Show that the Fourier sine transform of the function 0 < x≤1 {:- 1-x 1 ≤ x ≤ 2 2≤x is given by 2 sin (w(1 cos w)) w²
Show that the Fourier sine transform of the function 0 < x≤1 {:- 1-x 1 ≤ x ≤ 2 2≤x is given by 2 sin (w(1 cos w)) w²
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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Question
![**Title: Fourier Sine Transform of a Piecewise Function**
**Objective:**
To show that the Fourier sine transform of the given piecewise function is
\[ \frac{2 \sin(w(1 - \cos w))}{w^2} \]
**Given Function:**
The function \( f(x) \) is defined as follows:
\[
f(x) =
\begin{cases}
x & \text{for } 0 < x \leq 1 \\
1 - x & \text{for } 1 \leq x \leq 2 \\
0 & \text{for } 2 \leq x
\end{cases}
\]
**Explanation:**
The piecewise function \( f(x) \) is defined in three intervals.
1. For the interval \( 0 < x \leq 1 \), \( f(x) = x \).
2. For the interval \( 1 \leq x \leq 2 \), \( f(x) = 1 - x \).
3. For \( x \geq 2 \), \( f(x) = 0 \).
**Goal:**
To find the Fourier sine transform of the above piecewise function and demonstrate that it is equal to:
\[ \frac{2 \sin(w(1 - \cos w))}{w^2} \]
This involves applying the definition of the Fourier sine transform, typically noted as:
\[ F_s(\omega) = \int_{0}^{\infty} f(x) \sin(\omega x) \, dx \]
for the piecewise function \( f(x) \).
This description is aimed to provide a clear step-by-step guide on how to approach deriving the Fourier sine transform of the given piecewise function.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F62ca1c58-ac8a-4380-bd93-744ecdfbeb4a%2Faeb57b1a-a0fd-4e1f-9a6d-8b9a11e94b5b%2Fqyhxsl_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Title: Fourier Sine Transform of a Piecewise Function**
**Objective:**
To show that the Fourier sine transform of the given piecewise function is
\[ \frac{2 \sin(w(1 - \cos w))}{w^2} \]
**Given Function:**
The function \( f(x) \) is defined as follows:
\[
f(x) =
\begin{cases}
x & \text{for } 0 < x \leq 1 \\
1 - x & \text{for } 1 \leq x \leq 2 \\
0 & \text{for } 2 \leq x
\end{cases}
\]
**Explanation:**
The piecewise function \( f(x) \) is defined in three intervals.
1. For the interval \( 0 < x \leq 1 \), \( f(x) = x \).
2. For the interval \( 1 \leq x \leq 2 \), \( f(x) = 1 - x \).
3. For \( x \geq 2 \), \( f(x) = 0 \).
**Goal:**
To find the Fourier sine transform of the above piecewise function and demonstrate that it is equal to:
\[ \frac{2 \sin(w(1 - \cos w))}{w^2} \]
This involves applying the definition of the Fourier sine transform, typically noted as:
\[ F_s(\omega) = \int_{0}^{\infty} f(x) \sin(\omega x) \, dx \]
for the piecewise function \( f(x) \).
This description is aimed to provide a clear step-by-step guide on how to approach deriving the Fourier sine transform of the given piecewise function.
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