Fourier Analysis Fourier Series C H A P T E R 1 1

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1 C H A P T E R Fourier Analysis 474 This chapter on Fourier analysis covers three broad areas: Fourier series in Secs...4, more general orthonormal series called Sturm iouville epansions in Secs..5 and.6 and Fourier integrals and transforms in Secs The central starting point of Fourier analysis is Fourier series. They are infinite series designed to represent general periodic functions in terms of simple ones, namely, cosines and sines. This trigonometric system is orthogonal, allowing the computation of the coefficients of the Fourier series by use of the well-known Euler formulas, as shown in Sec... Fourier series are very important to the engineer and physicist because they allow the solution of ODEs in connection with forced oscillations (Sec..3) and the approimation of periodic functions (Sec..4). Moreover, applications of Fourier analysis to PDEs are given in Chap.. Fourier series are, in a certain sense, more universal than the familiar Taylor series in calculus because many discontinuous periodic functions that come up in applications can be developed in Fourier series but do not have Taylor series epansions. The underlying idea of the Fourier series can be etended in two important ways. We can replace the trigonometric system by other families of orthogonal functions, e.g., Bessel functions and obtain the Sturm iouville epansions. Note that related Secs..5 and.6 used to be part of Chap. 5 but, for greater readability and logical coherence, are now part of Chap.. The second epansion is applying Fourier series to nonperiodic phenomena and obtaining Fourier integrals and Fourier transforms. Both etensions have important applications to solving PDEs as will be shown in Chap.. In a digital age, the discrete Fourier transform plays an important role. Signals, such as voice or music, are sampled and analyzed for frequencies. An important algorithm, in this contet, is the fast Fourier transform. This is discussed in Sec..9. Note that the two etensions of Fourier series are independent of each other and may be studied in the order suggested in this chapter or by studying Fourier integrals and transforms first and then Sturm iouville epansions.. Fourier Series Prerequisite: Elementary integral calculus (needed for Fourier coefficients). Sections that may be omitted in a shorter course:.4.9. References and Answers to Problems: App. Part C, App.. Fourier series are infinite series that represent periodic functions in terms of cosines and sines. As such, Fourier series are of greatest importance to the engineer and applied mathematician. To define Fourier series, we first need some background material. A function f () is called a periodic function if f ( ) is defined for all real, ecept

2 SEC.. Fourier Series 475 f() p Fig. 58. Periodic function of period p possibly at some points, and if there is some positive number p, called a period of f (), such that () f ( p) f () for all. (The function f () tan is a periodic function that is not defined for all real but undefined for some points (more precisely, countably many points), that is >, 3p>, Á.) The graph of a periodic function has the characteristic that it can be obtained by periodic repetition of its graph in any interval of length p (Fig. 58). The smallest positive period is often called the fundamental period. (See Probs. 4.) Familiar periodic functions are the cosine, sine, tangent, and cotangent. Eamples of functions that are not periodic are,, 3, e, cosh, and ln, to mention just a few. If f () has period p, it also has the period p because () implies f ([ p] p) f ( p) f (), etc.; thus for any integer n,, 3, Á f ( p), () f ( np) f () for all. Furthermore if f () and g () have period p, then af () bg () with any constants a and b also has the period p. Our problem in the first few sections of this chapter will be the representation of various functions f () of period p in terms of the simple functions (3), cos, sin, cos, sin, Á, cos n, sin n, Á. All these functions have the period p. They form the so-called trigonometric system. Figure 59 shows the first few of them (ecept for the constant, which is periodic with any period). cos cos cos 3 sin sin sin 3 Fig. 59. Cosine and sine functions having the period p (the first few members of the trigonometric system (3), ecept for the constant )

3 476 CHAP. Fourier Analysis (4) The series to be obtained will be a trigonometric series, that is, a series of the form a a cos b sin a cos b sin Á a a n (a n cos n b n sin n). a, a, b, a, b, Á are constants, called the coefficients of the series. We see that each term has the period p. Hence if the coefficients are such that the series converges, its sum will be a function of period p. Epressions such as (4) will occur frequently in Fourier analysis. To compare the epression on the right with that on the left, simply write the terms in the summation. Convergence of one side implies convergence of the other and the sums will be the same. Now suppose that f () is a given function of period p and is such that it can be represented by a series (4), that is, (4) converges and, moreover, has the sum f (). Then, using the equality sign, we write (5) f () a a (a n cos n b n sin n) n and call (5) the Fourier series of f (). We shall prove that in this case the coefficients of (5) are the so-called Fourier coefficients of f (), given by the Euler formulas () (6) (a) a p p a n p p f () cos n d (b) b n,, Á n p p f () sin n d. f () d n,, Á The name Fourier series is sometimes also used in the eceptional case that (5) with coefficients (6) does not converge or does not have the sum f () this may happen but is merely of theoretical interest. (For Euler see footnote 4 in Sec..5.) A Basic Eample Before we derive the Euler formulas (6), let us consider how (5) and (6) are applied in this important basic eample. Be fully alert, as the way we approach and solve this eample will be the technique you will use for other functions. Note that the integration is a little bit different from what you are familiar with in calculus because of the n. Do not just routinely use your software but try to get a good understanding and make observations: How are continuous functions (cosines and sines) able to represent a given discontinuous function? How does the quality of the approimation increase if you take more and more terms of the series? Why are the approimating functions, called the

4 ` ` SEC.. Fourier Series 477 partial sums of the series, in this eample always zero at and p? Why is the factor >n (obtained in the integration) important? E X A M P E Periodic Rectangular Wave (Fig. 6) Find the Fourier coefficients of the periodic function f () in Fig. 6. The formula is (7) k if f () b k if p and f ( p) f (). Functions of this kind occur as eternal forces acting on mechanical systems, electromotive forces in electric circuits, etc. (The value of f () at a single point does not affect the integral; hence we can leave f () undefined at and.) Solution. From (6.) we obtain a. This can also be seen without integration, since the area under the curve of f () between and p (taken with a minus sign where is negative) is zero. From (6a) we obtain the coefficients a, a, Á f () of the cosine terms. Since f () is given by two epressions, the integrals from to p split into two integrals: a n p p f () cos n d p c p (k) cos n d k cos n d d because sin n at,, and p for all n,, Á. We see that all these cosine coefficients are zero. That is, the Fourier series of (7) has no cosine terms, just sine terms, it is a Fourier sine series with coefficients b, b, Á obtained from (6b); b n p p f () sin n d p c p (k) sin n d k sin n d d Since cos (a) cos a and cos, this yields b n p c sin n k n p c cos n k n k k [cos cos (np) cos np cos ] ( cos np). np np Now, cos p, cos p, cos 3p, etc.; in general, p sin n k ` n d cos n k n p ` d. for odd n, cos np b for even n, and thus for odd n, cos np b for even n. Hence the Fourier coefficients b n of our function are b 4k p, b, b 3 4k 3p, b 4, b 5 4k. 5p, Á Fig. 6. Given function f () (Periodic reactangular wave)

5 478 CHAP. Fourier Analysis Since the are zero, the Fourier series of f () is a n (8) 4k p (sin 3 sin 3 5 sin 5 Á ). The partial sums are S 4k sin, p S 4k p asin sin 3b. 3 etc. Their graphs in Fig. 6 seem to indicate that the series is convergent and has the sum f (), the given function. We notice that at and p, the points of discontinuity of f (), all partial sums have the value zero, the arithmetic mean of the limits k and k of our function, at these points. This is typical. Furthermore, assuming that f () is the sum of the series and setting p>, we have Thus f a p 4k b k p a 3 5 Á b Á p 4. This is a famous result obtained by eibniz in 673 from geometric considerations. It illustrates that the values of various series with constant terms can be obtained by evaluating Fourier series at specific points. Fig. 6. First three partial sums of the corresponding Fourier series

6 SEC.. Fourier Series 479 Derivation of the Euler Formulas (6) The key to the Euler formulas (6) is the orthogonality of (3), a concept of basic importance, as follows. Here we generalize the concept of inner product (Sec. 9.3) to functions. T H E O R E M Orthogonality of the Trigonometric System (3) The trigonometric system (3) is orthogonal on the interval p (hence also on p or any other interval of length p because of periodicity); that is, the integral of the product of any two functions in (3) over that interval is, so that for any integers n and m, (a) (9) (b) p cos n cos m d (n m) p sin n sin m d (n m) (c) p sin n cos m d (n m or n m). P R O O F This follows simply by transforming the integrands trigonometrically from products into sums. In (9a) and (9b), by () in App. A3., Since m n (integer!), the integrals on the right are all. Similarly, in (9c), for all integer m and n (without eception; do you see why?) p p p cos n cos m d p cos (n m) d p sin n sin m d p cos (n m) d p sin n cos m d p sin (n m) d p Application of Theorem to the Fourier Series (5) We prove (6.). Integrating on both sides of (5) from to p, we get We now assume that termwise integration is allowed. (We shall say in the proof of Theorem when this is true.) Then we obtain p p p f () d c a a (a n cos n b n sin n) d d. n f () d a p d a n aa n p cos n d b n p sin n db. cos (n m) d cos (n m) d. sin (n m) d.

7 48 CHAP. Fourier Analysis The first term on the right equals pa. Integration shows that all the other integrals are. Hence division by p gives (6.). We prove (6a). Multiplying (5) on both sides by cos m with any fied positive integer m and integrating from to p, we have () We now integrate term by term. Then on the right we obtain an integral of a cos m, which is ; an integral of a n cos n cos m, which is a m p for n m and for n m by (9a); and an integral of b n sin n cos m, which is for all n and m by (9c). Hence the right side of () equals a m p. Division by p gives (6a) (with m instead of n). We finally prove (6b). Multiplying (5) on both sides by sin m with any fied positive integer m and integrating from to p, we get () p p f () cos m d c a a (a n cos n b n sin n) d cos m d. n p p f () sin m d c a a (a n cos n b n sin n) d sin m d. n Integrating term by term, we obtain on the right an integral of a sin m, which is ; an integral of a n cos n sin m, which is by (9c); and an integral of b n sin n sin m, which is b m p if n m and if n m, by (9b). This implies (6b) (with n denoted by m). This completes the proof of the Euler formulas (6) for the Fourier coefficients. Convergence and Sum of a Fourier Series The class of functions that can be represented by Fourier series is surprisingly large and general. Sufficient conditions valid in most applications are as follows. f() T H E O R E M f( ) f( + ) Fig. 6. eft- and right-hand limits ƒ( ), ƒ( ) _ of the function f () b if > if Representation by a Fourier Series et f () be periodic with period p and piecewise continuous (see Sec. 6.) in the interval p. Furthermore, let f () have a left-hand derivative and a righthand derivative at each point of that interval. Then the Fourier series (5) of f () [with coefficients (6)] converges. Its sum is f (), ecept at points where f () is discontinuous. There the sum of the series is the average of the left- and right-hand limits of f () at. The left-hand limit of f () at is defined as the limit of f () as approaches from the left and is commonly denoted by f ( ). Thus ƒ( ) lim h* ƒ( h) as h * through positive values. The right-hand limit is denoted by ƒ( ) and ƒ( ) lim h* ƒ( h) as h * through positive values. The left- and right-hand derivatives of ƒ() at are defined as the limits of f ( h) f ( ) h f ( h) f ( ) and, h respectively, as h * through positive values. Of course if ƒ() is continuous at, the last term in both numerators is simply ƒ( ).

8 SEC.. Fourier Series 48 P R O O F We prove convergence, but only for a continuous function f () having continuous first and second derivatives. And we do not prove that the sum of the series is f () because these proofs are much more advanced; see, for instance, Ref. 3C4 listed in App.. Integrating (6a) by parts, we obtain a n p p f () cos n d The first term on the right is zero. Another integration by parts gives a n f r() cos n n p f () sin n np p p f s() cos n d. n p The first term on the right is zero because of the periodicity and continuity of f r(). Since f s is continuous in the interval of integration, we have ƒ f s() ƒ M p np p f r() sin n d. for an appropriate constant M. Furthermore, ƒ cos n ƒ. It follows that ƒ a n ƒ n p p f s() cos n d p M d M n p n. Similarly, ƒ b n ƒ M>n for all n. Hence the absolute value of each term of the Fourier series of f () is at most equal to the corresponding term of the series ƒ a ƒ M a 3 3 Á b which is convergent. Hence that Fourier series converges and the proof is complete. (Readers already familiar with uniform convergence will see that, by the Weierstrass test in Sec. 5.5, under our present assumptions the Fourier series converges uniformly, and our derivation of (6) by integrating term by term is then justified by Theorem 3 of Sec. 5.5.) E X A M P E Convergence at a Jump as Indicated in Theorem The rectangular wave in Eample has a jump at. Its left-hand limit there is k and its right-hand limit is k (Fig. 6). Hence the average of these limits is. The Fourier series (8) of the wave does indeed converge to this value when because then all its terms are. Similarly for the other jumps. This is in agreement with Theorem. Summary. A Fourier series of a given function f () of period p is a series of the form (5) with coefficients given by the Euler formulas (6). Theorem gives conditions that are sufficient for this series to converge and at each to have the value f (), ecept at discontinuities of f (), where the series equals the arithmetic mean of the left-hand and right-hand limits of f () at that point.

9 48 CHAP. Fourier Analysis P R O B E M S E T. 5 PERIOD, FUNDAMENTA PERIOD The fundamental period is the smallest positive period. Find it for. cos, sin, cos, sin, cos p, sin p, cos p, sin p. 3. If f () and g () have period p, show that h () af () bg() (a, b, constant) has the period p. Thus all functions of period p form a vector space. 4. Change of scale. If f () has period p, show that f (a), a, and f (>b), b, are periodic functions of of periods p>a and bp, respectively. Give eamples. 5. Show that f const is periodic with any period but has no fundamental period. 6 GRAPHS OF p PERIODIC FUNCTIONS Sketch or graph f () which for p is given as follows if 9. f () b p if p. f () b cos if cos if p. Calculus review. Review integration techniques for integrals as they are likely to arise from the Euler formulas, for instance, definite integrals of cos n, sin n, e cos n, etc. FOURIER SERIES Find the Fourier series of the given function f (), which is assumed to have the period p. Show the details of your work. Sketch or graph the partial sums up to that including cos 5 and sin 5.. f () in Prob f () in Prob cos n, sin pn k sin n, f () ƒ ƒ f () ƒ sin ƒ, f () sin ƒ ƒ f () e ƒ ƒ, f () ƒ e ƒ f () ( p) f () ( p) p cos k, sin p k, cos pn, k CAS EXPERIMENT. Graphing. Write a program for graphing partial sums of the following series. Guess from the graph what f () the series may represent. Confirm or disprove your guess by using the Euler formulas. (a) (sin 3 sin 3 5 sin 5 Á ) (b) ( sin 4 sin 4 6 sin 6 Á ) 4 p acos cos cos 5 Á b (c) 3 p 4(cos 4 cos 9 cos 3 6 cos 4 Á ) 3. Discontinuities. Verify the last statement in Theorem for the discontinuities of f () in Prob.. 4. CAS EXPERIMENT. Orthogonality. Integrate and graph the integral of the product cos m cos n (with various integer m and n of your choice) from a to a as a function of a and conclude orthogonality of cos m

10 SEC.. Arbitrary Period. Even and Odd Functions. Half-Range Epansions 483 and cos n (m n) for a p from the graph. For what if f is continuous but f r df>d is discontinuous, >n 3 m and n will you get orthogonality for a p>, p>3, if f and f r are continuous but f s is discontinuous, etc. p>4? Other a? Etend the eperiment to cos m sin n Try to verify this for eamples. Try to prove it by and sin m sin n. integrating the Euler formulas by parts. What is the 5. CAS EXPERIMENT. Order of Fourier Coefficients. practical significance of this? The order seems to be >n if f is discontinous, and >n. Arbitrary Period. Even and Odd Functions. Half-Range Epansions We now epand our initial basic discussion of Fourier series. Orientation. This section concerns three topics:. Transition from period p to any period, for the function f, simply by a transformation of scale on the -ais.. Simplifications. Only cosine terms if f is even ( Fourier cosine series ). Only sine terms if f is odd ( Fourier sine series ). 3. Epansion of f given for in two Fourier series, one having only cosine terms and the other only sine terms ( half-range epansions ).. From Period p to Any Period p Clearly, periodic functions in applications may have any period, not just p as in the last section (chosen to have simple formulas). The notation p for the period is practical because will be a length of a violin string in Sec.., of a rod in heat conduction in Sec..5, and so on. The transition from period p to be period p is effected by a suitable change of scale, as follows. et f () have period p. Then we can introduce a new variable v such that f (), as a function of v, has period p. If we set () (a) p p v, so that (b) v p p p then v corresponds to. This means that f, as a function of v, has period p and, therefore, a Fourier series of the form () f () f a p vb a a n (a n cos nv b n sin nv) with coefficients obtained from (6) in the last section (3) a p p f a p vb dv, a n p p b n p p f a p vb sin nv dv. f a p vb cos nv dv,

11 484 CHAP. Fourier Analysis We could use these formulas directly, but the change to simplifies calculations. Since (4) v p, we have dv p d and we integrate over from to. Consequently, we obtain for a function f () of period the Fourier series (5) f () a a n aa n cos np b n sin np b with the Fourier coefficients of f () given by the Euler formulas ( p> in d cancels >p in (3)) () a f () d (6) (a) (b) a n b n f () cos np d f () sin np d n,, Á n,, Á. Just as in Sec.., we continue to call (5) with any coefficients a trigonometric series. And we can integrate from to or over any other interval of length p. E X A M P E Periodic Rectangular Wave Find the Fourier series of the function (Fig. 63) if f () d k if p 4,. Solution. From (6.) we obtain a k> (verify!). From (6a) we obtain a n Thus a n if n is even and if f () cos np d k cos np d k np sin np. a n k>np if n, 5, 9, Á, a n k>np if n 3, 7,, Á. From (6b) we find that b n for n,, Á. Hence the Fourier series is a Fourier cosine series (that is, it has no sine terms) f () k k p acos p 3 cos 3p 5 cos 5p Á b.

12 SEC.. Arbitrary Period. Even and Odd Functions. Half-Range Epansions 485 f() k f() k k Fig. 63. Eample Fig. 64. Eample E X A M P E Periodic Rectangular Wave. Change of Scale Find the Fourier series of the function (Fig. 64) k if f () c k if p 4,. Solution. Since, we have in (3) v p> and obtain from (8) in Sec.. with v instead of, that is, the present Fourier series g(v) 4k p asin v 3 sin 3v 5 sin 5v Á b Confirm this by using (6) and integrating. f () 4k p asin p 3 sin 3p 5 sin 5p Á b. E X A M P E 3 Half-Wave Rectifier A sinusoidal voltage E sin vt, where t is time, is passed through a half-wave rectifier that clips the negative portion of the wave (Fig. 65). Find the Fourier series of the resulting periodic function if t, u(t) c E sin vt if t Solution. Since u when t, we obtain from (6.), with t instead of, and from (6a), by using formula () in App. A3. with vt and y nvt, a n v p p>v E sin vt cos nvt dt ve p p>v [sin ( n) vt sin ( n) vt] dt. a v p p>v E sin vt dt E p If n, the integral on the right is zero, and if n, 3, Á, we readily obtain p p v, p v. a n ve p c cos ( n) vt cos ( n) vt p>v ( n) v ( n) v d E ( n)p acos p n If n is odd, this is equal to zero, and for even n we have cos ( n)p b. n a n E p a n n b E (n )(n )p (n, 4, Á ).

13 486 CHAP. Fourier Analysis In a similar fashion we find from (6b) that b and for n, 3, Á E> b n. Consequently, u(t) E p E sin vt E p a cos vt # 3 3 # 5 cos 4vt Á b. u(t) / ω / ω Fig. 65. Half-wave rectifier t Fig. 66. Even function. Simplifications: Even and Odd Functions If f () is an even function, that is, f () f () (see Fig. 66), its Fourier series (5) reduces to a Fourier cosine series (5*) f () a a a n cos np ( f even) n with coefficients (note: integration from to only!) (6*) a f () d, a n f () cos np d, n,, Á. Fig. 67. Odd function If f () is an odd function, that is, f () f () (see Fig. 67), its Fourier series (5) reduces to a Fourier sine series (5**) f () a b n sin np ( f odd) with coefficients n (6**) b n f () sin np d. These formulas follow from (5) and (6) by remembering from calculus that the definite integral gives the net area ( area above the ais minus area below the ais) under the curve of a function between the limits of integration. This implies (7) (a) g () d g () d for even g (b) h () d for odd h Formula (7b) implies the reduction to the cosine series (even f makes f () sin (np>) odd since sin is odd) and to the sine series (odd f makes f () cos (np>) odd since cos is even). Similarly, (7a) reduces the integrals in (6*) and (6**) to integrals from to. These reductions are obvious from the graphs of an even and an odd function. (Give a formal proof.)

14 SEC.. Arbitrary Period. Even and Odd Functions. Half-Range Epansions 487 Summary Even Function of Period. If f is even and p, then f () a a a n cos n with coefficients a p p f () d, a n p p f () cos n d, n Odd Function of Period p. If f is odd and p, then n,, Á with coefficients f () a n b n sin n b n p p f () sin n d, n,, Á. E X A M P E 4 Fourier Cosine and Sine Series The rectangular wave in Eample is even. Hence it follows without calculation that its Fourier series is a Fourier cosine series, the b n are all zero. Similarly, it follows that the Fourier series of the odd function in Eample is a Fourier sine series. In Eample 3 you can see that the Fourier cosine series represents u(t) E>p E sin vt. Can you prove that this is an even function? Further simplifications result from the following property, whose very simple proof is left to the student. T H E O R E M Sum and Scalar Multiple The Fourier coefficients of a sum f f are the sums of the corresponding Fourier coefficients of f and f. The Fourier coefficients of cf are c times the corresponding Fourier coefficients of f. E X A M P E 5 Sawtooth Wave Find the Fourier series of the function (Fig. 68) f () p if p and f ( p) f (). f () Fig. 68. The function f(). Sawtooth wave

15 488 CHAP. Fourier Analysis y 5 S S S 3 S y Fig. 69. Partial sums S, S, S 3, S in Eample 5 Solution. We have f f f, where f and f p. The Fourier coefficients of f are zero, ecept for the first one (the constant term), which is p. Hence, by Theorem, the Fourier coefficients are those of, ecept for, which is p. Since is odd, a for n,, Á a n, b n n, and f a Integrating by parts, we obtain b n p f b n p p f () sin n d p p sin n d. c cos n n p n p cos n d d cos np. n Hence b, b, b 3 3, b 4 4, Á, and the Fourier series of f () is f () p asin sin 3 sin 3 Á b. (Fig. 69) 3. Half-Range Epansions Half-range epansions are Fourier series. The idea is simple and useful. Figure 7 eplains it. We want to represent f () in Fig. 7. by a Fourier series, where f () may be the shape of a distorted violin string or the temperature in a metal bar of length, for eample. (Corresponding problems will be discussed in Chap..) Now comes the idea. We could etend f () as a function of period and develop the etended function into a Fourier series. But this series would, in general, contain both cosine and sine terms. We can do better and get simpler series. Indeed, for our given f we can calculate Fourier coefficients from (6*) or from (6**). And we have a choice and can take what seems more practical. If we use (6*), we get (5*). This is the even periodic etension f of f in Fig. 7a. If we choose (6**) instead, we get (5**), the odd periodic etension f of f in Fig. 7b. Both etensions have period. This motivates the name half-range epansions: f is given (and of physical interest) only on half the range, that is, on half the interval of periodicity of length. et us illustrate these ideas with an eample that we shall also need in Chap..

16 SEC.. Arbitrary Period. Even and Odd Functions. Half-Range Epansions 489 f() () The given function f() f () (a) f() continued as an even periodic function of period f () k E X A M P E 6 (b) f() continued as an odd periodic function of period Fig. 7. Even and odd etensions of period Triangle and Its Half-Range Epansions Find the two half-range epansions of the function (Fig. 7) / Fig. 7. The given function in Eample 6 Solution. (a) Even periodic etension. From (6*) we obtain a c k > d k ( ) d d k, k if f() e k ( ) if. > a n c k > cos np d k > ( ) cos np d d. We consider a n. For the first integral we obtain by integration by parts > cos np d np sin np > np > sin np d Similarly, for the second integral we obtain np sin np n p acos np b. ( ) cos np > d np ( ) sin np > a np a b sin np b np sin np d > n p acos np cos np b.

17 49 CHAP. Fourier Analysis We insert these two results into the formula for. The sine terms cancel and so does a factor. This gives a n Thus, a n 4k n p a cos np cos np b. a 6k>( p ), a 6 6k>(6 p ), a 6k>( p ), Á and a n if n, 6,, 4, Á. Hence the first half-range epansion of f () is (Fig. 7a) f () k 6k p a cos p 6 cos 6p Á b. This Fourier cosine series represents the even periodic etension of the given function f (), of period. (b) Odd periodic etension. Similarly, from ( 6** ) we obtain (5) b n 8k n p sin np. Hence the other half-range epansion of f () is (Fig. 7b) f () 8k p a sin p sin 3p 3 5 sin 5p Á b. The series represents the odd periodic etension of f (), of period. Basic applications of these results will be shown in Secs..3 and.5. (a) Even etension (b) Odd etension Fig. 7. Periodic etensions of f() in Eample 6 P R O B E M S E T. 7 EVEN AND ODD FUNCTIONS Are the following functions even or odd or neither even nor odd?. e, e ƒ ƒ, 3 cos n, tan p, sinh cosh. sin, sin ( ), ln, >( ), cot 3. Sums and products of even functions 4. Sums and products of odd functions 5. Absolute values of odd functions 6. Product of an odd times an even function 7. Find all functions that are both even and odd. 8 7 FOURIER SERIES FOR PERIOD p = Is the given function even or odd or neither even nor odd? Find its Fourier series. Show details of your work. 8.

18 SEC.. Arbitrary Period. Even and Odd Functions. Half-Range Epansions (b) Apply the program to Probs. 8, graphing the first few partial sums of each of the four series on common aes. Choose the first five or more partial sums until they approimate the given function reasonably well. Compare and comment.. 4. Obtain the Fourier series in Prob. 8 from that in Prob f () ( ), p f () >4 ( ), p HAF-RANGE EXPANSIONS Find (a) the Fourier cosine series, (b) the Fourier sine series. Sketch f () and its two periodic etensions. Show the details f () cos p ( ), p f () ƒ ƒ ( ), p Rectifier. Find the Fourier series of the function obtained by passing the voltage v(t) V cos pt through a half-wave rectifier that clips the negative half-waves. 9. Trigonometric Identities. Show that the familiar identities cos 3 3 and sin 3 3 sin 4 cos 4 cos sin 3 can be interpreted as Fourier series epansions. Develop cos 4.. Numeric Values. Using Prob., show that 9 6 Á 6 p. 4. CAS PROJECT. Fourier Series of -Periodic Functions. (a) Write a program for obtaining partial sums of a Fourier series (5) f () sin ( p) 3. Obtain the solution to Prob. 6 from that of Prob. 7.