Question

Let $$ f(x)= \begin{cases}A x+B, & -\pi \leq x<0, \\ \cos x, & 0 \leq x \leq \pi .\end{cases} $$ For what values \(A\) and \(B\) does the Fourier series of \(f\) converge uniformly to \(f\) on all of \([-\pi, \pi]\) ?

Short answer

For uniform convergence on \([-\pi, \pi]\), \( A = 0 \) and \( B = 1 \).

Step-by-step solution

Determine Conditions for Uniform Convergence

For the Fourier series to converge uniformly to the function \( f(x) \) on the interval \([-\pi, \pi]\), \( f(x) \) must be continuous. Therefore, we need to ensure that \( f(x) \) is continuous at \( x = 0 \), where the piecewise definition changes. This means \( A(0) + B = \cos(0) \). Since \( A(0) = 0 \), this simplifies to \( B = 1 \).

Ensure Continuity at x = 0

We have already determined that \( f(x) \) is continuous at \( x = 0 \) if \( B = 1 \). Now, we need to verify if there are any other points where continuity could be affected, specifically the endpoints \( x = -\pi \) and \( x = \pi \). Since the definition of \( f(x) \) includes these points within each piece’s respective domain, additional continuity conditions are not required because they don’t influence the uniform convergence on the closed interval.

Confirm Continuity on (-π, π) with A and B

Next, confirm the condition for \( A \) by considering the smooth transition from \( A x + B \) to \( \cos x \). Since the derivatives must also match at \( x = 0 \) for smoothness, calculate from the left (\( Ax + B \)) and right (\( \cos x \)) limits of the derivative at \( x = 0 \). The derivative from the left is \( f'(0^-) = A \) and from the right, it is \( f'(0^+) = -\sin(0) = 0 \). Therefore, \( A = 0 \).

Key concepts

Uniform Convergence

Uniform convergence is a concept that ensures the Fourier series of a function, like our piecewise function \(f(x)\), behaves nicely over the entire interval \([-, ]\). When we talk about convergence, it's not just enough for the sequence of functions to converge at each point, but it needs to converge uniformly. This means the speed of convergence should be consistent across all parts of the interval.

For uniform convergence, the function \(f(x)\) must be continuous on the interval or meet certain conditions at points of discontinuity. In our exercise, ensuring uniform convergence requires that \(f(x)\) be continuous at the point \(x = 0\), where the function changes its definition. By ensuring this, we can say the Fourier series converges uniformly to \(f(x)\). To achieve this, we determined that \(A = 0\) and \(B = 1\), maintaining continuity and satisfying the uniform convergence requirement.

Piecewise Functions

Piecewise functions are a type of function defined by multiple sub-functions, each applying to a certain interval of the independent variable's domain. In our case, the function \(f(x)\) is divided into two parts: one part where \(f(x) = Ax + B\) for \(- \leq x < 0\), and another where \(f(x) = \cos x\) for \(0 \leq x \leq \).

The key to handling piecewise functions is ensuring a smooth transition between their defined intervals. This generally means matching the values and slopes at the points where the pieces meet. For our exercise, continuity at \(x = 0\), where the piecewise definition switches, is essential. By doing so, we ensure that the Fourier series can represent the entire function smoothly without abrupt changes, facilitating uniform convergence.

Function Continuity

Continuity in mathematics describes a situation where a function does not have abrupt changes or gaps. A continuous function can be drawn without lifting the pen from the paper. For a piecewise function like \(f(x)\), which switches its rule, ensuring continuity at the point of change is crucial.

To achieve this in our exercise, we needed to ensure that both parts of the piecewise function meet at \(x = 0\) without a break or jump. This achieved by setting \(A = 0\) and \(B = 1\), making \(Ax + B\) becomes \(x + 1\) at \(x = 0\), matching the value of \(\cos(0)\) which is also 1. Achieving this continuity not only facilitates uniform convergence but also ensures that the function remains well-defined and predictable across its entire domain.