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Homework 14

# Homework 14 - miller(dk1n645 HWlél ~ harnrick M(57395 1...

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Unformatted text preview: miller (dk1n645) - HWlél. ~ harnrick M (57395) - 1 This print-out should have. 20 questions. Multipiewchoice questions may continue on 2. intervelof cgce 2: [—1, l) the next column or page m find all choices before answering. [-2 2] 3. interval of cgce 001 (part 1 of 2) 10.0 points For the series 4~ converges only at a: m 0 0° (-1)” ' Z , .rc'”, 5. interval of cgce : (—1, 1] n + 8 nxl 7 (i) determine its radius of convergence, R. 6. interval of che :_— (m2, 2] 1m. 1. R = "g y 100100111125 @156 m 1 Deter - a - e radius of convergence; R, of the series 00?, ' 00 3. R z 0 2 mm _ :E XW Viva}; , __ (n+4)! “ﬁg/w" "“131 ‘ 1' QM 3' 1. = “M. (EKW f ”:1: V1? 3! 5. R m (~oo,oo) R 1 W” in” I ”’0 “”5 ' 1 (”M rerun .r H... _L._ 002 (part 2 of 2) 10.0 points 2' R m 23; ‘i‘fvw (t ,‘c—Nﬂww Mao“ one o, oo . (ii) Determine the interval of convergence Q R m of the series. H a} i 'e F” 4. R f 1. interval convergence 2 (-»1,1] - .\ ‘ 5. R = O O‘ 2. interval convergence 2 (43,8) -\ 005 10.0 points 3. converges only at :r m 0 / /—~\ Find the interval of convergence of the se— /' 4. interval convergence = (“1,1) rice I .. __.-» 0° W n \*" 5. interval convergence = [~1, 1) . Z( ”)3 6. interval convergence m {—8,8) 1. interval of cgoe m (—2, 2] (”818] V W 2. interval of cgce m [——2, 2) 7. interval convergence 003 10.0 points Find the interval of convergence of the se» 3. interval of cgce m (“1, 1) nos 00 ..... > Z \/ 2n 3:”. 353 (. converges only at :r; = 0 " If; 71—1 tervel of cgce : (m1, 1) 5. interval of cgee = (ME, 1] miller (dkinlﬁe‘é) ~ HW14 ~ hamrick w (57395) 2 6. intervaiof cgce = [—1,2) 1. interval convergence 2 [——1, 1] 2. interval convergence II 7‘ V8 e 006 10.0 points Find the interval of convergence of the 3' interval convergence m [ml/521%) power series 4. converges only at :1: == 0 00 8n 2 3n8+2\$n' 5. interval convergence = [—1/6,1/6] an 8 8 [interval convergence m [—6, 6] 1. interval} : [_g, g] “""‘W_k 7. interval convergence x {ml, 1) . 1 1 ' nterval : [733].“ 8. interval convergence 2 [——G, 6) goes part 1 of 2) 10.0 points For the series 2 (3: (M1)“, nml § . 8 8 3. interval m l“? E) 4. interval I! I 90 .93. 5/ 5. interval = [w8, 8) _ 1 1 (1) determine its radius of convergence, R. 6. interval m mu, —~« 8 8 1. R ‘: 1 007' (part 1 of 2) 10.0 points For the series 2. R = .31. C) an n :3 m e) PU ll C23 nml (1) determine its radius of convergence, R. 4 I? z 00 1. R = 0 5. R m 0 2. R = 1 _-~. I] Glgpert 2 of 2) 10.0 points 3. R = (moo, 00) an...“ 1 (ii) Determine the interval of convergence 4. R = 6 ofthe series, ~43 C K+l '4 3 g]? m 6 @tervai convergence 2: (“4, 2] . 008 {part 2 Of 2) 10'0 points 2. interval convergence m [—2, 4] (ii) Determine the interval of convergence of the series. 3. converges only at a: m —1 milier (dkm645) —~ HW14 m iiarnrick m (57395) 3 4. interval convergence 2 (—4, 2) ’9‘}! 013 10 0 points 5. interval convergence 3 (—oo, 00) If the series 2 Cam” LWMWW 6. intervai convergence m {—2, 4) ”no . . “ . converges when 2: x ——3 and diverges when 011 10-0 pornts :1: —~ 4, which of the following series converge Determine the interval of convergence of the series 1 . interval convergence 2. interval convergence ’ervel convergence 4. interval convergence rib-IUT 5. series converges only at a: m — 012 Determine the intervai of convergence of the inﬁnite series Zl ”Mirth? km 0 10.0 points (8m+5)k. 1. interval conv. 3. interval conv. 4. interval conv. 5. interval conv. without further restrictions on {en}? ( I) Dell/U333}?- IS yet/X WCwJ/LWLAULKD“ Off u‘l“ 0 00 Z Ewen?“ ” c w v ‘ n: fad/W “X MW B DO 3% ﬁfE—HX“ v ‘7 f) I 7; cni Cw Oils}; :3 1’ 00 h ()9 )3,” l 7?; __ i Q E Q15 5975 {“1" we f Fol nzﬂ “” wjd 1 BandConly T" Q ﬂqui grow we LCW ,, W / 2.Aonly #126“ (—13 N v 0995,45 3. .33 only G:_/.,..;T*.” 4. C only Silk/L W 5. A and C only \Wwe Jar/.2 v \$ \ qr”, ri or 7 6. all of them is 7. A and B only @me of them 014 10.0 points Compare the radius of convergence, R1, of the series 71:0 with the radius of convergence, R2, of thgl 1‘4 series _L R e W miller (dkm645) — H‘Wlé — harnrick ._ (57395) ‘ i 1 4 »w[7-£-\€; ): ha. 7 .. 3 when F +1 00 {V X) . “TI. _ n ‘ I, In"; . ‘.\ ...—. —-"‘ . ﬁlimoo an I “ 5' 1 m) m 12:13“ :— E'wf}%“\""V‘/L 3’ ,n20_w 0") ‘W ad“ «\hf“ 1 ..... n);,-"--«“‘:_ ’L _ ("fl-Ev} {L5}; 1. Rl : 2R2 m g oo tn “gm >+ mgr-J: L 2 100331113” Z ”3”“; W W“ , z = a x -, >4. 2 R: 3R2 5 00 t“ 1‘. 33‘ ig‘ﬁw 3. fﬁi) : 1113 '+' “jg-T; (”W 32R1=Rg=5 nzon 00 in K 4R1=R2=5 4f(t)=ZO-m—n3nA R — R __ 1 0° W "a 1— 2mg 5f(t)mh*13+zlég/r- ”2 .— f 1 00 we 6212:sz t” 7%) 1 9 5 6- m m —- Z “3?. C" nzl 015 10.0points ‘ ‘ 017 10.0 points Find a power series representation for the . . , ' End a. power serles representatlon centered function . . . 1 at the orlgln for the functron N E“ f (3‘) ﬂ 6 3 . .. . _ 1; \$3 \ fC‘I?) w my {itw .__\$ J K ‘\ ./ ‘1ng) w in .l.” n n (f? e x) "“ \1/ 00 6. Km) 2 — Z 5an m3 ”:0 018 10.0 noints 016 10.0 points Fin‘i a. Power series representation for the f t' Find a power series representation for the um; 1011 function 1 + 5g 2 1 f( ) r": . f(Z) n 1 _— 52 W. as miller (dkm64-5) m- HWM — hamriok — (57395) 5 (Hint: remember properties of logs.) 1. ﬁg) m i ﬁzz” 2. f(m) = Emirw 2. m) m i gigjlzzn em) = go ﬂmznw 5. Hz) a: i %¥\$32n—1 6. f(:1:) : 2 it): \$n+3 -l\ m 019 10.0 points (7—07 4“”): QM r a} , 6% “a . - 1_ .1 z: m*<w7c7m‘/3 1-1—va m 5w??ofg-l cm qr. Jaw-vi. -: S A W a Evaluate the deﬁnite integral 13/ s “7' my \ W; f(m) : / E1 49:8. ’1— .j \ ‘L 0 ‘3} XMC mgﬂx ‘AWWW as apOWer series. “w 4“ : so we [A— Kati-2. \f‘ 1 f( ) i0: \$412 W K M - m = M ‘ J X 471. + 2 beam-)4 1r», / 911:” “m":g 4m, J: 13:4 {JV-Lt?— a 00 4n+2 a m "-‘ . = Z 4. 2 _, fate nut) 71+ v f ~("5é'ja !! M8 E \ “or ”(3 ‘3‘ 12.30 4n+2 V110 o0 _1nm4n 4, f(;1;) x 2 L1)?"— n30 0° £134” 5. “55) x Z: 74}; 020 10.0 points / Find a power series representation §or the (a?) function f(3:) : Eton—1:3 on(—1,i). f 1. re) m 2 (”ﬁrm n=G (n+1 ...
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