Question

Assume \(f\) satisfies the assumptions of Dirichlet's Theorem. Determine the following limits: (a) \(\lim _{n \rightarrow \infty} \frac{1}{\pi} \int_{-\pi}^{\pi} f(t) \sin n t d t\) (b) \(\lim _{n \rightarrow \infty} \frac{1}{\pi} \int_{-\pi}^{\pi} f(t) \sin \left(n-\frac{1}{2}\right) t d t\) (c) \(\lim _{n \rightarrow \infty} \frac{1}{\pi} \int_{-\pi}^{\pi} \frac{f(t)}{t} \sin \left(n-\frac{1}{2}\right) t d t\)

Short answer

All the given limits evaluate to zero as \( n \to \infty \).

Step-by-step solution

Understanding the Setup

We need to analyze the given limits involving integrals of the function \( f(t) \) multiplied by sine functions over the interval \([-\pi, \pi]\). The function \( f \) possesses the properties given by Dirichlet's Theorem, which generally implies that \( f \) is piecewise continuous on \([-\pi, \pi]\).

Step (a): Evaluating the Limit Involving \( \sin nt \)

The limit in part (a) is \[ \lim _{n \rightarrow \infty} \frac{1}{\pi} \int_{-\pi}^{\pi} f(t) \sin n t \, dt. \]For functions satisfying Dirichlet's theorem, trigonometric integrals such as this one tend to zero as \( n \to \infty \). This is because the oscillations of \( \sin nt \) average out to zero due to its periodic nature, assuming \( f(t) \) does not introduce non-vanishing terms in the integral.

Step (b): Evaluating the Limit Involving \( \sin(n-1/2)t \)

The limit in part (b) is \[ \lim _{n \rightarrow \infty} \frac{1}{\pi} \int_{-\pi}^{\pi} f(t) \sin \left(n-\frac{1}{2}\right) t \, dt. \]Similar to part (a), the oscillatory nature of \( \sin\left(n-\frac{1}{2}\right)t \) causes the integral to average out to zero. Hence, this limit is also zero.

Step (c): Evaluating the Limit with \( \frac{f(t)}{t} \sin(n-1/2)t \)

The limit in part (c) is \[ \lim _{n \rightarrow \infty} \frac{1}{\pi} \int_{-\pi}^{\pi} \frac{f(t)}{t} \sin \left(n-\frac{1}{2}\right) t \, dt. \]This type of limit where \( \frac{f(t)}{t} \) is incorporated, is known from Fourier analysis and convergence results of Dirichlet's theorem to converge to zero due to orthogonal properties and oscillatory behavior.

Key concepts

Dirichlet's Theorem

Dirichlet's Theorem is an essential concept in mathematical analysis, especially when dealing with Fourier series. This theorem provides conditions under which the Fourier series of a function converges to the function itself. In simple terms, if a function \( f \) is periodic, has a finite number of discontinuities, and a finite number of maximum and minimum values, it meets the requirements of Dirichlet's Theorem.

This theorem helps us determine the behavior of integrals involving trigonometric functions. Particularly, it tells us that for functions that meet these criteria, the oscillations of sine and cosine terms cancel out their effects over a period, leading to the convergence of integrals to zero as the frequency increases. This is particularly useful when evaluating limits that involve oscillatory behavior, as seen in the original exercise where the function \( f(t) \) is considered under these conditions.

Trigonometric Integrals

Trigonometric integrals play a crucial role when working with Fourier series. In the context of the exercise, we have integrals of the form \( \int_{-\pi}^{\pi} f(t) \sin nt \, dt \). The key point about these integrals is their tendency to converge to zero.

• The sine function \( \sin nt \) oscillates with increasing frequency as \( n \) grows.
• When the product \( f(t) \sin nt \) is integrated over an entire period, the negative and positive areas created by these oscillations often cancel each other out.
• This causes the integral to converge to zero when \( f \) fulfills the properties outlined by Dirichlet's Theorem.

This convergence is crucial because it means the Fourier coefficients for non-constant terms diminish, providing smoother approximations of \( f(t) \) over the interval \([-\pi, \pi]\). This explains why the limits evaluated in the exercise tend towards zero.

Piecewise Continuous Functions

The function \( f \) in the exercise is assumed to be piecewise continuous on the interval \([-\pi, \pi]\). A function is piecewise continuous if it is continuous within each subinterval and has a finite number of discontinuities where the left and right limits exist and are finite.

• This property is crucial in the context of Fourier analysis, as most functions in real-world applications are not perfectly continuous but are piecewise continuous.
• The continuity conditions allow the use of Dirichlet's Theorem, which tells us about the convergence properties of the Fourier series.
• A piecewise continuous function can still be represented by a Fourier series, allowing us to evaluate limits of integrals effectively, as demonstrated in the exercise.

Understanding piecewise continuity helps in determining whether the results of Fourier series analysis, like those involving limits and integrals, are applicable.

Limit Evaluation

Limit evaluation in this context involves understanding how trigonometric integrals behave as the frequency of the sine function increases.

• The problems in the exercise demonstrate limits of the form \( \lim_{n \to \infty} \int f(t) \sin(nt) \ dt \), leading to zero due to the reasons grounded in Dirichlet's Theorem.
• As \( n \) becomes very large, the rapid oscillation of \( \sin(nt) \) causes any non-pathological function \( f(t) \) that satisfies Dirichlet's conditions to result in the integral averaging to zero over the interval \([-\pi, \pi]\).
• These evaluations are simplified through the use of techniques like recognizing symmetry, periodicity, and cancellation effects stemming from the orthogonality of sine functions.

Limit evaluation is a critical step in understanding the convergence behavior in Fourier series, and this insight helps us see how higher harmonics affect the culmination of such integrals.