Fourier Series Chapter 3 of Coleman

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1 Fourier Series Chapter 3 of Coleman Dr. Doreen De eon Math 18, Spring 14 1 Introduction Section 3.1 of Coleman The Fourier series takes its name from Joseph Fourier ( ), who made important contributions to the study of trigonometric series after preliminary investigations by eonhard Euler, Jean le Rond d Alembert, and Daniel Bernoulli. In 187, Fourier published his initial results dealing with solving the heat equation in a metal plate, for which he introduced the Fourier series. His later results were published in 18. Before Fourier s work, no solution to the heat equation was known in the general case, although solutions were known for special cases (e.g., if the heat source was a sine or cosine wave). Fourier s idea was to model a complicated heat source as a combination of sine and cosine waves, and write the solution to the problem as a superposition of the corresponding eigenfunctions. This superposition is known as the Fourier series. You might notice that the above more or less follows our formal solution procedure using separation of variables. If you recall, we obtained our solutions by assuming that the series representation converged and could be differentiated and integrated termwise. The goal in this chapter is to determine what properties a function f(x) must satisfy in order for this to be true. In other words, we wish to determine the following: Given a function f(x) on x, is it possible to find constants a n and b n such that ( f(x) a + a n cos nx + b n sin nx ) on x? If this series exists, it will be called the Fourier series of f(x) on x. Properties of Sine and Cosine Section 3. of Coleman Periodicity From your trigonometry class (oh so long ago), you know that the period of sin px and cos px is p. What does this mean? It means that ( sin px + ) p ( sin px and cos px + ) cos px. p 1

2 In general, we have the following definition. Definition. If there exists a number T for which a function f(x) satisfies f(x + T ) f(x) for all x in the domain of f, then we say that f is periodic with period T. The smallest positive period of f is called its fundamental period. Example. Determine the period of cos nx. The period is T n n n. If n 1, then, the period is. Similarly, for any n, the period of cos nx multiple of the fundamental period is also a period). The same, then, is true for sin nx. This means that any finite series of the form F N (x) a + N ( a n cos nx + b n sin nx ) is (since any integer must be a function that is periodic. It stands to reason, then, that if the infinite series ( a + a n cos nx + b n sin nx ) converges to a function f(x), then f(x) has period. Symmetry From trigonometry, we know that: the graph of sin x is symmetric about the origin; and the graph of cos x is symmetric about the y-axis. The same, of course, is true for sin nx nx and cos. Can we obtain information about the symmetry of a function without graphing it? Yes. Definition. A function f(x) is even if its graph is symmetric with respect to the y-axis; that is, if f( x) f(x) for all x in the domain of f. f(x) is odd if its graph is symmetric about the origin; that is, if f( x) f(x) for all x in the domain of f. We can use this to more easily determine if a function is odd or even.

3 Examples. (1) x i, i, ±, ±4,... are all even functions (note the even powers) () x i, i 1, ±3, ±5,... are all odd functions (note the odd powers) (3) Is the function f(x) x sin x even, odd, or neither? f( x) ( x) sin( x) x ( sin x) x sin x f(x). Therefore, f(x) x sin x is odd. We can generalize this to the following. Proposition 1. et h(x) f(x)g(x). Then, 1. h(x) is odd if one of f(x) and g(x) is odd and the other is even.. h(x) is even if both of f(x) and g(x) are either odd or even. Proof. 1. Without loss of generality, suppose that f(x) is even and g(x) is odd. Then h( x) f( x)g( x) f(x)( g(x)) (since f is even and g is odd) f(x)g(x) h(x). Therefore, h(x) is odd, as claimed.. First, assume that both f(x) and g(x) are even. Then h( x) f( x)g( x) f(x)g(x) h(x). Therefore, h(x) is even. Now, assume that both f(x) and g(x) are odd. Then h( x) f( x)g( x) ( f(x))( g(x)) f(x)g(x) h(x). Therefore, h(x) is even. 3

4 Are there any implications for integration? Yes (or I probably would not ask the question). Proposition. (i) If f(x) is even over x, then f(x) dx (ii) If f(x) is odd over x, then f(x) dx f(x) dx. f(x) dx. Proof. We will prove the first statement. The second statement is proven in the text. Suppose that f(x) is even over the interval x. Then for all x [, ], f( x) f(x). Therefore, f(x) dx f(x) dx + f( u) du + f(u) du + f(u) du + f(x) dx + f(x) dx. f(x) dx f(x) dx (u substitution, with u x) f(x) dx (since f is even) f(x) dx f(x) dx (let x u and substitute) Orthogonality We actually discussed this informally, without actually defining orthogonality. We will now be a bit more rigorous. Definition. et f and g be two functions integrable on a x b. Their inner product on a x b is the real number < f, g > b Orthogonality is a characteristic of the inner product. a f(x)g(x) dx. Definition. et f and g be two functions integrable on a x b. Then f and g are orthogonal if their inner product is. 4

5 Thus, we can define the orthogonality of a set of functions. Definition. A set of functions f n (x)} that is integrable on [a, b] is said to be orthogonal if the functions satisfy < f n, f m > whenever m n. This means that the functions f n (x) are pairwise orthogonal. We have already seen that the set of functions given by 1, sin nx nx }, cos is orthogonal, since sin nx mx, sin sin nx mx sin dx, m n,, m n. cos nx sin nx mx, cos, cos mx 1, sin nx 1, cos nx cos nx sin nx sin nx cos nx mx cos dx, m n,, m n. mx cos dx for all m and n. dx for all n. dx for all n. 3 The Fourier Series Section 3.3 of Coleman Supose that we have a function f(x) that is continuous on x and that we would like to represent by the series ( f(x) a + a n cos nx + b n sin nx ). We can formally determine expressions for a, a n, and b n in the following way. To determine b n, multiply both sides of the equation by sin mx to x from to. Then, we obtain f(x) sin mx dx a sin mx dx + a n cos nx and integrate both sides with respect mx sin dx + b n sin nx mx sin dx a 1, sin mx + a n cos nx mx, sin + b n sin nx mx, sin b m b m 1 f(x) sin mx We can determine a and a n similarly. Thus, we obtain the following definition. dx. 5

6 Definition. Given a function f(x) on x, the Fourier series of f on x is given by F (x) a + a n cos nx + b n sin nx, (1) where the coefficients are given by provided these integrals exist. a n 1 b n 1 f(x) cos nx f(x) sin nx dx, n, 1,,..., dx, n 1,,..., Examples. Determine the Fourier series of f(x) on the given interval. (1) f(x) x + 1 on x a 1 (x + 1) dx 1 ( x + x ). a n b n 1 (x + 1) cos nx dx x cos nx dx + 1 cos nx dx (x + 1) sin nx dx x sin nx dx + 1 cos nx dx (since x cos nx is odd) sin nx dx 4 x sin nx dx + (since x sin nx is even) 4 [ 1n x cos nx + 1 ] cos nx dx n 4 [ 1n ] cos n 4 [ 1n ] ( 1)n b n 4 n ( 1)n+1. Therefore, we obtain F (x) 1 + ( 1) n+1 4 sin nx. n 6

7 () f(x), if 1 x,, if < x 1 a a n 1 1. b n f(x) dx dx + 1 dx f(x) cos nx dx cos nx dx + n sin nx f(x) sin nx dx sin nx dx + n cos nx (1 cos( n)) n (1 cos n) n n (1 ( 1)n ) n (( 1)n 1), if n is even, ( ). n, if n is odd cos nx dx sin nx dx Therefore, we obtain or F (x) 1 + F (x) 1 + n (( 1)n 1) sin nx, 4 sin(n 1)x. (n 1) 4 The Fourier Series, Continued Section 3.4 of Coleman In this section, we will begin discussion of the convergence of the Fourier series. 7

8 Definition. Given f(x) and a point x x, not necessarily in the domain of f, we define f(x +) lim f(x) x x + f(x ) lim f(x). x x One more definition before we get to the all-important convergence theorem. Definition. Given a function f(x) with domain a x b, we say that f is piecewise continuous on a x b if 1) f has a finite number of discontinuities on a x b. ) At each point of discontinuity x, with x (a, b), f(x +) and f(x ) both exist and are finite. 3) f(a+) and f(b ) both exist and are finite. If f (x) is also piecewise continuous on a x b, then we say that f is piecewise smooth there. Examples. (1) All continuous functions are piecewise continuous. () All polynomials are piecewise smooth. 3x 6x + 3, < x < 3, (3) f(x) 5, x 3, 5x + x, 3 < x < 7. This function is piecewise smooth, since f (3+) and f (3 ) both exist. A graph of this function is below. 8

9 (4) f(x) x 3 ( is piecewise continuous, ) but it is not piecewise smooth on any interval containing x since f (x) 3 x 1 3. From Calculus, we know that piecewise continuous functions are integrable, and if f(x) is piecewise continuous on a x b with discontinuities at x 1, x,..., x n 1 (and possibly at x a and x n b), then b n xi f(x) dx f(x) dx. a x i 1 This is all we need to know for the big theorem. i1 Theorem 1. If f(x) is piecewise smooth on x, then its Fourier series F (x) converges on x, and (1) F (x) f(x) for all x on x where f is continuous. () F (x) 1 (f(x+) + f(x )) for all x in < x < where f is discontinuous. (3) F ( ) F () 1 (f( +) + f( )). The proof of this theorem is given in Section 3.5. It can also be proven (see Appendix A) that the series converges uniformly, and so we may integrate it term-by-term. Examples. Sketch the function and three periods of the graph of its Fourier series. (1) f(x) x, x, x, x (a) f(x) (b) three periods of F (x) Figure 1: The graphs of f(x) and its Fourier series, F (x). 9

10 () f(x) x 3, x (a) f(x) Figure : The graphs of f(x) x 3 (b) three periods of F (x) and its Fourier series, F (x). Note that the Fourier series ignores points of discontinuity. For example, if we were to change the function in Example (1) to x, < x, x, x <, f(x).5, x,, x, then the Fourier series for f(x) would be the same as the graph in Table 1(b). From earlier, we discussed the fact that the Fourier series has period, but the domain of the function f(x) is x. We can, thus, see that the Fourier series is the graph of f(x) extended periodically. Definition. Given a function f(x) defined on a x b, the function g(x) defined by g(x) f(x), g(x + T ) g(x), a < x < b, and for all x, where T b a is called the periodic extension of f (of period T ). With this in mind, we can restate Theorem 1 as follows. Corollary 1. If f(x) is piecewise smooth on x, and f p (x) is the periodic extension of f(x), then the Fourier series F (x) converges for all x and F (x) f p(x+) + f p (x ) for all x. 1

11 5 The Fourier Series Proof of Pointwise Convergence Section 3.5 of Coleman In this section, Theorem 1 is proved. We will skip the proof in this class, but for those with the math background (at least Math 111, although Math 171 is better), you may want to review it. The Gibbs Phenomenon One naturally expects that the more terms we use in the finite series approximation to the Fourier series, the closer we will get to the function f(x) (where f is continuous). However, for many functions, this is not the case. We get what is known as the Gibbs phenomenon. Although the phenomenon was discovered by Henry Wilbrahim in 1848, it was rediscovered by J. Willard Gibbs, for whom it is named, in Consider a simple square wave f(x) First, we will determine the Fourier series for f(x)., if x,, if x. a 1 f(x) dx 1 ( ) dx + dx. a n 1 f(x) cos nx dx 1 ( ) dx + cos nx dx 1 1 sin nx n. b n 1 f(x) sin nx dx 1 ( ) dx + sin nx dx 1 1 cos nx n 11

12 1 (1 cos n) n 1 n (1 ( 1)n ) 1 n, n odd, n even. So, the Fourier series of f(x) is given by F (x) sin(n 1)x. (n 1) We may approximate the Fourier series by using truncated series of the form F N (x) 4 + N 1 sin(n 1)x (n 1) for various values of N. In the figures below, we plot the truncated Fourier series for N 5, N 5, and N 15. Figure 3: The Gibbs phenomenon for the square wave example. The Gibbs phenomenon shows that if f has jump discontinuities, then its Fourier series will not converge uniformly to f. 1

13 6 Fourier Sine and Cosine Series Section 3.6 of Coleman We also encountered the following questions when we were looking at solution by separation of variables: Given a function f(x) on x, can we find constants A n, n, 1,,..., or B n, n 1,,..., such that either f(x) A + A n cos nx or on x? f(x) B n sin nx The answer will be similar to the previous section, wherein if f(x) is piecewise smooth, we first extend it to a piecewise smooth function on [, ] and then find the Fourier series of this. Question: How do we extend f from x to x? Answer: We know that if g(x) is an even function on x, then its Fourier series is really a cosine series, and if h(x) is an odd function over x, then its Fourier series is really a sine series. So, if we want to expand f in a cosine series, we will want to extend f to an even function on x, and if we want to expand f in a sine series, we will want to extend f to an odd function on x. Definition. Given f(x) on x, the even function f(x), if x, g(x) f( x), if x, is called the even extension of f to x. The odd function f(x), if x, h(x) f( x), if x, is called the odd extension of f to x. Note: If f is piecewise smooth, then so are g and h. We now wish to determine the coefficients for each of these series. If we have an odd function h(x), then its Fourier series will be the sine series given by H(x) b n sin nx. (Why?) Then, from our work with Fourier series, we know that b n 1 h(x) sin nx dx h(x) sin nx dx ( since h(x) sin nx is even ). 13

14 We can determine the contants a and a n, n 1,,..., for the cosine series for the even function g(x) similarly: This gives us the following definition. a 1 a n 1 g(x) dx g(x) dx. g(x) cos nx g(x) cos nx Definition. Given a function f(x) on x, the series F c (x) a + a n cos nx, where a n dx dx. is called the Fourier cosine series of f on x. The series F s (x) b n sin nx, where b n is called the Fourier sine series of f on x. f(x) cos nx f(x) sin nx dx, dx, Convergence properties of the Fourier cosine and sine series follow. Theorem. If f(x) is piecewise smooth on x, then the Fourier sine series of f(x) converges to: (i) f(a) if f(x) is continuous at a (, ); (ii) 1 (f(a ) + f(a+)) if f(x) has a jump discontinuity at a (, ). (iii) 1 (f(+) + f( )) at x and x. If f(x) is continuous and f() f(), then the Fourier sine series of f(x) converges to f(x) on [, ]. Theorem 3. If f(x) is piecewise smooth on [, ], then the Fourier cosine series of f(x) converges to: (i) f(a) if f(x) is continuous at a (, ); (ii) 1 (f(a ) + f(a+)) if f(x) has a jump discontinuity at a (, ). (iii) 1 (f(+) + f( )) at x and x. If f(x) is continuous, then the Fourier cosine series of f(x) converges to f(x) on [, ]. 14

15 Examples. Calculate the Fourier sine and Fourier cosine series of f(x). (1) f(x) 1, if x <,, if x 4. Fourier cosine series: 4 a f(x) dx 4 1 ( 1 dx Therefore, the Fourier cosine series of f(x) is 4 ) dx a n f(x) cos nx 4 4 dx 1 ( cos nx ) 4 dx nx sin n 4 n sin n. F c (x) 1 + n sin n (n 1)x cos. 4 Fourier sine series: 4 Therefore, the Fourier sine series of f(x) is b n f(x) sin nx 4 4 dx 1 ( sin nx ) 4 dx nx cos n 4 (cos n ) n 1 ( 1 cos n ). n F s (x) ( 1 cos n n ) sin nx 4. 15

16 () f(x) x 1, x Fourier cosine series: a (x 1) dx 1 (x 1). a n (x 1) cos nx dx x cos nx dx nx x sin n n ( ) + cos nx n ( ) (cos n 1) n 4 (n) (( 1)n 1). Therefore, the Fourier cosine series of f(x) is If we observe that F c (x) 4 (n) (( 1)n 1) So, let n k 1. Then, we have F c (x) If we then let k n, we obtain F c (x) 8 cos nx dx sin nx dx n 4 (n) (( 1)n 1) cos nx. (n), if n is odd,, if n is even. sin nx 8 (k 1)x cos. ((k 1)) k1 8 (n 1)x cos. ((n 1)) cos nx dx 16

17 Fourier sine series: b n n (x 1) sin nx dx x sin nx dx sin nx dx nx x cos + cos nx n dx 4 ( ) cos n + sin nx + n n n 4 cos n + (cos n 1) n n (cos n + 1) n n (( 1)n + 1) Therefore, the Fourier sine series of f(x) is F s (x) n (( 1)n + 1) sin nx. cos nx sin nx dx We note that b n 4 n, if n is even,, if n is odd. Therefore, if we let k n, we can rewrite the Fourier sine series of f(x) as F s (x) Now, let k n to obtain F s (x) k1 k1 4 kx sin k sin kx. k sin nx. n Example. Sketch the function f(x) and its Fourier sine and Fourier cosine series, where 1, if x 3, f(x), if 3 < x 5. 17

18 (a) f(x) (b) three periods of F c (x) (c) three periods of F s (x) Figure 4: The Fourier cosine and Fourier sine series. Differentiation and Integration of Fourier Series Term by Term Differentiation Fourier series Suppose f(x), defined on the interval [, ], is piecewise smooth on [, ] and the Fourier series of f, F (x), is continuous (including the endpoints). Then its Fourier series can be differentiated term by term. For the Fourier series of f to be continuous, it is sufficient to assume that f(x) is continuous and f( ) f(). Fourier cosine series: Suppose f(x) is defined on the interval [, ]. Then the Fourier cosine series of f can be differentiated term by term if f is continuous on [, ] and f is piecewise smooth on [, ]. Fourier sine series: Suppose f(x) is defined on the interval [, ]. Then the Fourier sine series of f can be differentiated term by term if f is continuous on [, ], f is piecewise smooth on [, ], and f() f(). Term by Term Integration Fourier series: Suppose f(x), defined on the interval [, ], is piecewise smooth on [, ]. Then the Fourier series of f can be integrated term by term and the obtained series is convergent to the integral of f(x). 18

19 7 Completeness Section 3.7 of Coleman The following just gives the idea of completeness. If we have a piecewise smooth function f(x) on x, we can find constants b n, n 1,,..., such that f(x) b n sin nx, where the equality above is in the sense previously defined. Therefore, the functions sin nx, n 1,,..., essentially span the space of piecewise smooth functions on x. We say that the set of functions sin nx } sin x } x, sin,... is complete in this space of functions. We also know that these functions form an orthogonal set. Therefore, () is a complete orthogonal set in this space. () Similarly, the set of functions cos nx } n 1, cos x } x, cos,... (3) also forms a complete orthogonal set in the space of piecewise smooth functions on x. Note that both () and (3) are sets of eigenfunctions for the eigenvalue problem y + λy, with y() y() and y () y (), respectively. Notes. (i) There are other sets of eigenfunctions that also form a complete orthogonal set for the space of piecewise smooth functions on x. (ii) Not all sets of eigenfunctions have this property. For example, in Exercise 1.7, problem 16(a), we saw that the eigenvalue problem has as eigenfunctions sin And, we can show that the set sin y + λy y() y () (n 1)x, n 1,,.... } (n 1)x also forms a complete orthogonal set in the space of piecewise smooth functions on x. 19

20 As another example, in the class notes for Section 1.7 (see pp of Introduction to Partial Differential Equations (PDEs)), we saw that the eigenvalue problem has as eigenfunctions cos And, we can show that the set cos y + λy y () y() (n 1)x, n 1,,.... } (n 1)x also forms a complete orthogonal set in the space of piecewise smooth functions on x.