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Chapter 10: Problem 28

A function is given on an interval \(0

Short Answer

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Question: Sketch the even (g(x)) and odd (h(x)) extensions of a piecewise function f(x) on the interval \(0<x<2\), find its Fourier cosine and sine series, plot partial sums of each series, and investigate the dependence on n of the maximum error on [0, 2] for each series. Answer: To find the even (g(x)) and odd (h(x)) extensions of the function, we need to extend the function periodically and create even and odd versions by mirroring it over the y-axis and origin, respectively. Afterward, we can find the Fourier cosine and sine series by computing their coefficients using the given formulas and constructing the series. We can then plot partial sums of the series and analyze the maximum error dependency on n by computing the error functions and finding their maximum absolute values as n increases.

Step by step solution

01

1. Analyzing f(x)#

First, let's make sense of the given function. It is a continuous piecewise-defined function with two parts: 1. \(f(x)=x\) when \(0<x<1\) 2. \(f(x)=0\) when \(1<x<2\) With L = 2, we can imagine a graph with a straight line connecting (0, 0) and (1, 1), then a horizontal line at y=0 from (1, 0) to (2, 0).
02

2. Finding g(x) and h(x)#

To find the even extension g(x) and the odd extension h(x), we first define the original function for multiple periods. Extend f(x) periodically with period L (in our case, 2). Then, we construct the even and odd extensions as follows: - g(x) is an even function, so it's symmetrical about the y-axis. We can mirror f(x) on the interval [-2, 2] to obtain g(x). - h(x) is an odd function, so it's symmetrical about the origin. To construct h(x), we need to set \(h(0)=0\) and then mirror the function f(x) on the interval [-2, 2] while flipping its sign.
03

3. Fourier Cosine (FCN) and Sine (FSN) series#

Now, let's find the Fourier cosine (FCN) and sine (FSN) series for f(x). The Fourier series is given by: $$ f(x) \approx a_0 + \sum_{n=1}^{\infty} [a_n \cos(\frac{n\pi x}{L}) + b_n \sin(\frac{n\pi x}{L})] $$ To find the coefficients a0, an, and bn, we'll use the formulas: $$ a_0 = \frac{1}{L}\int_{0}^{L} f(x)dx $$ $$ a_n = \frac{2}{L}\int_{0}^{L} f(x)\cos(\frac{n\pi x}{L})dx $$ $$ b_n = \frac{2}{L}\int_{0}^{L} f(x)\sin(\frac{n\pi x}{L})dx $$ We can then construct the FCN and FSN series by summing the a_n terms with the cosine function and b_n terms with the sine function, respectively.
04

4. Plotting partial sums#

Using the FCN and FSN formulas derived in step 3, we can approximate the function by plotting a few partial sums of each series (say, up to N = 5): 1. FCN: Sum the cosine terms up to N = 5. 2. FSN: Sum the sine terms up to N = 5.
05

5. Investigating dependence on n of the maximum error#

Finally, let's investigate how the maximum error on the interval [0, L] depends on n for each series (FCN and FSN). To do this, we can: 1. Compute the error function for each series, which is the difference between the original function f(x) and the approximated series (FCN or FSN). 2. Find the maximum absolute value of the error function for increasing values of n up to a certain value (say N = 10). 3. Examine the trend of the maximum errors as n increases.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

even extension
An even extension of a function is a technique to create a new function that has even symmetry. For a function to be even, it must satisfy the condition \( g(-x) = g(x) \). This means the graph of the function is mirrored around the y-axis. To find the even extension of a given function, you repeat the pattern of the function from the positive side into the negative side without altering the shape, just reflecting it.
  • For the given function \( f(x) = x \) when \( 0 < x < 1 \) and \( f(x) = 0 \) when \( 1 < x < 2 \), we reflect the graph across the y-axis.
  • This reflection creates a function that is x for \(-1 < x < 1\) and zero for \(1 < x < 2\) and \(-2 < x < -1\).
This technique ensures that the resulting function is "complete" with even symmetry over the interval \([-2, 2]\). This is particularly useful in Fourier series, as it allows us to use cosine terms, which are naturally even functions.
odd extension
To create an odd extension of a function, we generate a function that has odd symmetry. An odd function satisfies the condition \( h(-x) = -h(x) \). This requires mirroring the function around the origin and also flipping the sign. In essence, whatever happens on the positive side of the x-axis is inverted on the negative side.
  • In the case of the given function \( f(x) = x \) for \( 0 < x < 1 \), we extend it as \( h(x) = -x \) for \( -1 < x < 0 \).
  • For the interval \( 1 < x < 2 \), since \( f(x) = 0 \), this naturally extends as \( h(x) = 0 \) for \( -2 < x < -1 \).
This odd extension is crucial when obtaining the Fourier sine series, as sine terms are inherently odd functions.
cosine series
The Fourier cosine series is used to represent functions using only cosine terms. Since cosine functions are even, this series is naturally paired with the even extension of a function. The general form of a Fourier cosine series is given by:\[f(x) \approx a_0 + \sum_{n=1}^{\infty} a_n \cos\left(\frac{n\pi x}{L}\right)\]
  • Here, \(a_0\) is the zeroth coefficient, calculated as \( \frac{1}{L}\int_{0}^{L} f(x)dx \).
  • Each \(a_n\) is found using \( a_n = \frac{2}{L}\int_{0}^{L} f(x)\cos\left(\frac{n\pi x}{L}\right)dx \).
This series is particularly efficient when the function itself is or can be made into an even function by extending it appropriately.
sine series
The Fourier sine series expresses a function using only sine terms, which are odd functions. This series is particularly fitting when a function possesses odd symmetry as obtained from its odd extension. The Fourier sine series takes the form:\[f(x) \approx \sum_{n=1}^{\infty} b_n \sin\left(\frac{n\pi x}{L}\right)\]
  • The coefficients \( b_n \) are found using \( b_n = \frac{2}{L}\int_{0}^{L} f(x)\sin\left(\frac{n\pi x}{L}\right)dx \).
Using a sine series is ideal when the initial function is odd or has been extended to have odd symmetry, allowing us to leverage these naturally odd sine functions for constructing a series representation.

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Most popular questions from this chapter

Let an aluminum rod of length \(20 \mathrm{cm}\) be initially at the uniform temperature of \(25^{\circ} \mathrm{C}\). Suppose that at time \(t=0\) the end \(x=0\) is cooled to \(0^{\circ} \mathrm{C}\) while the end \(x=20\) is heated to \(60^{\circ} \mathrm{C},\) and both are thereafter maintained at those temperatures. (a) Find the temperature distribution the rod at any time \(t .\) (b) Plot the initial temperature distribution, the final (steady-state) temperature distribution, and the temperature distributions at two repreprentative intermediate times on the same set of axes. (c) Plot u versus \(t\) for \(x=5,10,\) and \(15 .\) (d) Determine the time interval that must elapse before the temperature at \(x=5 \mathrm{cm}\) comes (and remains) within \(1 \%\) of its steady-state value.

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

Find the solution \(u(r, \theta)\) of Laplace's equation $$ u_{r r}+(1 / r) u_{r}+\left(1 / r^{2}\right) u_{\theta \theta}=0 $$ outside the circle \(r=a\) also satisfying the boundary condition $$ u(a, \theta)=f(\theta), \quad 0 \leq \theta<2 \pi $$ on the circle. Assume that \(u(r, \theta)\) is single-valued and bounded for \(r>a\)

Suppose that we wish to calculate values of the function \(g,\) where $$ g(x)=\sum_{n=1}^{\infty} \frac{(2 n-1)}{1+(2 n-1)^{2}} \sin (2 n-1) \pi x $$ It is possible to show that this series converges, albeit rather slowly. However, observe that for large \(n\) the terms in the series (i) are approximately equal to \([\sin (2 n-1) \pi x] /(2 n-1)\) and that the latter terms are similar to those in the example in the text, Eq. (6). (a) Show that $$ \sum_{n=1}^{\infty}[\sin (2 n-1) \pi x] /(2 n-1)=(\pi / 2)\left[f(x)-\frac{1}{2}\right] $$ where \(f\) is the square wave in the example with \(L=1\) (b) Subtract Eq. (ii) from Eq. (i) and show that $$ g(x)=\frac{\pi}{2}\left[f(x)-\frac{1}{2}\right]-\sum_{n=1}^{\infty} \frac{\sin (2 n-1) \pi x}{(2 n-1)\left[1+(2 n-1)^{2}\right]} $$ The series (iii) converges much faster than the series (i) and thus provides a better way to calculate values of \(g(x) .\)

Consider the conduction of heat in a rod \(40 \mathrm{cm}\) in length whose ends are maintained at \(0^{\circ} \mathrm{C}\) for all \(t>0 .\) In each of Problems 9 through 12 find an expression for the temperature \(u(x, t)\) if the initial temperature distribution in the rod is the given function. Suppose that \(\alpha^{2}=1\) $$ u(x, 0)=50, \quad 0
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