Suppose that \(f(x)\) and its derivative \(f^{\prime}(x)\) are both expanded in Fourier series on \((-\pi, \pi) .\) Call the coefficients in the \(f(x)\) series \(a_{n}\) and \(b_{n}\) and the coefficients in the \(f^{\prime}(x)\) series \(a_{n}^{\prime}\) and \(b_{n}^{\prime}\) Write the integral for \(\left.a_{n} \text { [equation }(5.9)\right]\) and integrate it by parts to get an integral of \(f^{\prime}(x) \sin n x\). Recognize this integral in terms of \(b_{n}^{\prime}\) [equation \(\left.(5.10) \text { for } f^{\prime}(x)\right]\) and so show that \(b_{n}^{\prime}=-n a_{n}\). (In the integration by parts, the integrated term is zero because \(f(\pi)=f(-\pi)\) since \(f\) is continuous- sketch several periods.). Find a similar relation for \(a_{n}^{\prime}\) and \(b_{n} .\) Now show that this is the result you get by differentiating the \(f(x)\) series term by term. Thus you have shown that the Fourier series for \(f^{\prime}(x)\) is correctly given by differentiating the \(f(x)\) series term by term (assuming that \(f^{\prime}(x)\) is expandable in a Fourier series).

Step by step solution

01

Define the Fourier Series Coefficients

The Fourier series for a function on (-\pi, \pi) is given by \(f(x) = a_0 + \sum_{n=1}^{\infty} [a_n \cos(nx) + b_n \sin(nx)]\). The coefficients are defined as: \(a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos(nx) \, dx\) and \(b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(nx) \, dx\).

02

Apply Integration by Parts

Start with the integral for \(a_n\): \(a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos(nx) \, dx\). Integrate by parts: \(u = f(x), \, dv = \cos(nx) \, dx, \, du = f'(x) \, dx, \, v = \frac{\sin(nx)}{n}\). Using integration by parts formula \(\int u \, dv = uv - \int v \, du\), we have: \(a_n = \frac{1}{\pi} \left[ f(x) \frac{\sin(nx)}{n} \right]_{-\pi}^{\pi} - \frac{1}{\pi} \int_{-\pi}^{\pi} \frac{\sin(nx)}{n} f'(x) \, dx\). The boundary term here is zero because \(f(\pi)=f(-\pi)\), so we get \(a_n = -\frac{1}{n\pi} \int_{-\pi}^{\pi} f'(x) \sin(nx) \ dx\).

03

Relate Integral to Fourier Coefficient \(b_{n}^{\prime}\)

Recognize the integral \(\int_{-\pi}^{\pi} f'(x) \sin(nx) \, dx\) corresponds to \(b_{n}^{\prime}\), which is the Fourier coefficient of \(f'(x)\). Hence \(a_n = -\frac{1}{n} b_{n}^{\prime}\). Rearranging gives \(b_{n}^{\prime} = -na_n\).

04

Find Relation for \(a_{n}^{\prime}\) and \(b_{n}\)

Using similar steps, consider the term \(b_n: b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(nx) \, dx\). Applying integration by parts: \(u=f(x)\), \(dv=\sin(nx) dx\), \(du=f'(x)dx\), \(v=-\frac{\cos(nx)}{n}\). We obtain: \(b_n = \frac{1}{\pi} \left[-f(x) \frac{\cos(nx)}{n} \right]_{-\pi}^{\pi} + \frac{1}{\pi} \int_{-\pi}^{\pi} \frac{f'(x) \cos(nx)}{n} \, dx\). The boundary term is zero. Thus \(b_n = \frac{1}{n\pi} \int_{-\pi}^{\pi} f'(x) \cos(nx) \ dx\) = \(\frac{1}{n} a_{n}^{\prime}\) which rearranges to \(a_{n}^{\prime} = nb_n\).

05

Verify by Term by Term Differentiation

Differentiate the Fourier series term by term: \(f(x) = a_0 + \sum_{n=1}^{\infty} \left[ a_n \cos(nx) + b_n \sin(nx) \right]\). Derivative: \(f^{\prime}(x) = \sum_{n=1}^{\infty} \left[ -na_n \sin(nx) + nb_n \cos(nx) \right]\). Comparing with the Fourier series for \(f^{\prime}(x)\), the coefficients match the relations previously derived: \(a_{n}^{\prime}= nb_n\) and \(b_{n}^{\prime} = -na_n\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Fourier Coefficients

In Fourier series analysis, Fourier coefficients are crucial. They break down a complex function into simple sine and cosine components. For a function defined on \((-\pi, \pi)\), the series is: \( f(x) = a_0 + \sum_{n=1}^{\infty} [a_n \cos(nx) + b_n \sin(nx)] \). Each coefficient, \( a_n \) and \( b_n \, \), represents a specific frequency component. Coefficients are computed as: \[ a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos(nx) \, dx \] and \[ b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(nx) \, dx \]. These integrals measure the interaction of \( f(x) \) with \( \cos(nx) \) and \( \sin(nx) \). Higher \( n \) values correspond to higher-frequency components.

Integration by Parts

Integration by parts is a vital technique in calculus, especially for Fourier series. It transforms products of functions into simpler integrals. The formula is: \[ \int u \, dv = uv - \int v \, du \]. For the Fourier coefficient \( a_n \), start with \(: a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos(nx) \, dx \). Setting \( u = f(x) \) and \( dv = \cos(nx) \, dx \, \), then \( du = f\'(x) \, dx \) and \( v = \frac{\sin(nx)}{n} \. \) Applying integration by parts, we get: \[ a_n = \frac{1}{\pi} \left[ f(x) \frac{\sin(nx)}{n} \right]_{-\pi}^\pi - \frac{1}{\pi} \int_{-\pi}^{\pi} \frac{\sin(nx)}{n} f\'(x) \, dx. \] The boundary term is zero because \( f(\pi)=f(-\pi) \), simplifying to: \[ a_n = -\frac{1}{n\pi} \int_{-\pi}^{\pi} f\'(x) \sin(nx) \ dx. \]

Term-by-Term Differentiation

When differentiating a Fourier series term-by-term, each term in the series is separately differentiated. The original Fourier series for \( f(x) \) is \( f(x) = a_0 + \sum_{n=1}^{\infty} \left[ a_n \cos(nx) + b_n \sin(nx) \right] \. \) Differentiating term-by-term, we obtain: \[ f\'(x) = \sum_{n=1}^{\infty} \left[ -na_n \sin(nx) + nb_n \cos(nx) \right]. \] These new expressions match the Fourier coefficients for the derivative of \ f(x) \, confirming the correctness of the term-by-term differentiation. This process validates the relationships: \( a_{n}^{\prime}= nb_n \) and \( b_{n}^{\prime} = -na_n \). It ensures that the Fourier series for \( f\'(x) \) accurately represents the function's derivative.

Derivative of Fourier Series

Understanding the derivative of a Fourier series entails differentiating every sine and cosine term. The original series is \( f(x) = a_0 + \sum_{n=1}^{\infty} \left[ a_n \cos(nx) + b_n \sin(nx) \right]. \) The derivative involves: \[ f\'(x) = \sum_{n=1}^{\infty} \left[ -na_n \sin(nx) + nb_n \cos(nx) \. \] The coefficients for the series of \( f\'(x) \) can be seen as \( a_{n}^{\prime}- nb_n \) and \( b_{n}^{\prime} = -na_n \. \) This method confirms the results obtained by integration by parts: \( a_{n}^{\prime}= nb_n \) and \( b_{n}^{\prime} = -na_n \. \) This agreement ensures that differentiating the series term-by-term properly represents the original function's derivative, providing a deeper understanding of the function's behavior.