Question

Let \(g(x)= \begin{cases}\cos x, & -\pi

Short answer

(a) Calculate Fourier coefficients. (b) Use \(a = 0\) for uniform convergence of \(h(x)\).

Step-by-step solution

Understanding the Piecewise Function

Given the function \( g(x) \) which is defined piecewise:\[ g(x) = \begin{cases} \cos x, & -\pi < x < 0 \ \sin x, & 0 < x < \pi \end{cases} \]We need to calculate its Fourier series. Notably, the function is defined differently in the intervals \(-\pi, 0\) and \(0, \pi\). This will affect the calculation of coefficients.

Calculate Fourier Coefficients

The Fourier series is expressed as:\[ f(x) = a_0 + \sum_{n=1}^{\infty} \left( a_n \cos(nx) + b_n \sin(nx) \right) \]First, calculate \(a_0\):\[ a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} g(x) \ dx = \frac{1}{\pi} \left( \int_{-\pi}^{0} \cos x \ dx + \int_{0}^{\pi} \sin x \ dx \right) = 0 \]Now, calculate \(a_n\):\[ a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} g(x) \cos(nx) \ dx \]Split the integral:\[ a_n = \frac{1}{\pi} \left( \int_{-\pi}^{0} \cos x \cos(nx) \ dx + \int_{0}^{\pi} \sin x \cos(nx) \ dx \right) \]Each integral results in solving the integral involving trigonometric identities, simplifying it. The resulting expression leads to conditions where these terms mostly cancel or simplify based on trigonometric orthogonality.Similarly, calculate \(b_n\):\[ b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} g(x) \sin(nx) \ dx \]Apply a similar splitting and trigonometric simplification.After calculation, the coefficients yield a representation converging pointwise to function segments.

Integrate to Find h(x)

Define the function \( h(x) \) as:\[ h(x) = \int_{-\pi}^{x} g(t) \ dt + a \sin \frac{x}{2} \]We need to find \( h(x) \) over the interval \([-\pi, \pi]\). Break the integral into two parts based on where \(g(t)\) changes:\[ h(x) = \int_{-\pi}^{0} \cos t \ dt + \int_{0}^{x} \sin t \ dt + a \sin \frac{x}{2} \]Compute these integrals separately, ensuring you integrate the function correctly over its defined intervals, and combine them for the given range of \( x \).

Determine Uniform Convergence with Fourier Series

Now that we have the Fourier series representation of \(g(x)\) and the integration yielding \(h(x)\), we set the uniform convergence condition:The convergence of the Fourier series uniformly on \([-\pi, \pi]\) requires continuous derivatives as implied by strong Dirichlet conditions.Since \(a \sin \frac{x}{2}\) is smooth, we need the Fourier series of the integral to converge uniformly for \(h(x)\).Check for convergence through adjusting \(a\): if anti-symmetric components cancel effectively, \(a = 0\) achieves this as \(h(x)\) continuity and convergence may hinge on eliminating bias.

Key concepts

Piecewise Functions

Piecewise functions are mathematical constructs where different expressions or rules define the function over different intervals of its domain. This can be particularly useful when modeling phenomena that behave differently over distinct phases or boundaries. In the exercise, we have a function \( g(x) \) which is piecewise defined as:

• \( \cos x \) when \(-\pi < x < 0\)
• \( \sin x \) when \(0 < x < \pi\)

Notice how the trigonometric functions are used differently in each segment. This division means that each piece needs to be considered separately when performing operations like integration or finding a Fourier series. To manage this, we analyze each interval individually and later combine the results. When integrating or finding a Fourier series, we must integrate over each specific interval and sum the results to achieve a comprehensive solution. Piecewise functions like this showcase the beauty of adapting mathematical models to represent diverse real-world scenarios.

Trigonometric Integrals

Trigonometric integrals are integrals involving trigonometric functions such as \( \sin \) or \( \cos \). In the context of Fourier series, these integrals are crucial because the Fourier coefficients involve integrating products of trigonometric functions over specific intervals. The formula for integrals of the form:

• \( \int \cos(x) \cos(nx) \, dx \)
• \( \int \sin(x) \cos(nx) \, dx \)

require specific techniques such as trigonometric identities and properties of orthogonal functions.
In calculating coefficients for a Fourier series, we integrate over pieces defined by the piecewise function. For example, to calculate \( a_n \), we split the integral at the transition points:

• \( \int_{-\pi}^{0} \cos x \cos(nx) \, dx \)
• \( \int_{0}^{\pi} \sin x \cos(nx) \, dx \)

Each integral is computed using trigonometric identities, often simplifying through known integrals of sine and cosine products. These integrals play a pivotal role in expanding a piecewise function as a Fourier series, breaking it down into a sum of simpler sine and cosine waves.

Uniform Convergence

Uniform convergence is a concept concerning the way sequences of functions converge to a limiting function. Specifically, it means that the sequence converges at the same rate over the entire domain. In simpler terms, a function \(h(x)\) defined as a series converges uniformly if the difference between the function and the finite partial sums becomes arbitrarily small uniformly for all points in its domain as more terms are added.
For the Fourier series of \(h(x)\) to converge uniformly, certain conditions must be satisfied. The function must be continuous, and typically its derivatives must exist and be bounded. In the provided problem, the task is to find values of \( a \) such that the Fourier series of \( h(x) = \int_{-\pi}^{x} g(t) \, dt + a \sin \frac{x}{2} \) converges uniformly over \([-\pi, \pi]\). Crucially:

• A continuous \( h(x) \) ensures better chances of uniform convergence.
• When \( a = 0 \), the anti-symmetric components might cancel effectively, leading to smoother convergence.

This concept helps in ensuring that the approximation by the Fourier series is valid through the entire range rather than just pointwise, making it more reliable for practical calculations and interpretations.