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H ANOI U NIVERSITY OF S CIENCE AND T ECHNOLOGY S CHOOL OF A PPLIED M ATHEMATICS AND I NFORMATICS N GUYEN T HI T HU H UONG AND T RAN M INH T OAN Lecture on M ATH M ULTIPLE I NTEGRAL , I NTEGRAL THAT DEPENDS ON A PARAMETER , L INE I NTEGRAL , S URFACE I NTEGRAL , F IELD T HEORY AND S ERIES Summary, Examples, Exercises and Solutions Ha Noi - 2008 C ONTENTS Contents Chapter Multiple Integral Double Integral 1.1 Calculation of a double integral in Cartesian coordinate system 1.2 Change of variables in double integrals, polar coordinate 1.3 Applications of double integrals 1.4 Exercises 1.5 Solutions Triple Integral 2.1 Calculation of a triple integral in Cartesian coordinate system 2.2 Change of variables in triple integrals 2.3 Calculate the triple integrals in cylindrical coordinate 2.4 Calculate the triple integrals in spherical coordinate 2.5 Exercises 2.6 Solutions Chapter Integrals that depend on a parameter 5 12 14 17 20 20 22 22 24 24 25 29 29 29 29 32 32 33 33 34 35 39 Line integral of the first kind 39 The definite integrals that depend on a parameter 1.1 Definition 1.2 Properties The generalized integarls that depend on a parameter 2.1 The uniformly convergent integrals 2.2 Properties 2.3 Euler’s integrals Exercises Solutions Chapter Line integral 1 CONTENTS 1.1 Definition 1.2 Calculation formulae Line integral of the second kind 2.1 Definition 2.2 Calculation formulae 2.3 Theorem of four equivalent propositions 2.4 Area of a plane domain Exercises Solution Chapter Surface integral 39 39 41 41 41 45 46 46 49 53 Surface integral of the first kind 1.1 Definition 1.2 Calculation formulae Surface integral of the second kind 2.1 Definition 2.2 Calculation formulae 2.3 Stokes’ formula Exercises Solution Chapter Field theory 53 53 53 54 54 55 58 59 60 63 Scalar field Vector field Exercises Solution Chapter Series 63 64 66 66 69 Number series 1.1 Definition 1.2 Convergent criterion 1.3 Exercises 1.4 Solution Function series 2.1 Function sequence 2.2 Function series 2.3 Power series 2.4 Exercises 2.5 Solution Fourier series 69 69 70 74 75 78 78 78 80 82 83 85 CONTENTS 3.1 3.2 3.3 Decomposition theorem Exercises Solution 85 89 89 CONTENTS CHAPTER M ULTIPLE I NTEGRAL §1 D OUBLE I NTEGRAL 1.1 Calculation of a double integral in Cartesian coordinate system Consider the integral I= f ( x, y)dxdy (1.1) D (The Corollary of Fubini’s theorem) Suppose that D = [ a, b] × [c, d] and f : D → R is a continuous function on D Then I= f ( x, y)dy = dx a b d d b c c f ( x, y)dx dy a  a ≤ x ≤ b If D is described as follows: D = ,  ϕ( x ) ≤ y ≤ ψ( x ) where y = ϕ( x ), y = ψ( x) are continuous and have continuous derivatives on [ a, b] ψ( x ) b then I = a   ϕ( x )  f ( x, y)dy dx or ψ( x ) b I= f ( x, y)dy dx a ϕ( x ) (1.2) Chapter Multiple Integral  c ≤ y ≤ d If D is described as follows: D = ,  ϕ(y) ≤ x ≤ ψ(y) where x = ϕ(y), x = ψ(y) are continuous and have continuous derivatives on [c, d] then ψ(y) d I= f ( x, y)dx dy c (1.3) ϕ(y) Example 1.1 Calculate the double integral x2 ydxdy, I= D where D = [0, 1] × [0, 2] Solution We have I= x ydxdy = D = x2 x ydy = dx 0 y2 2 dx x3 = x2 dx = 2 3 x3 + xy dxdy where D is bounbed Example 1.2 Calculate the double integral I = by the curves y = x2 and y = √ D x Solution We have the region D = ≤ x ≤ 1, x2 ≤ y ≤ √ x (Figure 1.1) y y = x2 y= √ x O Figure 1.1 x Double Integral Therefore √ I= x x2 = x + xy dy = dx √ y2 x y+x x x2 dx √ x3 x − x5 + x2 − x5 dx = 2 36 Example 1.3 Interchange the order of the following integrals: i) I = Solution 2x dx x f ( x, y)dy; i) We have D = e ii) I =      x = 0, x = y=x    y = 2x ln y dy f ( x, y)dx (Figure 1.2) y O x Figure 1.2 From above figure, we have y I= ii) We have D =          f ( x, y)dx + dy y/2 y = 1, y = e x=0 x = ln y (Figure 1.3) dy y/2 f ( x, y)dx Chapter Multiple Integral y e O x Figure 1.3 Hence e I= f ( x, y)dy dx ex 1.2 Change of variables in double integrals, polar coordinate In general case Put I = f ( x, y)dxdy D To calculate I, we can perform the tranformation   x = x (u, v)  y = y(u, v) (1.4) The two equations in (1.4) define a mapping which carries a point ( x, y) ∈ D ⊂ Oxy ′ (or inversion) to (u, v) ∈ D ⊂ Ouv We shall consider mapping for which the functions x = x (u, v), y = y(u, v) are continuous and have continuous partial derivatives on D Then I= D ∂x where J = ∂u ∂y ∂u f ( x (u, v), y(u, v)) | J | dudv, f ( x, y)dxdy = ∂x ∂v = D ( x, y) = ∂y D (u, v) ∂v D Double Integral Polar coordinate In this case we write r and ϕ instead of u and v and discrible the mapping by the two equations   x = r cos ϕ ; | J | = r ≥ (1.5) y = r sin ϕ Then I= f ( x, y)dxdy = DOxy f (r cos ϕ, r sin ϕ) rdrdϕ D Orϕ Example 1.4 Calculate I = D y dxdy, where the region D is bounded by x y = x, y = 2x, xy = 1, xy = ( x > 0) y Solution Because x > 0, put = u, xy = v (u > 0, v > 0) Therefore we perform the x tranformation √  √ v 1 v  √ √ − √ x = √ D ( x, y) u√ u √v u u =− =⇒ J = = √ √ v u  D (u, v) u y = v u √ √ u v  1 ≤ u ≤ The region D → D = 1 ≤ v ≤ Hence I= dv 1 u − du = u v u = y y = 2x y=x xy = xy = O Figure 1.4 x Number series 1.4 Solution Solution 6.1 n2 n 1 a) = = + ,n ≥ n! ( n − 1) ! ( n − 2) ! ( n − 1) ! ∞ ∞ ∞ n2 1 = ∑ + ∑ = 2e ∑ n =2 ( n − ) ! n =1 ( n − ) ! n=1 n! 1 √ −√ √ n+2+ n+1 n+1+ n √ 1 √ ⇒ S = −√ = 1− −√ Sn = √ n+2+ n+1 2+1 2+1 1 1 c) − = 2 2n − 2n + 4n − 1 1 1− ⇒S= Sn = 2n + 1 1 1 1 d) = − −2 + = n(n + 1)(n + 2) n(n + 1) (n + 1)(n + 2) n n+1 n+2 1 1 1− − + ⇒S= Sn = 2 n+1 n+2 ( n + 1) − n = arctg(n + 1) − arctg n = arctg e) arctg + ( n + 1) n 1+n+n π Sn = arctg(n + 1) − arctg ⇒ S = b) an = √ Solution 6.2 All of these series not satisfy necessary conditions n2 − 2n + √ = a) lim =0 n→∞ 5n2 + (−1)n n n n b) lim = e− = n→∞ n + π 2n = =0 c) lim arctg n→∞ n Solution 6.3 n+1 ∼√ ∼ (n → ∞), the series is convergent a) √ ln n n−1 nn−1 n2 √ √ (2 − a ) n + (2 − a ) n + √ ∼ b) n4 + 2n + − n4 + an = √ 2n2 n4 + 2n + + n4 + an If a = 2, un ∼ the series is convergent 2n a−2 , the series is divergent If a = 2, un ∼ 2n c) For sufficiently large number n: n3 e−n ≤ , the series is convergent n n ( n + 1) + a < 1, the series is convergent = d) lim n+1 = lim n→∞ n→∞ an n2 + 3n +1 75 76 Chapter Series ln2 n e) α > 2: an ≤ α−1 ≤ 1+ε ; (0 < ε < α − 2), the series is convergent n n ln2 α ≤ 2: an ≥ α−1 , the series is divergent n a n +1 a a f) lim = The series is convergent if a < e, is divergent if a > e = lim n n→∞ an n→∞ e 1+ n (2n + 1)!! 22n (n − 1)! a n +1 = lim 2(n+1) = < 1, the series is convergent g) lim n→∞ n→∞ an n! (2n − 1)!! n √ 1 = < 1, the series is convergent h) lim n an = lim − n→∞ n→∞ n 5e √ √ n < 1, the series is convergent = i) lim n an = lim n n n→∞ n→∞ 4n − 16 √ n+a n = e a−b j) lim n an = lim n→∞ n→∞ n + b If a > b then e a−b > and the series is divergent If a < b then e a−b < and the series is convergent If a = b, an = does not satisfy the necessary condition, the series is divergent ,x ≥ k) f ( x ) = x ln x (ln ln x )2 ∞ f ( x )dx = − ln ln x ∞ < +∞, the series is convergent ,x ≥ x ln p x  ∞ ∞  if p = ln ln x f ( x )dx = (ln x )1− p ∞   if p = 1− p The series is convergent if p > 1, is divergent if < p ≤ l) f ( x ) = Solution 6.4 ln n ln n a) lim is decreasing as n → ∞ because = 0; an = n→∞ n n f (x) = − ln x ln x ′ ; f (x) = < 0, ∀ x ≥ x x2 The series is convergent √ √ n n = 0; an = b) lim is decreasing as n → ∞ because n→∞ n + e n+e √ e−x x ′ ; f (x) = √ f (x) = < 0, ∀ x ≥ x+e x ( x + e )2 The series is convergent Function series 77 ∞ (−1)n ∞ (−1)n 5n + = , this series is convergent because both se+ ∑ ∑ n n2 + n n =1 n + n =1 n =1 ries in the righthand-side are convergent ∞ c) ∑ (−1)n Solution 6.5 ln n a) We choose < ε < α − 1, when n is large enough α ≤ α−ε , α − ε > so the series is n n convergent 1 b) For an arbitrary > ε > 0, we have for a large enough number n: ≥ ε , the p (ln n) n series is divergent 2n 2n c) as n → ∞ ∼ n + 2n 2n a ( n + ) 2n lim n+1 = lim = < 1, the series is convergent n + n→∞ an n→∞ 2n x2 d) ln(1 + x ) = x − + o ( x2 ) as x → 0, so 1 ∼ , as n → ∞ n − ln + n 2n the series is convergent √ √ a2 π e) sin(π n2 + a2 ) = (−1)n sin(π n2 + a2 − nπ ) = (−1)n sin √ n2 + a2 + n a π a2 π < √ is a positive < π, ∀n, when n is large enough sin √ n2 + a2 + n n2 + a2 + n sequence which converges to The original series is convergent √ √ f) {Sn }, Sn = (2 + 3)n + (2 − 3)n satisfy Sn+2 = 4Sn+1 − Sn , for all n ≥ By induction prove that Sn is divisible by 4, then it is even for all n √ √ √ Hence sin[π (2 + 3)n ] = − sin[π (2 − 3)n ] ∼ −π (2 − 3)n as n → ∞ ∞ √ √ ∑ π (2 − 3)n is convergent because < π (2 − 3) < 1, the original series is convern =0 gent g) α > 1: nα (ln n) β ≤ nα−ε where < ε < α − 1, the series is convergent 1 where < ε < − α, the series is divergent ≥ nα+ε nα (ln n) β α = 1, see (1.3.l) Summary the series is convergent if and only if α > or α = 1, β > 1; and is divergent if < α < or α = 1, < β ≤ < α < 1: h) lim √ n i) lim a n +1 an n→∞ n→∞ a n2 a2 = e− < 1, the series is convergent n→∞ n ( n + 1)2 = lim = 0, the series is convergent n→∞ 22n+1 an = lim cos 78 Chapter Series §2 F UNCTION SERIES 2.1 Function sequence Assume that f , f , , f n , is a sequence of functions defined in a set X ⊂ R x0 ∈ X is called a convergent point of the above sequence if { f n ( x0 )} is a convergent sequence in R A sequence { f n } is called uniformly convergent in a set X to a function f , denoted by X f n ⇒ f , if for an arbitrary positive number ε > 0, there exists a number n0 ∈ N such that for all n ≥ n0 , we have | f n ( x ) − f ( x )| < ε, ∀ x ∈ X The number n0 depends only on ε, does not depend on x Example 2.1 The sequence f n ( x ) = f ( x ) = because xn is uniformly convergent in [0, 1] to the function n | f n ( x ) − f ( x )| = xn ≤ n n for all x ∈ [0, 1] Then for an arbitrary ε > we can choose n0 = + ε X Proposition: f n ⇒ f if and only if lim max | f n ( x ) − f ( x )| = n→∞ x ∈ X 2.2 Function series Definition ∞ A function series is ∑ un ( x ), where un ( x ), n ≥ 1, are functions defined in a set X ⊂ R n =1 Denote by Sn ( x ) the n−th partial sum of the above function series ∞ The function series ∑ un ( x ) is called to converge at a point x0 if the sequence {Sn ( x0 )} n =1 converges, is called to converge in a set X if {Sn ( x )} converges for every point x ∈ X The ∞ set of all convergent points of ∑ un ( x ) is called the domain of convergence The limits S n =1 of the sequence {Sn } is called the sum of the function series ∞ It is convergent if and only if x > x n =1 n Then the domain of convergence of this series is (1, +∞) Example 2.2 We consider the function series ∑ ∞ The function series ∑ un ( x ) is called to converge uniformly to a function S in a set X if n =1 the sequence {Sn } converges uniformly to S in X Function series 79 Convergence criterion To prove uniform convergence, we often use the following criterion ∞ ∑ un ( x ) is uniformly convergent in X to a function S( x ) if and only if n =1 lim max |Sn ( x ) − S( x )| = n→∞ x ∈ X Weierstrass’ criteria: If for all x ∈ X, we have |un ( x )| ≤ an , ∀n ≥ ∞ ∞ and the number series ∑ an is convergent, then the function series ∑ un ( x ) conn =1 n =1 verges uniformly in X Example 2.3 ∞ cos nx converge uniformly in R due to Weierstrass’ criteria 2 n =1 n + x i) The function series ∑ Indeed, cos nx 1 ≤ ≤ , ∀x ∈ R 2 n +x n +x n ∞ is convergent n =1 n and the series ∑ (−1)n−1 n =1 n + x For each x ∈ R, the corresponding number series is convergent due to Leibnitz’s criteria Denote by S( x ), x ∈ R the sum of the number series For all x ∈ R, we have ∞ ii) Consider the function series ∑ |S( x ) − Sn ( x )| ≤ x2 1 ≤ n+1 +n+1 then =0 n→∞ n + ≤ lim max |Sn ( x ) − S( x )| ≤ lim n→∞ x ∈ X (−1)n−1 converges uniformly in R n =1 n + x ∞ Hence lim max |Sn ( x ) − S( x )| = and ∑ n→∞ x ∈ X Properties of uniformly convergent function series ∞ Given a function series ∑ un n =1 ∞ If {un }, n ∈ N, are continuous functions in the interval [ a, b] and ∑ un converges unin =1 formly in [ a, b] to S( x ), then S( x ) is continuous in [ a, b], then is integrable in this interval and b b ∑ un ( x ) S( x )dx = a ∞ a n =1 ∞ dx = ∑ n =1 a b un ( x )dx 80 Chapter Series If {un }, n ∈ N, are continuous functions together with their derivatives in the interval ∞ ∞ n =1 n =1 ( a, b) and ∑ un is convergent to S( x ), ∑ u′n converges uniformly in ( a, b), then S( x ) is differentiable in ( a, b) and S′ ( x ) = ∞ ∑ ′ un ( x ) ∞ = n =1 ∑ u′n ( x) n =1 2.3 Power series Power series, radial of convergence, domain of convergence A power series is a function series of the following form ∞ ∑ a n x n = a0 + a1 x + a2 x + + a n x n + n =1 ∞ Radial of convergence of a power series is a number such that ∑ an x n is absolutely conn =1 vergent when | x | < R and is divergent when | x | > R a If lim n+1 = ρ (or lim n | an | = ρ), then the radial of convergence is determined by n→∞ an n→∞    if < ρ < ∞   ρ R= if ρ = ∞     ∞ if ρ = The domain of convergence contains (− R, R), together with the end points x = R or x = − R if the power series converges at x = ± R respectively Properties of a power series ∞ The power series ∑ un converges uniformly in every closed interval [ a, b] ⊂ (− R, R) n =1 ∞ The sum of the power series ∑ un is continuous in its domain of convergence We can n =1 integrate or differentiate each terms of this series: ∞ ∑ un n =1 b ∞ = ∑ an nxn−1 n =1 ∞ ∑ un a ′ n =1 for all closed intervals [ a, b] ⊂ (− R, R) ∞ dx = ∑ n =1 a b an x n dx Function series 81 ∞ If ∑ un ( x ) also converges at x = ± R, then n =1 ∞ lim S( x ) ∑ un (± R) = x→± R n =1 ∞ xn Example 2.4 Consider the series ∑ n =1 n √ R = lim n n = 1, then radial of convergence is R = n→∞ ∞ is divergent x = 1, ∑ n =1 n ∞ (−1)n x = −1, ∑ is convergent n n =1 So the domain of convergence is [−1, 1) We can calculte the sum S( x ) of the observing series In (−1, 1), S( x ) is differentiable and ∞ ′ S ( x ) = ∑ x n −1 = 1−x n =1 Hence x x ′ S ( x ) = S (0) + S (t)dt = 0 dt = −ln(1 − x ) 1−t Because the series also converges at x = −1, then −ln2 = S(−1) = lim S( x ) = x →−1 ∞ (−1)n n n =1 ∑ As in the above example, we often use differentiation and integration to find the sum of a power series We differentiate or integrate a series to obtain a new series, which we can use some fundamental series expansions to calculate this Here are some fundamental series ∞ = ∑ x n = + x + x2 + + x n + (| x | < 1) 1−x n =0 ∞ = ∑ (−1)n x n = − x + x2 + + (−1)n x n + (| x | < 1) 1+x n =0 ∞ sin x = x3 (−1)n x2n+1 (−1)n x2n+1 = x − + + + ( x ∈ R) ∑ 3! (2n + 1)! n=0 (2n + 1) ! ∞ cos x = ∞ ex = x2 x4 (−1)n x2n (−1)n x2n = − + + + + ; ( x ∈ R) ∑ (2n)! 2! 4! ( 2n ) ! n =0 xn x2 xn = + x + + + + ( x ∈ R) ∑ n! 2! n! n =0 82 Chapter Series Example 2.5 Find the sum of the series + √ 2+ n+1 + + + √ ( 2) n ∞ We consider the function series ∑ (n + 1) x n =: f ( x ) n =0 Radial of convergence is R = 1, then for x ∈ (−1, 1), we can integrate each term of this series in the interval [0, x ]: x x = ∞ ∑ ( n + 1) t f (t)dt = ∞ n =0 ∞ n +1 ∑x ′ x 1−x 1+ √ = dt = ∑ x (n + 1)tn dt n =0 = n =0 Hence f ( x ) = n x , x ∈ (−1, 1) 1−x The finding sum is (1 − x )2 2+ n+1 + = f √ + + √ n ( 2) √ = 2(3 + 2) 2.4 Exercises Exercise 6.6 Find the domain of convergence of the following series lnn ( x + n1 ) a) ∑ √ x−e n =1 ∞ ∞ (−1)n+1 b) ∑ + n2 x n =1 cos nx 2nx n =1 ∑ c) ∞ ∞ d) ∞ e) xn + x2n n =1 ∑ (n + x )n x +n n =0 n ∑ ∞ f) ∑ tgn ( x + n ) n =1 Exercise 6.7 Examine the uniform convergence of the following function sequences and function series a) f n ( x ) = x n − x n+1 in the interval [0, 1] x b) f n ( x ) = sin in the interval [0, 1] and in R n ∞ c) ∑ (1 − x ) x n in the interval [0, 1] n =1 ∞ x2 in the interval [− a, a], ( a > 0) n2 ln n n =1 ∞ x e) ∑ 2n sin n in the interval [− a, a], ( a > 0) n =1 ∞ 2x + n in [−1, 1], (| a| < 1) f) ∑ an x+2 n =1 d) ∑ ln + Function series 83 Exercise 6.8 Find the convergent radial and domain of convergence of the following exponential series ∞ ∞ xn a) ∑ √ n n =1 ∞ d) 2n n! ∑ n =1 (2n)! b) ∞ ∑x n n =1 ∞ x2n e) ln(n + 1) x3n+1 ∑ (−2)n n + n =1 c) ∑ n =1 ∞ f) 1+ n n2 xn 2n−1 x n−1 √ 3n −1 n=1 (2n − 1) ∑ Exercise 6.9 Find the sum of the following series ∞ a) ∞ x n +1 (2n)!! n =1 ∑ ∞ ∑ d) b) (−1)n−1 x2n 2n − n =1 ∑ ∞ c) ∞ n x n −1 e) n =1 ∑ n ( n + 2) x n n =1 x4n+1 n=0 4n + ∑ ∞ f) (−1)n ∑ 4n2 − n =1 2.5 Solution Solution 6.6 √ a) Domain of determination: x > e n an = ln( x + ) → ln x > as n → ∞ then the series n is divergent at x > e The domain of convergence is ∅ b) x = 0, | an | = 1, the series is divergent ∞ (−1)n+1 ∞ (−1)n+1 = x = 0, ∑ ∑ x n=1 1x + n2 n =1 + n x } is a positive decreasing sequence when n is large enough which For each x, { + n x tends to 0, so the series is convergent The domain of convergence is R∗ n+x n 1 c) an = ∼ e x x , so the series is convergent iff x > Domain of convergence is x n n n (1, +∞) ∞ n cos nx , the series is is convergent because 2x > d) x > 0: ≤ ∑ x 2nx 2nx n =1 x ≤ 0, if the series is convergent at x then necessary condition leads to cos nx = ⇒ lim cos nx = 0, n→∞ n→∞ 2nx lim this is impossible Domain of convergence is (0, +∞) | x |n 1 n as n → ∞; < so the series is convergent ∼ 2n |x| |x| 1+x | x |n ∼ | x |n as n → ∞; | x | < so the series is convergent | x | < 1: | an | = + x2n | x | = 1, | an | = 0, the series is divergent Domain of convergence is R\{±1} e) | x | > 1: | an | = 84 f) Chapter Series n → tg x as n → ∞ n π π If tg x < ⇔ − + kπ < x < + kπ , the series is convergent 4 π If tg x = ⇔ x = ± + kπ : an → e±2 = as n → ∞, the series is divergent If tg x > 1, the series is divergent π π Domain of convergence: − + kπ, + kπ ; (k ∈ Z) 4 | an | = tg x + Solution 6.7 a) f ( x ) = 0, ∀ x ∈ [0, 1]; | f n ( x ) − f ( x )| = x n (1 − x ) ≤ nn ≤ → as n → ∞ n + n+1 ( n + 1) [0,1] fn ⇒ f b) f ( x ) = 0, ∀ x ∈ R; | f n ( x ) − f ( x )| = sin x n [0,1] x x For all x ∈ [0, 1]: sin ≤ ≤ → as n → ∞ f n ⇒ f n n n n x ≥ sin = sin as n → ∞ f n does not converge uniformly For x ∈ R: max sin n n x ∈R to f in [0, 1] c) Sn ( x ) = x − x n+1 → x if ≤ x < 1, and Sn (1) → as n → ∞ The function f ( x ) = if x = 1; f ( x ) = x if ≤ x < is not continuous in [0, 1] then Sn does not converge uniformly The series does not converge uniformly too d) ln(1 + x ) ≤ x, ∀ x ≥ 0; then ln + ∞ a2 ∑ x2 n ln2 n ≤ x2 n ln2 n ≤ a2 n ln2 n ; ∀ x ∈ [− a, a] is convergent then use Weierstrass’ criteria, the series converges uniformly n ln2 n in [− a, a] n ∞ n x ; ∑ is convergent then use Weierstrass’ criteria, the series e) 2n sin n ≤ a 3 n =1 converges uniformly in [− a, a] ∞ 2x + 2x + n ∈ [−1, 1] for all x ∈ [−1, 1], then an f) ≤ | a|n , the series ∑ | a|n is conx+2 x+2 n =1 vergent then the function series converges uniformly in [−1, 1] n =1 Solution 6.8 a) R = 1, domain of convergence is [−1, 1) b) R = 1, domain of convergence is (−1, 1) 1 c) R = , domain of convergence is − , e e e Fourier series 85 d) R = +∞, domain of convergence is (−∞, +∞) 1 , domain of convergence is − √ ,√ e) R = √ 3 2 √ √ √ 3 f) R = , , domain of convergence is − 2 Solution 6.9 x a) S( x ) = x (e − 1), ∀ x ∈ R b) R = 1, ∀ x ∈ (−1, 1): S( x ) = x arctg x 1+x + arctg x c) R = 1, ∀ x ∈ (−1, 1): S( x ) = ln 1−x 1+x d) R = 1, ∀ x ∈ (−1, 1): S( x ) = (1 − x )3 e) R = 1, ∀ x ∈ (−1, 1): S( x ) = x2 − 3x ( x − 1)3 (−1)n x2n x2 + = − arctg x 2x n=1 (2n + 1)(2n − 1) ∞ (−1)n π = lim S( x ) = − ∑ 2 x →1 n=1 4n − ∞ f) S( x ) = ∑ §3 F OURIER SERIES 3.1 Decomposition theorem Assume that f ( x ) is a 2π periodic funtion and integrable in the closed interval [−π, π ] Its Fourier series is a trigonometric series ∞ a0 + ∑ an cos nx + bn sin nx n =1 whose coefficients are a0 = π bn = π π f ( x )dx; −π π −π an = π π −π f ( x ) sin nxdx, n ≥ f ( x ) cos nxdx, n ≥ 86 Chapter Series Theorem 6.1 Assume that f : R → R is a 2π periodic function that satisfies one of the following conditions: i) f is piecewise continuous function and its derivatives is piecewise continuous; ii) f is piecewise monotonous and is bounded Then, the Fourier series of f ( x ) converges at every points and its sum S( x ) coincides with f ( x ) at every continuous points of f At discontinuous point c of f ( x ), we have f ( c + 0) + f ( c − 0) S(c) = If f ( x ) is an odd function then an = 0, ∀n ≥ 0, bn = π π f ( x ) sin nxdx, n ≥ If f ( x ) is an even function then bn = 0, ∀n ≥ 1, a0 = π π f ( x )dx; an = π π f ( x ) cos nxdx, n ≥ Example 3.1 Find the Fourier series of the 2π periodic function f ( x ), f ( x ) = x for x ∈ (−π, π ) f ( x ) is bounded and is an increasing function in every intervals (−π + 2kπ, π + 2kπ ), then it can be decomposed into Fourier series We calculate the coefficients Because f ( x ) = x in (−π, π ) is an odd function then an = π bn = π π −π π f ( x ) cos nxdx = 0, n ≥ x − cos nx x sin nxdx = π n π π + cos nx dx n = (−1)n+1 ; (n ≥ 1) n Hence for x = (2n + 1)π, ∞ f (x) = (−1)n+1 sin nx n n =1 ∑ Theorem 6.2 If f ( x ) is 2l periodic function which also satisfies one of the conditions mentioned in the above theorem in the interval [−l, l ], then f ( x ) can be decomposed into Fourier series ∞ nπx nπx a0 + ∑ an cos + bn sin l l n =1 Fourier series 87 whose coefficients are a0 = l an = bn = l l f ( x )dx −l l l f ( x ) cos nπx dx, n ≥ l f ( x ) sin nπx dx, n ≥ l −l l −l Example 3.2 Find the Fourier series of the 2l periodic function f ( x ) that f ( x ) = x in ( a, a + 2l ) f ( x ) is bounded and is an increasing function in every intervals ( a + 2kl, a + 2(k + 1)l ), then it can be decomposed into Fourier series We calculate the coefficients Because the integral of a periodic function in every interval whose length is equal to the periodic is the same then a0 = l an = l l a+2l f ( x )dx = l −l a+2l a a l2 πna nπx sin + 2 cos l nπ l l n π 2l πna = sin ,n ≥ nπ l bn = l = a+2l a xdx = 2( a + l ) a xl nπx nπx dx = sin x cos l l nπ l 2l = a+2l f ( x )dx = l a+2l a l − nπ a+2l sin a nπx dx l a+2l a nπx − xl nπx x sin dx = cos l l nπ l a+2l a l + nπ a+2l cos a nπx dx l πna −2l cos ,n ≥ nπ l Hence for x = a + 2nl, f ( x ) =( a + l ) + =a + l + 2l π 2l π ∞ nπx nπa nπx nπa cos − cos sin sin n l l l l n =1 ∑ ∞ nπ sin (a − x) n l n =1 ∑ 88 Chapter Series Now we consider a function which satisfies one of the conditions in the first theorem in a closed [ a, b] To expand f ( x ) into a Fourier series, we construct a periodic function g( x ) whose periodic is larger or equal b − a such that g( x ) = f ( x ), ∀ x ∈ [ a, b] If g( x ) can be expanded into Fourier series then its sum coincides with g( x ), and also f ( x ), at every continuous points in [ a, b] If g( x ) is an odd function then its Fourier series is sum of sine functions If g( x ) is an even function then its Fourier series is sum of cosine functions Example 3.3 Find the expansion of the function f ( x ) = x for < x < into Fourier series of cosine functions and into Fourier series of sine functions To expand f ( x ) into Fourier series of cosine functions we construct an even function g( x ), which is periodic function and g( x ) = x for < x < 2, l = g( x ) is even then bn = 0, n ≥ a0 = xdx = 2; 2 2x nπx 2 nπx nπx dx = sin dx − sin an = x cos nπ nπ 0  0 if n is even = 2 (cos nπ − 1) = − n π if n is odd n2 π Hence for < x < 2, f (x) = − ∞ (2n + 1)πx cos 2 n=0 (2n + 1) π2 ∑ To expand f ( x ) into Fourier series of sine functions we construct an odd function h( x ), which is periodic function and h( x ) = x for < x < 2, l = h( x ) is odd then an = 0, n ≥ bn = =− nπx 2x nπx x sin dx = − cos nπ cos nπ = nπ f (x) = + nπ (−1)n+1 Hence for < x < 2, π nπ ∞ 2 (−1)n+1 nπx ∑ n sin n =1 cos nπx dx Fourier series 89 3.2 Exercises Exercise 6.10 Find the Fourier expansion of the following functions a) f ( x ) is a periodic function with T = 2π and f ( x ) = | x | in the interval [−π, π ] π−x in the interval (0, 2π ) b) f ( x ) is a periodic function with T = 2π, and f ( x ) = c) f ( x ) is a periodic function with T = 2π and f ( x ) = sin ax in the interval (−π, π ), a = Z Exercise 6.11 Decompose the following functions into Fourier series of cosine and sine functions  0 if ≤ x ≤ h a) f ( x ) = in the interval [0, π ] 1 if h < x ≤ π    if ≤ x ≤  x b) f ( x ) = if < x < in the interval (0, 3)    3 − x if ≤ x ≤ 3.3 Solution Solution 6.10 π ∞ cos(2n + 1) x , ∀ x ∈ [−π, π ] a) f ( x ) = − ∑ π n=0 (2n + 1)2 ∞ sin nx , ∀ x ∈ (0, 2π ) b) f ( x ) = ∑ n n =1 c) f ( x ) = sin aπ ∞ (−1)n n sin nx, ∀ x ∈ (−π, π ) ∑ 2 π n =1 a − n Solution 6.11 ∞ sin(nh) ∞ cos(nh) + (−1)n+1 π−h sin nx and f ( x ) = − cos nx a) f ( x ) = ∑ ∑ π n =1 n π π n =1 n 2nπ ∞ 2nπx and + − cos ∑ cos π n =1 n 3 2nπ 2nπx nπ ∞ + sin sin f ( x ) = ∑ sin 3 π n =1 n b) f ( x ) =