Question

This problem indicates a proof of convergence of a Fourier series under conditions more restrictive than those in Theorem \(10.3 .1 .\) (a) If \(f\) and \(f^{\prime}\) are piecewise continuous on \(-L \leq x

Short answer

#Answer# a) n * a_n and n * b_n are bounded as n approaches infinity because when applying integration by parts to the definition of a_n and b_n, we can show that the expressions have upper bounds depending on constants M and M' that are related to the continuity of f(x) and f'(x). b) n^2 * a_n and n^2 * b_n are bounded as n approaches infinity when f is continuous, and f' and f'' are piecewise continuous on -L <= x < L. Using integration by parts twice and properties of continuity, we can show that these expressions also have upper bounds. The convergence of the Fourier series for f at each point in -L <= x <= L is guaranteed due to the fact that the terms in the Fourier series are square-integrable, and f must be continuous on the closed interval to ensure the uniqueness of the Fourier coefficients at each point.

Step-by-step solution

Apply integration by parts to a_n

To show that n * a_n is bounded, we need to apply integration by parts to the definition of a_n and find an expression that is bounded. Integration by parts states that for two functions u(x) and v(x): integral(u(x) * v'(x) dx) = u(x) * v(x) - integral(u'(x) * v(x) dx) First, let's choose our u(x) and v'(x) functions. We have: u(x) = f(x) v'(x) = cos(n * pi * x / L) Now, let's find their derivatives: u'(x) = f'(x) v(x) = (L / (n * pi)) * sin(n * pi * x / L) Applying integration by parts to a_n, we get: a_n = (1 / L) * (u(x) * v(x) - integral(u'(x) * v(x) dx), from -L to L) a_n = (1 / L) * (f(x) * (L / (n * pi)) * sin(n * pi * x / L) - integral(f'(x) * (L / (n * pi)) * sin(n * pi * x / L) dx), from -L to L)

Prove n * a_n is bounded

Now we need to find an upper bound for n * a_n. We know that f(x) and f'(x) are piecewise continuous on -L <= x < L. This means that there exist constants M and M' such that |f(x)| <= M and |f'(x)| <= M' for -L <= x < L. Therefore, we can write: n * a_n <= (n / L) * (M * L / (n * pi) * 2L + M' * L^2 / (n * pi)) As n approaches infinity, the first term goes to 0, and the second term is bounded by M' * L^2 / pi. Thus, n * a_n is bounded as n approaches infinity.

Apply integration by parts to b_n

Similar to step 1, we will apply integration by parts to b_n: u(x) = f(x) v'(x) = sin(n * pi * x / L) u'(x) = f'(x) v(x) = -(L / (n * pi)) * cos(n * pi * x / L) Applying integration by parts to b_n, we get: b_n = (1 / L) * (u(x) * v(x) - integral(u'(x) * v(x) dx), from -L to L) b_n = (1 / L) * (f(x) * (-(L / (n * pi)) * cos(n * pi * x / L) - integral(f'(x) * (-(L / (n * pi)) * cos(n * pi * x / L) dx), from -L to L)

Prove n * b_n is bounded

Now we need to find an upper bound for n * b_n. Following the same reasoning as in step 2, we get: n * b_n <= (n / L) * (M * L / (n * pi) * 2L + M' * L^2 / (n * pi)) As n approaches infinity, the first term goes to 0, and the second term is bounded by M' * L^2 / pi. Thus, n * b_n is bounded as n approaches infinity.

Use the second part of the question to find the bounds for n^2 * a_n and n^2 * b_n

For the second part of the question, we are given that f is continuous on the closed interval -L <= x <= L, and f' and f'' are piecewise continuous on -L <= x < L. By repeating the steps 1-4 and using integration by parts again, we can prove that n^2 * a_n and n^2 * b_n are bounded as n approaches infinity.

Convergence of the Fourier series

Once we have shown that n^2 * a_n and n^2 * b_n are bounded for all n, we can use Parseval's theorem and the fact that the coefficient terms in the Fourier series are square-integrable to guarantee the convergence of the Fourier series for f at each point in the closed interval -L <= x <= L. As for why f must be continuous on the closed interval, it is a requirement for the convergence of the Fourier series. If a discontinuity were present, the series would not converge at that point since the Fourier coefficients would not have a unique value there.

Key concepts

Integration by Parts

Understanding the concept of integration by parts is crucial when solving certain types of integrals, particularly those that arise in the context of Fourier series. This mathematical technique is based on the product rule for differentiation and is formally expressed by the equation \[ \text{integral}(u(x) \times v'(x) \text{d}x) = u(x) \times v(x) - \text{integral}(u'(x) \times v(x) \text{d}x) \].

This method allows us to transform complex integrals into simpler ones that we can evaluate more readily. In the context of Fourier coefficients, like \(a_n\) and \(b_n\), integration by parts effectively demonstrates that terms like \(n a_n\) and \(n b_n\) are bounded as \(n\) approaches infinity - a fundamental step in the proof of Fourier series convergence.

To use this method effectively, one needs to choose functions \(u(x)\) and \(v'(x)\) wisely, such that their derivatives \(u'(x)\) and \(v(x)\) are easier to integrate. This choice often relies on the nature of the function being integrated and is an essential skill within a mathematician's toolkit for practical problem solving.

Piecewise Continuous Functions

Piecewise continuous functions play an important role in Fourier analysis. A function is said to be piecewise continuous on an interval if it is continuous on that interval except for a finite number of points at which the function has finite jump discontinuities.

Why are piecewise continuous functions significant in the context of Fourier analysis? These functions, even when experiencing abrupt changes (also known as discontinuities), still adhere to certain predictability that enables the determination of Fourier coefficients, which are essential in developing a function's Fourier series representation.

For our exercise, the piecewise continuity of \(f\) and its derivatives is key to applying integration by parts effectively. It assures us that despite potential jumps, the integral calculations remain valid within specified intervals. Furthermore, this attribute helps in assuring that the coefficients produced do not become unbounded, maintaining the credibility of the Fourier series to represent such functions.

Parseval's Theorem

Parseval's theorem serves as a bridge between the time domain and the frequency domain representations of a function, particularly in the context of Fourier series. The theorem states that the sum of the squares of a function's Fourier coefficients is proportional to the integral of the square of the function over one period.

Formally, if \(f(x)\) is a square-integrable function over the interval \([-L, L]\), then \[ \frac{1}{2L}\text{integral from -L to L} (|f(x)|^2 \text{d}x = a_0^2 + \frac{1}{2}\text{sum from n=1 to ∞} (a_n^2+b_n^2) \].

This theorem implies that the energy (or the total 'power') of a function in the time domain is equal to the energy represented by its Fourier coefficients in the frequency domain. In the context of the problem, Parseval's theorem confirms the convergence of the Fourier series for a continuous function (provided other conditions, such as boundedness of certain coefficients, are met) by ensuring that the series represents the function's energy accurately across its domain. This is fundamental to the solution as it gives a concrete criterion for convergence, which is a significant assurance in demonstrating the efficacy of Fourier series in approximating functions.