Teaching - Nguyen The Vinh UTC ď Chapter 1_Series tài liệu, giáo án, bài giảng , luận văn, luận án, đồ án, bài tập lớn v...
8 Series 8.1 Series Sum of series The series is an infinite sum u1 + u2 + + uk + = ∞ ∑ uk (8.1) k=1 The addends in this infinite sum are called the terms of the series and uk is called the general term If we assign to k some natural number, we get the related term of the series In (8.1) the k is called the index of summation and note that the letter we use to represent the index can be any integer variable i, j, l, m, n, The first index is for convenience, actually it can be any integer We can write (8.1) as ∞ ∑ uk = ∞ ∑ uk+1 = uk−1 = k=2 k=0 k=1 ∞ ∑ A number series is the series, whose terms are numbers In our course we consider the series of real numbers A functional series is the series, whose terms are functions of the variable x, i.e uk = uk (x), k = 1, 2, A geometric series is the series k a + aq + aq + + aq + = ∞ ∑ aq k (8.2) k=0 where each successive term is produced by multiplying the previous term by a constant number q (called the common ratio in this context) The harmonic series is the series ∑1 1 1 + + + + + = k k k=1 ∞ The sum of the first n terms Sn = n ∑ uk k=1 is called the nth partial sum of the series The partial sums S1 = u1 S2 = u1 + u2 (8.3) Sn = u1 + u2 + + un define the sequence of partial sums S1 , S , , S n , (8.4) Definition A series (8.1) is said to converge or to be convergent when the sequence (8.4) of partial sums has a finite limit If the limit of (8.4) is infinite or does not exist, the series is said to diverge or to be divergent When the limit of partial sums lim Sn = S n→∞ exists, it is called the sum of the series and one writes S= ∞ ∑ uk k=1 It is important not to get sequences and series confused! A sequence is a list of numbers written in a specific order while an infinite series is a limit of a sequence and hence, if it exists will be a single value Example The sum of the first n terms, i.e the n − 1st partial sum of the geometric series is Sn−1 = n−1 ∑ aq k = k=0 If |q| < 1, then a(1 − q n ) 1−q lim q n = n→∞ thus, a(1 − q n ) a aq n a = lim − lim = n→∞ n→∞ n→∞ 1−q 1−q 1−q 1−q lim Sn−1 = lim n→∞ So, if |q| < 1, then the geometric series converges and the sum is S= a 1−q If q > 1, then lim q n = ∞ n→∞ therefore, lim Sn−1 = ∞ n→∞ and the geometric series is divergent If q < −1, then lim q n does not exist n→∞ and hence, lim Sn−1 does not exist and the geometric series is divergent If n→∞ q = 1, then the n − 1st partial sum Sn = n−1 ∑ k aq = k=0 n−1 ∑ a = na k=0 and the limit lim Sn−1 = lim = na = ∞ If q = −1, then the S0 = a, n→∞ n→∞ S1 = a − a = 0, S2 = a − a + a = a, S3 = a − a + a − a = 0, We obtain the sequence of partial sums a, 0, a, 0, which has no limit Therefore, for q = ±1 the geometric series is divergent Conclusion If |q| < 1, then the geometric series (8.2) converges and if |q| ≥ then the geometric series diverges Example To find the nth partial sum Sn of the series ∞ ∑ k=1 k(k + 1) we use the partial fractions decomposition 1 = − k(k + 1) k k+1 We obtain Sn = n ∑ k=1 =1− 1 1 = + + + + k(k + 1) 1·2 2·3 3·4 n(n + 1) 1 1 1 1 + − + − + + − =1− 2 3 n n+1 n+1 The limit of this sequence, i.e the sum of this series ( ) S = lim − =1 n→∞ n+1 If we ignore the first term the remaining terms will also be a series that will start at k = instead of k = So, we can rewrite the original series (8.1) as follows, ∞ ∞ ∑ ∑ uk = u1 + uk k=1 k=2 We say that we’ve stripped out the first term We could have stripped out the first two terms ∞ ∞ ∑ ∑ uk = u1 + u2 + uk k=1 k=3 and first any number of terms respectively, ∞ ∑ uk = u1 + u2 + + um + k=1 ∞ ∑ uk = k=m+1 m ∑ k=1 uk + ∞ ∑ uk k=m+1 The first sum on the right side of this equality is the mth partial m ∑ uk k=1 sum of series (8.1) This is a finite sum, which is always finite Assuming that n > m, we can write the nth partial sum n ∑ k=1 uk = m ∑ n ∑ uk + uk k=m+1 k=1 or Sn = Sm + Sn−m where n ∑ Sn−m = uk k=m+1 Now, if Sn has the finite limit as n → ∞, then Sn−m must have also the finite limit Conversely, if Sn−m has the finite limit as n → ∞, then adding the finite sum Sm leaves the limit finite Similarly, Sn has the infinite limit or does not have the limit if and only if Sn−m has also the infinite limit or has no limit Conclusion Stripping out the finite number of terms from the beginning of the series leaves the convergent series convergent and divergent series divergent As well, adding the finite number of terms to the beginning of the series does not make the convergent series divergent and does not make the divergent series convergent 8.2 Necessary condition for convergence of series Suppose that the series (8.1) converges to the sum S, i.e lim Sn = S n→∞ The nth partial sum can be written Sn = n ∑ uk = k=1 n−1 ∑ uk + un k=1 or Sn = Sn−1 + un hence, un = Sn − Sn−1 The convergence of the series gives, since if n → ∞ then n − → ∞, lim un = lim Sn − lim Sn−1 = S − S = n→∞ n→∞ n→∞ We have proved an essential theorem, so called necessary condition for the convergence of the series Theorem If the series (8.1) converges, then the limit of the general term lim un = (8.5) n→∞ This theorem gives us a requirement for convergence but not a guarantee of convergence In other words, the converse is not true If lim un = the n→∞ series may actually diverge For example, the limit of the general term of the harmonic series (8.3) lim = k→∞ k but the harmonic series is divergent It will be a couple of subsections before we can prove this, so at this point the reader has just to believe this and know that it’s possible to prove the divergence In order for a series to converge the series terms must go to zero in the limit If the series terms not go to zero in the limit then there is no way the series can converge since this would contradict the theorem, i.e there holds Conclusion (the divergence test) If lim un ̸= then the series (8.1) n→∞ diverges For example the series ∞ ∑ k=1 is divergent because the limit of the constant term is that constant, lim = ̸= k→∞ 8.3 Convergence tests of positive series In Mathematical analysis there exist a lot of tests that give us the possibility to decide whether the series converges or diverges In this subsection we are going to consider the positive series, i.e the series (8.1), whose all terms are positive: uk ≥ 0, k = 1, 2, 8.3.1 Comparison test The nth partial sum of the series (8.1) is Sn = Sn−1 + un Since for any index n un ≥ 0, then Sn ≥ Sn−1 that means, the sequence of partial sums of the positive series is monotonically increasing We had the theorem in Mathematical analysis I, which stated that any bounded monotonically increasing sequence has the finite limit So, if we have succeeded to prove that the sequence of the partial sums of the positive series is bounded, we have proved the existence of the finite limit of the sequence of partial sums, that is, we have proved the convergence of the positive series The sequence S1 , S , , S n , has the finite limit means by the definition of the limit that for any ε > there exists the index N > such that for all n ≥ N |Sn − S| < ε This inequality is identical to the inequalities −ε < Sn − S < ε or S − ε < Sn < S + ε which means the sequence is bounded We have proved the following theorem Theorem The positive series (8.1) is convergent if and only if the sequence of its partial sums is bounded Suppose that we have another positive series ∞ ∑ vk (8.6) k=1 and we know whether it converges or diverges For instance we know that the geometric series (8.2) converges if |q| < and diverges if |q| ≥ We know that the harmonic series is divergent and we know that ∞ ∑ k=1 k(k + 1) is convergent Theorem (the comparison test) 1) If for any k = 1, 2, 3, uk ≤ vk then the convergence of the series (8.6) yields the convergence of the series (8.1) 2) If for any k = 1, 2, 3, uk ≥ vk then the divergence of the series (8.6) yields the divergence of the series (8.1) Proof 1) Denote the nth partial sums of the series (8.1) and (8.6) by Sn = n ∑ uk k=1 and σn = n ∑ vk k=1 respectively Since for any k = 1, 2, 3, uk ≤ vk , then Sn ≤ σn By the assumption the series (8.6) is convergent hence, by Theorem the sequence σ , σ2 , , σ n , , (8.7) is bounded by some constant σ, i.e σn ≤ σ But then Sn ≤ σ, which means that the sequence of the partial sums of the series (8.1) S1 , S , , S n , is bounded thus, by Theorem the series (8.1) is convergent 2) Next, let’s assume that (8.6) is divergent Because vk ≥ we then know that we must have lim σn = ∞ n→∞ The assumption uk ≥ vk yields Sn ≥ σn and by the limit theorem lim Sn ≥ lim σn n→∞ n→∞ which means that the sequence of the partial sums of the series (8.1) S1 , S , , S n , has no finite limit or the series (8.1) is divergent Example Prove that the series ∑ 1 1 + + + + = k k2 k=1 ∞ 1+ converges We know that the series ∞ ∑ k=1 ∑ 1 = k(k + 1) k=2 (k − 1)k ∞ converges For any k = 2, 3, it is obvious that 1 < k2 (k − 1)k and by Theorem the series ∞ ∑ k2 k=2 converges Adding the term to the beginning of the series preserves the convergence Example Prove that the series ∑ 1 1 √ + √ + √ + + √ + = k k k=1 ∞ diverges √ For any k ≥ there holds the inequality k ≤ k hence, 1 √ > k k The harmonic series (8.3) diverges thus, by Theorem the series given diverges also 8.3.2 D’Alembert’s test Sometimes the D’Alembert’s test is referred as the ratio test We consider again the positive series (8.1) Theorem (D’Alembert’s test) Suppose there exists the limit uk+1 lim =D k→∞ uk 1) If D < 1, then the series (8.1) converges 2) If D > 1, then series (8.1) diverges 3) If D = 1, then this test us inconclusive, because there exist both convergent and divergent series that satisfy this case Proof Suppose the limit D < and let q be a real number between D and 1, i.e D < q < By definition of the limit there exists N > such that for k ≥ N uk+1 −D q uk or uk+1 > quk > uk , i.e the terms of the series form an increasing sequence The limit of the increasing sequence cannot be zero thus, by divergence test the series (8.1) diverges ∞ ∑ Example Does the series converge or diverge? k! k=1 1 The ratio of two consecutive terms uk+1 = and uk = is (k + 1)! k! uk+1 uk 1 k! (k + 1)! = = = (k + 1)k! k+1 k! and the limit of this ratio =0 k→∞ k + D = lim Since D = 0, this series converges by the D’Alembert’s test ∞ ∑ converge or diverge? Example Does the series k2 k=1 10 Thus, at the points of discontinuity the Fourier series of the square-wave function converges to Since sin((2k + 1) · 0) = and sin((2k + 1)π) = for any integer k, then the direct computation also gives ∑ + sin((2k + 1)0) = k=0 (2k + 1)π ∞ and ∑ + sin((2k + 1)π) = k=0 (2k + 1)π ∞ Figure 8.4 shows the graphs of the partial sums S2n+1 = 2 + sin x + sin 3x + + sin(2n + 1)x π 3π (2n + 1)π for n = 0, 1, 2, y y 1 −π π −π x y x π x y −π π π −π x Figure 8.4 Partial sums of the Fourier series for the square-wave function 8.16 Fourier sine and cosine series of 2π-periodic functions In some of the problems that we encounter, the Fourier coefficients become zero after integration Finding zero coefficients in such problems can be 42 avoided because using knowledge of even and odd functions, a zero coefficient may be predicted without performing the integration Corollary If f (x) is even, then ∫a ∫a f (x)dx = −a and if f (x) is odd, then f (x)dx ∫a f (x)dx = −a Proof By the additivity property of the definite integral ∫a ∫0 f (x)dx = −a ∫a f (x)dx + −a f (x)dx and substituting in the first addend on the right side of this equality x = −t, we have dx = −dt If x = then t = and if x = −a, then t = a Thus, ∫a ∫0 f (x)dx = − −a ∫a f (−t)dt + a ∫a f (x)dx = ∫a f (−t)dt + f (x)dx and, denoting in the first addend the variable of integration by x again ∫a ∫a f (x)dx = −a [f (−x) + f (x)]dx Now, if f (x) is even, then f (−x) + f (x) = 2f (x) and if f (x) is odd, then f (−x) + f (x) = Recall from the course of Mathematical analysis three assertions about even and odd functions: • the product of two even functions in an even function; • the product of two odd functions is an even function; • the product of an even and an odd function is an odd function 43 Suppose that f (x) is an even 2π-periodic function Then for any integer k ≥ the product f (x) cos kx is even and the product f (x) sin kx is odd Now, Corollary simplifies the computation of Fourier coefficients: a0 = π ∫π f (x)dx ak = π ∫π f (x) cos kxdx, k≥1 and bk = 0, k≥1 Therefore, an even function f (x) has only cosine terms in its Fourier expansion ∞ a0 ∑ f (x) ∼ ak cos kx + k=1 This series is called a Fourier cosine series Next, suppose that f (x) is an odd 2π-periodic function Then for any integer k ≥ the product f (x) cos kx is odd and the product f (x) sin kx is even Again, Corollary simplifies the computation of Fourier coefficients: a0 = ak = 0, and bk = π k≥1 ∫π f (x) sin kxdx, k≥1 Thus, an odd function f (x) has only sine terms in its Fourier expansion f (x) ∼ ∞ ∑ bk sin kx k=1 and this series is called a Fourier sine series Example Find the Fourier coefficients and Fourier series of the function defined by f (x) = |x|, if − π < x ≤ π and f (x + 2π) = f (x) The graph of this 2π-periodic function is in Figure 8.5 44 y π −π −2π π x 2π Figure 8.5 The function is even and on the interval [0; π] the absolute value |x| = x We know, that the Fourier coefficients bk = for k ≥ Find a0 = π ∫π xdx = x2 π = and for k ≥ we integrate by parts ∫π 2 ak = x cos kxdx = x sin kx π π k = cos kx πk π π2 · =π π π − k ∫π sin kxdx = − , k 2 · (−1) πk − = = πk πk 0, if k is odd if k is even The Fourier cosine series expansion of the function given f (x) ∼ or π 4 − cos x − cos 3x − cos 5x − π 9π 25π ∞ π ∑ cos(2k + 1)x f (x) ∼ − π k=0 (2k + 1)2 The function is continuous and has in the interval (−π; π] one local maximum and one local minimum Hence, by Dirichlet’s convergence theorem the Fourier series converges to f (x) at any real x and now we may write for all real x ∞ ∑ cos(2k + 1)x π f (x) = − π k=0 (2k + 1)2 45 In particular |x| = ∞ π ∑ cos(2k + 1)x − π k=0 (2k + 1)2 for −π < x ≤ π Since f (0) = 0, then taking in the last equality x = 0, we get ∞ π 4∑ 0= − π k=0 (2k + 1)2 or ∞ 4∑ π = π k=0 (2k + 1) which gives ∞ ∑ k=0 π2 = (2k + 1)2 Let’s denote the sum of the convergent series S= ∞ ∑ k2 k=1 Writing 1 1 1 + + + + + + + = 16 25 36 ( 49 64 ) 1 1 1 π2 = 1+ + + + + 1+ + + + = + ·S 25 49 4 16 S = 1+ gives π2 ·S = or π2 Thus, the sum of the reciprocals of the squares of positive integers S= ∞ ∑ π2 = k2 k=1 The convergent Fourier series enable us to find a lot of sums of the kind Example Find the Fourier coefficients and Fourier series of the rectangular wave defined by { −1 if −π < x < f (x) = and f (x + 2π) = f (x) if < x < π 46 y −2π −π π 2π 3π x −1 Figure 8.6 The graph of this 2π-periodic function is in Figure 8.6 The function is odd hence, ak = for k = 0, 1, 2, and bk = π ∫π sin kxdx = − cos kx kπ π = − ((−1)k −1) = kπ { , kπ 0, if k is odd, if k is even The Fourier sine series expansion of the function given f (x) ∼ or 4 sin x + sin 3x + sin 5x + π 3π 5π ∞ ∑ sin(2k + 1)x f (x) ∼ π k=0 2k + By Dirichlet’ convergent theorem for any x ∈ ((2k − 1)π; 2kπ) the series converges to −1, for any x ∈ (2kπ; (2k + 1)π) the series converges to and at −1 + the points of jump discontinuity x = kπ the series converges to = 8.17 Fourier series of functions with whatever period If a function has period other than 2π, we can find its Fourier series by making a change of variable Suppose ( T f (x)) has period T , that is f (x + T ) = f (x) for all x Then the function f 2π · x has the period 2π because ( ) ( ) ( ) T Tx Tx f (x + 2π) = f +T =f 2π 2π 2π 47 ( ) The Fourier series of the function f T2πx is ( ) ∞ Tx a0 ∑ f ∼ + ak cos kx + bk sin kx 2π k=1 where a0 = π ( ∫π f −π Tx 2π ) dx and, for k = 1, 2, 3, ( ∫π ak = π f −π and bk = π ( ∫π f −π Tx 2π Tx 2π ) cos kxdx ) sin kxdx Tx 2πt 2π If we use the substitution t = , we have x = and dx = dt If 2π T T T T x = −π, then t = − and if x = π, then t = Therefore, the Fourier 2 series of the function f (t) with period T is a0 ∑ 2kπt 2kπt ak cos + + bk sin T T k=1 ∞ f (t) ∼ where T T a0 = ∫2 f (t)dt − T2 and, for k = 1, 2, 3, T ak = T ∫2 f (t) cos 2kπt dt T f (t) sin 2kπt dt T − T2 and T bk = T ∫2 − T2 48 2π is called the T angular frequency or simply the frequency of the function f (x) Denoting in the Fourier series expansion and in the formulas of Fourier coefficients the variable by x again, we have the Fourier series If T is the period of the function f (x), then the ratio ω = a0 ∑ f (x) ∼ + ak cos kωx + bk sin kωx k=1 ∞ where T a0 = ∫2 T f (x)dx − T2 and, for k = 1, 2, 3, T ak = T T ∫2 f (x) cos kωxdx and bk = − T2 T ∫2 f (x) sin kωxdx − T2 Of course, the Dirichlet’ convergence theorem is also valid for functions with period T Example Find the Fourier series expansion for { 0, if − < x < f (x) = and f (x + 4) = f (x) x, if ≤ x ≤ The graph of this function is shown in Figure 8.7 The function has the y −6 −4 −2 Figure 8.7 49 x 2π π period T = and the frequency ω = = Since the function equals to between −2 and 0, then a0 = ∫2 −2 f (x)dx = ∫0 −2 · dx + ∫2 xdx = x2 · 2 =1 Integration by parts gives ak = ∫2 x cos kπx dx = [0 kπx = x sin kπ = k2π2 kπx 22 + 2 cos k π 2 ] = ((−1)k − 1) As well, integrating by parts, we get bk = ∫2 x sin kπx dx = [0 kπx = − x cos kπ = − kπx + 2 sin k π 2 ] = 2 · (−1)k+1 (−1)k = kπ kπ So, the Fourier series expansion of the function is ] ∞ [ ∑ (−1)k − kπx (−1)k+1 kπx cos + sin f (x) ∼ + π k=1 k2π k Noticing, that the numerators of the coefficients of cosine functions are for even k-s and −2 for odd k-s, we may re-write this expansion as ∞ ∞ ∑ (2k + 1)πx ∑ (−1)k kπx cos − sin f (x) ∼ − 2 π k=0 (2k + 1) π π k=1 k 50 8.18 Fourier series of half range functions If the function with period T is even, then the Fourier coefficients for k = 0, 1, 2, 3, T ∫2 ak = f (x) cos kωxdx (8.38) T and for k = 1, 2, 3, T bk = T ∫2 f (x) sin kωxdx = (8.39) − T2 The Fourier series expansion of the even function is a0 ∑ ak cos kωx f (x) ∼ + k=1 ∞ (8.40) If the function with period T is odd, then for k = 0, 1, 2, the Fourier coefficients T ∫2 ak = f (x) cos kωxdx = (8.41) T − T2 and for k = 1, 2, T bk = T ∫2 f (x) sin kωxdx (8.42) and the Fourier series expansion of the odd function is f (x) ∼ ∞ ∑ bk sin kωx (8.43) k=1 ] ( If a function over half the range, say 0; T2 , instead of the full ( T isT defined ] range from − ; , it may be expanded in a series of cosine terms only or of sine terms only The series produced is then called a half range Fourier series ( ) The function given should be extended to the interval − T2 ; as an even or odd function This allows the expansion of the function in a series solely of cosines (even) or sines (odd) 51 ( ] Suppose the function f (x) is defined in the interval 0; T2 The even extension for this function is defined as ] ( { f (x), if x ∈( 0; T2 ] (8.44) φ1 (x) = f (−x), if x ∈ − T2 ; Now, if we define by φ1 (x + T ) = φ1 (x) the periodic extension of this even function over the whole number axis, we have the even periodic function, whose Fourier ( cosine ] series is (8.40) with coefficients computed by (8.38) In T the interval 0; (8.40) is also Fourier series for f (x) y y T x −T The function f (x) T x and its even extension φ1 (x) y −3T −T T 3T x The periodic extension of φ1 (x) Figure 8.8 ] ( The odd extension for function f (x) defined in the interval 0; T2 is ( ] { f (x), if x ∈( 0; T2 ] φ2 (x) = (8.45) −f (−x), if x ∈ − T2 ; The periodic extension of φ2 (x) defined by φ2 (x + T ) = φ2 (x) is an odd periodic function, whose Fourier ( sine ] series is (8.43) with coefficients computed by (8.42) In the interval 0; T2 (8.43) is also Fourier series for f (x) 52 y y −T T x T The function f (x) x and its odd extension φ2 (x) y −3T −T T 3T x The periodic extension of φ2 (x) Figure 8.9 Example Find the Fourier cosine series expansion for the function defined as f (x) = − x in the interval (0; 1] Since f (−x) = + x, then by (8.44) the even extension of this function in (−1; 1] is { − x, if x ∈ (0; 1] φ1 (x) = + x, if x ∈ (−1; 0] and the periodic extension over the whole number axis is defined by φ1 (x + 2) = φ1 (x) In Figure 8.10 is shown the graph of the function (red), the graph of its even extension (blue) and its periodic even extension (green) The period of 2π = π By the formula the periodic extension is and the frequency ω = 53 y y 1 x −1 1 x y −3 −1 Figure 8.10 (8.38) we compute first a0 = ∫1 ( ) x2 (1 − x)dx = x − =1 and next, integrating by parts ∫1 (1 − x) cos kπxdx = ak = [0 ] 1 1 = (1 − x) sin kπx − 2 cos kπx = kπ k π 0 { , if k is odd, = − 2 ((−1)k − 1) = k2π2 k π 0, if k is even Thus, the Fourier cosine series of the extension φ1 (x) is φ1 (x) ∼ ∞ ∑ cos(2k + 1)πx + 2 π k=0 (2k + 1)2 and in the interval (0; 1] ∞ ∑ cos(2k + 1)πx 1−x∼ + 2 π k=0 (2k + 1)2 54 x Since −f (−x) = −1−x, then by (8.45) the odd extension of this function in (−1; 1] is { − x, if x ∈ (0; 1] φ2 (x) = −1 − x, if x ∈ (−1; 0] and the periodic extension over the whole number axis is defined by φ2 (x + 2) = φ2 (x) y y 1 x −1 1 x y −3 −1 x −1 Figure 8.11 In Figure 8.11 is shown the graph of the function (red), the graph of its odd extension (blue) and its periodic odd extension (green) The period of the periodic extension is still and the frequency ω = π By (8.41) the Fourier coefficients ak = for k = 0, 1, 2, and by (8.42) (we integrate by parts again) ∫1 (1 − x) sin kπxdx = bk = [0 1 cos kπx − 2 sin kπx = −(1 − x) kπ k π 55 ] = kπ for k = 1, 2, 3, Hence, the Fourier sine series of the odd periodic extension φ2 (x) is ∞ ∑ sin kπx φ2 (x) ∼ π k=1 k and in the interval (0; 1] ∞ ∑ sin kπx 1−x∼ π k=1 k 56 ... go to zero in the limit then there is no way the series can converge since this would contradict the theorem, i.e there holds Conclusion (the divergence test) If lim un ̸= then the series (8.1)... has the finite limit as n → ∞, then Sn−m must have also the finite limit Conversely, if Sn−m has the finite limit as n → ∞, then adding the finite sum Sm leaves the limit finite Similarly, Sn has the. .. Theorem (the comparison test) 1) If for any k = 1, 2, 3, uk ≤ vk then the convergence of the series (8.6) yields the convergence of the series (8.1) 2) If for any k = 1, 2, 3, uk ≥ vk then