Chapter 7: Problem 23
Find the Fourier coefficients \(a_{0}, a_{k},\) and \(b_{k}\) of \(f\) on \([-\pi, \pi].\)\ $$f(x)=\left\\{\begin{array}{ll} 0 & \text { if }-\pi \leq x<0 \\ 1 & \text { if } 0 \leq x \leq \pi \end{array}\right.$$
Short Answer
Expert verified
\(a_0 = \frac{1}{2}, a_k = 0, b_k = \frac{2}{\pi k}\) for odd \(k\), 0 for even \(k\).
Step by step solution
01
Understand the Function
The function \(f(x)\) is a piecewise function. It equals 0 for \(-\pi \leq x < 0\) and equals 1 for \(0 \leq x \leq \pi\). The function is periodic with period \(2\pi\) on the interval \([-\pi, \pi]\).
02
Calculate the Average Value \(a_0\)
The coefficient \(a_0\) is given by the formula \(a_0 = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(x) \, dx\). We split this integral into two intervals: \([-\pi, 0]\) and \([0, \pi]\). \[a_0 = \frac{1}{2\pi} \left( \int_{-\pi}^{0} 0 \, dx + \int_{0}^{\pi} 1 \, dx \right) = \frac{1}{2\pi} \left( 0 + x \bigg|_{0}^{\pi} \right) = \frac{1}{2\pi} \cdot \pi = \frac{1}{2}.\]
03
Find the Cosine Coefficients \(a_k\)
The coefficient \(a_k\) is computed as \(a_k = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos(kx) \, dx\). Like \(a_0\), we split into two integrals:\[a_k = \frac{1}{\pi} \left( \int_{-\pi}^{0} 0 \cdot \cos(kx) \, dx + \int_{0}^{\pi} 1 \cdot \cos(kx) \, dx \right) = \frac{1}{\pi} \int_{0}^{\pi} \cos(kx) \, dx.\]Solving this integral gives:\[a_k = \frac{1}{\pi} \left( \frac{\sin(kx)}{k} \bigg|_{0}^{\pi} \right) = \frac{1}{\pi} \left( \frac{\sin(k\pi)}{k} - \frac{0}{k} \right) = 0,\]since \(\sin(k\pi) = 0\) for integer \(k\).
04
Find the Sine Coefficients \(b_k\)
The coefficient \(b_k\) is given by \(b_k = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(kx) \, dx\). Again, splitting, we get:\[b_k = \frac{1}{\pi} \left( \int_{-\pi}^{0} 0 \cdot \sin(kx) \, dx + \int_{0}^{\pi} 1 \cdot \sin(kx) \, dx \right) = \frac{1}{\pi} \int_{0}^{\pi} \sin(kx) \, dx.\]Solving this integral gives:\[b_k = \frac{1}{\pi} \left( -\frac{\cos(kx)}{k} \bigg|_{0}^{\pi} \right) = \frac{1}{\pi} \left( -\frac{\cos(k\pi) + 1}{k} \right) = \frac{2}{\pi k}(1 - (-1)^k).\]This simplifies to \(b_k = \frac{2}{\pi k}\) when \(k\) is odd and \(b_k = 0\) when \(k\) is even.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Piecewise Function
A piecewise function is a type of function characterized by different expressions, depending on the input value's interval. These are particularly useful when a single formula cannot describe all cases of a situation. In this scenario, the function is defined within two distinct intervals for the domain
Additionally, this piecewise behavior shows that our function is periodic with a period of \(2\pi\). Periodicity is important because it allows the use of Fourier Series to decompose or analyze the function.
- For \(-\pi \leq x < 0\), the function value is 0.
- For \(0 \leq x \leq \pi\), the function value is 1.
Additionally, this piecewise behavior shows that our function is periodic with a period of \(2\pi\). Periodicity is important because it allows the use of Fourier Series to decompose or analyze the function.
Fourier Coefficients
Fourier coefficients are essential for transforming a function into a series of sines and cosines. These coefficients – specifically \(a_0\), \(a_k\), and \(b_k\) – help in constructing a Fourier series that can express the original function. The Fourier coefficients are significant for converting complex mathematical functions into simpler trigonometric terms.
- Average Value \(a_0\): This term represents the average value of the function over a period and is calculated using \(a_0 = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(x) \, dx\). It gives an offset for the series and forms the basis.
- Cosine Coefficients \(a_k\): For symmetry with cosine functions, these are found by \(a_k = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos(kx) \, dx\). These coefficients reflect how much of each cosine wave is needed.
- Sine Coefficients \(b_k\): Analogous to the cosine coefficients, sine coefficients are calculated by \(b_k = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(kx) \, dx\). These coefficients represent the sine wave components of the function.
Trigonometric Integrals
Trigonometric integrals are a crucial part of finding Fourier coefficients, as they involve integrating functions multiplied by sine or cosine over a period. Here, each integral corresponds to assessing how much each trigonometric function resembles the target function over one complete cycle.
- Integrating Cosines: To find \(a_k\), you use \(\int_{0}^{\pi} \cos(kx) \, dx\). This integral focuses on cosine's contribution to reconstruction. In this scenario, it ultimately vanishes because \(\sin(k\pi) = 0\).
- Integrating Sines: For sine coefficients \(b_k\), you evaluate \(\int_{0}^{\pi} \sin(kx) \, dx\). This highlights how sine waves describe the original function's shape. It leads to non-zero outcomes, symbolizing the required sine components.