Math 656 Midterm Examination March 27, 2015 Prof. Victor Matveev
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1 Math 656 Mdtrm Examnatn March 7, 05 Prf. Vctr Matvv ) (4pts) Fnd all vals f n plar r artsan frm, and plt thm as pnts n th cmplx plan: (a) Snc n-th rt has xactly n vals, thr wll b xactly =6 vals, lyng n th crnrs f a rglar hxagn, wth vrtcs n a crcl f rads : j j 6 j / / 6 n 0,,, n ================================================================================= (b) = csh Infntly many vals, snc csh() has prd f π: csh w w w 0 44 w lg wlg lg l n Whr w sd lg and lg l n ln ================================================================================= ) (pts) Stch th mag f a sqar dfnd by vrtcs =, =0, = and =+ ndr th mappng w. Hnt: trat ths mappng as a sqnc f smpl transfrmatns.. njgat. Sqar th sqar. Rtat by π/4 Parablas
2 Stp : njgatn (flp arnd th R() axs) Stp : Sqar: lns alng axs rman lns, tw thr lns bcm parablas: v y y y A parabla v v x x x A parabla Stp : Rtatn by π/4: w xp 4 v =============================================================================== ) (pts) Us an apprprat mthd t calclat ach ntgral vr th ndcatd cntr d 4 4 (a) Only tw snglarts ar wthn th crcl, s w can dfrm th cntr t cnvrt t t tw ntgrals arnd ach snglarty, and thn s th achy Intgral Frmla fr ach: cs d cs cs d f ( ) f' 0 g( ) 4 g( ) sn cs cs cs 8 ============================================================================== (b) d Intgral vr a crcl f rads (hnt: fnd a cpl dmnant trms n th Larnt srs) O( ) O ( ) O( ) O( ) O( ) d O ( ) d
3 d = sm-crcl n th rght half-plan f rads cntrd at th rgn and (c) csh lg cnnctng pnt t pnt +. Intgrand s nt analytc anywhr, s all w can d s paramtr: / / csh lg csh lg csh d d d / / / / d d / / / / ========================================================================== lg d 4) (pts) Fnd an ppr bnd fr 4 cnnctng pnt = t pnt = (assm 0 arg < π)., whr th ntgratn cntr s a straght ln y. snc y[0, ] ln. lg ln r ln Snc [ /, ] and arg [0, /] Nt: a strngr bnd lg s als crrct, bt nds prf lg 4 d ln 4. Lngth f cntr: L =========================================================================== 5) (pts) Fnd th frst thr dmnant trms n th Taylr srs fr fnctn =; ndcat whr th fll srs wld cnvrg. f( ) nar pnt lg f O O( ) ( ) ( ) lg lg ( ) O( ) O( ) O( ) O( ) ( ) O(( ) ) Narst snglarty s a pl at lg()=, crrspndng t =xp(-), thrfr th rads f cnvrgnc s
4 ====================== Pc any tw prblms btwn 6, 7, 8 ====================== 6) (pts) Spps a gvn Larnt srs c has a maxmal dman f cnvrgnc dscrbd by 0 r R. Ds th prncpl part f ths srs cnvrg anywhr tsd ths rng? What c? =0 abt th pstv-pwr part, Any Taylr srs cnvrgs n sm ds (nt a rng), thrfr:. Th pstv-pwr part s a Taylr srs and thrfr cnvrgs n a ds R. Th prncpal part s a Taylr srs n (.g., Taylr srs arnd ) and thrfr als has t cnvrg n a ds, crrspndng t r r Nt that th ntrsctn f ths tw dmans gvs th rng r R, xplanng why Larnt srs cnvrgs n a rng. =============================================================================== 7) (pts) Fnd all srs rprsntatns cntrd at = fr fnctn f( ), and ndcat thr rspctv dmans f cnvrgnc. Nt: partal fractns ar nt ndd n ths prblm Frst, lt s shft th argmnt: f( ) Larnt srs ( lcal Larnt srs): factr t (+) LS LS Larnt srs : factr t : 0 =
5 8) (pts) Ma a rgh stch f th dman f cnvrgnc f th srs 0 lg Rat tst ylds th cndtn lg ln r lnr r lg lnr Thrfr, ths dman ls wthn a rng sctr, tchng ts bndars at {=0; r= } and {r= (ln r=0); =} r = = rad r = r = = rad
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