9.2-9.3 solutions - Now we want to sum the alternating...

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Unformatted text preview: Now we want to sum the alternating series 1 1 1 1 “33+???“ of reciprocals of odd cubes. Having used a Fourier series of t4 in Problem 24 to evaluate 2(1/n4), it is natural to look at a Fourier series of t3 . Let f(t) be the period 2% function with f(t) = t3 if ~7r < t < n. We calculate the Fourier coefficients of fit), and get 1 3 1 3 a=—— tdt=0, a=— tcosntdt=0 0 71' [In n 72' f” 2m: I’lzfl'z —6 cosmr—6 1127!2 —2 sinmr 2 b" = if t3sinm‘dt = — ( ) 4 ( ) = 2[£3‘L) 7r It 117: n n °° ,H sinnt °° ,,+ sinnt ’3 = 27132 (“1) I n ‘1ZZ("1) 1 n3 - (graphbelow) ":1 "=1 If we substitute 1‘ = ”/2 and use Leibniz's series 2(—1)"+1/n = 7r/4 of Problem 17 we find that Section 9.2 EP2—71 There 15 no value of t whose substitution 1n the Fourier series of fit) = t3 yields the series 2(1 / n 3) containing the reciprocal cubes of both the odd and even integers. Indeed, the summation in "closed form" of the series °° 1 =1 1 1 1 2—3 +—+3—3+4—3+5—3+--- is a problem that has challenged many fine mathematicians since the time of Euler. in modern times (by R. Apery in 1978) has it been shown that this sum is an irratio. ; V number. For a delightful account of this work, see the article "A Proof that Euler ' ' . . An Informal Report" by Alfred van der Poorten in the The Mathematical Intelligencer, Volume 1 (1979), pages 195—203. 3. SECTION 9.3 FOURIER SINE AND COSINE SERIES 1. a0 = Zjfll dt = 2, a" = 3—J”cosnt dt = 2sinn7r = 0 7r 0 71' 0 7171' Cosine series: f (t)=1 b : M ~71 7r 2‘7r in I: t b = if sinntdt= ———2(1 WSW) =———[1— (—1)"] n7! "7! Sine series: f (t) :(sint + isin 3t + ; sin 5t + ; sin 7t + n) See the graph at the top of the next page. EP2—72 Chapter 9 2 . nm 4 — 4 cos mr — 2n7r s1n nrr a = (1—t)cos———d = ————————————— " 2 2 2 o nit 8 M 1 37:1 1 5m 1 7m —2— cos—+7cos—+ ——cos———+ 7cos—- +- 3 2 52 2 7 2 2 _ . : (l—t)sinn—:£dt— 2nfl(1+cosn7r) 251nm: __ _[1+(_21),,] n27r2 nn' . . [sin m sin 2m sin 3m sin 4m Sine serles: f (t) + + + + 6 8 \\\\\ Section 9.3 Ep2_73 II 2 2 5. a0 gfl ldt = 3 —2x/3/mr if n=2,s,14,... mtt 2 2n7r 2 ,1 a = 2} COS—dt = —(sin——-—sinfl] = +2x/3/n7t if n=4,10,16,--- : " 3 1 3 rm 3 3 0 otherwise Cosine series: f (t) = -.__ 3 72' -cos——————cos———+—cos—————cos—— 2 3 4 3 8 3 10 3 12J§[1 2m] 4m1 8m 1 10m 0 for n even 2 2 . mrt 2 mt 21m +2/n7r if n=1,7,13,--~ b J- sm—dt . 1 ~4/n7r 1f n=3,9,15,-~ +2/mr if n=5,11,17,~- 2 71' 3 7. a0 = if t(7r—t)dt = 7’ 0 EP2-74 Chapter 9 2 2[mr cos mt + mt — 23in mr] 2 a = — t7r—t cosntdt = —.._.———————-———-—- = ——- 1+ —1 ” ,, ”LN > "a” "A ()1 . . ”2 [c0521 cos4t cos6t c058t Cosme senes: t =———-4 + + + +--- f() 6 22 42 62 82 ) 7T2 1: —7r It 371 Sn 2 2—2 —-2 . b" = 3E t(7r-t)sinntdt = w = ~43[1-(—1)"] 71' n 71' 71'" 8 . sin 3t sin 5t sin 7t Sine series: t = — smt + + + + --- f() A 33 53 73 ] 9. a0 = 3} sintdt = 34—, 7T 0 71' 2[1+cosn7r] _ 2[1+("1)"] if >1 ”(l—n2) * ”(l—n2) " II a ll 2 . ——Ln smtcosntdt = 7r 2 . a‘ —f smtcostdt = 0 71' 2 4 (cosZt cos4t cos6t cosSt ] + + +--- C ' 's: t =~—— + osme sene f () 77 2 3 15 35 63 7! See the graph at the top of the next page. Section 9.3 EP2-75 l3. b" = 3f sintsinntdt = _2__sm_mzr_ = 0 if n >1 7! 7f (1 ‘ n ) 2 . 2 15. b = - Sll’l tdt = l 1 fl 1: Sine series: f (t) = sint (graph below) 11. In order to satisfy the endpoint conditions x(0) = x(7t) = 0 we substitute the ' a _ 17 °° . 4 ‘ . . = x(t) = 2b,, sm nt and 1 = — 2 sm nt (from Example 1 1n Sectlon 9.1) into n" ”=1 7! nodd n L 19- differential equation x” + 2x :1 . This gives n ":1 n=1 °° . °° . 4 sinnt —E nzbnsmnt+2§ bnsmnt = ~ 2 71' "odd We therefore choose b" = 4/ 7m(2 —- n2) for 11 odd, b" = 0 for n even. This gi formal series solution " x0) = 1: 334,2. 2 i[sin,_.smi_§3151_§171_..., ”nodd "(z—n) 7r EP2-76 Chapter 9 [ . sin2t sin 3t sin 4! t = 2 smt- + — +... 2. 3 4 and 2 7:2 cos 2! cos 3! cos 4! t = ——4 cost— + +--- 1 L 4 9 1.6 J if—7r<t<7r. 19. a0 = —1—J2 —dt = 0, a,, = z-l—JZ Lcosntdt = 0 __ M! b" = _J‘:; —sinnt ttd s_______inmt— mrcosmz _( 1) n n+l % = M (—7r<t<7r) (graph below) n=l n 21. a0 = —- 7t 2 2_ . _ n a,. = l f tzcosntdt = W = 4( 21) 7’ ” nit n 1 2 , b = — t Slnntdt = 0 II n [I 2 __+4Z(—1)"ncosnt (~7r<t<7r) "=1 ”2—3t2 ”2 (—1)'ncosnt (_ -1)"+lzcosnt = —-—— —+4 12 12 4 "Z:____ =§——“ See the graph at the top of the next page. EP2-70 Chapter 9 13. 15. l7. l9. In order to conditions x(0) = x(1) = 0 we substitute the sine series in __ n+1 ' x(t) = 26 ( 1) sm nrrt (from Example 1 in Sectlon 9 3 With ,4 n L=1)i1m , m x'+x=t. This gives 1 ”“ sinmrt 1 mm+2bn sinnm= —Z(—:-)——-—-——. "=1 n: -l n We therefim: b_=2(-l)'”' /7rn(1— nznz) . This gives the formal series solution x(t)= ;ZZ—————— ('4); an: um?— —i) of our endpoint value problem. In order to satisfy the endpoint conditions x'(0) = x’(2) = 0 we substitute the cosine au 71 4 cos nt series x(t)=—+ a cosnt and t=————— 2 z 2 ”nodd "2 (from Example 1 in Section 17:1 9.3, with L = 7!) into the differential equation x” + 2x = t . This gives 7r 4 cos nt «3 c0 —zn2a” cos nt +a0 + 22a" cosnt = —-—— 2 n=| u=l 2 71' n odd n We therefore choose a0 = 7r/ 2 , a" = O for n > 0 even, and a" = 4/ 7rr12(n2 — 2) for n odd. This gives the formal series solution 7r+ 4 cosnt 7r 4( cos3t cos5t cos7t ) cost+ + +--- x(t): — M=—+—— + 4 Humane? —2) 4 72' 63 575 2303 of our endpoint value problem. Suggestion: Substitute u = —t in the left-hand integral. The first termwise integration yields t (-1)"cosnt+ 2 2:—_—'— C" and substitution of t= 0 gives CI = 222(4)”+l / n2 = n2 / 6, so ":1 t2 =22(—1)"cosnt+ _7r_2 +6 n=| A second termwise integration gives Section 9.3 EP2-7‘ 21. 23. ’4 E ,,=, n4 12 3’ and substitution of t= 0 yields C3 = 22(4)" / 12“. 11:1 We want to calculate the coefficients in the period 4L Fourier sine series m mrt Ft = b '— () gnSIHZL which agrees with fit) if 0 < t < L. Then b 2 _:J mrt 2 £1 mtt II =2L —dt —— 21.—t --— f(t)Sln2L +2L f( t)sin2Ldt. The substitution u = 2L ——t yields t ._ b, =1——:J f(t )sin%df ——[: f(u)sin£7flL—u)du t 1” L rear f(t )sinE—72—dt— ( L—_) L f(u)sin%7—r—u—du Now it is clear that 2 nm tcos—dt b,,= Ll: f () 2L if n is odd, whereas bn = 0 if n is even. __ (n—1)/2 ‘7 b" = ~2— tsinn—dt = —iz—[ZSin—n—fl-—n7rcosl—1£) = §£_1)2— for n odd 1 7t 0 2 7m 2 2 7m 4, 82 t l . 3t 1 . 5t 1 . 7t f(t)= sin———sm——+—sm n————sm—-+- -- 2 32 2 52 2 72 2 EP2-78 Chapter 9 SEC APP ...
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