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 \(x .\) $$ f(x)=\left\\{\begin{array}{lll}{0,} & {-1 \leq x<0,} & {f(x+2)=f(x)} \\\ {x^{2},} & {0 \leq x<1 ;} & {f(x+2)=f(x)}\end{array}\right. $$

Short answer

Question: Find the Fourier series and error term of the following function: Function: \(f(x) = 0\) for \(-1 \leq x < 0\) and \(f(x) = x^2\) for \(0 \leq x < 1\) Answer: The Fourier series of the function is as follows: $$ s_n(x) = \frac{1}{6} + \sum_{n=1}^{\infty} \left[ \frac{((-1)^n - 1) (\pi^2 n^2 - 6) }{6\pi^2n^2} \cos(n\pi x) + \frac{2}{1 - (-1)^n}\cdot\frac{1}{n\pi} \sin(n\pi x) \right] $$ The maximum error term for \(n = 10, 20, 40\) is \(\left|e_{n}(x)\right| \leq \frac{1}{2}\). However, no value of \(n\) exists where the error term is \(\left|e_{x}(x)\right| \leq 0.01\) for all \(x\), as the error term converges to \(\frac{1}{2}\) when approaching discontinuity points.

Step-by-step solution

Given that the function is periodic with \(f(x + 2) = f(x)\), we can determine that this is a half-wave periodic function with the period \(T = 2\). #Step 2: Find the Fourier coefficients a_n and b_n#

As it is a half-wave periodic function, the Fourier coefficients are calculated as follows: $$ a_0=\frac{1}{T}\int_{-T/2}^{T/2} f(x) dx $$ $$ a_n=\frac{2}{T}\int_{0}^{T/2}f(x)\cos(\frac{n\pi x}{T/2})dx $$ $$ b_n=\frac{2}{T}\int_{0}^{T/2}f(x)\sin(\frac{n\pi x}{T/2})dx $$ #Step 3: Calculate a_0, a_n and b_n#

Now, we will compute the values of \(a_0\), \(a_n\), and \(b_n\). For \(a_0\): $$ a_0 = \frac{1}{2}\int_{-1}^{1} f(x) dx = \frac{1}{2}\int_{-1}^{0} 0 dx + \frac{1}{2}\int_{0}^{1} x^2 dx = \frac{1}{6} $$ For \(a_n\): $$ a_n = \frac{2}{2}\int_{0}^{1}x^2 \cos(n\pi x) dx = \int_{0}^{1} x^2\cos(n\pi x) dx $$ For \(b_n\): $$ b_n = \frac{2}{2}\int_{0}^{1}x^2 \sin(n\pi x)dx = \int_{0}^{1} x^2\sin(n\pi x) dx $$ To compute \(a_n\) and \(b_n\), we have to integrate the given functions. $$ a_n = \frac{((-1)^n - 1) (\pi^2 n^2 - 6) }{6\pi^2n^2} $$ $$ b_n = \frac{2}{1 - (-1)^n}\cdot\frac{1}{n\pi} $$ #Step 4: Write down the Fourier series#

Now we can write down the Fourier series of the function \(f(x)\): $$ s_n(x) = \frac{1}{6} + \sum_{n=1}^{\infty} \left[ \frac{((-1)^n - 1) (\pi^2 n^2 - 6) }{6\pi^2n^2} \cos(n\pi x) + \frac{2}{1 - (-1)^n}\cdot\frac{1}{n\pi} \sin(n\pi x) \right] $$ #Step 5: Find the maximum error term for n=10, 20, 40 #

We have \(e_n(x)=f(x)-s_n(x)\), and we need to find the largest value of the error term \(\left|e_{n}(x)\right|\) for \(n=10, 20\), and \(40\). Since the Fourier series converges to the average of the jumps at the discontinuities and the given function has discontinuities at the end points, we can use the formula of the maximum error term as follows: $$ \left|e_{n}(x)\right| \leq \frac{1}{2}\left|f(1)-f(-1)\right| = \frac{1}{2} \cdot 1 = \frac{1}{2} $$ Hence, for all values of \(n=10, 20\) and \(40\), the maximum error term, \(\left|e_{n}(x)\right| \leq \frac{1}{2}\). #Step 6: Find the smallest n for which the error term is less than or equal to 0.01 #

The question asks to find smallest \(n\), such that \(\left|e_{x}(x)\right| \leq 0.01\) for all \(x\). However, as x approaches the discontinuity points, the error term converges to the half of the jump of the function at the discontinuity, which in this case, as shown in step 5, would always be \(\frac{1}{2}\). Therefore, a value of \(n\) where the error term is less than or equal to \(0.01\) for all \(x\) does not exist in this case.

Key concepts

Understanding Periodic Functions

Periodic functions are the bread and butter of Fourier analysis. They repeat their values in regular intervals or periods. Think of the motion of a pendulum or the cycles of day and night; they all have a pattern that repeats over time. The mathematical definition is just as straightforward: a function f(x) is said to be periodic with period T, if for all values of x, it satisfies the condition that \(f(x+T) = f(x)\).

In the given exercise, the function \(f(x)\) is defined piecewise, with different expressions in different intervals, but it is made periodic by extending it outside the original interval with the property \(f(x+2) = f(x)\), making 2 the period. When dealing with such functions, it's crucial to analyze behavior within one period before extending it to the entire real line through periodicity.

Demystifying Fourier Coefficients

Fourier coefficients are essentially the 'DNA' of a periodic function. They uniquely describe the function's behavior and form the building blocks of the Fourier series. These coefficients, \(a_{n}\) and \(b_{n}\), are found using the inner product of the function with sine and cosine functions—a process akin to projecting the function onto the different frequencies of the sine and cosine waves.

To calculate them, integrals are used over one period of the function, as shown in the given textbook solution, where coefficients are computed for a half-wave rectified function. These coefficients tell us how much each 'harmonic' (sine or cosine wave with a particular frequency) contributes to the makeup of our original function. So, by finding the Fourier coefficients, we can reconstruct the periodic function using a sum of sines and cosines, which is what the Fourier series is all about.

Decoding the Error Term in Fourier Series

When we approximate a function using a Fourier series, there’s naturally a discrepancy between the original function and its Fourier representation. This difference is known as the error term, \(e_n(x) = f(x) - s_n(x)\), where \(s_n(x)\) is the partial sum of the first 'n' terms of the Fourier series. The goal often is to make the error term as small as possible; in other words, to get the approximation as close to the original function as we can.

In the solution provided in the textbook, finding the least upper bound for the error term involves considering the discontinuities of the function, which is where the largest errors usually occur. The maximum error is related to the jump discontinuity of the function—a concept known as the Gibbs phenomenon. However, for some functions or points of discontinuity, as suggested by the solution's step 6, the error does not become smaller than a certain bound regardless of how many terms of the series we include. It's a profound insight into the limitations of Fourier series when it comes to approximating functions, especially at points of discontinuity.