1. Find the first four nonzero terms in the MacLaurin series for the function f(x) = e-² DA. 1 + x² + OB. 1+ 22 + OC. none of these OD. 1-² + DE.-x²-x³-22 ++ 24 x6 2 3! - - 26 3!
1. Find the first four nonzero terms in the MacLaurin series for the function f(x) = e-² DA. 1 + x² + OB. 1+ 22 + OC. none of these OD. 1-² + DE.-x²-x³-22 ++ 24 x6 2 3! - - 26 3!
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![**Problem Statement:**
1. Find the first four nonzero terms in the Maclaurin series for the function \( f(x) = e^{-x^2} \).
**Options:**
- **A.** \( 1 + x^2 + \frac{x^4}{2} + \frac{x^6}{3!} \)
- **B.** \( 1 + \frac{x^2}{2} + \frac{x^4}{4!} + \frac{x^6}{6!} \)
- **C.** none of these
- **D.** \( 1 - x^2 + \frac{x^4}{2} - \frac{x^6}{3!} \)
- **E.** \( -x^2 - x^3 - \frac{x^4}{2} - \frac{x^6}{3!} \)
**Explanation:**
To solve this problem, we use the Maclaurin series expansion formula. The Maclaurin series for \( e^x \) is given by:
\[ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \]
For \( f(x) = e^{-x^2} \), we replace \( x \) with \(-x^2\):
\[ e^{-x^2} = 1 + (-x^2) + \frac{(-x^2)^2}{2!} + \frac{(-x^2)^3}{3!} + \cdots \]
Thus, the first four nonzero terms are:
1. \( 1 \)
2. \( -x^2 \)
3. \( \frac{x^4}{2} \)
4. \( -\frac{x^6}{3!} \)
The correct answer is **D.** \( 1 - x^2 + \frac{x^4}{2} - \frac{x^6}{3!} \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffb2133c9-e1e5-4d56-9c72-044227328930%2F9982d215-51a5-4d15-97c1-36b2c84ef427%2Fx8egf5d_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
1. Find the first four nonzero terms in the Maclaurin series for the function \( f(x) = e^{-x^2} \).
**Options:**
- **A.** \( 1 + x^2 + \frac{x^4}{2} + \frac{x^6}{3!} \)
- **B.** \( 1 + \frac{x^2}{2} + \frac{x^4}{4!} + \frac{x^6}{6!} \)
- **C.** none of these
- **D.** \( 1 - x^2 + \frac{x^4}{2} - \frac{x^6}{3!} \)
- **E.** \( -x^2 - x^3 - \frac{x^4}{2} - \frac{x^6}{3!} \)
**Explanation:**
To solve this problem, we use the Maclaurin series expansion formula. The Maclaurin series for \( e^x \) is given by:
\[ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \]
For \( f(x) = e^{-x^2} \), we replace \( x \) with \(-x^2\):
\[ e^{-x^2} = 1 + (-x^2) + \frac{(-x^2)^2}{2!} + \frac{(-x^2)^3}{3!} + \cdots \]
Thus, the first four nonzero terms are:
1. \( 1 \)
2. \( -x^2 \)
3. \( \frac{x^4}{2} \)
4. \( -\frac{x^6}{3!} \)
The correct answer is **D.** \( 1 - x^2 + \frac{x^4}{2} - \frac{x^6}{3!} \).
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