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# Chapter 2 (maths 3)

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### Chapter 2 (maths 3)

1. 1. CHAPTER 2 FOURIER SERIESPERIODIC FUNCTIONS A function f (x ) is said to have a period T if for all x, f ( x + T ) = f ( x) , where T is apositive constant. The least value of T>0 is called the period of f (x) .EXAMPLES We know that f (x ) = sin x = sin (x + 4 π ) = … Therefore the function has period 2 π , 4π , 6 π , etc. However, 2 is the least value and therefore is the period of f(x). Similarly cos x is a periodic function with the period 2 π and tan x has period π .DIRICHLET’S CONDITIONS A function f (x ) defined in c ≤ x ≤ c+2l can be expanded as an infinite trigonometric a nπx nπxseries of the form o + ∑ a n cos + ∑ bn sin , provided 2 l l 1. f (x) is single- valued and finite in (c , c+2l) 2. f (x) is continuous or piecewise continuous with finite number of finite discontinuities in (c , c+2l). 3. f (x) has no or finite number of maxima or minima in (c , c+2l).EULER’S FORMULAS If a function f (x) defined in (c , c+2l) can be expanded as the infinite trigonometric ao ∞ nπx ∞ nπxseries 2 + ∑ an cos n =1 l + ∑ bn sin n =1 l then c + 2l 1 nπx an = l ∫ c f ( x) cos l dx, n ≥ 0 c + 2l 1 nπx bn = l ∫c f ( x) sin l dx, n ≥ 1[ Formulas given above for a n and bn are called Euler’s formulas for Fourier coefficients]
2. 2. DEFINITION OF FOURIER SERIES ao ∞ nπx ∞ nπx The infinite trigonometric series 2 + ∑ an cos n =1 l + ∑ bn sin n =1 l is called theFourier series of f (x) in the interval c ≤ x ≤ c+2l, provided the coefficients are given by theEuler’s formulas.EVEN FUNCTION If f (x) = φ (x) in (-l , l) such that φ (− x) = φ (x) , then f (x ) is said to be an evenfunction of x in (-l , l). φ1 ( x ) in (−l ,0) If f ( x) =  φ 2 ( x) in (0, l )Such that φ1 (− x) = φ 2 ( x) or φ 2 ( − x ) = φ1 ( x ) , then f (x) is said to be an even function of x in(-l , l).EXAMPLE y = cos x , y = x 2 are even functions.ODD FUNCTION If f (x) = φ (x) in (-l , l) such that φ (− x) = - φ (x) , then f (x) is said to be an oddfunction of x in (-l , l). φ1 ( x ) in (−l ,0) If f ( x) =  φ 2 ( x) in (0, l )Such that φ1 (− x) = - φ 2 ( x ) or φ 2 ( − x ) = - φ1 ( x ) , then f (x) is said to be an odd function of x in(-l , l).EXAMPLE y = sin x , y = x are odd functions.FOURIER SERIES OF EVEN AND ODD FUNCTIONS 1. The Fourier series of an even function f (x ) in (-l , l) contains only cosine terms 2
3. 3. (constant term included), i.e. the Fourier series of an even function f (x) in (-l , l) is given by ao nπx f (x) = 2 + ∑a n cos l , nπx l 2 where a n = ∫ f ( x ) cos dx. l 0 l 2. The Fourier series of an odd function f (x) in (-l , l) contains only sine terms, i.e. the Fourier series of an odd function f (x ) in (-l , l) is given by nπx f (x) = ∑b n sin l , nπx l 2 where bn = ∫ f ( x) sin l dx. l 0PROBLEMS1. Find the Fourier series of period 2l for the function f (x ) = x(2l – x) in (0 , 2l). Deduce 1 1 1the sum of f (x) = 2 − 2 + 2 − 1 2 3Solution: ao ∞ nπx ∞ nπx Let f (x ) = 2 + ∑a n =1 n cos l + ∑ bn sin n =1 l in (0 , 2l) …………(1) nπx 2l 1 an = l0∫ x(2l − x) cos l dx 2l   nπx   nπx   nπx    sin   − cos   − sin  1 2  l  − (2l − 2 x ) l  + (−2) l  , = (2lx − x ) l  nπ   n 2π 2   n 3π 3           l   l2   l3  0 using Bernoulli’s formula. 2 1 = [ − 2l cos 2nπ − 2l ] = − 4l 2 n 2π 2 n 2π 2l 1 x3  2l 1 4 a o = ∫ x(2l − x )dx = lx 2 −  = l 2 . l0 l 3 0 3 3
4. 4. nπx 2l 1 bn = l0∫ x(2l − x) sin l dx =0 Using these values in (1), we have 2 2 4l 2 ∞ 1 nπx x (2l - x) = l − 2 3 π ∑n n =1 2 cos l in (0, 2l) ……………..(2) 1 1 1 The required series 2 − 2 + 2 − … ∞ can be obtained by putting x = l in the Fourier 1 2 3series in (2). x = l lies in (0 , 2l) and is a point of continuity of the function f (x) = x(2l – x). ∴ [ Sum the Fourier series in (2) ] x =1 = f(l) ∞ 2 2 4l 2 1 i.e. l − 2 3 π ∑n n =1 2 cos nπ = l(2l - l) 4l 2  1 1 1  l 2 i.e.. -  − 2 + 2 − 2 + ...∞  = π2  1 2 3  3 ∴ 1 1 1 π2 − 2 + 2 − …∞ = 12 2 3 122. Find the Fourier series of period 2 π for the function f (x) = x cos x in 0 < x < 2 π .Solution: ∞ ∞ ao Let f (x ) = 2 + ∑ an cos nx + ∑ bn sin nx n =1 n =1 .……..…………(1) 2π 1 an = π ∫ x cos x cos nxdx 0 2π 1 = 2π ∫ x[ cos(n + 1) x + cos(n − 1) x]dx 0 1  sin( n + 1) x cos(n + 1) x  2π  sin( n − 1) x cos(n − 1) x  2π  =  x. +  +  x. +  , 2π  n +1 (n + 1) 2  0  n −1 ( n − 1) 2  0    if n ≠ 1 =0, if n ≠ 1 ao = 0 4
5. 5. 2π 2π 1 1 an = ∫ x cos xdx = 2π ∫ x(1 + cos 2 x)dx 2 π 0 0 2π 1 x sin 2 x cos 2 x  2 =  +x +  = π. 2π 2 2 4 0 2π 1 bn = π ∫ x cos x sin nxdx 0 2π 1 = 2π ∫ x[ sin(n + 1) x + sin(n − 1) x]dx 0 1  − cos(n + 1) x sin( n + 1) x  2π  − cos(n − 1) x sin( n − 1) x  2π  =  x. +  +  x. +  , 2π  n +1 (n + 1) 2  0  n −1 (n − 1) 2  0    if n ≠ 1 1 1  1 1  2n =− − = − + =− 2 , if n ≠ 1 n +1 n −1  n + 1 n − 1 n −1 2π 2π 1 1 b1 = π ∫ x cos x sin xdx = 0 2π ∫ x sin 2 xdx 0 2π 1   − cos 2 x  sin 2 x  1 =  x +  =−2 2π   2  4 0Using these values in (1), we get ∞ 1 n f(x) = π cos x − sin x − 2 ∑ sin nx n = 2 , 3,... n − 1 2 23. Find the Fourier series expansion of f (x) = sin ax in (-l , l).Solution: Since f (x) is defined in a range of length 2l, we can expand f (x ) in Fourier series ofperiod 2l. Also f ( − x) = sin[a(-x)] = -sin ax = - f (x) ∴ f (x) is an odd function of x in (-l , l). Hence Fourier series of f (x ) will not contain cosine terms. ∞ nπx Let f (x ) = ∑b n =1 n sin l ………………….(1) 5
6. 6. 1   nπ  nπ l    = ∫ cos l − a  − cos l + a  xdx l 0      l   nπ   nπ   1  sin  l − a  x sin  l + a  x  =    −    l nπ nπ −a +a   l l   0 1  nπ  1  nπ  = sin  − a l − sin  + a l nπ − la  l  nπ + la  l  = 1 nπ − al {− (−1) n sin al} − nπ 1 al {(−1) n sin al} +  1 1  = (−1) n +1 sin al  +   nπ − al nπ + al  (−1) n +1 2nπ sin al = n 2π 2 − a 2 l 2Using these values in (1), we get ∞ (−1) n +1 n nπx sin ax = 2π sin al ∑ sin n =1 n π −a l 2 2 2 2 l4. Find the Fourier series expansion of f (x) = e − x in (−π , π ) . Hence obtain a series forcosec πSolution: Though the range (−π , π ) is symmetric about the origin, e − x is neither an even functionnor an odd function. ∞ ∞ ao∴ Let f (x) = 2 + ∑ an cos nx + ∑ bn sin nx n =1 n =1 ..…..…………(1)in (−π , π ) [ the length of the range is 2π ] 6
7. 7. π 1 ∫π e cos nxdx −x an = π− π 1  e −x  =  2 ( − cos nx + n sin nx )  π n +1  −π =− 1 {e −π (−1) n − e π (−1) n } π ( n + 1) 2 2( −1) n = sinh π π (n 2 + 1) 2 sinh π ao = π π 1 ∫π e sin nxdx −x bn = π− π 1  e −x  =  2 ( − sin nx − n cos nx )  π n +1  −π =− n {e −π (−1) n − e π (−1) n } π ( n 2 + 1) 2n(−1) n = sinh π π (n 2 + 1)Using these values in (1), we get sinh π 2 sinh π ∞ (−1) n 2 sinh π ∞ (−1) n ne−x = π + π ∑ n 2 + 1 cos nx + π n =1 ∑ n 2 + 1 sin nx n =1 in (−π , π )[ Sum of the Fourier series of f ( x )] x =0 = f (0), [Since x=0 is a point of continuity of f(x)] sinh π  ∞ (−1) n i.e., 1 + 2∑ 2 −0  = e =1 π  n =1 n + 1  −1 ∞ (−1) ni.e., π cos ech π = 1 + 2  + 2∑ 2  2  n=2 n + 1 2 ∞ (−1) ni.e., cos ech π = ∑ π n=2 n 2 + 1 7
8. 8. HALF-RANGE FOURIER SERIES AND PARSEVAL’S THEOREM (i) The half range cosine series in (0 , l) is ao ∞ nπx f (x) = 2 + ∑a n =1 n cos l l 2 l∫ where a o = f ( x )dx. 0 nπx l 2 an = ∫ f ( x) cos l dx. l 0 (ii) The half range sine series in (0 , l) is ∞ nπx f (x) = ∑b n =1 n sin l , nπx l 2 where bn = ∫ f ( x) sin l dx. l 0 (iii) The half range cosine series in (0 , π ) is given by ∞ ao f (x) = 2 + ∑a n =1 n cos nx π 2 where a o = ∫ f ( x )dx. π 0 π 2 π∫ an = f ( x ) cos nxdx. 0 (iv) The half range sine series in (0 , π ) is given by ∞ f (x) = ∑b n =1 n sin nx , π 2 where bn = ∫ f ( x) sin nxdx. π 0 8
9. 9. ROOT-MEAN SQUARE VALUE OF A FUNCTIONDefinition c +2 l 1 ∫y 2 If a function y = f (x ) is defined in (c , c+2l), then dx is called the root mean- 2l csquare(R.M.S.) value of y in (c , c+2l) and is denoted by y. c + 2l 1 2 Thus y = ∫y 2 dx. 2l cPARSEVAL’S THEOREM If y = f (x ) can be expanded as a Fourier series of the formao ∞ nπx ∞ nπx2 + ∑ an cos n =1 l + ∑ bn sin n =1 l in (c , c+2l), then the root-mean square value y of y = f (x)in (c , c+2l) is given by 1 2 1 ∞ 1 ∞ y = a o + ∑ a n + ∑ bn 2 2 2 4 2 n =1 2 n =1PROOF ao ∞ nπx ∞ nπx f (x) = 2 + ∑ an cos n =1 l + ∑ bn sin n =1 l in (c , c+2l) ....……………….(1)∴ By Euler’s formulas for the Fourier coefficients, c + 2l 1 nπx an = l ∫ c f ( x) cos l dx, n ≥ 0 ..…………………(2) c + 2l 1 nπx bn = l ∫ c f ( x) sin l dx, n ≥ 1 …....……………..(3)Now, by definition, c + 2l c + 2l 1 1 ∫ [ f ( x)] 2 ∫ y dx = 2 y = 2 dx 2l c 2l c c + 2l 1 a ∞ nπx ∞ nπx  = ∫ f ( x)  o + ∑ a n cos + ∑ bn sin dx, using (1) 2l c  2 n =1 l n =1 l  ao1 c + 2l  ∞ a 1 c + 2l nπx  ∞ bn 1 c + 2l nπx  =  ∫ f ( x)dx  + ∑ n  ∫ f ( x) cos dx  + ∑  ∫ f ( x) sin dx  4  l c  n =1 2  l c l  n =1 2  l c l  ∞ ∞ ao an bn = .a o + ∑ .a n + ∑ .bn , by using (2) and (3) 4 n =1 2 n =1 2 9
10. 10. ∞2 ∞ 2 2 ao an b = +∑ +∑ n . 4 n =1 2 n =1 2EXAMPLES1. Find the half-range (i) cosine series and (ii) sine series for f (x ) = x 2 in (0 , π )Solution: (i) To get the half-range cosine series for f (x ) in (0 , π ), we should give an evenextension for f (x) in ( − π , 0). i.e. put f (x) = ( − x ) 2 = x 2 in ( − π , 0)Now f (x) is even in ( − π , π ). ∞ ao ∴ f (x) = 2 + ∑a n =1 n cos nx ………………….(1) π 2 an = π ∫ f ( x) cos nxdx. 0 π 2 2 π∫ = x cos nxdx 0 π 2   sin nx   − cos nx   − sin nx  = x2   − 2 x  + 2  π  n   n 2   n 3  0 4 4(−1) n = .π (−1) n = ,n ≠ 0 πn 2 n2 π π 2 2 2 2 2 ao = ∫ f ( x)dx = π ∫ x dx = 3 π π 0 0∴ The Fourier half-range cosine series of x 2 is given by π2 ∞ (−1) n x2 = + 4∑ 2 cos nx in (0 , π ). 3 n =1 n (ii) To get the half-range sine series of f (x ) in (0 , π ), we should give an odd extensionfor f (x) in (- π , 0). i.e. Put f (x ) = - ( − x ) 2 in (- π , 0) = - x 2 in (- π , 0)Now f (x) is odd in (- π , π ). 10
11. 11. ∞ ∴ f (x) = ∑b n =1 n sin nx ……………….(2) π π 2 2 bn = ∫ f ( x) sin nxdx = ∫ x 2 sin nxdx π 0 π 0 π 2   cos nx   sin nx   cos nx  = x2  −  − 2 x −  + 2  π  n   n 2   n 3  0  2 π 2   (−1) + 3 {(−1) − 1}  n +1 2 = n π n n   2 π 2 4    − , if n is odd = π  n n 3  − 2π , if n is even  nUsing this value in(2), we get the half-range sine series of x 2 in (0 , π ).2. Find the half-range sine series of f (x) = sin ax in (0 , l).Solution: We give an odd extension for f (x) in (-l , 0). i.e. we put f (x) = -sin[a(-x)] = sin ax in (-l , 0) ∴ f (x) is odd in (-l , l) ∞ nπx Let f (x ) = ∑b n =1 n sin l nπx l 2 bn = ∫ sin ax. sin l dx l 0 1   nπ  nπ l    = ∫ cos l − a  x − cos l + a  x dx l 0      l   nπ   nπ   1  sin  l − a  x sin  l + a  x  =    −    l   nπ   nπ     l − a  + a      l  0 1 1 = (−1) n +1 sin ( nπ − al ) − sin ( nπ + al ) nπ − al nπ + al 11
12. 12. 1 1 = (−1) n +1 sin al + ( −1) n +1 sin al nπ − al nπ + al 2nπ = (−1) n +1 sin al. 2 2 n π − a 2l 2Using this values in (1), we get the half-range sine series as ∞ (−1) n +1 .n nπx sin ax = 2π sin al ∑ 2 2 sin n =1 n π − a l 2 2 l3. Find the half-range cosine series of f (x ) = a in (0 , l). Deduce the sum of 1 1 1 2 + 2 + 2 + ∞ . 1 3 5 Solution: Giving an odd extension for f (x) in (-l , 0), f (x ) is made an odd function in (-l , l). nπx ∴ Let f(x) = ∑b n sin l ..……………(1) nπx l 2 bn = ∫ a sin dx l 0 l l  nπx   − cos l  = 2a   nπ l    = 2a nπ 1 − ( − 1) n { }    l 0   4a  , if n is odd =  nπ 0,  if n is even Using this value in (1), we get 4a ∞ 1 nπx a= ∑5 n sin l in (0 , l ) π n =1,3, Since the series whose sum is required contains constant multiples of squares of bn , we applyParseval’s theorem. l 1 1 ∑ bn = l ∫ [ f ( x)] dx 2 2 2 0 12
13. 13. ∞ 1 16a 2 1 i.e. . 2 π2 ∑ ( 2n − 1) n =1, 3, 5 2 = a2 ∞ 8a 2 1 i.e. π2 ∑ ( 2n − 1) n =1 2 = a2 ∞ 1 π2 ∴ ∑ ( 2n − 1) 2 8 . n =1 =4. Expand f (x) = x - x 2 as a Fourier series in -1 < x < 1 and using this series find the r.m.s. value of f (x ) in the interval.Solution: The Fourier series of f (x ) in (-1 , -1) is given by ∞ ∞ ao f (x) = 2 + ∑ an cos nπx + ∑ bn sin nπx n =1 n =1 .………………(1) 1 1 a o = ∫ f ( x)dx = ∫ ( x − x 2 ) dx 1 1 −1 −1 1  x2 x3  1 1  1 1 =  −  =  − − +   2   3  −1  2 3   2 3  −2 ao = ..........................(2) 3 1 1 ∫1 f ( x) cos nπx dx = −∫1( x − x ) cos nπx dx 1 an = 2 1− 1   sin nπx   − cos nπx   − sin nπx  = ( x − x 2 )   − (1 − 2 x )  2  + (−2) 3    n   n   n  −1 − cos nπ 3 cos nπ = − n2 n2 4 cos nπ an = − ……………….(3) n2 13
14. 14. 1 1 ∫1 f ( x) sin nπx dx = −∫1( x − x ) sin nπx dx 1 bn = 2 1− 1   − cos nπx   − sin nπx   cos nπx  = ( x − x 2 )   − (1 − 2 x )   + (−2) 3 3   nπ  nπ  n π  −1 2 2    − 2 cos nπ 2 cos nπ 2 cos nπ = − + 3 3 n 3π 3 nπ nπ n +1 2(−1) bn = ..........................( 4) nπSubstituting (2), (3), (4) in (1) we get 1 ∞ 4(−1) n +1 ∞ 2(−1) n +1 f (x) = − + ∑ cos nπx + ∑ sin nπx 3 n =1 n 2 n =1 nπWe know that r.m.s. value of f(x) in (-l , l) is 1 2 1 ∞ 1 ∞ a o + ∑ a n + ∑ bn 2 2 2 y = ……………….(5) 4 2 n =1 2 n =1From (2) we get −2 2 4 ao = ⇒ ao = .………………..(6) 3 9From (3) we get 4( −1) n +1 2 16 an = 2 ⇒ an = 4 ………………..(7) n nFrom (4) we get 2(−1) n +1 2 4 bn = ⇒ bn = 2 2 ..………………(8) nπ nπSubstituting (6), (7) and (8) in (5) we get 1 1 ∞  16 4  + ∑ 4 + 2 2  2 y = 9 2 n =1  n nπ 5. Find the Fourier series for f (x) = x 2 in − π < x < π . Hence show that 1 1 1 π4 + 4 + 4 + =14 2 3 90Solution: The Fourier series of f (x ) in (-1 , 1) is given by π2 ∞ 4(−1) n f (x) = 3 + ∑ n 2 cos nx n =1 14
15. 15. The co-efficients a o , a n , bn are 2π 2 4(−1) n ao = , an = , bn = 0 3 n2Parseval’s theorem is (a ) π ∞ 1 1 2 1 ∫ [ f ( x)] dx = ao + ∑ 2 2 2 n + bn 2π −π 4 2 n =1  ao 2 1 ∞ 2  ( ) π ∫ [x ] + ∑ a n + bn  2 2 dx = 2π  2 ∴ −π  4  2 n =1   π  x5   π 4 1 ∞ 16  i.e.,   5  = 2π   + ∑ 4   −π  9 2 n =1 n  2π 5 2π 5 ∞ 16 i.e., − =π∑ 4 5 9 n =1 n 8π 4 ∞ 16 =∑ 4 45 n =1 n ∞ 1 π4 i.e., ∑ n 4 = 90 n =1 1 1 1 π4 i.e., + 2 + 2 + ∞ = 12 3 5 90HARMONIC ANALYSIS The process of finding the Fourier series for a function given by numerical value isknown as harmonic analysis. In harmonic analysis the Fourier coefficients ao , a n , and bn of thefunction y = f (x) in (0 , 2 π ) are given by a o = 2[mean value of y in (0 , 2 π )] a n = 2[mean value of y cos nx in (0 , 2 π )] bn = 2[mean value of y sin nx in (0 , 2 π )](i) Suppose the function f (x) is defined in the interval (0 , 2l), then its Fourier series is, ao ∞ nπx ∞ nπx f (x) = 2 + ∑a n =1 n cos l + ∑ bn sin n =1 l and now, a o = 2[mean value of y in (0 , 2l)]  nπx  a n = 2 mean value of y cos in (0 , 2l )  l  15
16. 16.  nπx  bn = 2 mean value of y sin in (0 , 2l )  l (ii) If the half range Fourier sine series of f (x) in (0 , l) is, ∞ nπx f (x) = ∑b n =1 n sin l , then  nπx  bn = 2 mean value of y sin in (0 , l )  l (iii) If the half range Fourier sine series of f (x) in (0 , π ) is, ∞ nπx f (x) = ∑b n =1 n sin l , then bn = 2[ mean value of y sin nx in (0 , π )](iv) If the half range Fourier cosine series of f (x) in (0 , l) is, ao ∞ nπx f (x) = + ∑ a n cos , then 2 n =1 l a o = 2[mean value of y in (0 , l)]  nπx  a n = 2 mean value of y cos in (0 , l )  l (v) If the half range Fourier cosine series of f (x) in (0 , π ) is, ao ∞ nπx f (x) = 2 + ∑a n =1 n cos l , then a o = 2[mean value of y in (0 , π )] a n = 2[ mean value of y cos nx in (0 , π )] .EXAMPLES1. The following table gives the variations of a periodic function over a period T. 0 T T T 2T 5T T x 6 3 2 3 6 f (x) 1.98 1.3 1.05 1.3 -0.88 -0.25 1.98 2πxShow that f (x ) = 0.75 + 0.37 cos θ +1.004 sin θ , where θ = TSolution: Here the last value is a mere repetition of the first therefore we omit that value andconsider the remaining 6 values. ∴ n = 6. 16
17. 17. 2πx Given θ= ..………………..(1) T T T T 2T 5T π 2π ∴ when x takes the values of 0, , , , , θ takes the values 0, , , 6 3 2 3 6 3 3 4π 5ππ, , . (By using (1)) 3 3Let the Fourier series be of the form ao f ( x) = + a1 cos θ + b1 sin θ , ………………(2) 2 ∑y where a o = 2 ,  n     ∑ y cos θ  a1 = 2 ,  n     ∑ y sin θ  b1 = 2 , n=6  n    θ y cos θ sin θ y cos θ y sin θ 0° 1.98 1.0 0 1.98 0 π 1.30 0.500 0.866 0.65 1.1258 3 2π 3 1.05 -0,500 0.866 -0.525 0.9093 π 1.30 -1 0 -1.3 0 4π 3 -0.88 -0.500 -0.866 0.44 0.762 5π 3 -0.25 0.500 -0.866 -0.125 0.2165 4.6 1.12 3.013 ∑y a o = 2  = 1.5, a1 = 2 ∑ y cos θ = 0.37  6  6   2 b1 = ∑ y sin θ = 1.00456 6Substituting these values of a o , a1 , and b1 in (2), we get ∴ f (x) = 0.75 + 0.37 cos θ + 1.004 sin θ2. Find the Fourier series upto the third harmonic for the function y = f (x) defined in (0 , π ) from the table x 0 π 2π 3π 4π 5π π 6 6 6 6 6 17
18. 18. f (x) 2.34 2.2 1.6 0.83 0.51 0.88 1.19 Solution: We can express the given data in a half range Fourier sine series. f ( x) = b1 sin x + b2 sin 2 x + b3 sin 3 x ..………………...(1) x y = f(0) sin x sin 2x sin 3x y sin x y sin 2x y sin 3x 0 2.34 0 0 0 0 0 0 30 2.2 0.5 0.87 1 1.1 1.91 2.2 60 1.6 0.87 0.87 0 1.392 1.392 0 90 0.83 1 0 -1 0.83 0 -0.83 120 0.51 0.87 -0.87 0 0.44 -0.44 0 150 0.88 0.5 -0.87 1 0.44 0.76 0.88 180 1.19 0 0 0 0 0 0 4.202 3.622 2.25  ∑ y sin x  1 Now b1 = 2   = [ 4.202] = 1.40   6  3   ∑ y sin 2 x  1 b2 = 2   = [ 3.622] = 1.207   6  3   ∑ y sin 3 x  1 b3 = 2   = [ 2.25] = 0.75   6  3  Substituting these values in (1), we get f (x) = 1.4 sin x + 1.21 sin 2x + 0.75 sin 3x3. Compute the first two harmonics of the Fourier series for f(x) from the following data x 0 30 60 90 120 150 180 f (x ) 0 5224 8097 7850 5499 2626 0Solution: Here the length of the interval is π . ∴ we can express the given data in a half rangeFourier sine series i.e., f ( x) = b1 sin x + b2 sin 2 x ………………………(1) 18
19. 19. x y sin x sin 2x 0 0 0 0 30 5224 .5 0.87 60 8097 0.87 0.87 90 7850 1 0 120 5499 0.87 -0.87 150 2626 0.5 -0.87  ∑ y sin x  Now b1 = 2  = 7867.84   6    ∑ y sin 2 x  b2 = 2   = 1506.84   6   ∴ f (x) = 7867.84 sin x + 1506.84 sin 2x4. Find the Fourier series as far as the second harmonic to represent the function given in the following data. x 0 1 2 3 4 5 f (x ) 9 18 24 28 26 20Solution: Here the length of the interval is 6 (not 2 π ) i.e., 2l = 6 or l = 3 ∴ The Fourier series is ao πx 2πx πx 2πx f ( x) = + a1 cos + a 2 cos + b1 sin + b2 sin …………………..(1) 2 3 3 3 3 πx 2πx πx πx 2πx 2πx y cos y sin y cos y sin x 3 3 y 3 3 3 3 0 0 0 9 9 0 9 0 1 π 3 2π 3 18 9 15.7 -9 15.6 2 2π 3 4π 3 24 -12 20.9 -24 0 3 π 2π 28 -28 0 28 0 4 4π 3 8π 3 26 -13 -22.6 -13 22.6 5 5π 3 10 π 3 20 10 -17.4 -10 -17.4 125 -25 -3.4 -19 20.8 19
20. 20.  ∑ y  2(125) Now a o = 2 = = 41.66,  6  6   2 πx a1 = ∑ y cos = −8.33 6 3 2 πx b1 = ∑ y sin = −1.13 6 3 2 2πx a2 = 6 ∑ y cos 3 = −6.33 2 2πx b2 = 6 ∑ y sin 3 = 6.9Substituting these values of a o , a1 , b1 , a 2 and b2 in (1), we get 41.66 πx 2πx πx 2πx f ( x) = − 8.33 cos − 6.33 cos − 1.13 sin + 6.9 sin 2 3 3 3 3COMPLEX FORM OF FOURIER SERIES ∞ The equation of the form f ( x) = ∑c e n = −∞ n inπx lis called the complex form or exponential form of the Fourier series of f (x) in (c , c+2l). Thecoefficient c n is given by c + 2l 1 ∫ f ( x )e −inπx l cn = dx 2l cWhen l = π , the complex form of Fourier series of f (x) in (c , c+2 π ) takes the form ∞ f ( x) = ∑c e n = −∞ n inx , where c + 2π 1 ∫ f ( x )e −inx cn = dx. 2π cPROBLEMS1. Find the complex form of the Fourier series of f (x) = e x in (0 , 2).Solution: Since 2l = 2 or l = 1, the complex form of the Fourier series is ∞ f ( x) = ∑c e n = −∞ n inπx 20
21. 21. 2 1 c n = ∫ f ( x)e −inπx dx 20 2 1 = ∫ e x e −inπx dx 20 2 1  e ( 1−inπ ) x  =   2  1 − inπ  0 = 1 2(1 − inπ ) {e 2(1−inπ ) − 1} = (1 + inπ ) {e ( cos 2nπ − i sin 2nπ ) − 1} 2 2(1 + n π 2 2 ) = (e 2 − 1)(1 + inπ ) 2(1 + n 2π 2 )Using this value in (1), we get  e 2 − 1  ∞ (1 + inπ ) inπx  2  ∑ (1 + n 2π 2 ) e ex =     n =−∞2. Find the complex form of the Fourier series of f (x) = sin x in (0 , π ).Solution: Here 2l = π or l = π 2 . ∴ The complex form of Fourier series is ∞ f ( x) = ∑c e n = −∞ n i 2 nx …………………..(1) π 1 c n = ∫ sin xe −i 2 nx dx π 0 π 1  e −i 2 nx  =  { − i 2n sin x − cos x}  π 1 − 4n 2 0 = 1 π ( 4n − 1) 2 [ − e i 2 nx − 1 = − ] 2 π ( 4n 2 − 1)Using this value in (1), we get 2 ∞ 1 sin x = − ∑ 4n 2 − 1 .e i 2nx π n =−∞ in (0 , π )3. Find the complex form of the Fourier series of f (x) = e − ax in (-l , l).Solution: 21
22. 22. Let the complex form of the Fourier series be ∞ f ( x) = ∑c e n = −∞ n inπx l l 1 c n = ∫ f ( x)e −inπx l dx 2l −l l 1 = ∫ e − ax e −inπx l dx 2l −l l 1 = ∫ e −( al +inπ ) x / l dx 2l −l l 1  e −( al +inπ ) x l  =   2l  − ( al + inπ ) l  −l = 1 2( al + inπ ) [ e −( al +inπ ) − e ( al +inπ ) ] = 2 1 ( al + inπ ) [ e al (−1) n − e − al (−1) n ] [ e ± inπ = cos nπ ± i sin nπ = (−1) n ] sinh al (−1) n = al + inπ sinh al.( al − inπ ) (−1) n = a 2 l 2 + n 2π 2Using this value in (1), we have ∞ (−1) n ( al − inπ ) inπx l e − ax = sinh al ∑ 22 2 2 e n = −∞ a l + n π in (-l , l)4. Find the complex form of the Fourier series of f (x) = cos ax in (- π , π ), where a is neither zero nor an integer.Solution: Here 2l = 2 π or l = π . ∴ The complex form of Fourier series is ∞ f ( x) = ∑c e n = −∞ n inx ………………….(1) 22
23. 23. π 1 ∫π cos ax.e − inx cn = dx 2π − π 1  e −inx  =  2 { − in cos ax + a sin ax}  2π  a − n 2  −π = 1 2π ( a − n 2 ) 2 [ e −inπ ( − in cos aπ + a sin aπ ) − e inπ ( − in cos aπ − a sin aπ ) ] 1 = (−1) n 2a sin aπ 2π ( a − n ) 2 2Using this value in (1), we get a sin aπ ∞ (−1) n inx cos ax = π ∑ n = −∞ a2 − n2 e in (- π , π ). UNIT 2 PART – A 1. Determine the value of a n in the Fourier series expansion of f ( x) = x 3 in − π < x < π . Ans: f ( x) = x 3 is an odd function. ∴ an = 0 2. Find the root mean square value of f ( x) = x 2 in the interval (0 , π ) . Ans: RMS Vale of f ( x ) = x 2 in (0 , π ) is π π π 1  x5  ∫ [x ] 2 1 2 2 1 y = dx = ∫ x 4 dx =   π 0 π 0 π 5   0 1 π 5  π 4 = = π5  5  3. Find the coefficient b5 of cos 5 x in the Fourier cosine series of the function f ( x ) = sin 5 x inthe interval (0 , 2π )Ans: Here f ( x) = sin 5 xFourier cosine series is 23
24. 24. ∞ ao f (x) = 2 + ∑a n =1 n cos nx , where π π 2 2an = ∫ f ( x) cos nx dx = π ∫ sin 5 x cos nx dx π 0 0 π 2 = 2π ∫ [ sin(5 + n) x + sin(5 − n) x] dx 0 π − 1  cos(5 + n) x cos(5 − n) x  = + =0 π  5+n  5 − n 0  cos x, if 0 < x < π4. If f ( x) =  and f ( x) = f ( x + 2π ) for all x, find the sum of the Fourier 50, if π < x ≤ 2πseries of f (x ) at x = π .Ans: Here π is a point of discontinuity.∴ The sum of the Fourier series is equal to the average of right hand and left hand limit of thegiven function at x = π . f (π − 0) + f (π + 0)i.e., f (π ) = 2 cos π + 50 49 = = 2 25. Find bn in the expansion of x 2 as a Fourier series in (−π , π ) .Ans: bn = 0Since f ( x) = x 2 is an even function in (−π , π ) .6. If f (x) is an odd function defined in (-l , l) what are the values of a 0Ans: a0 = 0a n = 0 since f (x) is an odd function.7. Find the Fourier constants bn for x sin x in (−π , π ) .Ans: bn = 0Since f ( x) = x sin x is an even function in (−π , π ) . 24
25. 25. 8. State Parseval’s identity for the half-range cosine expansion of f (x ) in (0 , 1).Ans: 1 2 ∞ a0 2 ∫ [ f ( x)] dx = + ∑ an 2 2 0 2 n =1where 1 a 0 = 2 ∫ f ( x) dx 0 1 a n = 2 ∫ f ( x) cos nx dx 09. Find the constant term in the Fourier series expansion of f ( x ) = x in (−π , π ) .Ans: a 0 = 0 since f (x ) is an odd function in (−π , π ) .10. State Dirichlet’s conditions for Fourier series.Ans:(i) f (x) is defined and single valued except possibly at a finite number of points in (−π , π ) .(ii) f (x) is periodic with period 2 π .(iii) f (x) and f ′(x) are piecewise continuous in (−π , π ) . Then the Fourier series of f (x ) converges to (a) f (x) if x is a point of continuity f ( x + 0) + f ( x − 0) (b) if x is a point of discontinuity. 211. What you mean by Harmonic Analysis?Ans: The process of finding the Fourier series for a function given by numerical value isknown as harmonic analysis. In harmonic analysis the Fourier coefficients ao , a n , and bn of thefunction y = f (x) in (0 , 2 π ) are given by a o = 2[mean value of y in (0 , 2 π )] a n = 2[mean value of y cos nx in (0 , 2 π )] bn = 2[mean value of y sin nx in (0 , 2 π )] 25
26. 26.  2x 1 + π , − π < x < 0 12. In the Fourier expansion of f ( x) =  in (−π , π ) . Find the value of bn , 1 − 2 x , 0 < x < π  π the coefficient of sin nx.Ans: Since f (x) is an even function the value of bn = 0. 2( − x) 2x  In − π ≤ x ≤ 0 i.e., 0 ≤ − x ≤ π , f (− x ) = 1 − π = 1 + π = f ( x ) 13. What is the constant term and the coefficient of cos nx, a n in the Fourier expansion off ( x) = x − x 3 in (-7 , 7)?Ans:Given f ( x) = x − x 3 f ( − x ) = − x + x 3 = −( x − x 3 ) = − f ( x )The given function is an odd function. Hence a 0 and a n are zero.14. State Parseval’s identity for full range expansion of f (x ) as Fourier series in (0 , 2l).Ans: c + 2l 2 2 2 1 ∞ ∞ ∫ [ f ( x)] dx. = ao + ∑ a n + ∑ bn . 2 2l c 4 n =1 2 n =1 2 where c + 2l 1 nπx an = l ∫ c f ( x) cos l dx, n ≥ 0 c + 2l 1 nπx bn = l ∫ c f ( x) sin l dx, n ≥ 115. Find a Fourier sine series for the function f (x ) = 1; 0 < x < π .Ans: ∞The Fourier sine series of f ( x) = ∑ bn sin nx …………………….(1) n =1 26
27. 27. π 2 bn = π ∫ f ( x) sin nx dx 0 π π 2  − cos nx  2 = ∫ sin nx dx =  π 0 π  n 0  =− 2 nπ ( (−1) n − 1) bn = 0, when n is even 4 = , when n is odd nπ ∞ 4 ∴ f ( x) = ∑ . sin nπ n =1, 3, 5, nπ 0 0< x<π16. If the Fourier series for the function f ( x ) =  is sin x 0 < x < 2π − 1 2  cos 2 x cos 4 x  1 1 1 1 π −2f ( x) = +  + +  + sin x Deduce that − + − ∞ = . π π  1.3 3.5  2 1.3 3.5 5.7 4Ans: π Putting x = we get 2  π  −1 2  1 1 1  1 f = + − + − +  ∞ +  2  π π  1.3 3.5 5.7  2 −1 2  1 1 1  1 0= + − + − + ∞ + π π  1.3 3.5 5.7  2 1 1 1 π −2 − + − ∞ = .1.3 3.5 5.7 417. Define Root mean square value of a function?Ans: c +2 l 1 ∫y 2 If a function y = f (x ) is defined in (c , c+2l), then dx is called the root mean- 2l csquare(R.M.S.) value of y in (c , c+2l) and is denoted by y. 27
28. 28. c + 2l 2 1 Thus y = ∫y 2 dx. 2l c18. If f ( x) = x 2 + x is expressed as a Fourier series in the interval (-2 , 2), to which value thisseries converges at x = 2.Ans: Since x = 2 is a point of continuity, the Fourier series converges to the arithmetic mean off (x) at x = -2 and x = 2 f (2) + f (−2) 4 − 2 + 4 + 2 i.e., = =4 2 219. If the Fourier series corresponding to f ( x ) = x in the interval (0 , 2π ) isa0 ∞ + ∑ (a n cos nx + bn sin nx), without finding the values of a 0, a n , bn find the value of2 n =1 2 ∞a0 + ∑ (a n + bn ). 2 2 2 n =1Ans:By using Parseval’s identity, 2 2π 2π a0 ∞ 1 1  x3  8 + ∑ (a n + bn ) = ∫ x 2 dx =   = π 2 . 2 2 2 π 0  3  π  0 3 n =120. Find the constant term in the Fourier series corresponding to f ( x) = cos 2 x expressed in theinterval (−π , π ) .Ans: Given f ( x) = cos 2 x π π π 1 1  1 + cos 2 x  1 sin 2 x  Now a 0 = ∫π cos x dx = π −∫π  2 dx = π  x + 2  0 = 1 2 π −     PART B1. (i) Express f ( x) = x sin x as a Fourier series in 0 ≤ x ≤ 2π . 28
29. 29. 2l  πx 1 2πx 1 3πx  (ii) Show that for 0 < x <l, x =  sin − sin + sin  . Using root mean square p l 2 l 3 l  1 1 1value of x, deduce the value of 2 + 2 + 2 + 1 2 32. (i) Find the Fourier series of periodicity 3 for f ( x ) = 2 x − x 2 in 0 < x < 3. (ii) Find the Fourier series expansion of period 2 π for the function y = f (x) which is definedin (0 , 2π ) by means of the table of values given below. Find the series upto the third harmonic. x 0 π 2π π 4π 5π 2π 3 3 3 3 f (x ) 1.0 1.4 1.9 1.7 1.5 1.2 1.03.(i) Find the Fourier series of periodicity 2 π for f ( x) = x 2 for 0 < x < 2 π . l 4l  πx 1 3πx  (ii) Show that for 0 < x <l, x = −  cos + 2 cos +  . Deduce that 2 π2  l 3 l  1 1 1 π4 + 4 + 4 + = .14 3 5 96 l − x, 0 < x ≤ l4. (i) Find the Fourier series for f ( x) =  . Hence deduce the sum to infinity of 0, l ≤ x ≤ 2l ∞ 1the series ∑ (2n + 1) n =0 2 . (ii) Find the complex form of Fourier series of f ( x ) = e ax (−π < x < π ) in the form sinh aπ ∞ a + in inx π ∞ (−1) ne ax = π ∑ (−1) n −∞ a2 + n2 e and hence prove that =∑ 2 a sinh aπ −∞ n + a 2 .5. Obtain the half range cosine series for f ( x) = x in (0 , π ).6. Find the Fourier series for f ( x) = cos x in the interval (−π , π ) . 1 1 π37. (i) Expanding x(π − x) as a sine series in (0 , π ) show that 1 − + 3 + = . 33 5 32 (ii) Find the Fourier series as far as the second harmonic to represent the function given in thefollowing data. x 0 1 2 3 4 5 29
30. 30. f (x ) 9 18 24 28 26 208. Obtain the Fourier series for f (x ) of period 2l and defined as follows  L + x in ( − L,0) f ( x) =   L − x in (0, L) 1 1 1 π2Hence deduce that + 2 + 2 + = . 12 3 5 89. Obtain the half range cosine series for f ( x) = x in (0 , π ). 1 in (0, π )10. (i) Find the Fourier series of f ( x ) =  2 in (π ,2π ) (ii) Obtain the sine series for the function  l  x in 0 ≤ x ≤ 2  f ( x) =  l − x in l ≤ x ≤ l   211. (i) Find the Fourier series for the function 0 in (−1, 0) f ( x) =  and f ( x + 2) = f ( x ) for all x. 1 in (0, 1) (ii) Determine the Fourier series for the function πx, 0 ≤ x ≤1 f ( x) =  π (2 − x ), 1 ≤ x ≤ 212. Obtain the Fourier series for f ( x ) = 1 + x + x 2 in (−π , π ) . Deduce that 1 1 1 π2 + 2 + 2 + = .12 2 3 613. Obtain the constant term and the first harmonic in the Fourier series expansion for f (x)where f (x) is given in the following table. x 0 1 2 3 4 5 6 7 8 9 10 11 f (x) 18. 18. 17. 15. 11. 8.3 6. 5. 6. 9. 12. 15.7 0 3 4 0 4 0 7 6 0 614. (i) Express f ( x) = x sin x as a Fourier series in (−π , π ). 30