is given by:
- \( f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots \)
Calculus
Calculus, the mathematical study of change, is divided into two main branches: differential calculus and integral calculus. Differential calculus concerns the concept of the derivative, which is the measure of how a function changes as its input changes. This branch of calculus is essential for constructing Taylor series.
- It provides the tools needed to create derivatives, which are pivotal in determining the polynomial form in a Taylor series.
- Calculus aids in understanding the behavior and rate of change in functions, which directly influences how functions are represented in series.
Derivatives
Derivatives are a core concept in calculus, serving as the building blocks for functions like the Taylor series. They represent the rate of change of a quantity and are fundamental in understanding how a function behaves. Deriving the Maclaurin series requires calculating derivatives at a specific point (often zero).
- The first derivative, \( f'(x) \), represents the slope of the function at a point.
- The second derivative, \( f''(x) \), indicates the curvature of the function.
- Higher-order derivatives continue to reveal more about the function's behavior.
Mathematical Expansion
Mathematical expansion, like the Maclaurin series, allows a function to be expressed as an infinite sum of terms. This provides a simplified polynomial approximation of the function that is easier to work with. The expansion process breaks down complex functions into a series of simple, easy-to-calculate terms based on derivatives.
- In a Maclaurin series, which is a specific type of Taylor expansion centered at zero, each term is built using the derivatives of the function evaluated at zero.
- These expansions lead to series like \( 1 + x^2 + \frac{1}{2}x^4 + \cdots \), replacing the original function with an infinite sum of polynomial terms.