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0910F-108-02-midterm2-solutions

# 0910F-108-02-midterm2-solutions - MATH 108 section 02...

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Unformatted text preview: MATH 108, section 02, Midterm #2, November 10, 2009. During this exam, you are allowed to use the following formulas if you ﬁnd them useful. 1 (_1)n+1 1 / a: sin(n7ra:)da: = , / :1: cos(mrm)dx = 0 0 mr 1 1 1 l / sin(n7rx)2dm = 5, / cos(mm:)2dx = 5, (both if n > 0) o 0 (*1)” — 1 712%2 ’ 1. Find the Fourier series of f (as) = 008(17) — cos(2:r). Justify your answer. 1% (3 a Foww MS airemﬂa. W6“ I I Ca» LL 9’6” SSWL‘HH‘IU Lillie/r E X Peaking A fuuoiien . M a Pam.» W R mega t“ a sfzu‘fwz .0456 / fixi‘ ‘8 { co\$(n3:_x) / 534014 l'rx L— 73/ “laws/“‘21; éwkot 1“".1-r LN, 11"— LzsLS I3 de ﬁftg‘ Cosé‘klcl “(L €vh£kxi . 964+ 4%) f5 «av/wk with“, M \$1ng 46193 62:) 2. Find the Fourier series of f(\$)={1’ ”3““; f(\$+27r)=f(m)- 0, OSm<7r; To What limit does the series converge at a: = 7r and at :L' = 37r/2? l n ‘Qﬁcasc x) X ’ T? g COS (n903x Tr l ’\ “'ﬁ Cn>63 ‘ o ’ ’ﬁ SWIUDOI V1 47 :— sinccﬁ -3MQMU ; 0 HT 0 Cake-2. is dx : I W 47 b , l 1? , 0 . m ~ :1} S ”95%) smﬁnxk‘x :— l g s'waHc‘x 4f 17 —T 14“?— FGW W173?) ' ~l)“’/ ~ ’5; + 2“ LV “Tr )3\V\(le) 0r) f? yea were”; i r; é: (—2.3 WSMQMDO -0 (2mm u V av 7-TC :5 as Fol/5b 6‘: JéCM%WAﬂ ) \$4: ’V‘L WES await/7&3 ¢ ﬂ”- ”gag “w:- ‘900 + ’w‘" +‘k2u { (’ Ref No») oLv;oU§ MW‘LDS *FVDM ’{ZQ/ 3944,45 ) 3. The sawtooth wave function is by deﬁned by ac, ~L < :10 < L; ﬁx): 0, :13=—Lor:1:=L; f (x — 2L), otherwise. Its Fourier series is 2L °° —1 ”+1 — ( ) MW). 71' n L n=1 Use this information to ﬁnd a formula for 7r of the form 7r : Z: ai Where the ca do not include 7r. ‘Ld‘s wk L12— ami oval. 30 ob ._ AM 1—, 10m =— i Cw!) ml Tr Z S'V‘C 2. lit-1:! V} 09 w“ \$5 Tl‘ :— H Z 9;") SMCL‘; Lari 3 'nsz All NJH n W mums M ”#7 WW ask—Qﬁﬁqmg Read flit z? va—HMH smCZZ—r > A H“ WW” 0 ‘41 m rs um 30 0° ’2 Side/gs SM s’owVMa’ w; {Hf Aramaic Cmqé'm 4. Use separation of variables to replace the PDE tum + {nut = 0 by a pair of ODE. [e/+ “(79+) ,_ XOOTC’E 3 0:. t “xx +xu€ 7' t XNT +XXT) ass-D Xl‘ ” ' T} 33‘? 2'1: SmcL h I&+«Lm1-g,iL MM‘SLS «L7 m 2; ml (zK [maLF/wﬁmia w? “I”; an ‘6 19974 534a W5+ (ﬁve! «a Cowman ““374“er 4 s \ __,, An 71.. X1 / A ; —I 94X - :27: Mm X" axco T +>x£T~ZO 5. Consider the heat conduction problem um=ut, O<x<7r, t>0, u(0,t) = 7r, u(7r,t) = 0, t > 0, u(ac, 0) = sin(:c) + 7r + 7rsin(3m) — m, 0 < m < 7r. Do all necessary steps. (You can, as always, use the back of the page if you lack space.) . u MWL»)\$.) flunk MMA+ {gkﬂ ALL M X 7/0 :7 XC13= “X419 Mg 9(”4"“) \$4643» 5 4t. lei— v 361% Xéoreo a baa/Maw =9 Mao. ”MAL—i=1“— VKK=Vf V :0 M “-044 x 1: vwxpw) var/f) :o 34‘” vast): “xx-[L 7T=Mqt> ; I: 6'=\/Ur,{) “41:47? e7 M, M M “36" pad 5‘; funk i5 lchﬂo ’ W", ”I" B + A<o m ; 1 .7» rue. a» we. in “9191/2, (M 4W?” File: 50 W “NW“ )(x 7 E W f/o): Lt /0 , u 2 ’0 WCD/t)‘: NLRI‘H‘DO _ V090) )‘>0 (9% A 4 F2) X + If X : 52404 +Tr53'1 a» 0..)((o\=_ c‘ ‘I ' t ;> (1:0 or 593% 16/ :7 0:.X(1r\—:—- Cz_s)hCW\ OF “)LL, batik “3w WW/Mb rf=ww ‘9»— M n Pbsw “34¢an S‘Wk “0‘1“" X0977? ‘ 75% 97f eigﬁm‘kws '33“ch :6sz Le \m/‘Aalé e. v1 an 7“ L’“ ; 2A5" A .. \Nxx fl.) \NO" L XT XT T‘ArknTt/O 97 Taﬂ’ @ L “’7 X 71 - /r\ (50:0 Fwomu schism M- 3M1 2~ 7 T WY: 01.?“ SIMCVWA e-“ 00 \Ne, Map. “.09.“ 9 “’2 c“ wm(x,f) :1 f1" 14 ANA NJ at - (D ) W 5g:4(rn)<)b/ I 6 5mm) +TI'5ML3X); Z "‘ ‘7) X13) ‘1 C3 Wan (cl, "ﬁfe... mix,“ f((1')) V‘- r) x/a);o. 94917 st- Xcm to . ...
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0910F-108-02-midterm2-solutions - MATH 108 section 02...

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