Chapter 10 Fourier Series

Transcription

1 Chapter Fourier Series. Periodic Functions and Orthogonality Relations The differential equation y +β 2 y = F cosωt models a mass-spring system with natural frequency β with a pure cosine forcing function of frequency ω. If β 2 = ω 2 a particular solution is easily found by undetermined coefficients or by using aplace transforms) to be y p = F β 2 ω 2 cosωt. If the forcing function is a linear combination of simple cosine functions, so that the differential equation is y +β 2 y = N F n cosω n t where β 2 = ωn 2 for any n, then, by linearity, a particular solution is obtained as a sum N F n y p t) = β 2 ωn 2 cosω n t. This simple procedure can be extended to any function that can be represented as a sum of cosine and sine) functions, even if that summation is not a finite sum. It turns out that the functions that can be represented as sums in this form are very general, and include most of the periodic functions that are usually encountered in applications. 723

2 724 Fourier Series Periodic Functions A function f is said to be periodic with period p > if ft+p) = ft) for all t in the domain of f. This means that the graph of f repeats in successive intervals of length p, as can be seen in the graph in Figure.. y p 2p 3p 4p 5p Fig.. An example of a periodic function with period p. Notice how the graph repeats on each interval of length p. The functions sint and cost are periodic with period 2π, while tant is periodic with period π since tant+π) = sint+π) cost+π) = sint cost = tant. The constant function ft) = c is periodic with period p where p is any positive number since ft+p) = c = ft). Other examples of periodic functions are the square wave and triangular wave whose graphs are shown in Figure.2. Both are periodic with period 2. y y a) Square wave swt) Fig..2 t t b) Triangular wave twt)

3 . Periodic Functions and Orthogonality Relations 725 Since a periodic function of period p repeats over any interval of length p, it is possible to define a periodic function by giving the formula for f on an interval of length p, and repeating this in subsequent intervals of length p. For example, the square wave swt) and triangular wave twt) from Figure.2 are described by if t < swt) = ; swt+2) = swt). if t < t if t < twt) = ; twt+2) = twt). t if t < There is not a unique period for a periodic function. Note that if p is a period of ft), then 2p is also a period because ft+2p) = ft+p)+p) = ft+p) = ft) for all t. In fact, a similar argument shows that np is also a period for any positive integer n. Thus 2nπ is a period for sint and cost for all positive integers n. If P > is a period off and there is no smaller period then we say P is the fundamental period of f although we will usually just say the period. Not all periodic functions have a smallest period. The constant function ft) = c is an example of such a function since any positive p is a period. The fundamental period of the sine and cosine functions is 2π, while the fundamental period of the square wave and triangular wave from Figure.2 is 2. Periodic functions under scaling If ft) is periodic of period p and a is any positive number let gt) = fat). Then for all t g t+ p ) p)) = f a t+ = fat+p) = fat) = gt). a a Thus If ft) is periodic with period p, then fat) is periodic with period p a. It is also true that if P is the fundamental period for the periodic function ft), then the fundamental period of gt) = fat) is P/a. To see this it is only necessary to verify that any period r of gt) is at least as large as P/a, which is already a period as observed above. But if r is a period for gt), then ra is a period for gt/a) = ft) so that ra P, or r P/a.

4 726 Fourier Series Applying these observations to the functions sint and cost with fundamental period 2π gives the following facts. Theorem. For any a > the functions cos at and sin at are periodic with period 2π/a. In particular, if > then the functions cos nπ t and sin nπ t, n =, 2, 3,... are periodic with fundamental period P = 2/n. Note that since the fundamental period of the functions cos nπ nπ t and sin t is P = 2/n, it follows that 2 = np is also a period for each of these functions. Thus, a sum an cos nπ t+b nsin nπ t) will be periodic of period 2. Notice that if n is a positive integer, then cosnt and sinnt are periodic with period 2π/n. Thus, each period of cost or sint contains n periods of cosnt and sinnt. This means that the functions cosnt and sinnt oscillate more rapidly as n increases, as can be seen in Figure.3 for n = 3. π π π π a) The graphs of cost and cos3t. b) The graphs of sint and sin3t. Fig..3 Example 2. Find the fundamental period of each of the following periodic functions.. cos2t 2. sin 3 2 t π) 3. +cost+cos2t 4. sin2πt+sin3πt Solution.. P = 2π/2 = π. 2. sin 3 2 t π) = sin3 2 t 3 2 π) = sin 3 2 tcos 3 2 π cos 3 2 tsin 3 2 π = cos 3 2t. Thus, P = 2π/3/2) = 4π/3.

5 . Periodic Functions and Orthogonality Relations The constant function is periodic with any period p, the fundamental period of cost is 2π and all the periods are of the form 2nπ for a positive integer n, and the fundamental period of cos2t is π with mπ being all the possible periods. Thus, the smallest number that works as a period for all the functions is 2π and this is also the smallest period for the sum. Hence P = 2π. 4. The fundamental period of sin2πt is 2π/2π = so that all of the periods have the form n for n a positive integer, and the fundamental period of sin3πt is 2π/3π = 2/3 so that all of the periods have the form 2m/3 for m a positive integer. Thus, the smallest number that works as a period for both functions is 2 and thus P = 2. Orthogonality Relations for Sine and Cosine The family of linearly independent functions, cos π t, cos 2π t, cos 3π t,..., sin π t, sin 2π t, sin 3π t,... } form what is called a mutually orthogonal set of functions on the interval [, ], analogous to a mutually perpendicular set of vectors. Two functions f and g defined on an interval a t b are said to be orthogonal on the interval [a, b] if b a ft)gt)dt =. A family of functions is mutually orthogonal on the interval [a, b] if any two distinct functions are orthogonal. The mutual orthogonality of the family of cosine and sine functions on the interval [, ] is a consequence of the following identities. Proposition 3 Orthogonality Relations). et m and n be positive integers, and let >. Then cos nπ cos nπ sin nπ cos nπ tdt = sin nπ tdt = ) mπ tsin tdt = 2) tcos mπ tdt = tsin mπ tdt =, if n = m, if n = m., if n = m, if n = m. 3) 4)

6 728 Fourier Series Proof. For ): cos nπ tdt = nπ sin nπ t For 3) with n = m, use the identity to get cos nπ = 2n+m)π = sinnπ sin nπ)) =. nπ cosacosb = 2 cosa+b)+cosa B)), mπ tcos tdt = 2 n+m)π sin t+ cos n+m)π 2n m)π t+cos n m)π ) t dt ) n m)π sin t =. For 3) with n = m, use the identity cos 2 A = +cos2a)/2 to get = cos nπ mπ tcos tdt = +cos 2nπ ) 2 t dt = 2 cos nπ t ) 2 dt = t+ ) 2nπ sin 2nπ t =. The proof of 4) is similar, making use of the identitiessin 2 A = cos2a)/2 in case n = m and sinasinb = 2 cosa B) cosa+b)) in case n = m. The proof of 2) is left as an exercise. Even and Odd Functions A function f defined on a symmetric interval [, ] is even if f t) = ft) for t, and f is odd if f t) = ft) for t. Example 4. Determine whether each of the following functions is even, odd, or neither.. ft) = 3t 2 +cos5t 2. gt) = 3t t 2 sin2t 3. ht) = t 2 +t+ Solution.. Since f t) = 3 t) 2 +cos5 t) = 3t 2 +cos5t = ft) for all t, it follows that f is an even function.

7 . Periodic Functions and Orthogonality Relations Since g t) = 3 t) t) 2 sin3 t) = 3t+t 2 sin3t = gt) for all t, it follows that g is an odd function. 3. Since h t) = t) 2 + t) + = t 2 t + = t 2 + t + = ht) t = t t =, we conclude that h is not even. Similarly, h t) = ht) t 2 t+ = t 2 t t 2 + = t 2 +) t 2 + =, which is not true for any t. Thus h is not odd, and hence it is neither even or odd. The graph of an even function is symmetric with respect to the y-axis, while the graph of an odd function is symmetric with respect to the origin, as illustrated in Figure.4. ft) f t) ft) t t t t f t) = ft) a) The graph of an even function. b) The graph of an odd function. Fig..4 Here is a list of basic properties of even and odd functions that are useful in applications to Fourier series. All of them follow easily from the definitions, and the verifications will be left to the exercises. Proposition 5. Suppose that f and g are functions defined on the interval t.. If both f and g are even then f +g and fg are even. 2. If both f and g are odd, then f +g is odd and fg is even. 3. If f is even and g is odd, then fg is odd. 4. If f is even, then 5. If f is odd, then ft)dt = 2 ft)dt =. ft)dt. Since the integral of f computes the signed area under the graph of f, the integral equations can be seen from the graphs of even and odd functions in Figure.4.

8 73 Fourier Series Exercises 9. Graph each of the following periodic functions. Graph at least 3 periods. 3 if < t < 3. ft) = 3 if 3 < t < ; ft+6) = ft). 3 if 2 t < 2. ft) = if t ; ft+4) = ft). 3 if < t < 2 3. ft) = t, < t 2; ft+2) = ft). 4. ft) = t, < t ; ft+2) = ft). 5. ft) = sint, < t π; ft+π) = ft). if π t < 6. ft) = ; ft+2π) = ft). sint if < t π t if t < 7. ft) = ; ft+2) = ft). if t < 8. ft) = t 2, < t ; ft+2) = ft). 9. ft) = t 2, < t 2; ft+2) = ft). 7. Determine if the given function is periodic. If it is periodic find the fundamental period... sin2t 2. +cos3πt 3. cos2t+sin3t 4. t+sin2t 5. sin 2 t 6. cost+cosπt 7. sint+sin2t+sin3t Determine if the given function is even, odd, or neither.

9 .2 Fourier Series ft) = t 9. ft) = t t 2. ft) = sin 2 t 2. ft) = cos 2 t 22. ft) = sintsin3t 23. ft) = t 2 +sint 24. ft) = t+ t 25. ft) = ln cost 26. ft) = 5t+t 2 sin3t 27. Verify the orthogonality property 2) from Proposition 3: cos nπ tsin mπ tdt = 28. Use the properties of even and odd functions Proposition.4) to evaluate the following integrals. a) c) e) π π π π tdt tsintdt cos nπ tsin mπ tdt b) d) f) π π π π t 4 dt tcostdt t 2 sintdt.2 Fourier Series We start by considering the possibility of representing a function f as a sum of a series of the form ft) = a 2 + a n cos nπt +b nsin nπt ) )

10 732 Fourier Series where the coefficients a, a,..., b, b 2,..., are to be determined. Since the individual terms in the series ) are periodic with periods 2, 2/2, 2/3,..., the function ft) determined by the sum of the series, where it converges, must be periodic with period 2. This means that only periodic functions of period 2 can be represented by a series of the form ). Our first problem is to find the coefficients a n and b n in the series ). The first term of the series is written a /2, rather than simply as a, to make the formula to be derived below the same for all a n, rather than a special case for a. The coefficients a n and b n can be found from the orthogonality relations of the family of functions cosnπt/) and sinnπt/) on the interval [, ] given in Proposition 3 of Sect... To compute the coefficienta n forn =,2, 3,..., multiply both sides of the series ) by cosmπt/), with m a positive integer and then integrate from to. For the moment we will assume that the integrals exist and that it is justified to integrate term by term. Then using ), 2), and 3) from Sect.., we get Thus, ft)cos mπ tdt = a 2 + cos mπ tdt } } = [a n cos nπ mπ tcos tdt +b n sin nπ }} if n = m = if n = m a m = or, replacing the index m by n, a n = tcos mπ tdt } } = ft)cos mπ tdt, m =, 2, 3,... ] = a m. ft)cos nπ tdt, n =, 2, 3,... 2) To compute a, integrate both sides of ) from to to get ft)dt = a dt+ [a n cos nπ 2 }} tdt +b n sin nπ ] tdt }}}} =2 = = = a. Thus, a = ft)dt. 3)

11 .2 Fourier Series 733 Hence, a is two times the average value of the function ft) over the interval t. Observe that the value of a is obtained from 2) by setting n =. Of course, if the constant a in ) were not divided by 2, we would need a separate formula for a. It is for this reason that the constant term in ) is labeled a /2. Thus, for all n, the coefficients a n are given by a single formula a n = ft)cos nπ tdt, n =,, 2,... 4) To compute b n for n =, 2, 3,..., multiply both sides of the series ) by sinmπt/), with m a positive integer and then integrate from to. Then using ), 2), and 4) from Sect.., we get ft)sin mπ tdt = a 2 + sin mπ tdt } } = [a n cos nπ mπ tsin tdt +b n sin nπ mπ ] tsin tdt = b m. }}}} = if n = m = if n = m Thus, replacing the index m by n, we find that b n = ft)sin nπ tdt, n =, 2, 3,... 5) We have arrived at what are known as the Euler Formulas for a function ft) that is the sum of a trigonometric series as in ): a = a n = b n = ft)dt 6) ft)cos nπ tdt, n =, 2, 3,... 7) ft)sin nπ tdt, n =, 2, 3,... 8) The numbers a n and b n are known as the Fourier coefficients of the function f. Note that while we started with a periodic function of period 2, the formulas for a n and b n only use the values of ft) on the interval [, ].

12 734 Fourier Series We can then reverse the process, and start with any function ft) defined on the symmetric interval [, ] and use the Euler formulas to determine a trigonometric series. We will write ft) a 2 + a n cos nπt +b nsin nπt ), 9) where the a n, b n are defined by 6), 7), and 8), to indicate that the right hand side of 9) is the Fourier series of the function ft) defined on[, ]. Note that the symbol indicates that the trigonometric series on the right of 9) is associated with the function ft); it does not imply that the series converges to ft) for any value of t. In fact, there are functions whose Fourier series do not converge to the function. Of course, we will be interested in the conditions under which the Fourier series of ft) converges to ft), in which case can be replaced by =; but for now we associate a specific series using Equations 6), 7), and 8) with ft) and call it the Fourier series. The mild conditions under which the Fourier series of ft) converges to ft) will be considered in the next section. Remark. If an integrable function ft) is periodic with period p, then the integral of ft) over any interval of length p is the same; that is c+p c ft)dt = p ft)dt ) for any choice of c. To see this, first observe that for any α and β, if we use the change of variables t = x p, then β α ft)dt = β+p α+p fx p)dx = etting α = c and β = gives so that c+p c ft)dt = = c c p c+p ft)dt = ft)dt+ ft)dt+ β+p p α+p c+p c+p c+p fx)dx = ft)dt ft)dt ft)dt = β+p p α+p ft)dt, ft)dt. which is ). This formula means that when computing the Fourier coefficients, the integrals can be computed over any convenient interval of length 2. For example,

13 .2 Fourier Series 735 a n = ft)cos nπ tdt = 2 ft)cos nπ tdt. We now consider some examples of the calculation of Fourier series. Example 2. Compute the Fourier series of the odd square wave function of period 2 and amplitude given by t <, ft) = ; ft+2) = ft). t <, See Figure.5 for the graph of ft). y t Fig..5 The odd square wave of period 2 Solution. Use the Euler formulas for a n Equations 6) and 7)) to conclude a n = ft)cos nπ tdt = for all n. This is because the function ft)cos nπ t is the product of an odd and even function, and hence is odd, which implies by Proposition 5 Part 5) of Section. that the integral is. It remains to compute the coefficients b n from 8). b n = ft)sin nπ tdt = ft)sin nπ tdt+ = )sin nπ tdt+ +)sin nπ tdt [ = nπ cos nπ ] [ ] } t nπ cos nπ t = nπ [ cos nπ))+ cosnπ))] = 2 nπ cosnπ) = 2 nπ )n ). Therefore, ft)sin nπ tdt

14 736 Fourier Series if n is even, b n = 4 nπ if n is odd, and the Fourier series is ft) 4 sin π π t+ 3 sin 3π t+ 5 sin 5π t+ 7 sin 7π ) t+. ) Example 3. Compute the Fourier series of the even square wave function of period 2 and amplitude given by t < /2, ft) = /2 t < /2, ; ft+2) = ft). /2 t <, See Figure.6 for the graph of ft). y t Fig..6 The even square wave of period 2 Solution. Use the Euler formulas for b n Equation 8)) to conclude b n = ft)sin nπ tdt =, for all n. As in the previous example, this is because the function ft)sin nπ t is the product of an even and odd function, and hence is odd, which implies by Proposition 5 Part 5) of Section. that the integral is. It remains to compute the coefficients a n from 6) and 7). For n =, a is twice the average of ft) over the period [, ], which is easily seen to be from the graph of ft). For n, a n = = /2 ft)cos nπ tdt ft)cos nπ tdt+ /2 /2ft)cos nπ tdt+ /2 ft)cos nπ tdt

15 .2 Fourier Series 737 = = = nπ /2 )cos nπ tdt+ [ nπ sin nπ t [ sin nπ 2 = 4 nπ sin nπ 2. Therefore, ] /2 /2 /2+)cos nπ tdt+ [ + nπ sin nπ t ) +sin nπ)+sin nπ 2 ] /2 /2 if n is even, a n = 4 nπ if n = 4m+ if n = 4m+3 4 nπ ) sin nπ 2 )cos nπ tdt } /2 ] [ nπ sin nπ t and the Fourier series is ft) 4 cos π π t 3 cos 3π t+ 5 cos 5π t 7 cos 7π ) t+ /2 ) sinnπ+sin nπ 2 )] = 4 π k= ) k 2k + 2k +)π cos t. Example 4. Compute the Fourier series of the even triangular wave function of period 2π given by t π t <, ft) = ; ft+2π) = ft). t t < π, See Figure.7 for the graph of ft). π 3π 2π π π 2π 3π Fig..7 The even triangular wave of period 2π. Solution. The period is 2π = 2 so = π. Again, since the function ft) is even, the coefficients b n =. It remains to compute the coefficients a n from the Euler formulas 6) and 7). For n =, using the fact that ft) is even,

16 738 Fourier Series a = π = π πft)dt 2 π ft)dt π = 2 π tdt = 2 t 2 π π π 2 = π. For n, using the fact that ft) is even, and taking advantage of the integration by parts formula xcosxdx = xsinx+cosx+c, a n = π = 2 π = 2 π π π = πft)cosntdt 2 ft)cosntdt π tcosntdt let x = nt so t = x n and dt = dx ) n π nπ x n cosx dx n = 2 n 2 π [xsinx+cosx]x=nπ x= = 2 n 2 π [cosnπ ] = 2 n 2 π [ )n ] Therefore, a n = if n is even, 4 n 2 π if n is odd and the Fourier series is ft) π 2 4 cost π 2 + cos3t cos5t cos7t ) = π 2 4 cos2k +)t π 2k +) 2. k= Example 5. Compute the Fourier series of the sawtooth wave function of period 2 given by ft) = t for t < ; ft+2) = ft). See Figure.8 for the graph of ft). Solution. As in Example 2, the function ft) is odd, so the cosine terms a n are all. Now compute the coefficients b n from 8). Using the integration by parts formula

17 .2 Fourier Series 739 y t Fig..8 The sawtooth wave of period 2 xsinxdx = sinx xcosx+c, b n = = 2 = 2 nπ ft)sin nπ tdt tsin nπ tdt dx xsinx nπ nπ = 2 n 2 π 2 = 2 n 2 π 2 [sinx xcosx]x=nπ x= let x = nπ t so t = nπ x and dt = ) nπ dx nπ = 2 n 2 π2nπcosnπ) = 2 nπ )n. xsinxdx Therefore, the Fourier series is ft) 2 sin π π t 2 sin 2π t+ 3 sin 3π t 4 sin 4π ) t+ = 2 π ) n+ sin nπ n t. All of the examples so far have been of functions that are either even or odd. If a function ft) is even, the resulting Fourier series will only have cosine terms, as in the case of Examples 3 and 4, while if ft) is odd, the resulting Fourier series will only have sine terms, as in Examples 2 and 5. Here are some examples where both sine an cosine terms appear. Example 6. Compute the Fourier series of the function of period 4 given by

18 74 Fourier Series 2 t <, ft) = ; ft+4) = ft). t t < 2, See Figure.9 for the graph of ft). 2 y t Fig..9 A half sawtooth wave of period 4 Solution. This function is neither even nor odd, so we expect both sine and cosine terms to be present. The period is 4 = 2 so = 2. Because ft) = on the interval 2, ), each of the integrals in the Euler formulas, which should be an integral from t = 2 to t = 2, can be replaced with an integral from t = to t = 2. Thus, the Euler formulas give a = 2 a n = 2 = 2 Therefore, 2 2 nπ tdt = 2 [ t 2 2 tcos nπ 2 tdt ] 2 = ; 2x 2dx cosx nπ nπ = 2 n 2 π 2 = 2 n 2 π 2 [cosx+xsinx]x=nπ x= let x = nπ 2x t so t = 2 nπ nπ = 2 2 n 2 π2cosnπ ) = n 2 π 2 )n ). Now compute b n : b n = 2 2 xcosxdx if n =, a n = if n is even and n 2, if n is odd. tsin nπ 2 tdt 4 n 2 π 2 let x = nπ 2x t so t = 2 nπ ) 2dx and dt = nπ ) 2dx and dt = nπ

19 .2 Fourier Series 74 = nπ 2x 2dx sinx 2 nπ nπ = 2 nπ n 2 π 2 xsinxdx = 2 n 2 π 2 [sinx xcosx]x=nπ x= = 2 n 2 π 2 nπcosnπ) = 2 nπ )n. Thus, b n = 2 )n+ nπ Therefore, the Fourier series is for all n. ft) 2 4 2k +)π π 2 cos t+ 2 2k +) 2 2 π k= ) n+ n sin nπ 2 t. Example 7. Compute the Fourier series of the square pulse wave function of period 2π given by t < h, ft) = ; ft+2π) = ft). h t < 2π, See Figure. for the graph of ft). 4π 2π h 2π 4π Fig.. A square pulse wave of period 2π. Solution. For this function, it is more convenient to compute the a n and b n using integration over the interval [, 2π] rather than the interval [ π, π]. Thus, a = π a n = π b n = π 2π h h ft)dt = π h dt = h π, cosntdt = sinnh πn, and sinntdt = cosnh, πn

20 742 Fourier Series and the Fourier series is ft) h 2π + π sinnh n cosnt+ cosnh n ) sinnt. If the square pulse wave ft) is divided by h, then one obtains a function whose graph is a series of tall thin rectangles of height /h and base h, so that each of the rectangles with the bases starting at 2nπ has area, as in Figure.. Now consider the limiting case where h approaches. The h 4π 2π h 2π 4π Fig.. A square unit pulse wave of period 2π. graph becomes a series of infinite height spikes of width as in Figure.2. This looks like an infinite sum of Dirac delta functions, which is the regular delta function extended to be periodic of period 2π. That is, ft) lim h h = δt 2nπ). n= Now compute the Fourier coefficients a n /h and b n /h as h approaches. a n = sinnh πnh and π b n = cosnh as h πnh Also, a /h = /π. Thus, the 2π-periodic delta function has a Fourier series n= δt 2nπ) 2π + π cosnt. 2) It is worth pointing out that this is one series that definitely does not converge for all t. In fact it does not converge for t =. However, there is a sense of convergence in which convergence makes sense. We will discuss this in the next section.

21 .2 Fourier Series 743 4π 2π 2π 4π Fig..2 The limiting square unit pulse wave of period 2π as h. Example 8. Compute the Fourier series of the function ft) = sin 2t + 5cos3t. Solution. Since ft) is periodic of period 2π and is already given as a sum of sines and cosines, no further work is needed. The coefficients can be read off without any integration as a n = for n = 3 and a 3 = 5 while b n = for n = 2 and b 2 =. The same results would be obtained by integration using Euler s formulas. Example 9. Compute the Fourier series of the function ft) = sin 3 t Solution. This can be handled by trig identities to reduce it to a finite sum of terms of the form sinnt. sin 3 t = sintsin 2 t = 2 sint cos2t) = 2 sint 2 sintcos2t = 2 sint 2 = 3 4 sint 4 sin3t The right hand side is the Fourier series of sin 3 t. 2 sin3t+sin t))) Exercises 7. Find the Fourier series of the given periodic function. if 5 t <. ft) = ; ft+) = ft). 3 if t < 5 2. ft) = 2 if π t < 2 if t < π ; ft+2π) = ft).

22 744 Fourier Series 4 if π t < 3. ft) = if t < π ; ft+2π) = ft). 4. ft) = t, t < 2; ft+2) = ft). Hint: It may be useful to take advantage of Equation ) in applying the Euler formulas. 5. ft) = t, π t < π; ft+2π) = ft). t <, 6. ft) = t < 2, ft+3) = ft). 2 t < 3, 7. ft) = t 2, 2 t < 2; ft+4) = ft). if t < 8. ft) = t 2 ; ft+2) = ft). if t < 9. ft) = sint, t < π; ft+π) = ft). if π t <. ft) = sint if t < π ; ft+2π) = ft). +t if t <. ft) = ; ft+2) = ft). t if t < +t if t < 2. ft) = ; ft+2) = ft). +t if t < tπ +t) if π t < 3. ft) = ; ft+2π) = ft). tπ t) if t < π 4. ft) = 3cos 2 t, π t < π; ft+2π) = ft) 5. ft) = sin t 2, π t < π; ft+2π) = ft) 6. ft) = sinpt, π t < π; ft+2π) = ft) where p is not an integer) 7. ft) = e t ; t < ; ft+2) = ft) Then, a n = e t cosnπtdt

23 .3 Convergence of Fourier Series 745 = +n 2 π 2et [cosnπt+nπsinnπt] = +n 2 π 2[e cosnπ e cos nπ)] and, b n = = e e ) ) n +n 2 π 2 e t sinnπtdt = 2 )n sinh) +n 2 π 2, = +n 2 π 2et [sinnπt nπcosnπt] = +n 2 π 2[e nπcosnπ) e nπcos nπ))] = e e ) nπ) ) n +n 2 π 2 Therefore, the Fourier series is = 2 )n nπ)sinh) +n 2 π 2. ft) sinh)+2sinh) ) n cosnπt nπsinnπt) +n 2 π 2..3 Convergence of Fourier Series If ft) is a periodic function of period 2, the Fourier series of ft) is ft) a 2 + a n cos nπt +b nsin nπt ), ) where the coefficients a n and b n are given by the Euler formulas a n = b n = ft)cos nπ tdt, n =,, 2, 3,... 2) ft)sin nπ tdt, n =, 2, 3,... 3) The questions that we wish to consider are ) under what conditions on the function ft) is it true that the series converges, and 2) when does it converge to the original function ft)? As with any infinite series, to say that c n = C means that the sequence of partial sums m c n = S m converges to C, that is,

24 746 Fourier Series lim S m = C. m The partial sums of the Fourier series ) are S m t) = a 2 + m a n cos nπt +b nsin nπt ). 4) The partial sum S m t) is a finite linear combination of the trigonometric functions cos nπt nπt and sin, each of which is periodic of period 2. Thus, S m t) is also periodic of period 2, and if the partial sums converge to a function gt), then gt) must also be periodic of period 2. Example. Consider the odd square wave function of period 2 and amplitude from Example 2 of Section.2 t <, ft) = ; ft+2) = ft). 5) t <, The Fourier series of this function was found to be letting = ) ft) 4 sinπt+ 3 π sin3πt+ 5 sin5πt+ 7 ) sin7πt+ 6) = 4 π k= etting t = in this series gives 4 π k= sin2k +)πt. 2k + 2k + sin2k +)π = 4 π =. Since f) =, it follows that the Fourier series does not converge to ft) for all t. Several partial sums S m are graphed along with the graph of ft) in Figure.3. It can be seen that the graph of S 5 t) is very close to the graph of ft), except at the points of discontinuity, which are, ±, ±2,... This suggests that the Fourier series of ft) converges to ft) at all points except for where ft) has a discontinuity, which occurs whenever t is an integer n. For each integer n, S m n) = 4 π [ m 2 ] k= k= 2k + sin2k +)πn = 4 [ m 2 ] π k= =. Hence the Fourier series converges to ft) whenever t is not an integer and it converges to, which is the midpoint of the jump of ft) at each integer.

25 .3 Convergence of Fourier Series 747 y t S t) = 4 π sinπt y t S 5 t) = 4 π sinπt+ 3 sin3πt+ 5 sin5πt) y t S 5 t) = 4 π 7 k= sin2k +)πt 2k+ Fig..3 Fourier series approximations S m to the odd square wave of period 2 for m =, 5, and 5. Similar results are true for a broad range of functions. We will describe the types of functions to which the convergence theorem applies and then state the convergence theorem. Recall that a function ft) defined on an interval I has a jump discontinuity at a point t = a if the left hand limit fa ) = lim t a ft) and the right hand limit fa + ) = lim t a + ft) both exist as real numbers, not ± ) and fa + ) = fa ).

26 748 Fourier Series The difference fa + ) fa ) is frequently referred to as the jump in ft) at t = a. For example the odd square wave function of Example.2.2 has jump discontinuities at the points ±n for n =,, 2,... and the sawtooth wave function of Example.2.5 has jump discontinuities at ±n for n =,, 2,... On the other hand, the function gt) = tant is a periodic function of period π that is discontinuous at the points π 2 +kπ for all integers k, but the discontinuity is not a jump discontinuity since, for example lim t π tant = +, and lim 2 t π +tant =. 2 A function ft) is piecewise continuous on a closed interval [a, b] if ft) is continuous except for possibly finitely many jump discontinuities in the interval [a, b]. A function ft) is piecewise continuous for all t if it is piecewise continuous on every bounded interval. In particular, this means that there can be only finitely many discontinuities in any given bounded interval, and each of those must be a jump discontinuity. For convenience it will not be required that ft) be defined at the jump discontinuities, or ft) can be defined arbitrarily at the jump discontinuities. The square waves and sawtooth waves from Section.2 are typical examples of piecewise continuous periodic functions, while tan t is not piecewise continuous since the discontinuities are not jump discontinuities. A function ft) is piecewise smooth if it is piecewise continuous and the derivative f t) is also piecewise continuous. As with the convention on piecewise continuous,f t) may not exist at finitely many points in any closed interval. All of the examples of periodic functions from Section.2 are piecewise smooth. The property of ft) being piecewise smooth and periodic is sufficient to guarantee that the Fourier series of ft) converges, and the sum can be computed. This is the content of the following theorem. Theorem 2. Convergence of Fourier Series) Suppose that ft) is a periodic piecewise smooth function of period 2. Then the Fourier series ) converges a) to the value ft) at each point t where f is continuous, and b) to the value 2 [ft+ )+ft )] at each point where f is not continuous. Note that 2 [ft+ )+ft )] is the average of the left-hand and right-hand limits of f at the point t. If f is continuous at t, then ft + ) = ft) = ft ), so that ft) = ft+ )+ft ), 7) 2 for anytwheref is continuous. Hence Theorem 2 can be rephrased as follows: The Fourier series of a piecewise smooth periodic function f converges for every t to the average of the left-hand and right-hand limits of f.

27 .3 Convergence of Fourier Series 749 Example 3. Continuing with the odd square wave function of Example, it is clear from Equation 5) or from Figure.3 that if n is an even integer, then lim t n + ft) = + and while if n is an odd integer, then lim t n ft) =, lim t n +ft) = and lim t n ft) = +. Therefore, whenever n is an integer, fn + )+fn ) 2 =. This is consistent with Theorem 2 since the Fourier series 6) converges to whenever t is an integer n because sin2k + )πn = ). For any point t other than an integer, the function ft) is continuous, so Theorem 2 gives an equality ft) = 4 sinπt+ 3 π sin3πt+ 5 sin5πt+ 7 ) sin7πt+. By redefining ft) at integers to be fn) = for integers n, then this equality holds for all t. Figure.4 shows the graph of the redefined function. Putting y t Fig..4 The odd square wave of period 2 redefined so that the Fourier series converges to ft) for all t. in the value of ft) for some values of t leads to a trigonometric identity 4 sinπt+ 3 π sin3πt+ 5 sin5πt+ 7 ) sin7πt+ = ft) = if < t <. etting t = /2 then gives the series

28 75 Fourier Series = π 4. Example 4. et f be periodic of period 2 and defined on the interval t < 2 by ft) = t 2. Without computing the Fourier coefficients, give an explicit description of the sum of the Fourier series of f for all t. Solution. The function is defined in only one period so the periodic extension is defined by ft) = t 2 for t < 2; ft+2) = ft). Both ft) and f t) are continuous everywhere except at the even integers. Thus, the Fourier series converges to ft) everywhere except at the even integers. At the even integer 2m the left and right limits are f2m ) = lim t 2m ft) = lim t ft) = lim t t+2) 2 = 4, and f2m + ) = lim t 2m +ft) = lim t +ft) = lim =. t +t2 Therefore the sum of the Fourier series at t = 2m is f2m )+f2m + ) 2 = 4+ 2 The graph of the sum is shown in Figure.5. = 2. y t Fig..5 The graph of sum of the Fourier series of the period 2 function ft) for Example 4. Example 5. The 2π-periodic delta function n= δt 2nπ) has a Fourier series δt 2nπ) 2π + cos nt 8) π n=

29 .3 Convergence of Fourier Series 75 that was computed in Section.2. See Figure.2 for the graph of n= δt 2nπ). The 2π-periodic delta function does not satisfy the conditions of the convergence theorem. Moreover, it is clear that the series does not converge since the individual terms do not approach. If t = the series becomes 2π + π [+++ ], while, if t = π the series is 2π + π [ + + ]. Neither of these series are convergent. However, there is a sense of convergence in which the Fourier series of n= δt 2nπ) "converges" to n= δt 2nπ). To explore this phenomenon, we will explicitly compute the n th partial sum Δ n t) of the series 8): Δ n t) = 2π + π [cost+cos2t+ +cosnt] = 2π [+2cost+2cos2t+ +2cosnt] Now use the Euler identity 2cosθ = e iθ +e iθ to get Δ n t) = [ +e it +e it +e i2t +e i2t + +e int +e int]. 2π etting z = e it, this is seen to be a geometric series of ratio z that begins with the term z n and ends with the term z n. Since the sum of a geometric series is known: a+ar+ar 2 + +ar m = a r m+ )/ r) it follows that Δ n t) = [ +e it +e it +e i2t +e i2t + +e int +e int] 2π = 2π [ z n +z n+ + +z n] = 2π z n[ +z + +z 2n] = z n z2n+ 2π z = 2π = 2π = 2π e int e in+)t e it = 2π e in+ 2)t e in+ 2)t e it/2 e it/2 sin n+ 2) t sin 2 t. = z n z n+ 2π z e in+)t e int e it e it/2 e it/2

30 752 Fourier Series For those values of t where sin 2 t =, the value of Δ nt) is the limiting value. The graphs of Δ n t) for n = 5 and n = 5 over period π t π] are shown in Figure.6. As n increases the central hump increases in height and becomes narrower, while away from the graph is a rapid oscillation of small amplitude. Also, the area enclosed by Δ n t) in one period π t π is always. This is easily seen from the expression of Δ n t) as a sum of cosine terms: π π Δ n t)dt = π π [+2cost+2cos2t+ +2cosnt] dt = 2π since π cosmtdt = for all nonzero integers m. Most of the area is concentrated in the first hump since the rapid oscillation cancels out the contri- π butions above and below the t axis away from this hump δ 5 t) 2 δ 5 t) π π Fig..6 The graphs of the partial sums δ nt) for n = 5 and n = 5. To see the sense in which Δ n t) converges to n= δt 2nπ), recall that δt) is described via integration by the property that δt) = if t =, and ft)δt)dt = f) for all test functions ft). Since we are working with periodic functions, consider the integrals over period, in our case [ π, π]. Thus let ft) be a smooth function and expand it in a cosine series ft) = a m cosmt. Multiply byδ n t) and integrate, using the orthogonality conditions to get π π As n this sum converges to f). Thus Δ n t)ft)dt = a + +a n. 9)

31 .3 Convergence of Fourier Series 753 π π lim Δ n t)ft)dt = f) = δt)ft) dt. n π π It is in this weak sense that Δ n t) under integration is the same. n= δt 2nπ). Namely, the effect Exercises 9. In each exercise, a) Sketch the graph of ft) over at least 3 periods. b) Determine the points t at which the Fourier series of ft) converges to ft). c) At each point t of discontinuity, give the value of ft) and the value to which the Fourier series converges.. ft) = 3 if t < 2 if 2 t < 4 ; 2. ft) = t, t, ft+2) = ft). 3. ft) = t, t 2, ft+2) = ft). 4. ft) = t, t, ft+2) = ft). 5. ft) = t 2, 2 t 2, ft+4) = ft). ft+4) = ft). 6. ft) = 2t, t <, ft+2) = ft). 2 if 2 t < 7. ft) = ; ft+4) = ft). t if t < 2 8. ft) = cost/2), t < π, ft+π) = ft). 9. ft) = cosπt, t <, ft+) = ft).. Verify that the following function is not piecewise smooth. t if t < ft) = if t <.. Verify that sint 2 sin2t+ 3 sin3t 4 sin4t+ = t 2, for π < t < π. Explain how this gives the summation

32 754 Fourier Series = π Use the Fourier series of the functionft) = t, π < t < π see Example 4 of Section.2) to compute the sum of the series Verify that π 2 ) n n 2 cosnπt = t 2, for t. 4. Using the Fourier series in problem 3, find the sum of the following series ) n+ n 2 and n Compute the Fourier series for the function ft) = t 4, π t π; ft + 2π) = ft) for all t. Using this result, verify the summations n 4 = π4 9 and ) n+ n 4 = 7π Fourier Sine Series and Fourier Cosine Series In Section.2 we observed see the discussion following Example 5) that the Fourier series of an even function will only have cosine terms, while the Fourier series of an odd function will only have sine terms. We will take advantage of this feature to obtain new types of Fourier series representations that are valid for functions defined on some interval t. In practice, we will frequently want to represent ft) by a Fourier series that involves only cosine terms or only sine terms. To do this, we will first extend ft) to the interval t. We may then extend ft) to all of the real line by assuming that ft) is 2 periodic. That is, assume ft+2) = ft). We will extend ft) in such a way as to obtain either an even function or an odd function. If ft) is defined only for t, then the even 2-periodic extension of ft) is the function f e t) defined by

33 .4 Fourier Sine Series and Fourier Cosine Series 755 f e t) = ft) if t, f t) if t <, ; f et+2) = f e t). ) The odd 2-periodic extension of ft) is the function f o t) defined by ft) if t, f e t) = f t) if t <, ; f ot+2) = f o t). 2) These two extensions are illustrated for a particular function ft) in Figure.7. Note that both f e t) and f o t) agree with ft) on the interval < t <. y t The graph of ft) on the interval t. y y t t The graph of f et), the even 2-periodic extension of ft). The graph of f ot), the odd 2-periodic extension of ft). Fig..7 Since f e t) is an even 2-periodic function, then f e t) has a Fourier series expansion f e t) a 2 + a n cos nπt +b nsin nπt ) where the coefficients are given by Euler s formulas Equations 6)-8) of Section.2) a = f e t)dt

34 756 Fourier Series a n = b n = f e t)cos nπ tdt, n =, 2, 3,... f e t)sin nπ tdt, n =, 2, 3,... Since f e t) is an even function, then f e t)sin nπ t is odd, so the integral defining b n is see Proposition 5 of Section.). Thus, b n = and we get the Fourier series expansion where a n = f e t) a 2 + a n cos nπt, f e t)cos nπt dt, for n. Since the integrand is even and f e t) = ft) for t, it follows that a n = 2 ft)cos nπt dt, for n. 3) If we assume that ft) is piecewise smooth on the interval t, then the even 2-periodic extension f e t) is also piecewise smooth, and the convergence theorem Theorem 2 of Section.3) shows that the Fourier series of f e t) converges to the average of the left-hand and right-hand limits of f e at each t. By restricting to the interval t, we obtain the series expansion ft) a 2 + a n cos nπt, for t, 4) where the coefficients are given by 3) and this series converges to the average of the left-hand and right-hand limits of f at each t in the interval [, ]. The series in 4) is called the Fourier cosine series for f on the interval [, ]. Similarly, by using the Fourier series expansion of the odd 2-periodic extension of ft), and restricting to the interval t, we obtain the series expansion ft) where the coefficients are given by b n = 2 b n sin nπt, for t, 5) ft)sin nπt dt, for n, 6)

35 .4 Fourier Sine Series and Fourier Cosine Series 757 and this series converges to the average of the left-hand and right-hand limits of f at each t in the interval [, ]. The series in 5) is called the Fourier sine series for f on the interval [, ]. Example. et ft) = π t for t π. Compute both the Fourier cosine series and the Fourier sine series for the function ft). Sketch the graph of the function to which the Fourier cosine series converges and the graph of the function to which to Fourier sine series converges. Solution. In this case, the length of the interval is = π and from 4), the Fourier cosine series of ft) is ft) a 2 + a n cosnt, with and for n, a = 2 π π ft)dt = 2 π π π t)dt = π, π π a n = 2 ft)cosntdt = 2 π t)cosntdt π π = 2 π n sinnt 2 π tcosntdt π = 2 [ t π n sinnt π π ] sinntdt n = 2 π πn 2 cosnt) = 2 πn2 cosnπ +) = 2 πn 2 )n ) if n is even, = if n is odd. 4 πn 2 Thus, the Fourier cosine series of ft) is ft) = π t = π cost+ ) π 3 2 cos3t+ cos2k )t+. 2k ) 2 This series converges to ft) for t π and to the even 2π-periodic extension f e t) otherwise, since f e t) is continuous, as can be seen from the graph of f e t) which is given in Figure.8. Similarly, from 5), the Fourier sine series of ft) is ft) b n sinnt

36 758 Fourier Series π 3π 2π π π 2π 3π Fig..8 The even 2π-periodic extension f et) of the function ft) = π t defined on t π. with for n ) b n = 2 π π ft)sinntdt = 2 π π π π t)sinntdt = 2 π n cosnt 2 tsinntdt π = 2 π n cosnt 2 [ tn π π cosnt + n = 2 n Thus, the Fourier sine series of ft) is ft) = π t = 2 sinnt n. π ] cosntdt This series converges to ft) for < t < π and to the odd 2π-periodic extension f o t) otherwise, except for the jump discontinuities of f o t), which occur at the multiples of nπ, as can be seen from the graph of f o t) which is given in Figure.9. At t = nπ, the series converges to. π 3π 2π π π 2π 3π π Fig..9 The odd 2π-periodic extension f et) of the function ft) = π t defined on t π.

37 .5 Operations on Fourier Series 759 Exercises. A function ft) is defined on an interval < t <. Find the Fourier cosine and sine series of f and sketch the graphs of the two extensions of f to which these two series converge.. ft) =, < t < 2. ft) = t, < t < 3. ft) = t, < t < 2 4. ft) = t 2, < t < if < t < π/2 5. ft) = if π/2 < t < π 6. ft) = t if < t, if < t < 2 7. ft) = t t 2, < t < 8. ft) = sint, < t < π 9. ft) = cost, < t < π. ft) = e t, < t <. ft) = 2/)t, < t <.5 Operations on Fourier Series In applications it is convenient to know how to compute the Fourier series of function obtained from other functions by standard operations such addition, scalar multiplication, differentiation, and integration. We will consider conditions under which these operations can be used to facilitate Fourier series calculations. First we consider linearity. Theorem. If f and g are 2-periodic functions with Fourier series and ft) a 2 + a n cos nπt +b nsin nπt )

38 76 Fourier Series gt) c 2 + c n cos nπt +d nsin nπt ), then for any constants α and β the function defined by ht) = αft)+βgt) is 2-periodic with Fourier series ht) αa +βc 2 + αa n +βc n )cos nπt +αb n +βd n )sin nπt ). ) This result follows immediately from the Euler formulas for the Fourier coefficients and from the linearity of the definite integral. Example 2. Compute the Fourier series of the following periodic functions.. h t) = t+ t, π t < π, h t+2π) = ht). 2. h 2 t) = +2t, t <, h 2 t+2) = ht). Solution. The following Fourier series were calculated in Examples 4 and 5 of Section.2: t = π 2 4 cos2k +)t π 2k +) 2 and t = 2 π k= ) n+ sin nπ n t. Since h is 2π-periodic, we take = π in the Fourier series for t over the interval t. By the linearity theorem we then obtain h t) = t+ t = π 2 4 π k= cos2k +)t 2k +) 2 +2 ) n+ sin nt. n Since h t) is piecewise smooth, the equality is valid here, with the convention that the function is redefined to be h t + )+h t ))/2 at each point t where the function is discontinuous. The graph of h t) is similar to that of the half sawtooth wave from Figure.9, where the only difference is that the period of h t) is 2π, rather than 4. The function h 2 t) = + 2t is a sum of an even function and an odd function 2t. Since the constant function is it s own Fourier series that is, a = 2, a n = = b n for n, then letting = in the Fourier series for t, we get h 2 t) = +2t = + 4 ) n+ sin nπt. π n

39 .5 Operations on Fourier Series 76 Differentiation of Fourier Series We next consider the term-by-term differentiation of a Fourier series. This will be needed for the solution of some differential equations by substituting the Fourier series of an unknown function into the differential equation. Some care is needed since the term-by-term differentiation of a series of functions is not always valid. The following result gives sufficient conditions for termby-term differentiation of Fourier series. Note first that if ft) is a periodic function of period p, then the derivative f t) where the derivative exists) is also periodic with period p. This follows immediately from the chain rule: Since ft+p) = ft) for all t, f t+p) = d dt ft+p) = d dt ft) = f t). Theorem 3. Suppose that f is a 2-periodic function that is continuous for all t, and suppose that the derivative f is piecewise smooth for all t. If the Fourier series of f is ft) a 2 + a n cos nπt +b nsin nπt ), 2) then the Fourier series of f is the series f t) nπ a nsin nπt + nπ b ncos nπt ) 3) obtained by term-by-term differentiation of 2). Moreover, the differentiated series 3) converges to f t) for all t for which f t) exists. Proof. Since f is assumed to be 2-periodic and piecewise smooth, the convergence theorem Theorem 2 of Section.3) shows that the Fourier series of f t) converges to the average of the left-hand and right-hand limits of f at each t. That is, f t) = A 2 + A n cos nπt +B nsin nπt ), 4) where the coefficients A n and B n are given by the Euler formulas A n = f t)cos nπt dt and B n = f t)sin nπt dt. To prove the theorem, it is sufficient to show that the series in Equations 3) and 4) are the same. From the Euler formulas for the Fourier coefficients of f,

40 762 Fourier Series A = f t)dt = ft) = f) f) = since f is continuous and 2-periodic. For n, the Euler formulas for both f and f, and integration by parts give A n = = = nπ b n, since f) = f), and f t)cos nπt dt nπt ft)cos + nπ ft)sin nπt dt B n = f t)sin nπt dt = nπt ft)sin nπ = nπ a n. ft)cos nπt dt Therefore, the series in Equations 3) and 4) are the same. Example 4. The even triangular wave function of period 2π given by t π t <, ft) = ; ft+2π) = ft), t t < π, whose graph is shown in Figure.7, is continuous for all t, and f π t <, t) = ; f t+2π) = ft), < t < π, is piecewise smooth. In fact f is an odd square wave of period 2π as in Figure.5. Therefore, f satisfies the hypotheses of Theorem 3. Thus, the Fourier series of f: ft) = π 2 4 cost π 2 + cos3t cos5t cos7t ) computed in Example 4 of Section.2) can be differentiated term by term. The result is f t) = 4 sint+ 3 π sin3t+ 5 sin5t+ 7 ) sin7t+,

41 .5 Operations on Fourier Series 763 which is the Fourier series of the odd square wave of period 2π computed in Eq ) of Section.2. Example 5. The sawtooth wave function of period 2 given by ft) = t for t < ; ft+2) = ft). see Example 5 of Section.2) has Fourier series ft) 2 sin π π t 2 sin 2π t+ 3 sin 3π t 4 sin 4π ) t+ = 2 π ) n+ sin nπ n t. Since the function t is continuous, it might seem that this function satisfies the hypotheses of Theorem 3. However, looking at the graph of the function see Figure.8) shows that the function is discontinuous at the points t = ±, t = ±3,... Term-by-term differentiation of the Fourier series of f gives 2 cos π t cos 2π t+cos 3π t cos 4π ) t+ = 2 ) n+ cos nπ t. This series does not converge for any point t. However, it is possible to make some sense out of this series using the Dirac delta function. Integration of Fourier Series We now consider the termwise integration of the Fourier series of a piecewise continuous function. Start with a piecewise continuous 2-periodic function f and define an antiderivative of f by gt) = t fx)dx. 5) Then g is a continuous function. It is not automatic that g is periodic. However, if it follows that gt+2) = t+2 g) = fx)dx = fx)dx =, 6) fx)dx+ t+2 fx)dx

42 764 Fourier Series = = t+2 t = gt), fx 2)dx since f is 2-periodic fu)du using the change of variables u = x 2 so that g is periodic of period 2. Therefore we have verified the following result. Theorem 6. Suppose that f is a piecewise continuous 2-periodic function. Then the antiderivative g of f defined by 5) is a continuous piecewise smooth 2-periodic function provided that 6) holds. Now we can state the main result on termwise integration of Fourier series. Theorem 7. Suppose that f is piecewise continuous and 2-periodic with Fourier series ft) a 2 + a n cos nπt +b nsin nπt ). 7) If gt) = t fx)dx and g) = fx)dx =, which is equivalent to a =, then the Fourier series 7) can be integrated term-by-term to give the Fourier series t gt) = fx)dx A 2 + nπ a nsin nπt nπ b ncos nπt ) 8) where A = tft) dt. 9) The integrated series 8) converges to gt) = t fx)dx for all t. Proof. Since f is assumed to be 2-periodic and piecewise continuous and a =, Theorem 6 and the convergence theorem shows that the Fourier series of gt) converges to gt) at each t. That is, gt) = A 2 + A n cos nπt +B nsin nπt ), ) where the coefficients A n and B n are given by the Euler formulas A n = gt)cos nπt dt and B n = gt)sin nπt dt. For n apply integration by parts using u = gt), dv = cos nπ tdt so that du = g t)dt = ft)dt, v = nπ nπ sin t. This gives

43 .5 Operations on Fourier Series 765 A n = gt)cos nπt dt = gt) nπt sin nπ = nπ nπ ft)sin nπt dt g t)sin nπt dt ). Hence, A n = nπ b n where b n is the corresponding Fourier coefficient for f. Similarly, B n = nπ a n and using integration by parts with u = gt), dv = dt so that du = g t)dt = ft)dt, v = t, A = gt)dt = tgt) tft)dt = tft)dt. Example 8. et ft) be the odd square wave of period 2: t <, ft) = ; ft+2) = ft), t <, which has Fourier series ft) 4 sin π π t+ 3 sin 3π t+ 5 sin 5π t+ 7 sin 7π ) t+ that was computed in Eq ) of Section.2. For t <, gt) = t fx)dx = t )dx = t and for t <, gt) = t fx)dx = )dx+ t dx = t. Thus, gt) is the even triangular wave function of period 2 shifted down by. That is, in the interval t gt) = t fx)dx = t with 2 periodic extension to the rest of the real line. See Figure.2. Theorem 7 then gives, for t, gt) = t = A π π cos π t 9π cos 3π t 25π cos 5π ) t = A 2 4 π 2 cos π t+ 9 cos 3π t+ 25 cos 5π ) t+

44 766 Fourier Series 3 2 t <, a) ft) = t <. y t y b) gt) = t fx)dx 2 3 t Fig..2 = A 2 4 π 2 This can also be written as 2k )π cos t. 2k ) 2 k= t = + A ) 4 2 π 2 2k )π cos t. 2k ) 2 k= The constant term +A /2 can be determined from Eq. 9): + A 2 = tft)dt = t dt = = 2. The constant term +A /2 can also be determined as one-half of the zero Fourier coefficient of t : + A 2 = t dt = 2 2. Exercises. et the 2π-periodic function f t) and f 2 t) be defined on π t < π by f t) =, π t < ; f t) =, t < π; f 2 t) =, π t < ; f 2 t) = t, t < π. Then the Fourier series of these functions are f t) π n=odd sinnt n,

45 .5 Operations on Fourier Series 767 f 2 t) π 4 2 π n=odd cosnt n 2 + ) n+sinnt n. Without further integration, find the Fourier series for the following 2πperiodic functions: a) f 3 t) =, π t < ; f 3 t) =, t < π; b) f 4 t) = t, π t < ; f 4 t) =, t < π; c) f 5 t) =, π t < ; f 5 t) = t, t < π; d) f 6 t) = 2, π t < ; f 6 t) =, t < π; e) f 7 t) = 2, π t < ; f 7 t) = 3, t < π; f) f 8 t) =, π t < ; f 8 t) = +2t, t < π; g) f 9 t) = a+bt, π t < ; f 9 t) = c+dt, t < π. 2. et f t) = t for π < t < π and f t+2π) = f t) so that the Fourier series expansion for f t) is f t) = 2 ) n+ sin nt. n Using this expansion and Theorem 7 compute the Fourier series of each of the following 2π-periodic functions. a) f 2 t) = t 2 for π < t < π. b) f 3 t) = t 3 for π < t < π. c) f 4 t) = t 4 for π < t < π. 3. Given that t = π 2 4 π n=odd cosnt, for π < t < π n2 compute the Fourier series for the function ft) = t 2 t 2 if < t < π sgnt = t 2 if π < t <. 4. Compute the Fourier series for each of the following 2π-periodic functions. You should take advantage of Theorem and the Fourier series already computed or given in Exercises 2 and 3. a) g t) = 2t t, π < t < π. b) g 2 t) = At 2 +Bt+C, π < t < π, where A, B, and C are constants. c) g 3 t) = tπ t)π +t), π < t < π. 5. et ft) be the 4-periodic function given by