Question

Let \(f \in E\) be an even function satisfying \(\int_{-\pi}^{\pi} f(t) d t=5\). Define the function \(F\) by $$ F(x)=\int_{-\pi}^{x} f(t) d t, \quad-\pi \leq x \leq \pi . $$ Let $$ F(x) \sim \frac{A_{0}}{2}+\sum_{n=1}^{\infty}\left[A_{n} \cos n x+B_{n} \sin n x\right] $$ denote the Fourier series of \(F\) and set $$ G(x)=\frac{A_{0}}{2}+\sum_{n=1}^{\infty}\left[A_{n} \cos n x+B_{n} \sin n x\right] $$ Calculate \(G(-\pi), G(\pi)\), and \(G(0)\).

Short answer

\( G(-\pi) = G(0) = G(\pi) = \frac{5}{2} \).

Step-by-step solution

Understand Even Function Integral

Since \( f \) is an even function, its integral from \( -\pi \) to \( \pi \) is given as \( \int_{-\pi}^{\pi} f(t) \, dt = 5 \). Even functions have symmetric properties, meaning \( \int_{-\pi}^{0} f(t) \, dt = \int_{0}^{\pi} f(t) \, dt = \frac{5}{2} \).

Define the Function F

The function \( F(x) \) is defined as the integral of \( f(t) \) from \( -\pi \) to \( x \): \( F(x) = \int_{-\pi}^{x} f(t) \, dt \), with \( -\pi \leq x \leq \pi \). This integral represents the accumulated area under the curve of \( f(t) \) from \( -\pi \) to \( x \).

Express the Fourier Series Components for F

The Fourier series for \( F(x) \) is given as \( F(x) \sim \frac{A_0}{2} + \sum_{n=1}^{\infty}\left[A_n \cos(nx) + B_n \sin(nx)\right] \). Note that the function \( F(x) \) derived from an even function \( f(t) \) will have even symmetry properties: \( F(-x) = -F(x) \).

Calculate Coefficients for G(x)

Since \( f(t) \) is even, \( B_{n} = 0 \) for all \( n \) due to the symmetry (sine terms integrate to zero over a symmetric interval). The constant term \( A_0 = \int_{-\pi}^{\pi} f(t) \, dt = 5 \). Thus, \( G(x) = \frac{5}{2} + \sum_{n=1}^{\infty} A_n \cos(nx) \).

Calculate G(-π), G(0), and G(π)

For \( G(-\pi) \), \( \, G(-\pi) = \frac{A_0}{2} = \frac{5}{2} \), since cosine terms for \( n \geq 1 \) evaluate to zero at \( x = -\pi \) due to the periodicity and symmetry. Similarly, at \( x = \pi \), \( G(\pi) = \frac{5}{2} \). For \( G(0) \), \( G(0) = \frac{5}{2} \) as cosine terms nullify due to the cosine of zero being \( 1 \) but effects cancel out since Fourier cosine terms rearrange back to fundamental value \( A_0/2 \).

Key concepts

Integral Transforms

Integral transforms are mathematical operations that convert a given function into another function. This is achieved through a process called integration, which helps analyze the properties of the original function in a different domain. One well-known integral transform is the Fourier Transform, but in the context of this exercise, we are dealing with the Fourier series as a form of representing functions.

Fourier series is a powerful tool used to approximate periodic functions by breaking them down into a series of sine and cosine terms. This representation is crucial for both theoretical and practical applications, as it allows for easier manipulation and analysis. In this problem, the function \(F(x)\) is derived through an integral that accumulates the values of an even function \(f(t)\) over an interval.

The integral transform process involves calculating definite integrals over specified intervals, converting the function into a new form that retains essential characteristics, such as periodicity and symmetry. By understanding the integral transform of \(F(x)\), you can explore its properties and representations in terms of its Fourier coefficients.

Even Functions

Even functions have a special symmetry about the y-axis. This means that for an even function \(f\), \(f(x) = f(-x)\) for all values of \(x\). This symmetry simplifies many calculations, especially integrals over symmetric intervals.

For example, given the integral \(\int_{-\pi}^{\pi} f(t) dt = 5\), the even property of the function allows us to assert that \(\int_{-\pi}^{0} f(t) dt = \int_{0}^{\pi} f(t) dt = \frac{5}{2}\). This split of the integral into two equal parts is possible because the function mirrors itself across the y-axis.

When dealing with Fourier series, even functions result in a simplification: the sine terms (which represent the odd components of the function) integrate to zero over symmetric intervals for even functions. Therefore, the Fourier series of an even function will only include cosine terms in its expansion, making the analysis and computation straightforward. This principle is evident in this exercise, where the derived function \(F(x)\) from an even \(f(t)\) only contains cosine terms in its Fourier series.

Fourier Coefficients

To express a function in its Fourier series, we decompose it into a series of sine and cosine terms with coefficients commonly referred to as Fourier coefficients. These coefficients determine the amplitude of the sinusoidal components that sum up to represent the original function. In this exercise, the Fourier series of \(F(x)\) is expressed as:

• \(\frac{A_0}{2}\) is the average value of the function over one period.
• \(A_n\) and \(B_n\) are the coefficients for the cosine and sine terms, respectively.

For an even function such as \(f(t)\), the sine terms \(B_n\) are zero because their contributions over symmetric intervals cancel out. Hence, the series only involves terms with \(A_n\), simplifying to:

• \(G(x) = \frac{A_0}{2} + \sum_{n=1}^{\infty} A_n \cos(nx)\)

In this problem, the value \(A_0\) is derived from the integral of the original function \(f(t)\) over the interval \([-\pi, \pi]\), which is equal to 5, making \(\frac{A_0}{2} = \frac{5}{2}\). Consequently, \(G(x)\) evaluates to constant values at \(x = \pm \pi\) and \(x = 0\) since higher-order cosine terms vanish due to periodicity properties, leaving \(G(x) = \frac{5}{2}\) at these points.