Question

Find the Fourier sine series. $$ f(x)=e^{x} ; \quad[0, \pi] $$

Short answer

Question: Find the Fourier sine series of the given function: \(f(x) = e^x\) on the interval [0, π]. Answer: The Fourier sine series of the given function is: $$ f_n(x) = \sum_{n=1}^{\infty} \left(\frac{1}{\pi n}\left(1 + \frac{(-1)^n -1}{2n^2}\right)\right) \sin(2nx) $$

Step-by-step solution

Define the function, domain, and relevant variables

We are given the function: $$ f(x) = e^x, \quad[0, \pi] $$ We are asked to find its Fourier sine series representation on the interval \([0, \pi]\). To find the Fourier sine series, we will use the general formula for Fourier sine series representation and calculate the coefficients required.

Determine the general formula for Fourier sine series

The general formula for the Fourier sine series is given by: $$ f_n(x) = \sum_{n=1}^{\infty} b_n \sin\left(\frac{n\pi x}{L}\right) $$ where \(b_n\) are the coefficients, and \(L\) is half of the interval length.

Find the coefficients required for the Fourier sine series using integration

To find the \(b_n\) coefficients, use this formula: $$ b_n = \frac{2}{L} \int_{0}^{L} f(x) \sin\left(\frac{n\pi x}{L}\right) dx $$ First, find the value of L, which is half of the interval length: $$ L = \frac{\pi - 0}{2} = \frac{\pi}{2} $$ Substitute the given function and the value of L into the \(b_n\) formula: $$ b_n = \frac{2}{\pi/2} \int_{0}^{\pi/2} e^x \sin\left(\frac{2n\pi x}{\pi}\right) dx $$ Simplify the expression: $$ b_n = \frac{4}{\pi} \int_{0}^{\pi/2} e^x \sin(2nx) dx $$ Then, compute the integral using integration by parts. Let \(u = e^x\) and \(dv = \sin(2nx)dx\). Differentiating u and integrating dv, we get \(du = e^xdx\) and \(v = \frac{-1}{2n} \cos(2nx)\). Apply integration by parts formula: $$ b_n = \frac{4}{\pi} \left(\left[uv\right]_{0}^{\pi/2} - \int_{0}^{\pi/2} v du\right) $$ Substitute u, v, and du: $$ b_n = \frac{4}{\pi}\left(\left[e^x \frac{-1}{2n} \cos(2nx)\right]_{0}^{\pi/2} - \frac{-1}{2n} \int_{0}^{\pi/2} e^x \cos(2nx) dx\right) $$ Calculate the integral again using integration by parts. Let \(u = e^x\) and \(dv = \cos(2nx)dx\). Differentiating u and integrating dv, we get \(du = e^xdx\) and \(v = \frac{1}{2n} \sin(2nx)\). Apply the integration by parts formula for the integral inside: $$ b_n = \frac{4}{\pi}\left(\left[e^x \frac{-1}{2n} \cos(2nx)\right]_{0}^{\pi/2} + \frac{1}{4n^2} \left(\left[uv\right]_{0}^{\pi/2} - \int_{0}^{\pi/2} v du\right)\right) $$ Substitute u, v, and du: $$ b_n = \frac{1}{\pi n} \left(-\frac{1}{2}\cos(n\pi)e^{\frac{n\pi}{2}} + \frac{1}{2}\cos(0) + \frac{(\sin(n\pi)-0)e^{\frac{n\pi}{2}}}{2n}-\frac{1}{2n^2}b_n\right) $$ Solve for \(b_n\): $$ b_n = \frac{1}{\pi n}\left(1 + \frac{(-1)^n -1}{2n^2}\right) $$ Note that for each \(n\) not equal to 0 the integral converges and the solution above becomes valid.

Plug the coefficient values into the Fourier sine series formula

Now, substitute the \(b_n\) values back into the general Fourier sine series formula: $$ f_n(x) = \sum_{n=1}^{\infty} \left(\frac{1}{\pi n}\left(1 + \frac{(-1)^n -1}{2n^2}\right)\right) \sin\left(\frac{n\pi x}{L}\right) $$

Provide the final equation for the Fourier sine series of the given function

Our final equation for the Fourier sine series of the given function is: $$ f_n(x) = \sum_{n=1}^{\infty} \left(\frac{1}{\pi n}\left(1 + \frac{(-1)^n -1}{2n^2}\right)\right) \sin(2nx) $$

Key concepts

Integration by Parts

Integration by parts is a powerful method for solving integrals, especially when dealing with products of functions. It's based on the product rule of differentiation and provides a formula to tackle the integral of a product of two functions. The formula is given by:

• \( \int u \, dv = uv - \int v \, du \)

Where:

• \( u \) is a function which is easy to differentiate.
• \( dv \) is a function which is easy to integrate.

In the context of finding Fourier sine series, particularly with functions like \( e^x \sin(x) \), integration by parts is used iteratively. In this exercise, we start by choosing \( u = e^x \) and \( dv = \sin(2nx) \, dx \). By differentiating \( u \) and integrating \( dv \), we gradually simplify the integral and find the coefficients needed for the series. This approach helps when direct integration is challenging due to complex products in the integrand.

Fourier Coefficients

Fourier coefficients are fundamental in expressing a function as a series of trigonometric terms. In Fourier sine series specifically, these coefficients, \( b_n \), are responsible for the weights applied to each sine term in the series.For the sine series, the coefficients \( b_n \) are determined by the formula:

• \( b_n = \frac{2}{L} \int_{0}^{L} f(x) \sin\left(\frac{n\pi x}{L}\right) \, dx \)

Where:

• \( L \) represents half the length of the interval in which we are interested.
• \( f(x) \) is the function we want to represent with the series.

Calculating these coefficients involves integrating the product of \( f(x) \) and the sine functions over the given interval. In the context of the exercise, \( b_n \) tells us how each sine wave contributes to approximating the original exponential function.

Trigonometric Series

Trigonometric series are a way of representing functions as sums of sine and cosine terms. They form the foundation for Fourier series, including the Fourier sine series we are addressing here.Within the interval \([0, \pi] \), any sufficiently "nice" function can be broken down into a sum of sines (as in this case) or sines and cosines. These series are instrumental in analyzing periodic behavior and solving differential equations.A trigonometric series like the one encountered in the exercise, composed solely of sine functions, is especially useful for problems with boundary conditions, where the function is zero at \( x = 0 \) or \( x = \pi \). The Fourier sine series gives an approximation of non-periodic functions over a finite interval by adjusting the sine wave's amplitude, frequency, and phase. This helps in capturing the function's details with growing accuracy as more terms are included.

Boundary Value Problem

Boundary value problems are a type of differential equation with solutions defined based on certain conditions set at the boundaries of the domain.In such problems, the function values or their derivatives are specified on the boundaries of the interval being examined. For example, you might be required to determine a function that satisfies a given differential equation on \([0, \pi] \) and also meets specific values or derivatives at \( x = 0 \) and \( x = \pi \).Fourier sine series are particularly suited for boundary value problems where function values vanish at both ends of the interval due to the fundamental properties of sine functions:

• Sine functions are naturally \(0\) at \(x=0\) and \(x=\pi\).
• Their series form does not include constant terms, making them perfect for intervals where the function naturally returns to zero at the boundaries.

In the exercise, finding the Fourier sine series helps solve these types of boundary problems efficiently, offering a way to express the solution in terms of basic trigonometric functions.