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: \( 1 + x^2 + \frac{1}{2}x^4 \).
Step by step solution
01
Understand the Maclaurin Series Definition
The Maclaurin series for a function is given by the formula \[ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots \].We need to find the derivatives of \( f(x) = e^{x^2} \) evaluated at \( x = 0 \) for the first few non-zero terms.
02
Calculate the Derivatives of the Function
Let's compute the necessary derivatives:- \( f(x) = e^{x^2} \)- \( f'(x) = 2xe^{x^2} \)- \( f''(x) = (4x^2 + 2)e^{x^2} \)- \( f'''(x) = (8x^3 + 12x)e^{x^2} \)- \( f^{(4)}(x) = (16x^4 + 48x^2 + 12)e^{x^2} \)
03
Evaluate Derivatives at x=0
Evaluate the derivatives at \( x = 0 \):- \( f(0) = e^{0} = 1 \)- \( f'(0) = 2 \cdot 0 = 0 \)- \( f''(0) = 2 \cdot e^{0} = 2 \)- \( f'''(0) = 0 \)- \( f^{(4)}(0) = 12 \cdot e^{0} = 12 \)
04
Construct the Maclaurin Series
Using the derivatives evaluated at \( x = 0 \), the first four nonzero terms of the Maclaurin series for \( e^{x^2} \) are:- The constant term: \( f(0) = 1 \)- Second term: \( \frac{f''(0)}{2!}x^2 = \frac{2}{2}x^2 = x^2 \)- Third term (zero, hence skipped)- Fourth term: \( \frac{f^{(4)}(0)}{4!}x^4 = \frac{12}{24}x^4 = \frac{1}{2}x^4 \)Thus, the series is \( 1 + x^2 + \frac{1}{2}x^4 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Derivative Evaluation
When studying functions and their approximations, derivative evaluation is essential. A derivative represents how a function changes as its input changes. In simple terms, it is the rate of change or the slope of the function at a particular point. When building a Maclaurin series, we rely on the derivatives of the function evaluated at the origin (i.e., when the input is zero) to construct terms in the series.
Here's how it helps:
Here's how it helps:
- First, we calculate the derivatives of the function one by one.
- We then substitute zero into each derivative (because we use Maclaurin, which is a special case of Taylor series centered at zero) to get the specific coefficients.
- Each coefficient derived from these derivatives will contribute to the term in the series.
Exponential Functions
Exponential functions are a special category of mathematical functions where a constant base is raised to a variable exponent. They appear frequently in mathematics due to their unique properties:
- One interesting property is that the derivative of an exponential function is proportional to the value of the function itself.
- This makes them perfect for modeling growth processes like population growth and radioactive decay.
- For our function \( f(x) = e^{x^2} \), the function itself and all its derivatives contain the exponential term \( e^{x^2} \).
Taylor Series Expansion
The Taylor series gives us a powerful way to approximate more complex functions by expressing them as infinite sums of polynomials. The Maclaurin series, a special type of Taylor series, simplifies this by centering at zero, making calculations a bit more straightforward.
- The series is constructed from the function's derivatives evaluated at a point of interest, usually zero.
- Using the formula \( f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \cdots \), we find polynomial terms that can effectively approximate the function near the center.
- The Maclaurin series for \( e^{x^2} \) focuses on derivative values at zero, generating nonzero terms in a step-by-step expansion.