Math 2142 Homework 2 Solutions. Problem 1. Prove the following formulas for Laplace transforms for s > 0. a s 2 + a 2 L{cos at} = e st.
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1 Mth 2142 Homework 2 Solution Problem 1. Prove the following formul for Lplce trnform for >. L{1} = 1 L{t} = 1 2 L{in t} = L{co t} = Solution. For the firt Lplce trnform, we need to clculte: e t 1 dt u e t dt e t u e u + e For ech fixed vlue of >, e u u. Therefore, the firt term drop out nd we re left with e / = 1/ required. For the econd Lplce trnform, we need to clculte: e t t dt u e t t dt To clculte u e t t dt, we ue integrtion by prt with u = t nd dv = e t. Therefore, du = dt nd v = e t /. u e t t dt = te t = ue u = ue u u u + 1 e t dt ( e t ) u ( e u ) 2 2 = u e u 1 2 e u By L Hopitl Rule, for ech fixed >, u/e u u. Furthermore, for ech fixed >, 1/e u u, Therefore, when we tke the limit u we re left with 1/ 2 re required. For the third Lplce trnform, we need to clculte: e t in t dt Rther thn deling with the limit form of thi integrl, I will do integrtion by prt leving the upper limit. Let u = e t nd dv = in t, o du = e t nd v = ( 1/) co t. e t in t dt = e t co t e t co t dt
2 Remember tht the firt term on the righthnd ide i n bbrevition for e t co t e t co t u e u co u + e co co u + 1 e u Since 1 co u 1, we hve 1/e u (co u)/e u 1/e u. Since >, lim 1/e u =, o by the Squeeze theorem lim (co u)/e u =. Therefore, nd we hve e t co t e t in t dt = 1 co u + 1 e u = 1 e t co t dt. (1) To clculte e t co t dt we ue prt gin with u = e t nd dv = co t, o du = e t nd v = (1/) in t. Therefore, e t co t dt = e t in t + Agin, the firt term on the righthnd ide i n bbrevition: e t in t e t in t dt. e t in t u e u in u e in in u e u Becue 1 in u 1, we hve 1/e u (in u)/e u 1/e u. For >, lim 1/e u = nd o by the Squeeze theorem, lim (in u)/e u =. Therefore, nd we hve e t in t e t co t dt = in u e u = e t in t dt. (2) Putting Eqution (1) nd (2) together nd doing ome lgebr, we hve which i the formul we re trying to prove. e t in t dt = 1 2 e t in t dt 2 ) (1 + 2 e t in t dt = e t in t dt = 1 2 e t in t dt = =
3 For the lt Lplce trnform, we need to clculte: e t co t dt We could repet imilr clcultion to thoe ued in the lt Lplce trnform, but it will be horter to ue the formul in Eqution (1) nd (2). Strting with Eqution (2) nd uing Eqution (1) to do ubtitution, we hve e t co t dt = e t in t dt = ( 1 ) e t co t dt = e t co t dt. We cn produce the formul for L{co t} from here with ome lgebr. e t co t dt = 2 e t co t dt 2 2 ) (1 + 2 e t co t dt = e t co t dt = 2 which i the formul we re trying to prove. Problem 2. Prove by induction on n N + tht e t co t dt = 2 L{t n } = n! n = Solution. For the be ce, we need to how tht L{t} = 1/ 2, which we did in Problem 1. For the induction ce, ume tht for fixed vlue of n nd we how tht L{t n } = n! n+1 L{t n+1 } = By the definition of the Lplce trnform, we hve L{t n+1 } = (n + 1)! n+2 t n+1 e t dt
4 Apply integrtion by prt to thi integrl with u = t n+1 nd dv = e t, o du = (n + 1)t n nd v = e t /. Then we hve L{t n+1 } = tn+1 e t + n + 1 t n e t dt (3) By the definition of the Lplce trnform, we know tht t n e t dt = L{t n }. Therefore, we cn ubtitute L{t n } into Eqution (3) nd ue the Inductive Hypothei. L{t n+1 } = tn+1 e t L{t n+1 } = tn+1 e t L{t n+1 } = tn+1 e t L{t n+1 } = tn+1 e t + n n n L{t n } (n + 1)! n+2 n! n+1 t n e t dt To finih the proof, we need to how tht the firt term on the righthnd ide of thee eqution i equl to. Writing thi term out in full, we hve t n+1 e t t n+1 e t u u n+1 e u = 1 u n+1 lim e u Applying L Hopitl rule n + 1 mny time how tht lim u n+1 /e u = nd o required to finih the induction ce. t n+1 e t Problem 3. Let f(t) be twice differentible function uch tht f i continuou on [, ) nd both f nd f hve exponentil order t. Fix contnt K, M nd uch tht both f(t) Ke t nd f (t) Ke t for t M. Prove tht for > = L{f (t)} = 2 L{f(t)} f() f (). Solution. A uggeted in the hint, let g(t) = f (t). From cl, we know tht for >, L{g(t)} = L{f (t)} = L{f(t)} f(). Applying the formul for the Lplce trnform of derivtive to g(t), we hve for > L{g (t)} = L{g(t)} g(). Subtituting in the fct tht g() = f () nd L{g(t)} = L{f(t)} f(), we hve L{f (t)} = L{g (t)} = ( L{f(t)} f() ) f () = 2 L{f(t)} f() f ().
5 Problem 4. Clculte the following Tylor polynomil. 4(). The 5th Tylor polynomil of ln x t x = 1. Solution. Let f(x) = ln x. Let c n denote the coefficient of the n-th term in the Tylor polynomil for f(x). f(x) = ln x f(1) = c =! = f (x) = 1 x f (1) = 1 c 1 = 1 1! = 1 f (x) = 1 x 2 f (1) = 1 c 2 = 1 2! = 1 2 f (x) = 2 x f (1) = 2 c 3 3 = 2 3! = 1 3 f (4) (x) = 2 3 f (4) (1) = 2 3 c x 4 4 = 2 3 = 1 4! 4 f (5) (x) = f (5) (1) = c x 5 5 = = 1 5! 5 Therefore, the 5th Tylor polynomil for ln x t x = 1 i T 5,1 ln x = + 1(x 1) 1 2 (x 1) (x 1)3 1 4 (x 1)4 + 1 (x 1)5 5 4(b). The 4th Tylor polynomil of x t x = 1. Solution. Let f(x) = x. Let c n denote the coefficient of the n-th term in the Tylor polynomil for f(x). f(x) = x f(1) = 1 c = 1! = 1 f (x) = 1 2 x 1/2 f (1) = 1 2 c 1 = 1/2 1! f (x) = x 3/2 f (1) = 1 4 c 2 = 1/4 2! f (3) (x) = x 5/2 f (3) (1) = 3 8 c 3 = 3/8 3! = 1 2 = 1 8 = 1 16 f (4) (x) = x 7/2 f (4) (1) = c 4 = 15/16 4! Therefore, the 4th Tylor polynomil for x t x = 1 i = T 4,1 x = (x 1) 1 8 (x 1) (x 1)3 5 (x 1)4 128
6 4(c). The 6th Tylor polynomil of co x t x =. Solution. Let f(x) = co x. f(x) = co x f() = 1 c = 1 f (x) = in x f () = c 1 = f (x) = co x f () = 1 c 2 = 1 2! f (x) = in x f () = c 3 = At thi point, the pttern of derivtive repet, o we hve f (4) () = co = 1, f (5) () = in = nd f (6) () = co = 1. Therefore, the 6th Tylor polynomil i T 6, co x = 1 1 2! x ! x4 1 6! x6 Problem 5. Prove tht the Tylor polynomil of degree 2n for co(x) t x = i T 2n (co(x)) = k= ( 1) k (2k)! x2k Solution. Let f(x) = co x. From the clcultion in Problem 4(c), the pttern of vlue of f (k) () i tht f (k) () = when k i odd nd f (k) () lternte between 1 nd 1 when k i even. Therefore, the Tylor polynomil i T 2n, co x = 1 1 2! x ! x4 1 6! x6 + + ( 1) n 1 (2n)! x2n = k= ( 1) k (2k)! x2k Problem 6. Let f(x) = 1/(1 x) = (1 x) 1. 6(). Prove by induction on k tht f (k) (x) = k! (1 x) (k+1) nd o f (k) () = k!. 6(b). Prove tht the n-th Tylor polynomil for f(x) t x = i T n f = 1 + x + x x n = k= x k Solution. For 6(), we proceed by induction. For the be ce k =, we hve f () (x) = f(x) = (1 x) 1 nd!(1 x) (+1) = (1 x) 1 which re equl. For the induction ce, ume tht f (k) (x) = k! (1 x) (k+1). We need to how tht f (k+1) (x) = (k +1)! (1 x) (k+2). f (k+1) (x) = d dx f (k) (x) = d dx k! (1 x) (k+1) = k! (k+1)(1 x) (k+1) 1 ( 1) = (k+1)!(1 x) (k+2).
7 For 6(b), ince f (k) () = k!, we hve T n, f = k= f (k) () x k = k! k= k! k! xk = k= x k Problem 7(). Ue Problem 6 to find the n-th Tylor polynomil for 1/(1 + x) t x =. Solution. Since 1/(1 + x) = 1/(1 ( x)), we hve by ubtitution tht 1 T n, 1 + x = ( x) k k= = 1 x + x 2 x 3 + x 4 ( 1) n x n = ( 1) k x k k= 7(b). Ue 7() to find the n-th Tylor polynomil for 1/(1 + x) 2 nd the degree 2n Tylor polynomil for 1/(1 + x 2 ), both lo t x =. Solution. Subtituting x 2 for x, we get the 2n-th Tylor polynomil 1 T 2n, 1 + x = ( 1) k x 2k = 1 x 2 + x 4 x ( 1) n x 2n 2 k= Differentiting 1/(1 + x) give 1/(1 + x) 2. Therefore, 1 T n, = d (1 + x) 2 dx T 1 n+1, 1 + x = d n+1 ( 1) k x k dx k= n+1 = k= n+1 d dx ( 1)k x k = ( 1) k kx k 1 Note tht in the lt um, the bottom index hifted to k = 1 becue the contnt term in the previou line dipper when you tke the derivtive. To implify thi expreion, note tht ( 1) k cn be rewritten ( 1) k+1 or ( 1) k 1. We hift indice follow: k=1 1 T n, (1 + x) = n+1 ( 1) k 1 kx k 1 = 2 k=1 ( 1) k (k + 1)x k k=
8 7(c). Ue 7(b) to find the degree 2n + 1 Tylor polynomil for rctn(x) t x =. Solution. Since rctn x = x 1/(1 + t2 ) dt, we cn find the (2n + 1)-t Tylor polynomil by integrting the formul for T 2n, 1/(1 + x 2 ). T 2n+1, rctn x = ( 1) k x2k+1 2k + 1 Problem 8. Conider the opertor defined by D(f) = 2f + f. 8(). Clculte D(x), D(x 2 ) nd D(3x 2 4x). Solution. We clculte follow. k= D(x) = 2(1) + x = 2 + x D(x 2 ) = 2(2x) + x 2 = 4x + x 2 D(3x 2 4x) = 2(6x 4) + 3x 2 4x = 3x 2 + 8x 8 8(b). Prove tht D i liner. Tht i, how tht D(αf + βg) = αd(f) + βd(g) for ny α, β R. Solution. D(αf + βg) = 2 d (αf + βg) + (αf + βg) dx = 2(αf + βg ) + αf + βg = (2αf + αf) + (2βg + βg) = α(2f + f) + β(2g + g) = αd(f) + βd(g) Problem 9. Conider the opertor defined by D(f) = f 2f + 3f. 9(). Clculte D(x), D(x 2 ) nd D(3x 2 4x). Solution. D(x) = 2(1) + 3x = 3x 2 D(x 2 ) = 2 2(2x) + 3x 2 = 3x 2 4x + 2 D(3x 2 4x) = 6 2(6x 4) + 3(3x 2 4x) = 9x 2 24x (b). Prove tht D i liner. Solution. D(αf + βg) = d2 d (αf + βg) 2 (αf + βg) + 3(αf + βg) dx2 dx = αf + βg 2(αf + βg ) + 3αf + 3βg = (αf 2αf + 3αf) + (βg 2βg + 3βg) = α(f 2f + 3f) + β(g 2g + 3g) = αd(f) + βd(g)
9 Problem 1. Conider the opertor defined by D(f) = e x + f. 1(). Clculte D(x 2 ), D(2x 2 ) nd 2D(x 2 ). Solution. D(x 2 ) = e x + 2x D(2x 2 ) = e x + 4x 2D(x 2 ) = 2(e x + 2x) = 2e x + 4x 1(b). Ue your nwer to 1() to explin why D i not liner. Solution. Since D(2x 2 ) 2 D(x 2 ), we cnnot pull multiplictive contnt out of D, which men D i not liner.
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