Chapter 10: Problem 28
A function is given on an interval \(0
Short Answer
Step by step solution
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
even extension
- For the given function \( f(x) = x \) when \( 0 < x < 1 \) and \( f(x) = 0 \) when \( 1 < x < 2 \), we reflect the graph across the y-axis.
- This reflection creates a function that is x for \(-1 < x < 1\) and zero for \(1 < x < 2\) and \(-2 < x < -1\).
odd extension
- In the case of the given function \( f(x) = x \) for \( 0 < x < 1 \), we extend it as \( h(x) = -x \) for \( -1 < x < 0 \).
- For the interval \( 1 < x < 2 \), since \( f(x) = 0 \), this naturally extends as \( h(x) = 0 \) for \( -2 < x < -1 \).
cosine series
- Here, \(a_0\) is the zeroth coefficient, calculated as \( \frac{1}{L}\int_{0}^{L} f(x)dx \).
- Each \(a_n\) is found using \( a_n = \frac{2}{L}\int_{0}^{L} f(x)\cos\left(\frac{n\pi x}{L}\right)dx \).
sine series
- The coefficients \( b_n \) are found using \( b_n = \frac{2}{L}\int_{0}^{L} f(x)\sin\left(\frac{n\pi x}{L}\right)dx \).