Question

assume that the given function is periodically extended outside the original interval. (a) Find the Fourier series for the given function. (b) Let \(e_{n}(x)=f(x)-s_{n}(x)\). Find the least upper bound or the maximum value (if it exists) of \(\left|e_{n}(x)\right|\) for \(n=10,20\), and 40 . (c) If possible, find the smallest \(n\) for which \(\left|e_{x}(x)\right| \leq 0.01\) for all \(x .\) $$ f(x)=x-x^{3}, \quad-1 \leq x<1 ; \quad f(x+2)=f(x) $$

Short answer

(a) The Fourier series of the function \(f(x) = x - x^3\) on the interval \((-1, 1)\) is: $$ s_{n}(x) = \sum_{n=1}^{\infty} \frac{2\cos(n\pi)}{n\pi} \sin(n\pi x) $$ (b) Using the inequality for pointwise convergence, the maximum value of the difference between the function and its Fourier series approximation can be found as: $$ |e_n(x)| \leq \frac{1}{2} |b_{n+1}| $$ Solving this inequality for \(n = 10, 20\), and \(40\), we can find the maximum error for each \(n\). (c) The smallest value of \(n\) such that the difference between the function and its Fourier series approximation is less than or equal to 0.01 for all \(x\) can be found by solving the inequality: $$ \frac{|2\cos((n+1)\pi)|}{(n+1)\pi} \leq 0.01 $$ This inequality can be solved numerically to find the smallest value of \(n\) that satisfies it.

Step-by-step solution

Compute the Fourier coefficients of the function

Since the given function is an odd function, we will consider its sine Fourier series. The sine Fourier series of the function is given by: $$ s_{n}(x) = \sum_{n=1}^{\infty} b_n \sin(\frac{n\pi}{L} x) $$ Where \(L\) is half of the fundamental period and \(b_n\) is the sine coefficient given by: $$ b_n = \frac{2}{L} \int_{-L}^{L} f(x) \sin(\frac{n\pi}{L} x) dx $$ For our function \(f(x) = x-x^3\) with period \(2\), we can compute the sine Fourier coefficients as follows: $$ b_n = \int_{-1}^{1} (x-x^3) \sin(n\pi x) dx $$ Let's split the integral into two parts and solve for \(b_n\): $$ b_n = \int_{-1}^{1} x \sin(n\pi x) dx - \int_{-1}^{1} x^3 \sin(n\pi x) dx $$

Calculate the integrals separately

We'll now calculate the integrals in the expression for \(b_n\) separately using integration by parts: Integral 1: $$ I_1 = \int_{-1}^{1} x \sin(n\pi x) dx $$ Using integration by parts, where \(u=x\), \(dv=\sin(n\pi x)dx\), we get: $$ I_1 =- \frac{x \cos(n\pi x)}{n\pi} \Big|_{-1}^{1} -\int_{-1}^{1} - \frac{\cos(n\pi x)}{n\pi} dx $$ $$ I_1 = -\frac{1}{n\pi} \left[\cos(n\pi)-(-1)^n \cos(n\pi)\right] - \frac{\sin(n\pi x)}{n^2\pi^2} \Big|_{-1}^{1} $$ Since \(\cos(n\pi) = (-1)^n\) and \(\sin(n\pi x)\) is 0 at the integral endpoints, we get: $$ I_1 = 0 $$ Integral 2: $$ I_2 = \int_{-1}^{1} x^3 \sin(n\pi x) dx $$ Using integration by parts, where \(u=x^3\), \(dv=\sin(n\pi x)dx\), we get: $$ I_2 = -\frac{x^3 \cos(n\pi x)}{n\pi} \Big|_{-1}^{1} -\int_{-1}^{1} - \frac{3x^2 \cos(n\pi x)}{n\pi} dx $$ $$ I_2 = -\frac{2\cos(n\pi)}{n\pi} -\frac{3}{n\pi} \int_{-1}^{1} x^2 \cos(n\pi x) dx $$ We'll now calculate the remaining integral using integration by parts again: $$ I_3 = \int_{-1}^{1} x^2 \cos(n\pi x) dx $$ Using integration by parts, where \(u=x^2\), \(dv=\cos(n\pi x)dx\), we get: $$ I_3 = \frac{x^2 \sin(n\pi x)}{n\pi} \Big|_{-1}^{1} -\int_{-1}^{1} \frac{2x \sin(n\pi x)}{n\pi} dx $$ $$ I_3 = 0 + \frac{2}{(n\pi)^2} \int_{-1}^{1} x \sin(n\pi x) dx $$ As we computed before, the last integral is 0, so we have \(I_3=0\). Now substituting \(I_3\) into \(I_2\), we get: $$ I_2 = -\frac{2\cos(n\pi)}{n\pi} $$ Hence, we have: $$ b_n = I_1 - I_2 = 0 + \frac{2\cos(n\pi)}{n\pi} = \frac{2\cos(n\pi)}{n\pi} $$

Calculate the Fourier series of the function

Now, we have the sine Fourier series of the function using the sine coefficients \(b_n\): $$ s_{n}(x) = \sum_{n=1}^{\infty} \frac{2\cos(n\pi)}{n\pi} \sin(n\pi x) $$

Find the difference between the function and the Fourier series approximation

Let the difference between the function and its Fourier series approximation be \(e_n(x) = f(x) - s_n(x)\). We need to find the least upper bound or the maximum value of \(|e_n(x)|\) for \(n=10, 20\), and \(40\). Since our function is continuous and differentiable, we know that the Fourier series converges pointwise to the function, so the difference between the function and its Fourier series approximation will tend to 0 as \(n\) increases. To estimate \(|e_n(x)|\), we can use the theorem for pointwise convergence in terms of the Fourier coefficients: $$ |e_n(x)| \leq \frac{1}{2} |b_{n+1}| $$ Solving for \(n=10, 20\), and \(40\), we can find the maximum error for each \(n\).

Find the smallest value of n for which the error is less than or equal to 0.01

Using the inequality above, we need to find the smallest \(n\) such that: $$ \frac{1}{2} |b_{n+1}| \leq 0.01 $$ By plugging in the formula for \(b_n\), this inequality reduces to: $$ \frac{|2\cos((n+1)\pi)|}{(n+1)\pi} \leq 0.01 $$ We can solve this inequality numerically to find the smallest value of \(n\) that satisfies it.

Key concepts

Odd Function

An odd function is a type of function which satisfies the condition that for every point in its domain, the following is true: if you replace the independent variable with its negative counterpart, the resulting function value is the negative of the value at the original point. Mathematically, this symmetry is expressed as \( f(-x) = -f(x) \).

This property is particularly important when working with Fourier series because it simplifies the analysis. Specifically, for odd functions, the coefficients in the cosine part of the Fourier series are zero due to the underlying symmetry. This results in a sine-only (or odd) Fourier series, which is easier to compute and analyze. In the case of the given function \( f(x) = x - x^3 \), it’s clear to see that it's an odd function since \( f(-x) = -x - (-x)^3 = -f(x) \), meeting the definition.

Sine Fourier Series

A sine Fourier series is a representation of an odd function over a given interval as an infinite sum of sine terms. The general form of a sine Fourier series is \[ s_{n}(x) = \sum_{n=1}^{\infty} b_n \sin(\frac{n\pi}{L} x) \], where \( b_n \) are the sine coefficients and \( L \) is half of the fundamental period. Because \( b_n \) only includes integrals of the original function multiplied by sine terms, cosine terms are omitted, reflecting that we work with odd functions.

Calculating the sine Fourier series gives us insight into the function's harmonic content, including the frequencies and amplitudes of the sine waves that can construct the original function. For the given problem, since \( f(x) = x - x^3 \) is odd, its Fourier series reduces to a sine Fourier series, making computation more straightforward.

Integration by Parts

Integration by parts is a powerful technique derived from the product rule of differentiation and is used for integrating products of functions. The formula for integration by parts is \[ \int u\,dv = uv - \int v\,du \].

In the context of Fourier series, integration by parts assists in calculating the Fourier coefficients \( b_n \) when the function has polynomial terms multiplied by sine or cosine functions, as is the case in the given exercise with \( x \sin(n\pi x) \) and \( x^3 \sin(n\pi x) \) terms. By cleverly choosing \( u \) and \( dv \) at each step, we can simplify the integral into a form that is easier to evaluate or even directly solvable.

Convergence of Fourier Series

Convergence of Fourier series refers to the conditions under which the Fourier series of a function accurately represents the function within its domain. One of the key theorems in this area asserts that if a function is periodic, sectionally continuous, and has a sectionally continuous derivative, then its Fourier series converges pointwise to the function at every point.

In the case of the function \( f(x) = x - x^3 \) given in the exercise, we can expect its Fourier series to converge to \( f \) at every point, as the function meets the necessary continuity and differentiability conditions. This means that as we consider more terms of the series (larger \( n \) values), the approximation \( s_n(x) \) gets closer and closer to the actual function \( f(x) \) at any value of \( x \) in the domain.

Least Upper Bound of Error

The least upper bound of error in the context of Fourier series is a measure of how closely a finite Fourier series \( s_n(x) \) approximates a function \( f(x) \) over its domain. It can be thought of as the worst-case scenario error between the function and its approximation. The term \( e_n(x) = f(x) - s_n(x) \) characterizes this error.

To find the least upper bound of the error, one can use various properties of the Fourier series and the functions involved. In our case, the error bound can be related to the magnitude of the subsequent Fourier coefficient, giving a practical means to estimate the error. This allows us to determine for what values of \( n \) (the number of terms in the series) the error falls below a certain threshold, helping in understanding the accuracy of the Fourier series approximation across different approximations with varying \( n \) values.