Fourier Series PDF

ESI. CP2. ANA3. Mrs Achour. Chapter 4: Fourier series. Introduction: In this chapter, we will learn how to …nd Fourier series to represent periodic functions as an in…nite series of sine and cosine terms. Let f be a real valued function de…ned on R,...

ESI. CP2. ANA3. Mrs Achour. Chapter 4: Fourier series. Introduction: In this chapter, we will learn how to …nd Fourier series to represent periodic functions as an in…nite series of sine and cosine terms. Let f be a real valued function de…ned on R, f is periodic () 9T > 0 = f (x + T ) = f (x) 8x 2 R, T is a period of f. If T is a period of f , then nT is also a period of f 8n 2 N. The smallest positive value (if it exists) is called the fundamental period f. The aim of this chapter is to expand a periodic function in Fourier series, we will examine the period 2 then a period T = 2l. 1 Trigonometric series 1.1 Regarding the period 2. De…nition 1 Let (an )n2N and (bn )n2N be two sequences of real numbers. We call a trigonometric series (of period 2 ), a series of functions on R with term un such as : un (x) = an : cos (nx) + bn : sin (nx) ; 8n 1 and a0 u0 (x) = 2 (an )n2N and (bn )n2N are called the coe¢ cients of the trigonometric series: a0 X In fact a trigonometric series is written as: + (an : cos nx + bn : sin nx) : 2 n 1 The theorems and results seen in the previous chapter can be used for the convergence of a trigonometric series. Theorem P 2 Let P (an )n2N and (bn )n2N be two sequences of real numbers. If jan j and jbn j are convergent then the trigonometric series: a0 X + (an : cos nx + bn : sin nx) 2 n 1 is normally convergent (hence uniformly convergent) on R and its sum is a continuous function. 1 Proof. In exercise. X cos nx sin nx Example 3 Give the nature of the series + 5 : n2 n2 n 1 Answer: It is a trigonometric series with: 1 an = ; 8n 1 and a0 = 0 n2 1 bn = 5 ; 8n 1 n2 It is enough to use the previous theorem in order to have the normal convergence of this one on R. Proposition 4 (Periodicity) If the trigonometric series: a0 X + (an : cos nx + bn : sin nx) ; 2 n 1 is convergent to a function f then this one is periodic of period 2. Proof. a0 X f (x) = + (an : cos nx + bn : sin nx) : 2 n 1 Then a0 X f (x + 2 ) = + (an : cos n (x + 2 ) + bn : sin n (x + 2 )) 2 n 1 a0 X = + (an : cos (nx + 2n ) + bn : sin (nx + 2n )) 2 n 1 a0 X = + (an : cos nx + bn : sin nx) 2 n 1 = f (x): We will therefore limit ourselves to working in an interval of length 2 as for example [0; 2 ] or [ ; ]: Proposition 5 (Expression of the coe¢ cients on [0; 2 ]) If the trigonomet- ric series: a0 X + (an : cos nx + bn : sin nx) 2 n 1 is uniformly convergent to a function f then its coe¢ cients are given by 8 > Z2 > > 1 > > a0 = f (x)dx > < Z2 0 1 and bn = f (x): sin nxdx; 8n 1: > > Z2 > > 1 0 > > a = f (x): cos nxdx; 8n 1 : n 0 2 Proof. We have a0 X f (x) = + (an : cos nx + bn : sin nx) 2 n 1 We multiply this equality by cos px then by sin px a0 X f (x): cos px = cos px + (an : cos nx: cos px + bn : sin nx cos px) 2 n 1 a0 X f (x): sin px = sin px + (an : cos nx sin px + bn : sin nx sin px) 2 n 1 The series obtained are also uniformly convergent because we multiplied by bounded functions (bounded independently of x). Now we integrate from 0 to 2 ; using the integrability theorem: 0 1 Z2 Z2 X Z2 Z2 a0 @an : cos nx: cos px:dx + bn : sin nx cos px:dxA f (x): cos px:dx = cos px:dx+ 2 0 0 n 1 0 0 0 1 Z2 Z2 X Z2 Z2 a0 @an : cos nx sin px:dx + bn : sin nx sin px:dxA f (x): sin px:dx = sin px:dx+ 2 0 0 n 1 0 0 Then it remains to use the trigonometric formulas. 1.2 Regarding the period T = 2l; l 6= 0 De…nition 6 Let (an )n2N and (bn )n2N be two sequences of real numbers. We call a trigonometric series (of period 2l), a series of functions on R with term un such as : n x n x un (x) = an : cos + bn : sin ; 8n 1 l l and a0 u0 (x) = 2 (an )n2N and (bn )n2N are called the coe¢ cients of the trigonometric series: A trigonometric series is generally written as: a0 X n x n x + an : cos + bn : sin : 2 l l n 1 Proposition 7 (Periodicity) If the trigonometric series: a0 X n x n x + an : cos + bn : sin ; 2 l l n 1 is convergent to a function f then this one is periodic of period 2l. 3 Proof. It is analogous to the case of the period 2. Proposition 8 (Expression of the coe¢ cients on [0; 2l]) If the trigonomet- ric series: a0 X n x n x + an : cos + bn : sin 2 l l n 1 is uniformly convergent to a function f then its coe¢ cients are given by 8 > Zl > > 1 > > a0 = l f (x)dx > < Zl l 1 n x and bn = f (x) sin dx; 8n 1: > > Z l l l > > 1 n x l > > a = f (x): cos dx; 8n 1 : n l l l Proof. It is analogous to the case of the period 2. 2 Fourier series expansion 2.1 De…nitions and parity De…nition 9 (For the period 2 ) Let f be a real valued function de…ned on R that has period 2 , integrable on any closed bounded interval of R (For example, we choose to work on [ ; ]). Set 8 > Z > > 1 > > a0 = f (x)dx > < Z 1 Z and b n = f (x) sin nxdx; ; 8n 1: > > 1 > > an = f (x) cos nxdx; 8n 1 > > : Then the trigonometric series a0 X n ot + (an : cos nx + bn : sin nx) = Ff (x) 2 n 1 is called the Fourier series associated to the function f. (an )n2N and (bn )n2N are called Fourier coe¢ cients. Remark 10 Notice that we are not saying f (x) is equal to its Fourier Series. Later we will discuss conditions under which that is actually true. Proposition 11 (Odd and even functions) Let f be a real valued function de…ned on R that has period 2 , integrable on any closed bounded interval of R. 1) If f is an even function, then its Fourier coe¢ cients verify: Z Z 2 2 a0 = f (x)dx; an = f (x) cos nxdx; 8n 1; bn = 0; 8n 1: 0 0 4 2) If f is an odd function, then its Fourier coe¢ cients verify: Z 2 an = 0; 8n 0; bn = f (x) sin nxdx; 8n 1: 0 Proof. 1) f is an even function f ( x) = f (x) 8x 2 R, so Z Z 1 2 a0 = f (x)dx = f (x)dx; 0 the function g = g(x) = f (x) cos nx is also even, hence Z Z 1 2 an = f (x) cos nxdx = f (x) cos nxdx; 8n 1 0 on the other hand the function h = h(x) = f (x) sin nx is odd ie h( x) = h(x), hence Z 1 bn = f (x) sin nxdx = 0; 8n 1: 2) It is evident that the same reasoning gives this case. Example 12 Let f be a real valued function de…ned on R of period 2 such that 1 if x 2]0; [ f (x) = 1 if x 2] ; 0[ Find the Fourier series Ff (x) associated to f. Answer: Step one: Sketch a graph of f over two periods (at least). Step two: f is integrable on any closed bounded interval of R, then Ff exists. Step three: f has period 2 and it is an odd function, we choose the interval [ ; ]. Step four: calculation of coe¢ cients Z 2 an = 0; 8n 0; bn = f (x) sin nxdx; 8n 1: 0 Then Z 2 2 1 2 (1 cos n ) bn = sin nxdx = : cos nx = ; 8n 1: n 0 n 0 X 2 X (1 cos n ) Step …ve: Ff (x) = bn : sin nx = : sin nx series ex- n n 1 n 1 pansion of sinuses. 5 Remark 13 In the previous example, we can improve the expression of bn , we n recall that cos n = ( 1) , we obtain: n 2 (1 ( 1) ) bn = ; 8n 1 8 n < 4 if n = 2p + 1; p 0 = (2p + 1) : 0 otherwise. 4X 1 4X 1 Then Ff (x) = : sin (2p + 1) x = : sin (2n + 1) x: 2p + 1 2n + 1 p 0 n 0 Remark 14 The previous example allows us to see that two di¤ erent functions can have the same Fourier series. De…nition 15 (For the period 2l) Let f be a real valued function de…ned on R that has period 2l, integrable on any closed bounded interval of R (For example, we choose to work on [ l; l]). The Fourier series for the function f is the trigonometric series a0 X n x n x n otée + an : cos + bn : sin = Ff (x) 2 l l n 1 (an ) and (bn )n2N are called Fourier coe¢ cients, they are given by : 8 n2N l > Z > > 1 > > a 0 = f (x)dx > < l Zl l 1 n x and bn = f (x) sin dx; 8n 1. > > Z l l l > > 1 n x l > > a = f (x) cos dx; 8n 1 : n l l l The parity proposition remains valid with the necessary modi…cations. 2.2 Dirichlet’s theorem and Parseval’s equality De…nition 16 A function f can be expanded in Fourier series on a set E R if f (x) = Ff (x) 8x 2 E: Remark 17 If it is asked to expand f into a Fourier series, it will be necessary to …nd the largest domain E such that f (x) = Ff (x) 8x 2 E: Theorem 18 (Dirichlet) Let f be a real valued function de…ned on R that has period 2l, integrable on any closed bounded interval of R , let x0 2 R such that: 1) lim+ f (x) and lim f (x) which exist and are …nite will be noted x!x0 x!x0 f x+ 0 = lim+ f (x) and f x0 = lim f (x): x!x0 x!x0 6  f (x) f x+ 0 f (x) f x0 2) lim and lim exist and are …nite. x!x0+ x x0 x!x0 x x0 1 Then Ff (x0 ) is convergent to f x+0 + f x0 ie: 2 1 Ff (x0 ) = f x+ 0 + f x0 : 2 In particular if f is continuous on E R and admits at every point of E a right and left derivative, then: Ff (x) = f (x); 8x 2 E: Example 19 Let f be a real valued function de…ned on R of period 2 such that f (x) = jxj ; 8x 2 [ ; ] 1) Sketch a graph of f over two periods. Calculate Ff (x). 2) Expand f in Fourier series. 3) Deduce the valuesof the following in…nite series: X 1 X 1 S1 = 2 ; S2 = (2n + 1) n2 n 0 n 1 Answer: 1) f is periodic with period 2 and its graph is given. Calculation of Ff (x). f is a 2 -periodic function which is integrable on [ ; ] (it is also contin- uous on R) then Ff exists. we choose to work in [ ; ], f is even, using the formulas for the Fourier coe¢ cients we have Z Z 2 2 a0 = f (x)dx; an = f (x) cos nxdx; 8n 1; bn = 0; 8n 1: 0 0 Z Z 2 2 2 x2 1 2 a0 = f (x)dx = xdx = = =. 2 0 0 0 Z ( 2 u = x ! u0 = 1 an = x cos nxdx; 8n 1, let’s do a IBP: 1. v 0 = cos nx ! v = sin nx 0 n 0 1 h i Z 2B B x sin nx 1 C an = @ n sin nxdxC A | {z 0 } n 0 =0 2 1 2 (cos n 1) = : cos nx = n n 0 n2 n 2 (( 1) 1) = : n2 7 Then a0 X X 2 (( 1) n 1) Ff (x) = + an : cos nx = + cos (nx) : 2 2 n2 n 1 n 1 We prefer to improve the expression of an in preparation for question 3): 8 < 4 2 if n = 2p + 1; p 0 an = (2p + 1) : 0 otherwise. 4 X cos (2p + 1) x Conclusion: Ff (x) = 2 ; ie : 2 (2p + 1) p 0 4 X cos (2n + 1) x Ff (x) = 2 cosine series expansion. 2 (2n + 1) n 0 2) To expand f into a Fourier series, let’s use Dirichlet’s theorem, since: f is 2 périodic (and locally integrable), f is continuous on R (according to the graph). f admits a derivative to the right and to the left of each point of R (in exer- cise calculate limits). 4 X cos (2n + 1) x Then f (x) = Ff (x) = 2 ; 8x 2 R 2 (2n + 1) n 0 We can also apply Dirichlet’s theorem only in [0; ] because f is an even 2 périodic function, we will have: 4 X cos ((2n + 1) x) 2 = x; 8x 2 [0; ]::::: (F) 2 (2n + 1) n 0 Therefore f is expanded in Fourier series on R. X 1 3) S1 = 2 ; replace x with 0 in (F). n 0 (2n + 1) 2 4 S1 = 0 = f (0) =) S1 = : 2 X 1 X 1 X 8 1 S2 = = 2 + 2 n2 (2p) (2p + 1) n 1 p 1 p 0 1X 1 X 1 S2 = 2 + 2 4 p (2p + 1) p 1 p 0 1 = S2 + S1 4 2 1 = S2 + : 4 8 2 2 1 We obtain S2 1 = =) S2 = : 4 8 6 8 De…nition 20 A function f is said of C 1 piecewise on an interval [a; b] if: i=n [1 1) [a; b] = [xi ; xi+1 ] (…nite union) with x0 = a and xn = b: i=0 2) 8i; f 2 C 1 (]xi ; xi+1 [), lim+ f 0 (x) and lim f 0 (x) exist and are …nite. x!xi x!xi+1 Corollary 21 Let f be 2l périodic, integrable on any closed bounded of R. If f is of the class C 1 piecewise on [ l; l]; then: 1 Ff (x) is convergent to f x+ + f x ; 8x 2 R 2 ie 1 f x+ + f x Ff (x) = ; 8x 2 R: 2 If moreover f is continuous on E R, then: Ff (x) = f (x); 8x 2 E. Example 22 Let the function 2 périodic, de…ned on R and given by: f (x) = x; 8x 2 [ ; [ 1) Draw the graph over 2 periods. Calculate Ff (x). 2) Expand f in Fourier series. Answer: 1) f is periodic with period 2 and its graph is given. Calculation of Ff (x). f is a 2 -periodic function which is integrable on [ ; ] (it is also contin- uous on R) then Ff exists. we choose to work in [ ; ], f is even, using the formulas for the Fourier coe¢ cients we haveCalcul de Ff (x). f is "odd", so : Z 2 an = 0; 8n 0; bn = f (x) sin nxdx; 8n 1: 0 Z ( 2 u = x ! u0 = 1 ie bn = x sin nxdx = 8n 1, let’s do a IBP: 1. v 0 = sin nx ! v = cos nx 0 n 0 1 Z 2@ x 1 bn = cos nx + cos nxdxA n 0 n 0 n n+1 2 2 ( 1) 2 ( 1) = ( cos n ) = = : n n n Then X X ( 1)n+1 Ff (x) = bn : sin nx = 2 sin nx: n n 1 n 1 9 2) Let us apply the Dirichlet corollary on [0; ], because f is 2 periodic and odd. f=]0; [ (x) = x is of the class C 1 (polynomial) and f 0=]0; [ (x) = 1: lim f 0=]0; [ (x) = 1 2 R and lim f 0=]0; [ (x) = 1 2 R: x!0+ x! f is therefore C 1 piecewise on [0; ]; hence: 1 Ff (x) is convergent to f x+ + f x ; 8x 2 R 2 In particular, f is continuous on R n f(2k + 1) ; k 2 Zg, then X ( 1)n+1 F(f )(x) = sin nx = f (x); 8x 2 R n f(2k + 1) ; k 2 Zg : n n 1 At points (2k + 1) : F(f )((2k + 1) ) = 0 (direct calculation); f ((2k + 1) ) = , the values are di¤ erent. f is therefore expended in Fourier series on E = R n f(2k + 1) ; k 2 Zg : Theorem 23 (Parseval equality) Let f be 2l périodic, integrable on any closed bounded of R. Let (an )n2N and (bn )n2N be the coe¢ cients of its Fourier series, they verify: X 1) a2n + b2n is a convergent in…nite series. n 1 Zl 1 a20 X 2 2) f 2 (x)dx = + an + b2n Parseval equality. l 2 l n 1 X sin (nx) Example 24 The trigonometric series p can it be the Fourier series n n 1 of a locally integrable function and 2 periodic on R? Answer: X1 NO, because according to Parseval’s theorem the in…nite series would be n n 1 convergent and this is not true. Example 25 Let f be a periodic function of period 2 such that 8 < 1 ( + x) if > < x < 0; f (x) = 2 > 1 : ( x) if 0 x : 2 1) Graphically represent the function f and theoretically justify the existence of Ff: 2) Determine the Fourier series of f and deduce its sum X ( 1)n X 1 3) Deduce the sums of in…nite series and : 2n + 1 n2 n 0 n 1 10 Answer: 1) Make the graph. f is a 2 -periodic function which is integrable on [ ; ] (it is therefore locally integrable on R) then Ff exists. 2) Calculation of Ff (x). we choose to work in [ ; ], f is odd, using the formulas for the Fourier coe¢ cients we have an = 0 8n 0: Z Z 2 1 and bn = f (x) sin (nx) dx = ( x) sin (nx) dx, 0 ( 0 x ! u0 = 1 u= Let’s do a IBP: 1 : v 0 = sin (nx) ! v = cos(nx) n 0 1 Z 1@ ( x) 1 bn = cos (nx) cos (nx) dxA n 0 n 0 1 1 1 = [sin(nx)]0 = : n n2 n X sin (nx) We obtain: Ff (x) = : n n 1 Let’s use the Dirichlet corollary: i) f ( ) = f ( ) and on [ ; ] f is discontinuous at 0 only; therefore the points of discontinuity on R are those of the form 2k , k 2 Z. ii) As f is odd and 2 periodic then it su¢ ces to restrict the work to [0; ], 1 iii) We have f is of the class C 1 on ]0; [; moreover lim+ f 0 (x) = lim = x!0 x!0 2 1 1 2 R and lim f 0 (x) = 2 R (we can not do the test in because f has 2 x! 2 no problem at this point), then f is C 1 piecewise on [0; ]. we then obtain: X sin (nx) F Ff (x) = = f (x) 8x 2]0; ] and Ff (0) = 0: n n 1 3) Deductions: X ( 1)n S1 =. 2n + 1 n 0 Let’s apply F for x = ; we have 2 p ( 1) if n = 2p + 1; p 0: sin n = : (To remember). 2 0 otherwise: X ( 1)p Then Ff = =f ie S1 = : 2 2p + 1 2 4 n 0 11  X 1 S2 = : Let’s use Parseval, as f is locally integrable on R odd and it is n2 n 1 2 périodic, we have Z X 1 f 2 (x)dx = b2n : n 1 Z X 2 2 Moreover f is even, so f 2 (x)dx = b2n ; which gives 0 n 1 Z X 1 1 2 ( x) dx = = S2 : 2 n2 0 n 1 2 We …nd S2 = : 6 2.3 Expanding a function de…ned on [a; b]. In this section, we would like to make expansions to a de…ned and locally inte- grable function on an interval [a; b]. Depending on the position of the interval, several cases can be given, we will simply suggest two of them. 2.3.1 f is a function de…ned and locally integrable on an interval [0; b]; b > 0. To make a Fourier series expansion for it, it su¢ ces to consider the extended function noted fe which will be de…ned on R, b periodic and equal to f on ]0; b] (or on [0; b[ or ]0; b[) and we apply Dirichlet to fe, then it will be necessary to restrict the results on [0; b]. To make a cosine series expansion for it, it su¢ ces to consider the extended function noted fe which will be de…ned on R, even, 2b periodic and equal to f on ]0; b] (or on [0; b[ or ]0; b[) and we apply Dirichlet to fe, then it will be necessary to restrict the results on [0; b]. To make a sine series expansion for it, it su¢ ces to consider the extended function noted fe which will be de…ned on R, odd, 2b periodic and equal to f on ]0; b] (or on [0; b[ or ]0; b[) and we apply Dirichlet to fe, then it will be necessary to restrict the results on [0; b]. Remark 26 If it is requested to do a Fourier series expansion, we may prefer to do a cosine or sine series expansion for ease of calculations. 2.3.2 f is a function de…ned and locally integrable on an interval [a; b]; a > 0. f (x); if x 2 [a; b]; Let g the function given by g(x) = 0; if x 2 [0; a[: The function g will be a de…ned and locally integrable function on the interval 12 [0; b], so we returned to the previous case, so it will be enough to work with the extended function of g then restrict the results to [a; b]. 1 Example 27 How can we proceed to expand the function g = g(x) = ( x) 2 on [0; ] in cosine series? ? It is not required to …nd expansion. Answer: (see previous example) It su¢ ces to consider the extension of g denoted ge which will be even 2 periodic and equal to g on ]0; ] (or [0; [ or ]0; [) and we apply Dirichlet to ge, then it will be necessary to restrict the results on [0; ] because this is the domain proposed for g. 13