Phil Old MATH II binder

| . ee t PhilLucht’sold Math IIBinder | ' Infinite Series ODE's Complex Variables | | Oo i | Infinite Series } |ContentsofthisSection. ah2niny |eo1,Theoremsoninfiniteseries. 2¢-The: meaning-of various kinds-of--convergencer 3.Interchanging: Limit with Integration. a.exampie given; FESR-application 4.Interchanging: Integration with Another Integration. 5+Facts abouthowyoucanfindoutwhetheraSUMisCarlsoninaparameter. 65Convergence: ofZaps 7.Convergence of: (a,<;x)= WCaje;x) 8.Some standard series.- “9.Sum-calcuiation- by-Mebhod--of- Piecewise-Contours- 1D.Howdoyouboundthisseriestal: ZOE ll.Asymptotic Series(Asymptotic expansions) _ 42y Infinite series asanalytic functions @ IB.Sung, Suma. s | ee f - |QD i.Reverences: nostofthese aregiven inBuckforrealvariables inhischapter on -+ convergence. Many however are generalized tocomplex variables _and.are stated inSpiegels book. Inthe library there are books specializing ininfinite series - somegiveing theory and advanced tests,-others stating various series. ~~ “2, Plan: below Istate innoparticular order anytheorems Imightbeabletouse. -—-Hopefully-all qualifiers are included. —-- _- -- _—--3+ PowerSeries Radius: Thedomain ofconvergence ofacomplex powerseries [email protected] ---—---Ry you mist-use the-various tests given below, Right atthe radius /z/=R,-you .— donot apriori know whether series converges ornot. when afunction f(z) is a —Yepreneiited imsone region bya series, and you write asexpansion ubout apoint _______ 8) thenseries converges incircle which isaslarge aspossible aboutawithouthitting asingularity of the function. ""Hr BimpleWivergence Test:ifa,dossnot“approach 0,thenseries mustdivérge. ~ " "5,Gauchy convergence: thepartial sums3,formasequence inthespace C.With-—-= usual-metricy thisspaceiscomplete. Thus,forinfinite series theconceptsofconvergence and Cauchy convergence coicide. —§.Ratiotest:let_L« lim/a(n+1)/a(n)/. IfLisgreaterthan1,seriesdiverges. TfLisTessthan1,series converges. IfI=l,noconclusion. (ifa(n)Werethe—— —- 00efficientsofapowerseriesinstead oftheactual-terms, thenthistestwould . tell you the radius ofconvergence: R=1/L.)Here, convergence isabsolute. 5 on ay fon eeee_7+RootTest:thistime,letL=lim(/a(n)/)® .Thentestissameasratiotest.~“Teisthis test which really shows that power series hasacircleofconvergence. -—.- If-ay-are_poner series caefficients, thenradina_is R=1/L.again. —— ~~Gs-Integral test: thistestonlyapplies toarealypositive definite-sequence a,. Tomniyy toa710@positive definite tailofaseries,ifyoulike.Testsaysthat “Sq converges ifIa(n)dn converges. NIso, sumdiverges ifititegral diverges. ~—- Sometimes integrals areeasier tocalculate thansuns. oe —— —9s-Raabe's Test: sometimes when Lal inthe ratio test and the test fails, RaabesFestwillsucceed. Itsaystodefine L'=lim(n(1~-/a(n+1)/a(n)/)) .THMHXEHS EMMELotccicnectueriiietieriseitecuet reettiiririectrelcvestiereccscet secre aigTESRSHI EASEAREANAAAAATIATR XBARHEREEERAET.Now,ifL'isgreater than1,series converges absolutely. IfLtisless than 1,series may diverge oritmay ~————— conditionally converge, --If L'=1,test-fails. - - oe ________10.. Gauss's Testi thisisamoresophisticated version ofRaabes test. Seepage142‘OfSpiegel fordetails. 11, Absolute versus Conditional Convergence: ifaseries converges absalutely "~~>“Shen itmustconverge. Thereverst~ofdourse isednotbetrue, Ifaseries—~ —sanverges, but does not absolutely converge,itiscalledconditionally _ convergent. Thus, aconvergent series iseither ACorCC. 12.Rearrangements: ifaseriesconverges absolutely, thenyoucanrearrange it @Howeveryou likesThus, iftwoserieses converge absolutely, sodoestheir _S sum and Cauchy product, see Buckp159todefine sumandproduct. Thisis_ the great strength ofthe concept of absolute convergence. Note that all power-series (including ofcourse hypergeometric function) converge_absolutely— inside their radius. ' a ( 13._Alternating Series Test: Lettheterms a,bereal andmonotoiée decreasinggatapproaching 0tatheLinit.ThentheAlternating signseriesconverges, butnoconclusion maybedrawnabouttheséries $a,itself. @ 14. The bracket theorem: this isaparticular form ofrearrangement where youInsert brackets asyoulikebutyoudonotchahge theorderonpaper. Then. clearly, the bracketsdodefinearearrangement, soweexpectbrackets to - have noeffect onanabsolutely convergent sequence. However, even ifseries ~ converges only—conditionally, brackets donot-hurt.- - ‘15.Cawohy product: evenifoneseries isACandsecondisoilyCC,theCatohyproduct converges. Ifboth are AC, then Cauchy product series isconvergent. 16, Double sums: Suppose two-series Aand Bare AC, ‘Then the double sum-series- -- isalsoAC,Youcanthenrearrenge asyoulike.Thus,theCauchy product is —~-glsoACy SewBuck for~more ‘details. 7~— ° ~~ I].Uniform convergence: consider aseries where each term carries anadditional parameser.like2.(Example:seriesoffunctions f,,(z)).Withthisseries-f, weassociate asequence ofpartial sums F,(z). Ifthis sequence converges~ uniformy omsome-domain ofz,thenseries "converges uniformly". Sée"page140°*—ornotes for meaning of“uniformly”. 1 18,Neiorstrauss Mtest: ifaseries f,(z) issuch that /f,(z)/ isless than. =-8convergent series ofrealsM,forall=inchosen region, thenseries- converges uniformly~in- that region, This isclearly the easiest test to - check for uniform convergence. . e_. 19,term-by-term caloulus: ifaseries isuniformly convergent, thenyoucan — -“Integrateordifferentiate termbyterm,buttherearecertain qualifiers. For integration tobe-defined, @function must be continuous, So ifthe terms inyour uniformly convergent series are continuous, you can prove - thatthe sum of the series is alto voHtiMidlis. The only othe® requiremént:is ; ofcourse that the integration domain (or contour) lie inside region of _ ___ uniform continuity inz. Also, inorder todifferentiate term bytern, you have toshow that the resultant series. inuniformly convergent before you are justified. “~~ 20,Comparison tests: seoSpiegel page141theorem 10. . Addendum onconvérgénce tests: “‘Tiiese tests often require somelimit L.For a_--.-Szample, intheratio fest,Eis thelimii.of thissequence: r,.=/a(n+1)/a(n)/.—_BUT, suppose this real sequence Tyhas nolimit !Ifthe sequence isbounded --- from.above, then itisbounded and must have-at-teast-one-ciuster-point, -but-if 7 ithastwocluster points, then there isnolimit .Example: r,=cos(PIn) =1,-l...In this-cawe the varioustestsmstberéformilated. Hereisthereformulation: ~ “fet Leaps limsup r,andLinf =liminf r,.Sven ifthere isaninfinity ofcluster _ -~-—- —pointsfor.sequence rn, these-t+wo limits-must exist ifY,isbounded. Or, perhaps 7 Limsup =infinity, The test then reads: ifLoup isless than 1;series converges. — if-tinfis-greater than-1, ther series diverges.Inwords, ifall¢luster points arebelow1,thenconvergence, andifallcluster points areabove1,then . --diverges.Sutifsomeareahoveandsomebelow,noconclusion, oOo w--Behm ane Lsup eye -~- --- Soe covdudiat. | Ve Ld>|SDdae _Sees sone \ - - tshoe \ = mal a |Convergence ofasequenceoffunctions£,(x)toafunction-£(x) +@_i.Weofcoursehaveinmindthetf(x)isthepartial sumofntermsinsomeseries —_ ~~ gothe question ofwhether the sequence converges isthe same asthe question “—~—“orWiiettigr’ theBéFISs converges. ae -ne -2. Think ofxasa-single real variable -for-illustratiom- There are basically-three— - - - kinds_of convergence: -pointwise,inthe-mean", anduniform convergence.——__ -— 3.Poitwise convergence. (Buck page 180) Letallthese functions fp(x) bedefined _ onsome domain D. Let Ebeany open piece ofthis domain. Pointwise convergence "~"""peans thatyoupickaparticular xinEandyoushowthatthesequence converges ~~ torthat particular x.You show that this istrue foranypoint xinE.Then _ yout séviés converges ‘pointivise. a - - —— — - ~~ This~ts-the-weakest'form ofconvergence>~ “As showmtin-the simple example on-~ -—— —~-page-182-0f- Buck,youcanhavepointwise~convergence, -but—the -supnorm-ft~te/sup--— ——can still_be_infinite.. Thisisbecause thetrouble canjamup.ggainst-oneof _.| _ your boundaries. - . a ©+Comergence intheMean.(Buck,page195)Thismeansthat/t-fy/,2 goestoo” Ie, the integral ofthe difference squared over the interval must vanish tohave ~~~ “"“““gonvergence inthemean. oo a 7 7 - Ivcanimagine thatyouCouldhave'f, cnvérging toTinthEiiéin,bitHOE-— +-pointwise-due to:someoutstending point(of-vero measure-under integration —— - This same outstanding point. would ruin uniform convergence also--beceuse the -——— ——--sup.wouldalways depend onthiaonebadpoint. So: ee —_ mean#pointwise To — ican3watPora TO eee -—— ——5,-Uniferm-Gonvergence, This iethe most importent-and significant convergence tdea.— ___ -—Unifrom.convergence.means( Buck, page-182).that /f.-fofsup goes.toO.Ie, =- _ the maximal difference between the two functions (atwhatever point. thatmaybe) must vanish. . ~~ a“Irfnconverges uniformly, thenithastheCauchyProperty /fp~te/eup..8 goes tozero ifm,n larger than some M.(etc) eo ~~~Mostimportantly: ifaseriesconverges uniformly, thenyoucanintegrate“ItGermbyterm. (page 187) ~ oF ~~Ipresumethatuniformconvergence allowsyoutointerchange theprocedureof‘summation ofseries with any- other procedure, such aslimit taking. @ oWrow Quornsquak - - b bb .acs SyaxXess) =oehaere - “a. Bane Soothe.duwiteonist.Quequichin as : wee ee —4.. ee LoL insLaeSag=YasiGog-- aH-@-Ra naval, examine amare To __ a_.- ee ee ee ee & : ==yatta paso). <Qaxlses) eel. -—--Meosugpest yosifislea dey2ee ©VRGa)FT SeONTREY net aes @Ww. --éXme ees) eweo ~--Que,ayoy)dlfunnce AhumsidanAO. UU |Mo“AeMwSAGD=$Gop)aor HenxeTakl, @ @osWwconnie . . : . 4 an _. — MeYER NO - _ ee ee ee Camm Wek FGs4) RRGsnye)—sencfoamnty, nate—Tab. (or cee eee en, _. \Qasoseg) —Yaaeates |Ce te — —b—-—. - ———---- - - a - - - ---- -. \YanWO)LRGSays] &Saxe. heehee Seago aee HO “<=Qahelé.. Oe . ©delRack”.QaMoshemangeedoeLakgrasof, -awh Wh Sano ce aMoe yanksSGyq) pavehoe2Gw)- Sronugie dh“Wronmn. wa eo©eewitan t=YamBa,mete a MNaivsy,.youJodeYoudmeamddeke graunside,oeo7-: Ho GYeRDae nL mye: oe ee TTRakAcomeyenbadandcect.chaansGokoceannmainHAfeeter -aeee wee ~——-©Voogtyparcor Urn, wey de ee wee -- ge. eye ee ee eee ee ee wee. FY Rae OG. tee eeeee — ee fee LL, Qik Reselencomer“\8e3) Foie VEN “FaYack9,gow e. awk.Du<©WX,chkakKy voporto case j_- WAVE AL- a a i. BAcuoo.sugptaennberry Eom WER)0ted ee ge ee ee ee 1oo hye UU Beye De OT ee ee KB ee. .3 _--Draw\ER\ (ae). =eo”HSAQ... a -nee Keh ~~Ee Ceghee eee GaRo:-Lee eee---~—washrowwineBK)=O ~93@-- oe4- st Ae sem oe ver 5> G veBO =GsyQ\GA% -Kime =ean 240 . SB nekened deee).snoanedthanding): @Yeas, amysowedarma A”Wooreowadh govern”cematerr [©Co)“awd —_e _ - - o~ *site : Waren, Sa ee Daa,uaemgply Masoud 4’ and emdudre Mowe noah. - ~ee @ ©Crga calle: - oo eS .~t tnN~lex - Cyrs =Cw kee8))“ - ‘(xy) SF4 (dN \ ox \ Aunt oe =7Ve wigteh¥). ye&)we) - ~ . - 7 oO... . - - os . - a - - BQ -- - - ee ey : . - : -- DreWeeweg YeJennvoteaaa=ei.. - Comment: inintegrals ofthis form, there would benoquestion about thevalidity ofthe "naive" result ifthe integral were finite. You could then take ytoinf _ _inside the integral, because itistrivial togetuniformity onafinite intersal. Therealquestion iswhathappens atthehighenedofthex-integration. Wee@ _seefromtheaboveexample thataslongash(x)decays faster thanxtyor ___ _are OK, The amount ofpower "borrowed" can beassmall (but finite) asneeded __ __tomake the argument work, , xe wee |weepee-.Rome AraFESRorale. _ a . @©Dardin DHSmoses wee el a BakAK paFay=3y“Ser we - Thisexample -@iffers fromtheprevious_one on.two.counts: well, mainly: the - ~minus sign- inthe denvm-which-means that. the."worst.case"x.willbe at¢xuy. *Te,even--borrowing- power, it—may—be. that.you.do_not-have_uniformity.. '- =©Bind,werAp_pisinacphes posks=wel2a Aeh ST Ryan Ey ee-2. S&S Lee eee ee =Bega ZuBOs) weAECQ)ted teL eee es _ RPV=ByKbaka _ e2 NBL wee wee ee ee ee dak) ~.Kn wee eee ee a ee de-we AY)=24Sdemde. ee x.§x1 --Wloxt —Gsts4S) weet =Gr) an ae a as) New Lede drSG5\worana %boVanifoswnscoaee -=Mer tf ee Wrasy wonsheaar ak.KeOenugh fn- a wenn ee, ©Wakpleintoradothach CoaneMaegece e ate we _ weeBo ea HKOCaieee ~2SaBN ( o “ an @Guabr pork veo FO=BYay_O)> raa. 2@y-9 -y a fe day ddd See ee ne tee Aee ee ee Well, no.mbretime fonthis.I.thinkthepoint.is:clean. If@2<) then.youare OK atthe.tighsend of x.integration. as:imilast example andthisds.whptis _important. Thecontribution fromthepole(haif acontour) willbeImF-¥ andsince.<¢~\ thietermis_helow wl.Thehighendpoint ‘contribution wis.ylprecisely, ahdibisthis.thing which mstbanish inFESRsonofixed pole. . or) a wee eee - * ’ - ee ee eG -oak r 1.IamfindingthatIfinaTlyHavetoworryabouttttiiss-~Thecontextisasfollows:” ~“Tnevaluating unknonw integrals ofspecial functions, astandard procedure isto elevate thespecial function intooneofitsintegral representations, andthen a to"do" the original integration bymoving ittothe right. Ofcourse this _________involued order. interchange_and_you have tomake sure the-interchange isallowed. ——"- ~=9,°qhe"precise theorem aboutorderInterchaigé isgiveninHickpage207.Themain aideaissimply stated: inorderthatinterchange beallowed, eachoftheinside integrals must beuniformly convergent inits"parameter" overtheentire ‘relevant a _range_ofintegration. Normally, theplace you.run_into trouble is.attheextremes____- ---—of-the-integrel -ranges usually whenitisinfinity. ""3, Hereisabeautifullysimpleillustrationoftheproblem.Ingeneral,whenyou_ tdoitone way, the thing converges, but when you gothe other way, itdivergbes. Butinthis example itconverges bothways,butgivesdifferent sanswers: 9 __ ae 20 a ~ee :-- -- — S-oh a NagGeeRSS Saye=A. 2one PEAS 3 - ee oe He .eee woe eee > cy 2 2Sane oe dakly SEOdx:SVee =axFaw=Me=ZO --——a: YBa,Y SSGay 2°. & en CeaSe a oyYah.: rySe N x fo,Wa: QdyyS Wk Usados camagec xe foe)ft - a ~—-———_____ <r ._ ~-Sdee -2ow weeMyeilaet Hereweseeveryclearlythat_oneofthe“internalintegrals" isnotmiformy —__ ®cnnyergent_over_the_range ofthe external-integration.-- .-_-. —~ — .- . -~Comments onUniform Convergences. . : @]«=4Mesorems iF(Jn).dsthesunnandofaseriesinnyandifF(j,n)-is theimit-- behavior asn-yo ,then fromtheformofF(j,n)- alone youcannot makeany |===eonelusions about.whetherseries-converges unityrmly:in-Je— ———-— | 7 eT wr mn eg Rew kT |~~Crarnge IPGE FgtLaeMem EGis)=ai,Uadagemauit. ee en 7~aawerns eeAY They .. one ~—-—Ae. Qa, LEG =PUM =Wigs, so.aaniea tawal, - _ surddonrendiy omnusaneit. ox,\S\>2,(Te,seanyu.Sexy.AN ce SOIT, LEP Ce) Womans oak Up) Bubsh ———— ogc ~ —- FSaQW =deananivost)VegLeandorardin, ueeae. ]——-->The-point-is,-obviousiy;-that the-lower termswhithyou"do-not showinFare --————possibly goingto“vause’non-uniform-vonvergence: : oa =- -2ezApplicution: ityouarewondering whether theaudition tneoren P=fPp—[email protected] ~uniformly’ cényergent’ as~/j/-»% 7—youwilllearnnothing bylooking “Stthe - --~targe /j/-benavior-of-the ‘summands ——~ ~~ > > OO ial a - -. 4 - DedeoatunGuilsencranadohwording, "Await!amasinkinite avin 5 DHoPagee FHZLFGowes |i Qasiona, SQ)aCodey joarumn, EGSASeitenInfiedoe.Qaasbion:a=bwin,ANSGad? ~eee ee -- we ee ew wee semi oo.2=ate e|DT Getewelool eee) ere ee cee Ter Sen om wan 2OS™ Lys 2DUT ae 2©Wedele_8Q= Zw LTeee ee eee eeeee ee ait2 Men.-\8Q\ =[atom <ZLAtal_¢ 57GS - ---;3 ---—- -3aay- fuses, Mags2)aZMon-YW)CEG)canna DreSH)@Gdn| ‘ JS ©6MeePocng! Shomer Gey FM =cbesateddis,amdrslpromdy Wun.REPL cee ._—oS7 oe ae . _ ==feoDasa: 8MO ALFG)Upsgeo Geeho OTT @Sa Cote ek - WSEQ cnCetera ancepeer)om. -4d. Lea +OZ EG)cmmnrgee. merloonatig. JomRhyeadcolastudady. | aan sanuotivne.iaSwede commen ye _- ~©ExamshtdyotkoWordBogpeden alah: Deereadbtde sommpSanede © ®GrobQedcanescamagm cmv someDnieemn beorngtiae oi? 2 ate =ZASTMee) ®t Be ACE) weCine (bene). Wyler -Racers el oeCRSRR)TO oo oo co pom se ( \.. .MMwandrole . RKeseaniarao..or . Beng|non)Bata=hlaa)Oo 2) (6gies - sell me(GateuttesweONCeSe ewe?Tye tiegangataarf? WaisrHt?”Teecmallerhed eee dene oe vys Qua, weed cnbis beange meY -- - eet Le eo - ~~ oT - Rindcoating Awoumpento: EB LF e°Yorke: WaSen telsadry waFoxMc)(ule,Swing,commana, alesisAe“saadgidatonnie.-~WocoeB=~MeratiostGeuncorstusines, laresias Le\. -=Sin[reQy-td]=doeLaC=ee} - YadwokonOia:VkteBG). 2 : -teG- Gy)LsbE. <3 a<=bg)._! Corduasin: QC SeASAsorogynces sianEN .e Qact)SL enesonhusion wuwnee JSaAeonRasteatoeMeares. 1Se)TonkBG)9)aalaslale~~eanvangasce— BakenieLe’chavious. Jaa. congamson /vshsqral Sest. -ORoua oeSe Wie\,Seoee C9 - APAL.B,comnangse super. -ReXo ne . Wvsi=4, oeaSSOL oe. . 4Rug<0,AO.=>oergpncty AQecQ=o nang, ©WecasewaormemosWRWivaiing GaI=\odog JOmag. ofhiS1.Moaotcanawas.Vaishane- Miadak SBdyGusder glo —— + ee a reea Ome bewee ee wo - G)ABaro Cia, a0 2%Lo. , J es phen apo nene arlorena: OxNel Soa a WA . ee ee - TSM hyo Rnera emp Ze - moe ne - -) .- ro Siaaa lielef es —ch ae ne soe”~ ~ mm . - z= <n — Ow yey Se wo 2h a ason Coe a ~ ee ee NL . QsLang.aevaravoid ah, .ee eeee \Ba)iaSboomdad. GWA ZO, |Consdunsion Jou. lala\_ahZA\od OG)>O Lo. {We20085. aay aCeeee ee a ~ 2pom) _ae =emvargaaiky .et pt ee / 4OM S35 AU =ANOVSK . s. bae 5 @Ya,ssx) =aia =alaeAw - cee Aes =-Bem) Pom) FOOL _ Am PG RC Gah _en) ow) . . . . . Se &e BHR We ro. LL Dory, Quasana commgaa cesolubily imanbice,2Some! w, b=Oo ohh - - Wadd Fedye,2)dadeWoeLH) a0Qoalenanasdad 6 . ; « to! |‘ ©f.me =© jel40 | |42.Aes =Te):G-2)> ica caoat =ces” Wet zKesa=mo.Basse) whew e zQe= =S(a,-\2) \ew o2OsF=F(ajeye\ 2) wey |2.es=Kye)Hoekace |°| if | -* q Calewlation ofsunsbypiecewise continuation ofcontours. r1,ThOrder“tdexplainwhatthistitlemeans,consiger-this sui—~ - -F9 ee - See om mao : ->‘Thisis-some function of-2,S(z). Suppose we-didnot-knowhowto-add upthis — ~~series. Gur-first approach would-be to-do aSWI-This yields: -—— ee — — _ Way Se + SEER — we -© x Wewould iike tosimply open thecontour and q- - —”FUWTA VeFLAGATIY: However, “Af2Isatallcopplex, oneofthetailswilldivergea 77 ‘aridyou"sre screwed. 7 7 7 —— ~~, $0,wetryapiecewise analysis. Breakthecontourintotwocontours C1andc2 7TS" dsstiowi. HuSE~| —_ oro ne i. -—-—SES4*S -Svs737 Jann -2e— —- -- -e - - Q& - a - 7SE a ‘ ~~ TT erie a ~ 7 ~ “7 ga -Ndue Sy@) —_ — ~~ See ee tan aOswe \ wee ee~isle gyn ~ iena aeee daZS--Y -ae =+r so = SC)Sh ern .2wee ee 3, Nonacaind.Dk ae‘Yaak —— _ ee OSS\dxSeay — _id 5 -eee eotie- - - =WHEL mL - ~ — a = - _ = ssA x a Van,S@e- SNaeLSS ~ayana @7 “Sale enaat ~Ete We Sai-s@% 3[S@= &@)| OS / . _____ Tims,weneed‘only,compute $;(x)sincewecanthengetSpfromthislittleformla, ~ feCalculation ofSl: 7 . ~_de GaloulstionofSis . eee— ee a a ere -~S@e-g\ eee QoS eT —radegpadeareoe“uk EDO ieO\ECT: — : ~ -LE(4Bos) —edn. alalalsnte.—. aan Zac. s =e JH _SWaa.2x9dacoes \atrnl Sldemally Therefore, when arg(z) Ispositive, weare jistified in“Potativig the"ecntour _ tothe vertical direction.—Thusr - _ . inktico ~oo _ Sy= teKmFete 2 8.TaoSaangevara taateeekQue oo ee _ - -- =Ye — =~ -- —- 8.=QaTE”te)°° aTeak(gSten)77 en Orne oe aene _ . . . Sa Ba SSoom7 8 a ee — a. . O.Wd, danger bomix Mominh= So) Gd:“apBEQoe2aordY : ». oswe5 ix[Quel td=BEYdxhorn)& ; , —ae- @Wo shane te AawH -e- 6 sCaebh Serr}Wa=t S ‘ feryas@ec-.. - 2 (ee ~ apeoswee Sets_ eeeee =FefedbaeoLi - — - = RE = TS -zS _-~i [=| -\iefi(Rar lee Se - wee eee GL . = eth --a Brsd— ——-~8.Waue Rewncamnpdale lakeSebe 2. Fate)_taefeove wo ee IT aoe -oe__. |S@)=e (-AneysBae AE - . = -_ F ~ - _- a eer ®qW I5.6)= KXCaedee)atEnef "JO. Now,animportant point: thebetafunction islikethegammafunction inthat ™~ attends tobesimple andrealif{tsargument isFéalafdpositive. “Ifthereal”———-= part’6fits”argullent isfosttive, then-beta tusasimpleintegral representation; —— ~-—— -—just-tikethe-gammas However;—as- expected; notive— that-our-S, will-be~"simplet- —- ~inthis sense only-4f arg(z)-is-positived, whereas S»issimple ifarg(z) isnegative. ---.—However, thebetafunction hasanalytic continuation toall_argument, soactually___ a _boththese functions aredefintd forall2,except where thebetamayhave . oo singularities. _ _ ee —_ Ti.Addthepieces together andwhatdoyougets t—C—=—sSs—s—sS~S~S ~ RR ae ve anSS 4)6O-WOR Te eS ee ig at 7 “TO =1) L > oogeSE He ene a cj fies GS 5rafel See eee ~a TOSne SRT _——_—_ ane WeWrakaA)SSE12 dabx=ARI. Van Re ee. 2=Cael tea w .(eazy =X+\ . - 2% . ee : 2-22 Nes =. - _ pte a)=Qa . a . @-Ya X] ye Be 7Ae a\® aee _ .8 —= WE KT ep -=~Yeele~-2)- —6 . 3. ly.Soaftermanyerrors whichIfinally corrected, wegetoutthedesired answer o thatwaKnewallal6Hg.Needless tdsay;Chisdecdmpdsition oftheSimpleyitsinex ~~" > fiinetiorr inquestton-here intotwomessypieces isnottooenlightening: - IVoTaso, andeseexoddans43> ~o eaaSeEEEbalaTo]teeS ec®p\--- Te ZyeCr |)vr Mne 2yeCT VA Amt “"~~" “ntact, thedecomposition isamazingly complicated involving aspecial function | ~"77 >WBSTgsoSpecial Gedoesht“evenHavearamér |~ TTT -~-Buty-the-method-is-whatts- important. -Suppose youhaveaconvergent sum - whichyoudo-not-know-the sum-ofy —Then-this method istheonlyalternative that —--——-Limwoft. 2... wet ~- 6 _15.Here istheinterpretatign ofwhat Ihave done here. Consider just the“upper contour. This upper contour actually generates onehalftheseries sumS?Seems that waysince itpicks uphalf theresidue ateach pole. Isthis Yeally true? No,becalse thisleavesittheTestOfthecontour Which ‘coers-along theFeal—— ~ - ~~axtirbetweerthe-poles? Thisintegralishighiy-"impropér" because: itcuts through~~ ————_—Lots-of-poles+ —You-might- think-that such-an integral would-be real,—being-& real_ _—_.- -_____contour_integral ofareal function, butthis isnotthecase. Consider; meen a —-- ot -—.-_AS =_Snx\, =eG) =Jo foe “V “16. Sowecontinue withtheinterpretation, Consider justtheuppercontour above “~The real axis. Doesthis upper contour integral converge for all2?Thatis, ""° fof2IRBLAS WheVALECiFELS (iiherethisSuiConvergées). Tthinktheanswerisyess --—————The~uppercontour ismade oftwo pieces: The "poles" piece-ts—just-half-the- —~ |—-———-~ original -sum-which weknow-converges. TheCauchy -piece-is-the-sum oflots of . little segements asve_have said.A typical segment ist ee ~. vme Net X m\ Wy - pee ee ee ~ .OSs 2Kae 2.Gare— Nek ont =olout — eee ( So-yes, thisCauchy-piece-elso converges because aissitting there. Infact;you knowthattheuppercontour converges because asyougoofftotherightalittle above the_real_axig, thetenfactorsimplyoscillates atfinitevalues,andthe2”e doesexpodecay. - Thus, both contours separately converge for zinthe unit circle, which isto say,for1n/z/smaller than0,ie, negative.7 17. Soéach contour weknéw separately converges. Weknow how toget one fromthe —- ~~other, so-the-wioleprobtemts-reduced “toevaluating the upper contour; ~Whem ~ - - arg(z) ispositive, -we~lnow-that -G€-vanishes-so we-can-replace theupper-contour- -—— _-witha semivertical_contour_end-we can evaluate that toget result. When-erg(z) .— isnegative, thensemiyertical contour diverges andGCdiverge, butweJmow.that ____ theuppercontour itselfisstillconverging. So,when2isinUHP,weidentify theupper contour with a¢ertain function beta(z), say. When 2goes tolower HP, :wedontknowwhattheuppercontour is,butweknowwhatbeta(z) is.Sincethe = “Upper contour converges forali2|and’sincecontour =beta(z) intheUEP,we . fustB86havecontéur(z) S“seta(z) intheLHP.Iftwo’analytic furctions agree inthe upper-half plane~, ‘tlerthey have tohave the same analytic vontinuation! - 18.-So once_more,-lets rewiewthegeneralmethod. .®- a)youwanttoevaluate ininfinite sum. Cantdoitdirectly. So, b)youSWTthéthingandbreakintotwocontours. Wherethesumconvérges, each ofthese contours going offtotheright will also converge. ¢)yourelatethelowercontour totheupper, soonlyhavetodotheupper. d)youfindaregion of"Zsuchthat’theupperquarter G¢vanishes. ~~ “—- =" a)Gharyourotate theUppercodtciy aidreplace withalaiom, Convergent - +~yveat-integralsIwok-upthe~integral andyouhavethe-upper(z) fonetion. -= - £)-get-Lower(z)},—edd- them,-and-youaredenel.—- ~— - ee Re - -- . :, ae . . : *, . C)~_197Expliéitstatement offhistechiigie - Be TS@y=Zeke |compu dozeRo Cems Gnome) =KEL ha eee —-Beraluidic inToothvenaletr, amdbeSeo. VaAso Ub OME ODAsal. TH) Wma) okostsugar ahaaamebeiseee ee Jes>OnSonaseetdamir._. ee . armyPLA Nod prod} =Shans~30=.oaSee) =ta@-+4a. ._ppoady ordurscoeguaber EC), __ an eee 2Waa @,Shessemeene SG=oo Acs)WGE-E,2). 0°.- e .tey= L@+h@). coe, a9. Apghusodeins _ _. Co 7=aAt. =aaaGuaioes 7 .a SLvba Ge) =Vonge _ JPRS ee \ =Wedeaeodadece!” (al)antene)hie)ala) (7_-.Wd ~-4—ee Rams-1BOme - -eB =lies We. ok -- See eee=Pand(T) a When \2\<1,ok200dace SC .ys? _. OY) man24is.comaee motbh. —eandulel—}anLE- (c)_2<one ess or =€ zg a- A\e>0) 5Aurasgan soGl=OY% 5 .soa e xAslidel WG-32) 21a) 2”4)ec : oe Pak+ie2) . . ee ee_. SNe)eteSastk® 1)©.om ont! sf : |MFSasshés)Whe#2)TCEth+7) e \ | mo _—— 8a arDasSHDICE) ~TIGION 3GePe Sap se Sorat ee | TS tk ke eS eke Yet —— mabe Ars WIE LUxkst%s | Oe ( “. Comment “inthis exemple, Ihave successfully replaced the undoable sum with-an integral. Unfortunately, this isanintegral which isnot.very easy todo. ItisdoableigIuseEulerexpansions forthetwogammafunctions, Butthisisa_ e lotofworks _| - - Itsuddently occurs tomethat there isanother possible way which is aninduction proof. Thisintegral business should onlybeusedasalaxtresort. _- -oe _- e _ @ Series:nowdoyououndatohSG et. Le 2as a aw <22 el re Oe © Gage ALSeS,baaLe © Cesitinquanti. GUE). GameteeShBR ry aa ae NTS se, Faomy690)unsambedomMemd>nh ese.Boksmagnundeaksacha,s_pouk th,vlKN: —@ MadatAMe&benecomawKS)ancea ae =ee a ——— we ee -- odl2\_< ) veeWANS KL, TB Glebe jorSuid1141, wecamMakemeDeagpanayleweDisk ee - = a a =~ SaquikWA8|Dror Wa! WBWerxsdst. | eee ee OB OO ee -- ee 2 oS _ MRL SKwe —©DawpewSkdao: Le - ee - :__—~|Givens. emmgiex. 4.wd &212S) Toweeail,: ———- -]-asa_gossisle_teinkomW.BorgesanongnsoOne - di RY eet {a=&ol.Qrarcror, wae _—a arorasea, Setsada anWy -oe - es cere: en i_. E.oye ¢SOO .EN KO 6 ;a a To==flame mee ON")cL Le ____. Asymptotic Series. — _-- S -- we eee ~~ivReferences: Smith "Math MethodsforSandEB" oe. Qe > ittakerane-Watson "bestyooK— =——-— __ __Jeffries bigkook eS ______2: Hereisthestandard exemple: 2 ee ee —6gg ee ¥ eo - t—fa— —- ——_—_— oT TTT TT_ =Daw VE WeeLodiiesCowyealt - eS oe x KS (amy x STS ik- ae 2 ea eee ———$- ay ge)——|ory _ WQS) =|eSoyWane sesete iy _ SuEES —————MawaQesaneonSecsegeds MaanMor_SoiKLenswasabi _. ——.__Sn()! Clearly, thisisthecatch. Theseriesdiverges, but-if_you-take-105_\ _._. —@—__*™=» theerrorislessthanthe105thterm,Theideaisthat_you neverreallyconsider theentire infinite series which actually doesnt_even_exist_(converges_-. nowhere). Ifyoucanshowanerrorcondition likethis,andifthefirst_few—— __terms temporarily converge, thenyoucanmakeagoodapproximation. withthese—- terms. Howsmall canyoumaketheerror forsomegiven_value ofx?Laok_where— — ; _____series has "minumum. _ _ ~ = SO _-______|A> MeDm |fo ny AN2osPW GKSPY a rn Se ee Sors pn em We aot n=ScMY eeDe eesa een neSefonnck]DOD wee,foeA -@— Anyway,—you-can-compute-the-turn-areund-point-to get-best-ideal error. Inpeactice~ —— ——-——you get within. desired-error-in saythree-termsy =———————-—_ =>— . w e 1.,Supposeyouhaveaninfinite seriesofanalytic functions f,(z)whichyou.add.upto.getsome.total function f(z). Suppose theseries converges uniformly. oversomeregion.in.the z=plane. ‘Theniftheindividual terms.are.analytic inthat. region, then the. function which4sthesummust.beanalytic inthat regions 2.Examples Consider theseries +242 tsa. =1/(-z). Thisxeviesconverges uniformly overtheinsideoftheunitdisc. Thua,inthe:interior ofthiskkk Sisc, thesumsimply cannot.haveanysingularitys. Nobranch points,nopoles. Thereason. is.that.theindividual ternsareanalyticintheentirez-plane!of _cowrae you.cannot..count. thebranch pointat221sincethis.in not.in the.disc. AsApplication: youarenot.goingtoget.aseries ofreggepolesterms whichproduce. thenormal threshold cuts, Reasonis: ifyouhavesuchaseries, thenthere can benobranch points intheconvergence region, _ - s @ Sorpe Suma, @ionotphevee ZW= ¥sz 2L(i) =*S rom Ay =b(vriKebei) a ¢ SW =E+” yt 4 ek, @®Cmeise ir eget sn. jae Se St (ARR BSR GEDeo:=Wagregecye ). $e)¥ eh Redbiz9&®ak S=Zo(aged regard) RokwoodaopoddSerue!Je » S=Ze(oded) =Ao]+aces bs.Aw =SHG) owadnAypeSatieCfadd) S S=Gen)|Foor +4] (v=LaAx8)A=LAB+ BtSr9] 4aSolu do| e s-=[AS8FGd] jewie SS Grunge Zr 9Axbe9y) DPS AS=GeGedy: O24 Gear ODE's Solution ofCertain Class ofNon-Linear Differential Bquations. Deo29,1994 ider: au eo 1.Consider: a £(u) Here isapossible way tosolve the thing: deey ay=few fky-seaeBey=By =yRqtO aSa a w+ 2yy2Fee ESSYAR +cet=FO)TA=F) FAIL >¥=ZVORO\TA, brMN2odSe daAKtLe A \on= \a= +8 = x(wA,8). =«= hraw *CSAS) Then you would invert this result toget u=u(x;A,B). Thus AandBarethe two undetermined constants which must becontrolled bytheboundary conditions which . accompany the original equation. Comments: Note that youcannot have anyfirst derivitave present intheoriginal equation orthismethodfails.Notealsothatyouhavetobeabletodotwoindefinite t)integrals. Ifthesecannot bedoneanalytically, youcannotgetaclosed formsolution atleast bythis method. Numerically ofcourse you areguaranteed asolution, Examle: Maya wu 9 Fl) =—bwur 4 SM dx waN (dn, a t w -\ tsevt=SNgee+8=25|G+ =*ire BNEMe ° 3QsreESM) >wey=Asirtoxrs) 7. This was alinear example. Now for anon-linear example: amie? fayKe™ =F)=Kea&=JakeA" % ond Fo Mm, hanaSlemdod ODES. -- @cumgovigue KE . ou. Gab Basco eer TT Bias pani we “©Goygassansh: rs a ee oDGa co OS, wane REND ot) cee Lot NelsQuad Sataeiece __@ Vaslaygeast Os)uw=mulaitnso |a—a _ Ge) want ~atu =o a © Momitig ted ke=OL ©Dayoan “tu+Gokulimeo ©@Algpigannsbue: —xQadula LeeGabe)glul=ob=o. 4 | \ue e OntheLimit-Point andLimit-Circle Business: Weyl's Theorem.QntheLinit-FeeeeeeyeeE—E—E_E—EEEE 1.Iamreading this inStakgold page 297. Lu=du LeSet. 2.Part Iofthe theorem isuncomplicated, The claim issimply this: if atanyparticular eigenvalue \y both solutions totheODE are of finite s-norm onthe interval ,then atany other eigenvalue, the same istrue. The proof, given onpage 298, istust anexercise in inequalities. 3.Part IIoftheofficial theorem says this: given any} notonthereal axis, there must beatleast one s-norm solution, Ie, either there is one solution, orthere are two solutions ofs-norm. Ifthere are two -solutions, then bypart Ithere are two finite s-norm solutions atall yoThus, theactual system itself ischaracterized tythisdichotomy: 4.Thedichotomy: oneofthese situations must prevail forthe stanilard | Sturm—Liouville Hermitian (self-adjoint) system with asingular point taken asone of the interval endpoints! a)forany, both solutions areoffinite s-norm (t&dq Cow) p)for any\,there exists onlyone(ormaybe no)solutions.of fs-norm, That is,ifforone you canfind two finite s-norm sols, then you greincatagory a)according topart I.Ontheother hand, suppose for 7 Some Athere isonly onesolution. Then atnoother®cantherebetwo solutions because then that would have tobealso true at. Thus, if you find a) where there isone solution, you know that atevery other there is either 1or NO solutions. - 5.Ithink this should bepresented asaseries ofcorrolaries. Well, weknow that ifyou areinsituation b), and for) off the real axis, thenknowthereisprecisely ones-norm solution. Ifyouthentake)ontothe real axis, then all you can say is: either there isone fans, orToneatall,awk QuaaaunlbeontPonsaN=walesigue. 6.The way you tent asingujar SLsystem tosee whicl case you have is simple: pick atrivial value of)andsolve forthetwosolutions. Then check their s-norms. Ifthey are both finite s-norm, you are a). Igone ofthem faile tobes-norm (orifboth fail), then you areb). 7.Wronskian facts: (time out) From Abels formula page 60, weknow that thelironekian oftwo solutions ofLu=0 (SXEXMXEXEN) =©exp(-m(x)). Therefore, ifW(x;) #0, then W(x) ¢0forallxininterval. Conversely, ifW(x,) =0atanyonepoint, then N(x) =0throughout interval. Inthis vahishing case, the two. solutions must bedependent. Inthe nonvanishing case, the solutions are independent. However, since n(x) isdetermined bydm/dx =a;/a), weseethat thewronskian can depend onx,inthis second case. 8,Another fact isthis: for S-L systems, a,' =a,.Inthis case, :m(x)=Inoe.Tms,thewronskianofanytwosolutionsisW=C/ao(x). @ Later, ag(x)iscalled —p(x). Still, Wisx-dependent ingeneral. |Qk, pOOW Gwyx) =C= onto | 9.Thobilinszr comjunct J(u,v) reduces to-p(x) W(u,v; x). Thus, itis this wronskian that you want tovanish atyour endpoints toget a trulyself-adjoint system.Ofcourse,ifp(b)=0,youhaveaslight |problem. Orifendpoint isatinfinity. 2 ( e Stillmoredataisneeded before wecandothecircle-point stuff. 10.Nuch use ismade ofaspecial-purpose “unmixed boundary condition" atthe engpoint a.Itis: | Buetu(a)+4ut(a) Ifome takes aboundary condition ofthis form ateach end, then easy toshow that thewronskian ofauuiwktm uandtheCC ofanother solution v,namely ¥,willvanish attheendpoints. Butthisis °thebidinear conjunct inthiscase,sothepointisthis:those ohspecial unmiaed boundary conditions give youfull sefl-adjointness.” ¢ Wehavesaidthateuu,¥)=Oatbothends.Weneednotassume that uafd vsatisfy lusO, These uand vare just two functions on the interval. The operator LisHermitian ifBC's are satisfied. 11,Bytheway,throughout allthis,pandaaresupposed tobereal,and the_weight sispositive reals@akWles)FO daceLnglyUNOrtyodoqendant: 12.Note that W(u,v) =0doesnotimply thatuandvaredependent? That is only true ifuandvaresolutions ofLu=0, Inwhat follows, weusually speak ofsolutions ofLu=shuwhich isadifferent equation. Iwant tothrow the idea that ifwronskian vanishes atthe ends then it vanishes throughout anduandvmist bedependent. Notgenerally true!Twwlunknyolbackr aolay pe)WO,Sjx),watWhasdlp. 13.IwishnowtopointoutequationGeos),Itsays:if)liesoffthe e real axis, then thewronskian objects p(a) W(u,uja) cannot bezero atboth ends atonce, Here, usolves Lu=s\u. However, these wronskian-like objects dohave clean limits even ifendpoints are singular. 14.Wenowstart part IIofWeyl's Theorem. Consider lu=Asu. Let9 :andWhetwoindependent soluttions chosen tomatch theveryspecialboundary conditions atregular endpoint a,seep299. You cancertainlyfindsuchaQandYbecause onlyoneBConeach, Clearly u=4@+mayisalso gsolution ofequation. Now, the big problem is:how do you putaboundary condition atb whichisasingularpoint?Youcannotsimply“doit”becausep(b)=0 lionployw(uibs)maybe,whichgoofsthingsup,ormaybesolutionudivergesatinf. Seibe) Phe trick isthis: rewrite the BCinits wronskian form, then take =o thelimit bygoes tob.Yougantake limit ofthewronsian object.SincebottQandsatisfytheusualwronskian=0 BCattheleft enda,weseefrom4.80athatingeneralniether&norWcould possibly satisfy ourdesired condition atendd.This was noted in13 above that both wronskians cannot vanish for the same solution, ‘Thus, any candidate which seeks tosatisfy our new condition atb must truly bealinear combination, ie,m#0. (Except ifX= real). Now all weare doing istaking over the old BCtothis singularcase. Thequestion nowis:whatchoices formallowutosatisfythas right-end boundary condtion? _ a 4| 6 15.Necontinue, Tofind theallowablevaluesofm,weinsertu«@+247 into theboundary condition atb,before taking limit. Weatonce find that: . a)if\offrealaxis,mliesonacertaincircle b)if\onrealaxis,thiscircle degenerates torealmaxis. 16.Notice that \=real isvery special because wehave been assuming that allother functions like pandqandsarereal. Thus, real means that solutions arereal andthis means that wronskians like W(4;A;x) =0.Ie,wronskinu offunction with itself, Go,h=auls qaningatins 1.Theexpression fortheradius when )not real isgiven in4.84. This shows clearly that,as youtakeb,tothelimit b,theradius ismonotonically decreasing! Thus, inthelimit, yougeteither a Circle ofviable values for m,xmoryou get exactly one allowed point value for m,inthe case that the circle shrinks toapoint. 18. What isthe bigadvantage ofhaving asolution solve this fancy voundary condition? The reason isshom inbottom ofpage 300. If xreal, then this magic solution (orclass ofsolutions) isof finite s-norm, This isthe big payoff. 19, But there iseven more payoff: ifyou get acircle inthe limit, then from 4.84 youseethat notonly isueQ§+mA offinite s-norm,e putitselfhasfinites-norm.Thus,ifyougotthecirclecaseby applying the magic boundary condition, then you have found two finite s-norm solutions. But bypart Iofthis theorem, you must bein situation a)of4.Sewaany sSondacea! ==Voua)" 20.Ontheother hand, ifyougetapoint, then r=0and4.84 says thatA must nothave finite s-nopm. Inthis case (point), there is iselyonefinites-normsolution. Butthisparagraph isonlyTt\aotreals 21._If\= real, wedoknowthatanymonrealaxis allows usQ+n¥ tosatisfy the fancy boundary condttion. Infact, any solution u Will satisfy this BC. Inthis case, weget noinformation onwhether any solutions are offinite s-norm .Maybe none, maybe one, maybe two. . Limit-Point Analysis. 1.Youhave asingular point b,butyouhave determined that youareinthe Limit point cads. What doyou donext? : : First, youwant toconstruct agreens function, Todothis, you chooseayontheleftbecausethissatisfies yourleftsideBC,and youchoose u=Q@+ma" onthoright, sayforbo. Thespecificgreens isthen-given in4,93. Youfind that this greens hasonly poles, hence usual discrete spectrum since wehave notyetgone singular. Wow, -asyoutake thelimit b,goes tob,youknow that mgoes toitslimit. Itmayhappen that asyoudothis, thepoles ofour meromorphic m(p) may move together toform aout. Ie, in thelimit, m,(h) mayhave poles sndouts. Itcould have only poles or only cuts. 2.Wedont really have tothink ofthis limiting procedure however. There isawell defined system for the desired greens function. Onthe left side you use some solution which satisfies your regular left endBC. Ontheright side, youuse the single finite s-norm solution which you know must exist. Thus, 4.94a specifies gcompletely. 3.Once youhave thegreens g,youcamtour integrate attofind your ‘completeness statement. Itwill have theform 4.95. 4.Conmentson thiscompleteness relation.IntheregularS-Ltheory, r) allyoursolutions areoffinite s-norm. Youimposetwounmixedpoundary conditions atregular points aandb.Thiscauses\to_ "quantize". Ie, only acertain denumerable set of)will give salutions tothe system. Ie, spectrum isdiscrete, eigenfunctions are orthogonal eté atc. Compare this tothe singular case. Inthe limit-circle case, there are finite s-norm solutions for all ).—te;—+herein-ne—quentinetion— (Re Inthe limit point case, there isexactly one finite s-norm solution for every \offtherealaxis.tgein,goudocobegetRtineticn,ir Gysolution W fact, 7ouosethet-old-nesudt thet—rigenveiner—met—be—neeky—Taie oxve,will Nevertheless, for5Teal,youmightstillgetthatquantisation. effect. Ithink you doé Infact, when =real, you get the mixed Qt,ox specturm situatiom, Thenthegcontour picks upallxthediscreteae eigenfunctions, anditpicksupthecontinuum ofpossiblynon A= finite s-norm eigenfunctions. dieSamesul Ifspectrum ispurelydiscrete, eigenfunctions willbeorthogonal. sabisty cS 5.Examples given.byStakgold oflimitpoint:aa S p305 Bessels functions on0to1;pure ddscrete spectrum7 p308 Mellin transform equation on0toinf: pure continuum (4800) widen. |p284. Fourier sine transform (4.58a): pure continuum other cosine and fourier integral transforms are done. pB94 exercise 24gives only example with mixed spectrum, e Orurs,whow 7ash)youcomutgatamsgenSumctin uw.SA.\aodhy ‘). , .4 — Tels,WhenASraah,yonwillgasaganSunetiind rs apouk:Lyyou.partner’ UCjA)hyppuad “alsodpsolveBau=o~ mo y byoudle: 4gna.youruln;A)" ‘ . Qwe, sprcbtuwo woslquentized. Vv . ': 1 o |Gomment onmachine solution tosingular eigenvalue problem.‘Ciont onmachine solution tosingular eigenvalue problem. 1.For\tobeatrueeigenvalue ,youru(x;A) mstsatisfy both 'boundary conditions, Inthe singular case atb,westill have 1 @boundary condition soconceptually the same asigegular case. ! Now, suppose you start with afunctign f(x) which istied down|attheregular endaeThenyou.pick somehandstartintegratingouttoward b.Ifthe)youpicked isaneigenvalue, thenyouknow Iyoursolution willsatisfy condition atb.utthenyouknowit iwill beoffinite s-norm. Thus, your solution should heel down toward the axis, ie, itmust approach zero. 2.Ifyour solution baws up,then youknow itgenotoffinite s-norm : and there fore itcannot besolving the condition atb.Therefore i your chosen )wasnotaneigenvalue. . i \ ': : 1 I rN |: 'A |4 : i boo . i | en e LimitCircleAnalysis 1.Taken from page 310 ofStakgold, 2.Inthe limit point case, weset upadefining system for the greens function byusing the single finite-e-norm solution onthe right.Forthecircle case,there isnounique suchsolution ontheright:the solution corresponding toany point-on-the-cirlce will do, So,, the idea isthis: pick aniceoeofftherealaxisandpickanyparticular solution "onthecircle", Thus, youwill haves ontheleft and uonthe right. These are independent, and wemay choose tonormalize sothat their product isthegreens atAe. 3.The scene suddenly shifts andweconsider the inhomo equation with soneftacked on: LyeV=sf. Howwhall woapply BC's tothis equation?The left end isusual unmixed BC. The right end isnow going toget p(b) W(v,4;d) =0asits condition. Here, uisone ofthe circle solutions chosen above, Given these two boundary conditions, the inhomo equation has only one solution: that shown in4.115. This one solution, called v,turns out tobeoffinite s-norm, 4.Now asaspecial case ofP3above, choose f=(h-ho) swwhere wis the solution tocome. Wehave just computed the greens upinPl, sothe solution given asintegral over greens isasin4,120. BUT thisisanintegralequationwithHilbert-Schmidt kernel.(ie, r) !meansbounded operator). Ifyouchoose Acreal,thenkernel is i symmetric, From previous theory, then, youlnow that theeigenvalueequation Lv=)sw_has 2pure discrete spectrum with orthogonal _Sigenfunctions, 5.Tosummarize, inthe circle-limit situation, the eigenvalue equation always gives you adiscrete specturm ofreal eigenvalues with usual orthogonality and completeness. Greens function iseasliy defined andthen constructed. Theactual results (greens; system oforthogonal functions) depends parametrically onwhich point you choose onthe circle. 6,Example: Bessel p313s Here, the points onthe ligit circle are paranetrixed byreal number A,Then 4.123 shows the A-dependent greens at\=0. The Bessel equation isregular at1,but singular at0,80atOweimpose thewronskial singular BC4.125 (which again isA-dependent). The differential system isshown equivalent toa real symmetric hsintegral system, soweknow eigenfunction data, If youchoose A=0,your eigenfunctions aretheTox) Ofcourse theboundary condttion w(1) =0puts especial, inside, Ie,you get quantization ofeigenvalues onthe real axis, 1.Other examples oflimit-ciréle: Hanekl transforms, Hermites, Legendres, aK-transform, etc. \ ¢ Application of All this Theory to the Radial ation. 1.Asshown say onpage 350 ofMessiah, A a~+Qn 4very] Haley=Egat) ROS oreCahde OIVONRO=Eele 2 a7 Qe)=Mba 4,vCo)amet WHAM SS[yuafae a=H. ame3 TheBCatr=0isshom. WO=o. 2. Inthis limit point, or isit limit circle? Cannot tiivially solve this thing even with E=0. However, wecan assume the potential V(r) drops off fast so that at large rthe equation is roughly: kt ~a3a]Q)=Eats | \awact)$ceEsogabe)=Axe Totest for finite-s-norm, weonly need the asymptotic behavior of the solutions anyway. Atour test point E=0, wefind that neither solution isfinite s-norm, Thus, Weare inzirgte limit—point case. 3.Aside:ithadtobelimit~point becausethatistheonlycasewhich @allows acontinuous segment ofthe spectrum. 4.Now,youcouldconstruct theE-greens betakinga¥(0)=0solution onthe left, and the unique decaying sofution onthe right (for B notonthereal axis. Ie,e“*T, )Then youcould contour integrate this greens toget the completeness relation and the spectrum, You would get the mixed result. The discrete eigenfunctions will beof finite snow, Ihave not proved this inthe theory, but seems that if-you-comsiderEjustaboveoneoftheserealeigenvalues, youdo know there isone finite s-norm solution. Asyou drop down onto the isolated eigenvalue, this remains true. Not so on the cut. In fact none ofthe eigenvalues on the cut will be of finite s-norm, . . oy! c ' . oS Discussion oftheGeneralSingularProblem, 1,Inall this, weare always restricting ourselves todifforential operators Lwhich are ofthe standard formally self-adjoint form. But, aspage 269 ofMishows, you can always cast your ODE in this form. Let xbethe variable, and (a,b) the range. 2.Now, apoint x=@ isasingular point if 1)o=infinity, plus ormins, or 2).p(c) =O Ineither case, you should armange tohave point ©beone ofyour | interval endpoints foryourscalar product. Usually, weletc=b. 3+Weyl's Theorem says: Look atyour equation. Pick asimple value ofeigenvalue ®forwhichyou can trivially solve the equation, Pind all solutions (two). Are ‘these solutions offinite s-norm over interval of interest? Ifthey both are, then you know that :foranyvalue of}, all solutions areoffinites-norm,Thismeansthatinregardstothesingular |point of interest, you are inthe "limit-circle case, Suppose one orboth ofyour aolutions are not offinite s-norm. Thenyoumustbeinthe"limit-point Case". Thenyouknowthatforany) offthereal axis, there is,onlp onesolution offinite s-norm, soak. e@ 4.Quickexample: consider page302Stagold, Legendre equation, butchooseinstead the interval ]toinf, soyou have two singular points, For X= 0you know that u=constant isasolution, But this solution is clearly not offinite s-norm onthe interval toinf. Therefore, the point infinity isalimit-point situation, I'm not sure about the class ofthe point z=l, have totest the other solution. 5.Greens functions: assume b=singular point. Then we defame the Greens function problem byLg~\sg =S(x-y) with these very important boundaryconditions: .1)Bg(g) =0.(Some simple condition atthe regular end. 2)the s-norm ofgmust befinite. You can imagine solving for this greens function. The contour-complete- ness relation is then valid, namely: A~-&Q) =- axBua -GD=-ZTrai)Soggy) However, ifequation has only limit-circle endpoints (singulaire), then the continuous spectrum isnot there and you have only discrete specturn and usual orthogonal functions ofweight 5. 6._Thus, weassociate continuous spectra (ormixed) with limit-point situations. 7.Thus,thereasonyouhavea“mixedspectrum" inthe0(2,1)UIRsis ethat the related differential system has alimit-point singularity ataeinfinity. In0(3), the differential system does not have this singular point, the other singular points are circl-limit cases, 80 you get apure discrete spectrum with orthogonal eigenfunctions. } : t / ® Reviews’ dS . . 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Orga - ©Taoubishabak Laohsu =2,“Naeampediale opreva debe. woQOeo, comvase oewx)=2dubans, Un=—CFBed/dn->)so — - as _- - —. w®.= So@lawa passagasWSGaofaae=th® bageery= <2ORO) yee’ ee _ __ dn~>- - 5 a ; Creaduia: soJonwaDrameReedinto(lis‘asqeton" ov.NuameLronoilde, Pda, - — _-=Saad &——Nlow;- if-you-take-a-Sturm-Liouville-or_regulartype-differential operator ;-you-generate the "singular" case.by_allowing one(orboth) oftheinterval endpoitts to-go to— __infinity, ormaybepvanished atoneendpoint. Thistakes youintotherealmof — _ __ ___ "singular" differential equations. Theanalysis ofSturmLiouville wasgeneralized _ byHermann Weyl, inparticular thetheorem onpage 297 ofStakgold. This gets you | “"~“4ntoallthisbusiness oflimit circle caseandlimitpointcase.Stakgold makesaveryrelevant cbservatisn “aitpage30hconcerning eigenfunction expansions. If,he says, you can equate @deltafunction toaIargecirlce integral ofyourGreens Function, -———~then itispossible-to write-the deltafunction asadiscrete-sum oftermsFelatifig — —---to-the-disereet part-of-the-spectrum,—and another-integral—over-some—terms-having tu~ dowiththecontinuous spectrum. Thecontinuaus partcomes from_cuts.in-the-greens~ function,thediscreet partcomesfrompoles. ThisiswhatVilenkin istaliing about—______whenhewrites the"expansion theorem" forF(x)ontheBluns ‘Thepoint above is,of _ course, that ifyouknow thedelta function, then youcanalways write aprojection/ expansion pairStakgoid doesthisforiotsofexamples, likeHankelTransforms, ete. nr a However, tamaLittleuncertain aboutStakgold’s analysis because henever @ wwe expettthegreat”circlegreensintegral tobeequaltothedeltafunction.“Iwould-dike toseea-chapter-in some-other book-on-this subjectsAs-references GHGiis - —- subjecty—Stakgold-gives-Hellwig-andNaimark,bothin-German-only, along-with—Titchmarsh .—_—Friedman, andYosida. — eee ee ee 7 —_ ns a Chapter 9 Regular Singular Points . . 7 apg. — 1+ -Buler's equation. This isawarmup_exersise. Euler's equation is_simply s _.Zw"+paw!+qw=0wherepandqareconstants. Abasisoftwoindependent solutions isgiven simply byw=2%where theaarethetwosolutions of ‘the indicial equation a%+(p-1)at+q=0 .Ifthese solutions happen tocoincide -s thenthe-second-sctutiow-ts-u@ing; Typically, ‘logarithms wtways' arisewhet“~~something coincides. Dee. Looe 2.Gircuit matrix: Given asecond-order DEthere will betwo solutions, Assume that these solutions are analytic inadisc except maybe atthe center. Without loss ~- ~~ ~~of-generality;—putthis-potential-singutarity-at-2=07 Then you can take “each — solution and analytically continue itone turnCCaboutthisorigin toget_ —Eno(possibly different) newsolutions. OFcourseeachofthesemstbe- linear_combination ofthe_original two..The2x2matrix whichdescribes thislinear combination is called the circuit matrix. 3.Simple branch point: asolution f(z) has asimple branch point at2-0 ifthe “tinction goes into amiltiple oFitself whenyouContinue around thebrarich ~——~ —-.-—Point once... - —~ - - — 4.Ageneral circuit theorem:-consider n—th-order—DE-#ith-n—indep-solutions.—— There will be annxn circuit matrix describing rotation about origin. The ~eigenvectors ofthin matrix give linsar comypinations ofthestarting solutions ~—-{used to_get_the mtrix)uhich gointothemselves undersuchacontinuations __ The eigenvalue isa. Then the “diagonal” solution has the form f(z) =2°g() --there.-b—=1na/2PIi— andwhere.g(s) has-nobranch-at 2=0-(single-valued),-——- @ Since every matrix has atleast one eigenvector, you are guaranteed to — -alvays"haveat~Teast-one-solution which-has‘a-simpre brancl point;~———-—— ~ 5.Circuit theorem forn=2:(Theorem 4).Yourcircuit matrix is2x2.There are-twoeigenvalues_correspondingtotwoexponentscalledaandb-Iftheseare- different,then the two solutions oan bechosen ashaving the form 29f(s) ———and-z>-g(2}, where~f-and-g-are-singte-valued-in-the-dincs—If-asb-you-might——~ ___still beable todiagonalize thecircuit matrix andhavebothsolutions of——_ "theform2°fand28g,Morelikely, youwillhaveoneoftheforme®fand——______the_otheroftheform2%f (h(z)+Inz)where£is.samef,hisSV._Again,.. — notice the log coming in inadegenerate case. 6.Regular Singular Point: Consider general equation w"+p(2P w'+q(2)W= 0. oO If atz=2l pandqarebothanalytic, then21-is8 called an‘érdinary points ~—— -—_—--If_at_z=21_p and/or_q are singular, then zl=singular point. Ifthe ___ singularity inpisatworst asimple pole, and inqatworst adouble pole, ——_———then-z1-= regular singular point. ——-——— ——-- ——— - (7 "7.Bignitromes ofr; rrwingular” atIf-2-0-is regular ‘singularpoint(we take 220only for convenience) then you canmultiply through by2°toget anequation resembling Euler's equation: 22w" +zP(z)w' +Q(z) w=0where ——_———_-P_and_Q_are_analytic. and may thus-be—power-seiies—expanded at2-0, The —— —- - significance of this fact is contained in the nemt theorem: .8.RegularSingularPointTheorem:FromP(z)andQ(z)computeP,andQ,,thetwo r) constants. Next, solve theEuler indicial equation fortwovalues ofv(see_.“________1_abowe,_replace p_with Py#tc,.). Then_at_least one_solution-hasthe-form. ——w(z) =2% (1l+cl2+02 0%+...). Ifthetworoots donotdiffer byinteger,~-————then-both indep-solutions have-this-formwhere-v takesonthe -tno-values. — -2- 9.Singular points atinfinity: This requires aspecial test. You change from 2totel/z .Then the DEisregular singular atinfinity ifthe transformed _ -equation isregular_singular attheorigin. Itturnsoutthatyou_cantell @ bydoing power series expanions ofpandq.inpowers of1/2=t.Then for -=-—— infinity toberegutar-singuir-point, p-must-have atleast afirst order zero int,and qmust have atleast aSarehdorder zero int. ae ~~""Suppose someDEisregular singular atinfinity. Expand pandqinthis- way eee ee — -aPeRhateeth tees -getea thee | —-—_—. —The-indicial_equation isthenv(v-1) +(2-pl)v +2=0,Solvethisforyour.—vand then by8equatiom will have one ortwo independent solutions ofthe form ov (1gaFT eadoeaes,)SOpendeing omvhether-difference-of - —_______——the-expoente-is-an-integer or-noty wee -- - 10.The“intecer difference theoren. Suppase yourtworootsaandbdiffer byan”~~.integer. Thenchoose aasthelarger root.Youmaythentakeasyour .independent solutions these forms: OO wl=2°(1+...usual) w2=2°(1 +4s.) +0wlIng “Ofcourse thetwopowerseries aredifferent, thisjustgivestheform.Note —— —— that-you-cen-choone-C-as-you likes —- ——--- "TLS Garionical“basig: forgot VOmeATLGH thatUTyourHootsaFeanequal“asd-aoWOT __.differbyinteger, thenyourtwostandard forms asin5willdiagonalize the little 2x2 circuit matrix and euch abasis is called cannonical. @ byRewien Equations: theseare(homogeneous linear)DE's having atworstregular ———~--——gingilay points intheextended complex plaié, i¢,imtluding infintty.—By-Uoing. linear fractional variable changes, youcanmovetheseregular singular pointswhereever you like. -— —-— —--- —For-Hi2-you -can-move-singular points. to-0-and-infinity. Then-equation - boils down toEuler equation with elementary solutions. (N= number ofsing — ~points-of-Fuchstan-emrationys — —_—-- —_—-: _______ ForN=3,youfirstmovepoints to0,1andinfinity, Thenyoucanintroduce 5parameters bylooking atthe singular points 0and 1.Thus, you fiddle a -——_______little_to_findthatp_and_qmusthavevery.specialform_in_terms ofthese_ 5constants. This leads to Riemanns DE. You can reparametrize interms 6 ——-the-six-exponents—{two at-each-regular-singulr point) andthen-the sum-of the six exponents =0togive only 5parameters, lvery Riemann DEcan be - “Reduced toahypergeoiistric form. ifthe6exponents takeoncertain values” ~~ --interms of3parameters, yougetexactly thehypergeomptirc with nochanges. _ ——-—_11._Apply.to hypergeometric equation: ateach_regular_point youcomputa the.roots by looking atvarious indicial equations: 2-0 vie0 _veel-o aiff=loa2 vése=a=B APE4cma~d ~ . int vlan vaeb_._ diff =aeb — - -- Glearly, theusualsolution F(a,b,0,2) corresponds toav=0.KEKKIXAXIHER BAXEENCSHEARIACIEREEBAKEREAEAL Te, this solution isthe 2-0 vle0 case. .Bateman-calls. this-ul.Its partner-is u5.So,2=0gives-ul-and u5.Then _ @ z=1 roots gives Batemans u2and u6. Finally, inf =2gives u3and ud. -— > Each pair’formss-viablé vannonical ‘basis ofsimple branch poirt-furctions. 7 ow jY Qoluyaie|Barada (=e _ .. @ormdm 5LL uo pS ae a ata a2 OtReagg |te =eee ~~ ®odePESeHtotnalangleyee . -0. SB - Cw es VO UO 57 @eyus @REO Q@lNeo TT 2 2RC)ateOt©ARERSIEY =LAT-eud) ,- _— en oo-=MeRO=6.QSSSeS eG_. Se a, Cn, --a a © »ObWedlu: SaQexgee TTT Se RLAGA AN 0 CBE GIT Da NEANECEA'T Seeae] ©Yen weyeoMS CL -Obses), LETE OTageMe y Bee POeOR“Be.-|.GeeCELE). =480).-- Bays NEote ot SE=a=s)-- . 2S= \-% bays AGA)+Ks)Raewo cond -9)0)GaTEtea] & Pesased5eHore-Be{peso26Joie thy sh 2 . a|code +saxhayg -$(pd-pu0-). - OK,$50,GPPE=\oeeye - @gerck ge07 Bo ) 2st as “TE cy5Ges) Sy- Ew LLOQ +Qu=0. Ce -WsGe”. .. ag sy WET..-QG)=AGHEE hee ~HA=- 3" Le - oo, -“-r- a avery. Sede le wt aS. Coe a /7 . ODE, e .ro ; OALat)=Sc-4), GanC=Yaeatdat re,CYu= Dactyae ak.Oye: orJRnweolohhunanaopuao,end.TetaSpecie Lowait Sdumile nosed wih vnvading,Qua! 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GaadrteakTidak dakodMieaWeRUSbool, \ToScakeGapanadanmennebehiond “A ‘ Complex Variables 1. . - N96 Contents ofComplex Variables Section. . e1,Wadcontaining severalbasictheoremg 7 a~aycontinuing” off“a7segmént_ ~~"——~ ~~ ———- -b) uniqueness @fa.continuation _ le; - ¢)meaning of.analyticity inclosed domain and boundary off-continuation aaa)afidlytic continuation aswitoWor-fumetions --—-- —©)-good example_of_ganma.function —.. 42... 8 - - f)Schwartz Reflection and1(s,t) application: real analyticity .~~~) hoh-analyticity off*(z) ahd~—r(a*), andlyticity orf*(g*)in —"~ ——-—--h)—Scwarts in-two-complex-variables 9. _ aaa - 2,Wad ofmatérial onCarlsGii“Theorem aid Phragsamlindelor -~~~ ~--f~e)logic.ofCarlson:theorem forour_purposes———_.. --__-_____- b)general reading conclusions “—~"~~e/YnotesPromTitchmarsh 1939-—~ —---——9)sapo-Acns. we ~---—_——_--d)-notes from-Hille-1962 9_____ .neBAe SumeAnemae! -- Bie abies ofspacialturievions toldsateTeuts - eee Aa GT —- —- o--eeeee -_——-- eeee _____b) eightpagesaboutsquarerootsofgammafunctions - -— —4,-Comparison of-adding twodiscs.andcomputing total.disc viareal_analyticity j~ - -5,The-meaning ofPrinciple Value =~ - -—§Deformation. ofcutsinfrominfinitywhenpossible._ - @ - 2bBeemple of-contour-pinchings — foe . ~= §+-How-to shiftthe_contourin.delta function integral andhavesame.thing..—-~+~Q-How-to.change variables. in_acomplex integration;.also, -how.tocomplex conjugate-———~9.- Wadon-the-mapping..2= ch(v). Related. details. 22 a ~ =—--LL. Solution tothe famous.tcutproblem.. = lk +---+--1g. Theorem aboutwhen_you-can-drop greatcircles. ..._ --cee —-—--1}. heinverse mapping theorem = ee --- 4 ee poumdous dL. Orem, iusecasoton r(2) alu alana Ans oONeebumSaw poste arm $@ =f@+ L\(z-o) +VerLO\@-a) +... 7ae Doles 9 ul Shed Chat Gho oon motein iu mga flow camer Gre omarera. pxnlodorra ohconway 2 PACXLN Quiaongarsod da dnolfoA Qoanby umecanacktpomy° : . lo} 6 Qoyyasonsers ofcsrQonpinng char oMarparte? on ge 0 a2 fOe@s fa ee ame = (m) a = a)(2a) /(m-\) Q@Q= 3G@)(a-a\ /@-r)!i Ses nA = Y= Ce) orET 2)= »)Ce~®)Ju’ go.Onenord. ae ~=é FO -2- *, abana Mom if (Ce) aamore, va2 =B__wroauweg ia . Qk: ao va.Oromo A, uso, onterd f=kdoOhotickdei Quis. tedaar bry olnaosuning Gas . 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RG =Za anim,anlage 2 =f@=\2a.(exunt eaa2\/ee —ME sy TS @bioel=feN=an(2)=? q'(@)=(2a, (Benet 22")-=a (2VJ/* q = ae := ann Sal Ba RYURVON —~ .’,S ..9-g a Yeohvowrar Qos,carSig9 __ Jas) =asta isdes Sao \(2,w)=2 Onn(=wyFae asal So olan =a *: ” (huss —- a” Gn=e = = =(Zam @)(uy) “=WekhakDacoleuneQuecibanvamialoda Agel a} GeiaAsoka. adyke. _ \ a. * ' ‘The Logical Structure ofCarlson's Theorem. @|1.First,defineasetoffunctionsofasinglecomplexvariablesuchthatif Ifunction obeysacertain condition (theCarlson Condition). Anyfunction which obeys this condition iscalled aCarlson Function, Or, you might say:thisfunction is“Carlson.” \ 2.TheCarlson Condition isthis:f(z)mustbeanalytic intheclosed RightHalfPlane,andasyougooffonsomerayinthisRHP,(f(z)|mustbelessthan aconstant times aay 3.Examples: f(z) =ef?I2 isnotCarlson because ifyougodown thimaginary | axis, z=-i/z/ andf(z) goes asef//%/which violates theCondition. Thus, axisra2,sinh(PIz) isnotCarlson. Ontheotherhand, e®”for/a PL isCarlson. Any polynomial inXf 2isCarlson. More special examples ere [discussed below. ' »estion: Suppose youhaveacertain function defined obonthepositive H integers. Can’you define aunique analytic continuation off the ‘integers? HIngeneral, theanswer 1sNO.Forexample, ifyour functionwasf(z)=U, | £2(2) «0+sin(PIz). HOWEVER, ifyouhave afunction @efined ontheintegers which isCarlson, then fa SClassOfVEFISON TUNCTIONS thereWOES_ExXist aUnique aMALy ontinuation_off theintegers! Thiswillbeadirect. consequenceofCarlson's Theorem below. / on>Parison s—Tneoreny -¢c)-ts-Cartson-ani-vantshes~on-the-positive—integersy ben theon.Carlson continuation of z)isf1(z).= 0 onside: e_example J . continuation offtheintegers. (If‘nottrue,youcouldfindtwodifferent Carlson continuations. Thedifference ofthesetwowouldalsobeCarlson, ha —__—____—_—_—_and-would_vanish_on_theintegers+But_by- thetheorem, such_a Carlson,funon aly] a. .*Gonelusions about Carlson's Theorem andRelated Theorems. - e 1,First,I.willquoteCarlson'sTheremdirectlyandaccuratelyfromT:piss 5.81." If£(z)ts-reguar andoftheform0(e\ZL),where k<W-y-tor} -~~ —_ \-n€z) %0,anid£(z)-S-0 for-z-S-0F4;273 7-7.ytherf(z)=O-kdentically. |--—~ 2.ThereaveSeveret-quéstivns hers.First, theconclusionis reattyonly£(z)=0 identivally forRe(zyyObecause f(z)hasmo-definition elsewheres Raof‘course - thieonly avalyvic-continuation of-f(z)-= 0totheleft-half=pianeis f(z)=0 ~ NotethéeqiaLity partof-Re(z) yO~s~Thishav-the “important implicstion that “+ yor museiveTude the“imaginary rays. = ~ 37°THE Pealquestiin ‘is¥whatisthewemming ofO-(...). According-to thefirst’page - "QEHASBOOK,CHEEmistwieanas72/9, /£(z}/KA o8/2/weresisan ~agpgoluts positive Constant’. Thisequality” should holdfor-/2/- larger~tntin ~soilsVelie. —SeeRS TasPly“Wiadguous:-Here tsaralmostgoodproofwhithshows -->@WGlterdstive waytoWRItEtinisveondtbiom = ne - e SimQMFL 2ka Al gkLN LDR.\A\9@ 12h 21 - we eee ee AL ee =e Dai che ee WE CO ‘heTAIS Btatenentof-the-theorem gibes-with-a-reduction-ef-one-of Hilles: theorems~ “With I-have done—in-somenotes-there;-exceptthere-you need-Re(z)7 ~tk. I -—aéveptT's-wtutement of-the theorem, ———- —- -- _—- ~- 5.HOWEVER; —“norreof-these guysHilleorTitch. gives-a proof-I likes Tietch is simply wrong, Ithinks “~And-Hilte-does-something more-generai-thatr Iwants-Z: - teally-should “look-for-another- book to-reference-on this. - oe a- _ ©MWowgeod 56Waleed Cdl - -First aquick reminder ofthemaximummodulus theorem—if-f(z) s-enalytic’ inside some_finite closed contour, thenitisofcourse bounded onthedomain_D_ which.includes the boundary. Thebig fact isthat the maximum of/f(z)/ actually must-"—-oveur"somewhere omthe-boundary curveitselfs —————————--—— ~ So,foropeners Titchinsection”5.6 generalizes justsiightly themax-mod theorem -—~by extending-it to-a case -where-f(z) fails to-be-andlytic atone-point onthe-— -boundary. .Not.toosignificant? Ah,butwhenthat.one pointbecomes infinity, this-—~Sheoren becomesverybaste!|Betterwork3rome —F7 C7 Rs po ne__(.%)AOE omtuticlylmee Lk _--eefake w—-BOLUS WaSiexseftodPh 2LLLL -- eee BerAwe)anebbic Ho,wee wil\u2l|S|caseC- -abn: WR.comaygoaas Fascheng,satsWeek.~Jpmandabaypntouaut,rn ar Se ee -- e_.|wetde)| Sm—-yrony, €20.,sherwellenwadclaraloyfesscamstoP -=LoarWOTSMchatouswoeSo ir PNT PRSOkcancemiiiier entaonteeelets weienr itone pak FR)=_V@ AG)weonokyhe msdeCSe,asin-EES, ——\B®\<Han antarsusdia TeaJE(SlGh dro. -- eee een er aey -n\h@) SAMWOlT ikWG) eal rn sow doo - . -Gerd ero, Bel Cn. _ +"AeeetrasASSepnegiatesayEe)isboundedintheighb ,P< - er CO7ESIMEITAINSRFtderste Poearr ‘ Wellnowforthefuture theorems-he sayshe-wont; ude-thié ‘fesuit?--- ° --~ “2Booms ©Gumi ©Keone imwidyed(Eel).——{) HELE owManspeoDive wo,pitt SS)MSHere0)hrB64,writ Uo.aunghe e(*) =Se es18 :. Qua:WOLSM omatD. er a-_oewee -Po e pe: e . Feree Fe) dow 4 - ee neSD ~ + Loeb entry cn ope odre = AROy Be Te glee eggsseSoegoe'«Pe aDAVOS HOLS BR LZ TS Yothoamun 20. Man xees arta seo . _ sXe) =cosltd\ 70-eremmer Wel -.- -- -=WreorR. gol eb ..- ©SayeNadine, yyented \re<REVSm. Urea e hod,WaduneDokmamare (2).YarOanwe Cnn okSeOuran. okumn JQ) SH.Mun FRSA owdankchur=>\PRIEM wise-4 -- a ery. eee 7 - Ge, M@LERO™: oe -Theproof.givenhereissimply nagood,though I_assume.the theorem.is correct. Thetheoremusesheavily thefactthate> oO,andgotstheconclusion,—BENEmMSxolvelel] owWade. _ ‘NOmatter how-omall'@ 48;~this isnottdboundataltén/t(z)/-. -Youcennet setepg=Ovbecgul thenproofgossnotgpthrélehs”SUOP!|# -- Nevertheless, I-willquotethenextimportant theorem -before-I-goes-into details .withtheindiestor function eure fm hm . 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ACS RS00LD ©WieompsaewithAeVEey A a - aA S Aconverse ofCarlson's Theorem. -- - $<Aconverse ofCarlson's Theorem. @ s-First werestate thetheorem: -9=——-=0+=n>- - (a) f(z)-ds-analytic -inRe(z)%-O-. -—- -—— me - —- {b)-there exists- anR-such that,for-all-/2/> R-with arg(z)-in RHPrange, --— -ff(z)/ <-exp(k/s/}- where keQe -- = = soe ee (c)£(z) isCarlson. ~ es ~ . ~ ~ fieppem saysi (a).and(b)BHCY ——~ 3)ThatTstheconversé? ~~.not.(ce)Snob. (avand.by .Re ee ae =tyfsnobve)and/ore (.not.b) -=- -- Suppose: (a)isstidi-true. Then-vnobve =Pwnobsbys—————— —- ~The: ddea-is that: ifyou-have aset-of-conditions-ayby-+.. which imply x,then : -- =notex says-that- at-least- oneof-those~conditions-is-false:— If-yow assume-that - ~all-the conditions-except one are-truey-then snobsx-~imphies-that—that one-~ ~—- unassumed-condition bs-falsoes ©=== --——- 3.Soif(a)istrue;thenf(z)isnot-Carlson’ means-there existsnoamRsuch e - -that for /2/> R/£(2?/z> exp(k/z/) -where ketry -Theeefore no-matter-what-R -- you choose, this-must befalses—-—-—- -- Te " /tla)/\ explk/2/) where KOWForall/2J>R~” _ -Inotherwords,thismugt_befalse: eee eee5a wees nee TT Joly RB Sy [i/Zexpla/a/) itiEy _— Iamgetting confused. Here istheorem Xwant toprove: 2 _ hoxexexiata anxixonch khatxLoxkokwRx/8/Rxwxn(k/a/ dexSexmayhex: Lora anyBythere exists apoint 2/asuch thet /t(21)/ 7) EIforall kv, Iftherewerenosuchpoint,,then/¢(z)/wouldbelossthanflgoratt/2/7k forallk4, and_then fwouldbeCarlson.ThisisOK,forget thelogics FumedionsAVepoorartialFoun. :i_i— eweew we -—- lel=©dite: —AfS)ad\“angoNorm3hMEG <C_.faBlosoneR : JovenJoakfmsemenang aye 2 en--@Cun: ABTSEWiadRyShowTWeydomeWe=Sak,Onak.\A(2a)<A-SI JorA=omypositive oossted. _ 5=BoTAP,dioneHeke960forASLdense,[RSeseMA),|-- so eh} —-—_- ha —------eel ee ---Ciew\a\ce8MAS&Guiaeel)=eh ee. Tron SLse =etl CRY. Ack gen. enSe A aa ee lef} —— le-@Mhaowm: Stoveaneotsave,. wae Za <a oly | vw 7\.Teyouhometouiostase kgtile, ==7~_oogel :_ 7_—OO Wiel ©ota le ite BY ko INCeeSLYWNAR ©7peeashat” Duis)<aBalaWeal .~ -wudDalFl oleJel wee - “Se qa:ata Cele aed Gy yaSA BaDebyBELweRUSE TEDaCe)St Crag:ones, a.eaMae|:Qa£V,drocse\iste). .ern One “ey ©Vasa(orsionNOFA,8<CEBw Md\M\ce* -Nearly.WeaFin|NCcea Fe r) - ote LS wene ee . —2— ©dadpeddein: MEYA"Cadane” WO<SYfeangeBRBAEIB1>&,ber=GUSAcoda, Wenwnregraarge 2”Goatwistsnae ©.-Rawk MaskWLBW ceBasm kL |ra ~- lela =~@Qurwa aWelsOM alett eiJakor,2. QeCeothmentahing), $1)co.shahCadsoyt 928 oe ee ee ee —--- ~ -AMe\- gyre e-—Prodsiattrenas vasaJundBeMec ew wk,~——ShowwrsomApeed. Wage —Rsucka(ok-eek)=Saso -Dak Warralso,Que, BSRosnds Mok_it] co Mkeo,—oe SOR] Cada 2 -— -—-— --- @-.--- --- - eee Lene Wooo:RowFm)bedaknadJo.MeO\te©__Ossuane ©gaero SG BaslR|eenSasa, Gu aAGoanJal,Soulets spdeGL GT haar.” Guin,eormaangea. alsa wlgrwag) EAL. Fa(sR, lt@l\.cge*Flas | oo! 2 comi ATWEASOaeGSE laity 4RISE, wejowibs Saathnandaath assign pauses ROT <DeedkEReawgisabsaad, undo, WAYR,Maw.WISZIP<LRSM <WE, J, pm soycalerectnde QuikE2Dncomenger!(Ey, ©erage isdMVE Zien Sage Coanais: tee |= - ooOQarmuggesnal ayeMen|oom eeee —-——Aarooh Bowaane igorSaba ea.2.§te—--| - @andar: vartockDockTalermdendie Grab ._ MWhsedusite aosedsackueflabtornsodorci. .Gawa——~ —opeows obswhan yooble+Gadecnn oceans toeAna, Wgaging,Drata we ©Siem aneitiesan oo ET a Hae ee og ope Banogen ins ~ON —-—-- en aeLoot TBE aquikas -SS==eekcanbeGnas. Alon use, aeeeee OO ©Oss: 8@2\wa- ~-(nats) T(a-%) - OF ne ©Ornappends adevanPoerjaadle)_ 20_feik i oSan (en ACee ae NM. —~—— Mon24,werWorst we eee eee Ler 1 eee eee OER TEPERRYS TVife-aat) © =Dsadsy,agecod --eeAp -——Aesng teagan ~A Ain Be.aig-. wt - _. Vousciskshamed. joomuddWaachon: oo -._&®= | \*.|e (RG@retD ge f=a Mow sredtey dodNeanedrere,ekeqaastch yerWearben~Sa ess — ~oe _@Luggase aawh aabalkrash.Yoro. , fee fs.\aeat~. aa Oo -oe . @\eack QName hoe _. .8@y=AGG) |\nes Be . NCEENIESre) Parker) _ ae . . cece oan “ _— ~~ ee at , ew Ranw) Thecutstructure ofthefunction: Seay=|ete |=rs aie PGRwy1(2) DFzdonotnavethepatience todrawallthespecialcases1haveworkedout.~“Instead;—tet us consider ~just two casesy “In-botir these-cases,-we-shatl~ _assume that both mand nare integers. However, the result isprobably _ ~ just the same formandnbothhalf integers. "~~”*«yTnthecagem.GT.nwegettwokindsofpictures depending onwhether the ~———ifference-is-even or-offs—Hamely, —— “se ann ee ig OS - ; Askat chsoomath mt annety! \= - aeae Sanstheisot ics om=stoneO82 Sona 2gunnnetball INx ~ —aeoeoe—.. -4 ange ee ==mee -Tren DT Nin attSi f= =00)3Sgpg 0 agagpageASE ~- t \ qa -——___! . - : ant. oearea am nr) a 3.Suppose nislarger thanm.Thenthepicture looks justthesameexcept every—--——where_you see alabel_mraplaceitwithn,andviesversa.Ihave,workedout. anexample ofthis but you can itanyway bysimply inverting the inside of @tebescket;Mpinversiondoes-not-change-theposition ofthebranch points, The xmarks poles, the omark geroes. —————A.Phasing, Fromexperience withsimpler functions, imagine starting onthefar___ _ right where we choose the phase at zero. Then move left. Asyou cross over each —- - geroy-you-gain-90°, -so-I imagine that- with each pole you lose-90%,—Seelabels- onthe pictures. Itwould appear that ifyou start with +1onthe far right,-—s——you-also MaveF1-onthefarleft,unlike thesimpler fanotion t2@sI}e> >a ——These phase claims grebornout_if youjusttake thelarge2limitof- thefunction. You get simply 1, This isthe same inall directions. Thus, you ~——-—must chave plus phase-in-the-negative-direction.- ao _ - : ‘ = RE =\Gat -- 5. Comparewithearlierexample, f@\=~+IFH!wo=ECe\=Vetpro Herewepicked upa(-)phase going fromfarright = ¢) —---49farslefte Sutin-the present example thieisnot-so: --2=X(@) _ — OO ~~ “st ar rr Crse7—s aesDeeg] GS GSOE V24%)PCR-%) Goat Gay cvak - so Toe emPeete) ) sensory - wee LE |ew(Sel =|paste ieaiom) | ---- 6,Application: . ~ ee beeae)eh)Eaeia{Ses aS e [232 atgce-dtoh Gaus =29@) —4-1-suspect. you can-make thisargum\nt: if-this little-relation-holds forlarge-—a aigsidesareanalytic}in sasidefromouts,thenrelation.quat. hold ‘Orali2.Hence: fA-aBeene EGwdarge) -a 4 :Pex m1 Q-~) @ DR LNGew ; fateh ne _mae 5a;“ i i iOe ON ON po ow a 1agpet.omyw? _ BRB Ricrhaterpnl wr | Poot eee ee ttl oN - _. _ i} -oeee - . - wee ee / Ash.estad|", ee wee Mom) PGS VO TY asepena, Gowvamwees, en Jeolen, ott Yates aggain ASLgointomne gale. - WSjotargan se:(CLinodoggfoi.pikie--ee eet aad. (Aleem nunatin, giesAsmat LeIBat —Ge MaMeike a eeaY20esvee a 8 BGS POYAS) anceGUN) _ Orc SP ee ae) PED ereyend JL a a i ee CO Eee . - =; \ Sorry) AIGA) - ATGray PQ aserGyyp) seiCy) DO,HKtearySeoadbesahesasle, AGE.satoaespeeee, Py [Ags-ov"lsh poeDatei| @ 2 , Boradyind.sN@d=Tigareeae Ypaollad _Pm)Pex) e.-.— Ligon tes ©OMeWakemyoonk Late. BehedMele AOobertonManone — aa ai tancsdQ obenoka:ee eer.saucalso_fagihhewn——— wo.Okas=nslatioso shorohayaa4A RES.Exrnody «daansdOEE wo=me+ a - a@Raminsoakjakconto ne- onctaet TSB, aginAtaoko.Bue,ypegiDkmafeckewthe > © OTT eee FPog oeF@\rnes, Jt,kG\=+ |SGan are a eo annecayan)\)ooent CyR_@)ao. ©OYA Necomeneeenodose qponaas: (REE =CNKR) ©Diggeremro.j molabkan£0.Qs oo JT es : a ——-—GLa) 2avinngSagentJING), Sesmsrovabdae, Sonnerootamodyata -- a -— - e@®Vasrmeaudiesameway, -- a Austindynes nce)=[Gena Hoe 8. —0Bmanontona.~ 'Ly _ Yea)=ae tat a dy [SeiehPatel peg ae~~—---alnsanya xeonSalt,lookBksisgnansegthe pan ennnemen«Dya,OsGaaraMeLJunction oo A aepee ollaevita sat tae) 2 eo - ee re a beeoboPat ck a2. BGA ee—Stay. saadagdockCNcrendastaben soignATD.PoaSogs a waaNe SIEM. Qaaxomgia gharSAT =iVocette ~pont emercnaletn meant ee _ee ert -—Sea ee CT Bee OL ee 1~ 1. Te ee] icc oo_ -ee A -— OnMa_tulfe, Shsrodd Shoreghana EV=asd| —— a uy —-Siaugleduasenknuibiay —- a @oR Mie cinekhowt Gna Og yeeeee tig,aad=.eel(e")|=BissConey ——Onur seyG@aaia lanai’_ row _ ase =.2ibn ee_.Hee ~RG)_arsolt udmaeado._.Vieterse omotabse, -_ ee —w@ohBAeGee)” ula Proedile andRanRb_ —Orn -prichis ieee . - -‘ a LL EE Cee Ti tmkak segue eetel— wee ele oleh OhOY . jew —ieg [SET oe ols Ai.rinfre (eri(ors)Co-afeee diag, SQ)=EREisinCay+GaZisien(b-5)" “Ra Gate deo _CaagtEe-t ee Ler cenynat [SicaSSepS| _.- _ -—— SS ese RHP (SS fomwoequeeseme @dk$@)= G2). Dow io daet@l= joaoMO™|, -aw(evton) 6 : re . 5 watetBasSeay 77Beieles)onal Qua: das.Co-2) =~dismi) (g-ay" se ~ 6 e Themeaningofprinciple values “ 1.Areasonable definition ofprinciple value might bethis: choose your cut insuch away that the "principle value function" takes real,positive values somewhere onyour selected sheet. The notion ofprinciple value does not force.you todraw your cutinany special way. 2.,Examplé: .for f(z) =In(z), wecandraw cutinlots ofwajs, Here are two standafd ways: ee ae e & . ir \ mF rAsad>0hoe Or se Ld ls ater e ) - 3/. any other position ofthe-cut 4salso acceptible. The important fact is that you want your-"image"-region- to-include the real axis. Otherwise you are not 6nthe “principle branch". ‘heHere is‘nother example: f(z)~= 2°-Consider z-planes shown inA)andB) above. Again, the point-here istodefine the principle branch -sothat -when 2takes real positive values, the image isreal. For-a=1/3: OS :{ | ' Cy $@\= €2) |Becareful. Inthis case youmst ‘define principle branch bysaying: iwhenzliesonthenegative realaxis,théfunction isrealpositive.|Agaisi, youcandrawthecutasyoulike. a>Yj :_ | 4 |. . :;oo:4 '-~ ~- neva .1 hs? is e |2e= \ete@ z=\e\e 1 . a . i@e-w;2-e=i\2e ? -z=(elo ) |= |~ ‘ iesw) - -% =\AC ' ~OSM 1 : }. |e 1 e Nowebonsdein. lowtoenmad ringswp OComa ornuk $@\2 Jawany =Jam , kQuoe aangitacky ok226% Nowy, 3)=,soad! noes& nual, BS SCe\= ga)=Tan) -[ace 3Mie\= EVAR QuaBATA iepoleakKeosoHReeeignr petroov.Yow. ee > | DWaianaskoDrasckegoishsok=|wee,gorcae eer Rakyorcandaspen Quiook.LanaiGuk“poioo"de Qhessenadvo sodaWee et ooNdeCsiSe7S*wandsgooake.Vee,ho ak"peaane Deragh® 2aSikaKKwadpoeOem nnyHangousk. * roomit Le)=GH) er uso legs. as , onSuns v . \Sosa ee meatareloscomag onalagen5yonaknonsense |~ee . a-b Grdwecan, FR\>t soodbeoe r~ovwrensYodraw&a: .| :ON: . Ves arlo+>1A,pollag'samas trgctn Voneun cos oe | : | » | 4. ~~Gramadaobgonkour giashivig) ——- woeee Sa Mook SO=Sete and _Gistane DennOte Qe ooa: a oo a po Wekpuescde, Sa ee weea a - ee ee <r Ar enyseamngin Nuart dnnenninilay\echviegpecnanlCase. weaaievidagolreeseenclqpendee wo OeMenge BeEyath Arman rgd . 4aeiconmRta i —ak -» pe = x x CNN er onBeCE>RS(KE-BYCRSH) x r oe — - ~ VanSah.yantak 4x 3 - x3 .-® QrneaYdOka2yay oe _ So dy. =e: ; oe -Sayeyatxt) *“Saxe yo= — ex: +— - beens) en -Sire x -b) -- Caa\(4-b) - we — 4G Sen {ot > SohZZE-DH4) G~a\(a-4)CE ae --“*. ae _.tay y _. . - shaos Mra One Be ne ee - moe ©. tosummarizes- weconsider -an--contour—integral-which—is-a function ofsome- parameter -- which parameter controls theposition.of apoléin.the integrand, Vehaveother knownpolesintheintegrand. Wewanttomowifthereareanysingularities inthis ______ external parameter. Even ifweadjust the parameter sothatitspoleliesright == onthe integration contour, wedonot get asingularity atthis value ofthe parameter ~ because wecansimply deform thecontour. But, iftheparameter varies insucha - ~waythatitspolepinchies thecontour ontosomeotherpole,thenwemaycompute 6 ~the Singular part“or the Integral bycomputing theresiaue’ oftneintegrand ~at . ~the pinetrpoint. - The problem ofdouble-contour integrals and pinching. oo 1,This subject comesx uponpage3-14ofBDWreview. Idontreallythinkit_ ehas much todowith pinches. oe . ""2,First,Iclaimtherearecertain instances where,adoubleintegral doesnot have anymeaning. For example, what isthis integral: x oy - — --< fag —-—- --oe Vayaea -—-- - Sy Thissimply doesnotmeananything! Youcannot havetheendpoint “of‘theouter ~""~" integration depending onthedummy variable ofaninside integration. “This is “~~~ just. plain nonsense. 7 — oe "~~ 3,Now consider thisdouble contour integral, withthecontours asshown: wee ee eeeSa\a gsshyne --Qe ic -aa -Hy, _ rd. - . BY. - - 74. Ng weeeSe OLS Thesolidlinesshowtheintegration contours. Thedottedlinesshowthelocusof -thepole1/(x-y). Inone”plane,“the~locatdoir-of this’polédepends onwide’youare— - onthe other integration contour,___ wee ee -- - 4.Now, weare going toconsitier closing "tomone-wr‘the‘otlierOfBOTAOFtise~ .contours. Iclaimthatinthisxespect, thecontour £2,depends onthevariable y.The reason isthis: ifyou change yasyou would change itinorder toclose down theC;-contoury youvare~going tohaye todeformthe"C, comtour 0avoid tlie” ~~poleatx-ywhichisimpinging uponit.Youcannot_leave.Co..where itwas.Thus,weshould really writeCo(y). However, ontheotherhand, ifyqu_were. totrytaclase theLycontour, you wouldnot have todeform C, atali because the pole aty-x always Steys away [email protected] Thus;€;*isin-this senseindependent-of x. “Therefore, theabove double integration written inthe.order shown makes complete sense. : -- aeeeeeeNeves qe - ;x=DeyROSE Tey =1. e@ OSQoQ G9)Gs) A Tey FL. 5.The’Contoii's takén‘intheoffieFOFUEYsiliply HavenoWeaning, justasin2above. ~~A-pinch-ts-suggested, butreallyitdoesntearlanything, ~"~ | $ ______Datta Hmction: howtoabit}thesontouandstillmakesenses ____4, consiger-the standard-reprasentationfor-the deltafunction, Theusual ___-contour teas-shomne Be= te Ze|— oe22 EYESte ——)——-- -- _ 2.Now, howcan youdeform this contour without changing anything? Hereis2. exampleofwhatyou(canghfdot ‘i-ian rT aoebaWee Fo =ysSE ——-- -- $0 eo oo ere ne_arelooking atthefinitelittlegraen_segnent_as wemoveitoff +othe right. Thus, weseek: . ee nn thn.— S77.Gna 7) rewery IP.____this-"function-"-of k-is-actualiy agenersiized-function-Like-the delta— — function, sowetreat_the limit_in thesenseofdistributions. Hereare some facts_I knowinthedistributional sense: reca}l that _. _ > abide: —— : ——— kde aaeBes a ee oN ee —~-sCeaska) =_9_-3, BREN SOT a Lg shest - ee eees =~pes:——sSeastess (oh5aSyee SRN ex | ig" yates eS eae v a ee= ERShe Wee Aba acl Se = TE HBALSYSvexKk 20” oo=“EAS VeK%X=OS__ Geronson cmanh - .— - onSEY te — Lin FEENOLO 2 ~ _-Sorece=\e) ———=FPyTS Doe eae eS ae ee --= : wOSK ~ x. = =Yon Se (Ce=) A%7s QuaMarie,roaddoa,kreudeila,Ja.eee -- _~eee ee eene ee DSK~ e\: St Q=O-.-. a — _-—SOXone ~—-——-80-tt-eams—that this-timit-onty- worksite himit-of the-psendofuncties ————~—--—distribution, ——— —_———-3;-Therefore;_we-arvabie toshowthatthe Tittle greensegment vanishes-"in ~~— a the-pseudodistribution- sense.I-dont~think there~is-any reatproblen-with =———=the"pseudo" heres weeeee “~~ 4,TOBvoid this possible contusion withthePseudo-visiniess, Lets~~ e.-+consider ‘instedd-the effect‘of“simply-ratsing-thw-wntine contourup— - tohigher level. Then-we-have two green segments and: ta =ete tatth— > jheoeikeoy ce -— \DO eSNOK =en(el S—ay S&Sye Be_ ae 7 a a te fit RYLL gz a-sieka-_—=(S2>\) oeere BE) aer---- Qa yer3r8te) Yeaut,$2.00L. -- sep kb 4 a eee ee 0 2.=tee CVS ROE S,TT 5.Therefore, thare isabsolutely.no_guestion butthat_you canreiseorlower—thecontour without changing anything. Basytoshowthisdirectly: Pee De sik i- pepe ee ee\eedy=S| awdCh) =BWC) @ Leto- =a .Ceviouis nddousecdvaiiolsth enfecrueonnyayelion —? ew Chom verde th.ya)xa xu\ 4aabe _ oe ax=a)ay. S00=SG) OOS LseS OG) teGea oo ee 23 eee LO sn . ug omar 4. -O==>_> = —<—. re oo rsQAte)SsSaysca.VA ne ae)Qu t=) azf@ —- a S|otte tethelGg| ~"@®Ree [aYEYOodee2Cu)lorQeomog: _ __ -Ts Baye —2— a Sa \ek an yeaSSako. 2Fen te) oo Gat 2=Bayi® soaE=Eate =AE 7 ——AWaser beens seebmiai ---=H. ~~veeayaR,SORTTee oe ae Soe 3) Ce eeeee ns Bee --Oe - ee | , icstructure of2=cha andg=chz. - ‘ @ 1.Comment: thisisoneofthemorecomplicated situations because yougeta “reflected cut" inyour briginal plane, which cutwanders through another cut. ~~" “Consider intwostagess — a — oe . a aeo =88@sImlexSeer\_ =daw we_eedent _2, First: -= wep -soo oe ee eae ee : is rn en 4 >a “"Here weshowthez-plene forthefunction w.Allpoirits onthis sheetofthez-plane —~ _ map into points which[ieoutsdie théunitcircle inthew-plané. “Anylifewhich ~~ - —“EFOSSSS ENEUMECircle“inthe‘w-plane;-must-crossthrough-the -cut~in ‘thez-planey — ~~~~~hs-en- examphe;-we-consider-this lines—-—--——..-- _ pone ete ——- — Lea _ oe oo eeepoeWwWLLLe-oe --Youhavetoworkonthedetailstoreallyseo©this,butthetoppictureshowsthe actual penetration ofthieunit circle. Aswetrytodraw theline straight offtothe left inthew-plane, wehave tomake adetour atwe-1inordertoknowwhathappens ‘inthe z-plane. ~~ TO 7 a ---~>3¥Ifwenowconsider-q-=-inw;-we~see that--there- will-be-a cutim-the-w-plene just -~~wherewe-havebeen-drawing-the-above_lines_justfor_fun. Thus,we-mayconsider. the—— —picture above_as_the_w-plane for. the function.q.>-ln(w). Thiscut_at_w=0to--inf is thenreflected intotheg-plane astheabove pictures show.So: w4 PTA : ae. -2- , oeNoticethatthechosensheetofthez-planemapsintotherighthalfq-planestrip._ hsSofarwehaveconsidered thefunction q=ch”4z. Thisisnotananalytic function of%because there are two cuts toworry about. fne ofthe cuts isobsious and 7 comesfromtheSQRT.Theotherismoresubtle andistheonewhosebothendsare” ~"~~“atSnfinity bitTonGifférent Sheets. Wenow wish toconsider thefunction” ~~~ ~ =~ afg) =chq.” ‘Thby function“ts-anelytie frthetintte-qplane, Nomatterhow=—— - - youwander-in-q, -you-wil-get something-senstbe~in 2.Here-are~-some-picturest- -=--fe +——— re en oo on eee —=>= aa —_- cee ah FL te a tr Ko FO g 1 ++ ------ eR +t. = =F --———-- i ee es — - weeee tyros a-o L Asqwandersthroughthe0toPIboxshown,2=chqwandersthroughtheentire @ «plone without obstruction. Thephase of2canbedefined, eg,as-PI¢ LE<1" “ >deaaapyparcnaaes .-.MosonWaomg2S -- Bromgg varsateded, deB=DV.Aaadyage Muaonog - - - - et -- QuewQen, 2@S2@ly| @) ~@'y (ow) yee oe Sa @ _(qosv">Csav"y7 a ®Wogan aedol“wes! sooarablogKeomeGaWo, Sigar. Aptiecrmn. enn ©),1)- a ©Quds:BN4"hoose.Soma,yp!+omyuy=rovevan,ees,Aepse.BmDendy, Sw woke THveYY.Tuswacansho ok = &Jou “rare: anne wee wene : ,y “oo ze )et CeeennokoSBAOR-RKanVaseameter wkGh!),selostayen QuashakByBawa comst. eseg BC On -— ee -Goadie Buphena someagaids doasemiship—_ i a -a- - ®Warskadohoa, 4Agana +hyprloolga: — , well -ps A —--.-f —, aZafe CS “he ayeos" i : : -@ - =-e\-f-> f- __~ a ee ee fe ae :SY, VU rn aoe“Bagariul Grecunmeespeak abdbaGinyedk. |” ©®NowwadsooralinOiSackwile. gids? aad. 2: Rd>o.: vido . . \.duv\. <7 itor —— . ar jotend) |<7 TO . ©Weds Yaweaow,.. -.\ealea\ =Js)=o. SO ewmogt Wow TVCRel =sola L. a - og(esSE) Sy" so) lage lorSWE >lagGeo\ov +detailsonomaaninetheneplabad boULVaha --- @obeys wg pateh._7a w . - - a ao Sigs — - . woke =og [fe oo - a TOTS . a a-- - :re aETgLaQeid .@Wofads -2he-ong(24) _a a-ee eee ~OrsoeQa). eH.--- ===Semen (8), e : “edeyeLe” oie.= sles 2BPa 2" 2 oe 2 te eka ete -®Prowegnee;wee oo eatenetn — ro a.“tyotOT Tee ehG-Se)= Aataich =sheSe ----oe Re, 2aehSor. ©NowJada: - ceeeeee eeeee ee BBR owss 2 |gS ettON eee SESE ae e wse,=221 She=.2 —-oes" 2 Mav pe Swe=at : ee.) je) _ «Orn asain woh.Coote imSd), - ee ‘e©Calder 2ecesex daisie). Waskdome, TeeSedesdeul LsSocost. Wnsa-plams once. Wao _ ete 8 © cesGeanisth =coselcosGell —e@ISKG ~ -+=sealde"-isiasiss” ______#1 =sense!dus" - «2 t+ LLL. 2 oe. Beceslshe OL - - - ——--— — a - - @Fasenn, waduntoatite ndaCWavicageass. Met onan -=eset keLe- . OY +eV ek 2 .one 2 OS Le wee = Ainenoneahcostae Baio).SlainSiWut sacnat,aoe anyperauat boneinteger Le ~ eeeee ee —-. eo: i eae en Be ERSwieoeensahiceegloneCakvlonehuat onanleasoa,deme OW Vetake, ween a cS cs 0 z= wse- dus%o = Smee RY LLL © Gie=+i) oeQaody soonest lala inMaem @NaoMk©=gor.nach+TeapawSess iewonrdanguorne +Seng-- RAKHrollLawad Asqyune + _- - wee©“)[focce spa 2Joteles conoush Qolescrashed - ZL. vgaDisssetuaLu- +45 aa Qs,EONSTECTS Jaye Ven Mrashedsars \| fo,- . odsugck \\eR Sco. - a * .Ness—yveneer aynAES \, _@ Caan :ae)2,)08. Sedssk AAay(edeo jh . Loe ean - sudaup. amas) =Or "Sas aleve VA, ekgbs\= =\1- oe Wes Nvccose =)MRsueSs\soQehkwas oxpesed. giesJiosealeeSangin ayeOkemamet, ®Gonsod sSon2emadr some FC). Ox 8-8, fodawona, prtorda2giey, SWOAM|QuonBWaoes - AySG)henwentegetz=i,Mono-oo WWmwasodrama 2). Novinsdan asay >-8pets A.SGass).leakSete Wak ;SMES _—__aus=Vises =Gore). : ©Agghasolvin: Sool ern Bqpen.os Sanwa.CadesauaadonyQ:3).,s0aeCOEHRAWHroan Gy om -- : a ne vy aaa i N- a - _ - a) -a- esOMe wmesn sor Ea). MaueBoe: _Ute oO a VITSIT 7 ___rn rs =Dedheck.\fayva-9hBustdoagenedhMe._Qewcgla, OaaeSoe,we she | a Saane=).as MN..Ww =ragecarYo.—| onigaeGAME Dagens. S@\2Nod)Yoweasanacho—.Lokes,OmeStal i" :a ---|-.- a @26nd" =oS Gwe. Ccraustaa YA-V,AG) Bere warywet “nnsandeat,OO oe | otPBaw Ge >otiSENS STDau AA ©©Gongles -ugg SOV=Bpod. Sead 2. - ae eats VPS pyle afang «GonMeV=TrynaQl’ (ote)Be , e Gays Sas Ieryeation Leo QpONEstorGr=BoC) —CoP Gaya La) Ne ESRe eeeeFe So =Sek SPY)Amutts$Q! ad ona,26,1296 \. Sodskiw soda wkerkdew2 —--- a ae @ooYuta WKlogyGlswe leg(oy-aNawDandy gosschte salen a2 He (8een GelsomOe FE Te Hao Sewm =.Veereason abanBadd, Or2)LeeinLoveeathandDuos_. -ongiit)<Q.80cuacoansh add artogah.-- eoeeSsenatninginoue. ol,04(2-a)i,owVURVUSA. 5. QualSomeua eae fey Se -298 ap.Rwamighs Ree a CROnSal fy aeIGee)|x} Q-) =e. fe LL we eeeee RT aSGI ACG Race). sit 2Que) Gaia eee[teGS, lo TT epTatars she] a . wee eee Re BR) CL LL C- aNOOOO. CtOKPEVOSUTOCSPOSOEU “TT [eee Gea ST _—_ Ce) =G@2) Gl .. eo eee eee @ - Sieico:Ft.(aw.rah——ana - a at —— positivewhen 2707VEYTyeae @o Qader AgMESL aRESYsalenBHTVary)SALSEO) | maa 7+ =~positive when2éeb—— 2=~--— ~dmgfre}-- OrsonATEL lean)OF~NetssMeeaayer. nn EY 7 ns ®Veradresaowe.now?-Apckionn. “singtheusualcutto-thenLeft. meunonie:ao SSH Ex Salle eeeaaenE nenaeee yee seawS&S. _bvame 0-<Jong.) |.6M. This- new function is-not strictly defined anywhere onthe-real~axis. Each -ofthe factors rulesout,onehalf.oftherealaxis. However, ifyouapproach therealaxisontheleftfrom either direction, you.easily get tiie same result, sothis function isuncut onthe -left. Moreover, itris-positive-and-reatr -—— >— sete ee ree: L - . a w= pee ow eR ee--+Joes ssaie Jve’= \\©ee Wo 7774.Therunes fsata—SSFUAINly TOOKtiie“saiies FOFComplex zyBOTH“are~~---unambiguous throughtheir definitions. You-must-keep_in_mind_that_the-standard.function— has its zcharacterized byarg(1-z) which must be-less than PIinmaghitude, whereas ~the""functionischaracterized byarg(z-1)whiehniustbelessthanPIinwagiitude. tdInfact, both these functions. agree. onthenegative realaxisin-signt_--.——_ ~~ ~ “Investigation “indetail Shows’that“theyar'éthesainéTusietTori~ ~a MSeet.-LXee =.Ai-? = lan StEdam rnyeot 5.Nowdefine theother ""funttion andseewhathappens: ©~ ee eee Ee ae Jen = 2_@ —usingcutcto=tbesright.meumonic (note_how_opposityetothe above meumonic for left going cut) _Again,thisfunction.is. defined tor_all_complex. xunambiguously. Locket positive =_real 2 : ;-eee ME} H----~- pe ——-—t at}eE-—_——. =i—aeee-——p_*#= asic. 2=\|.€ SAME __. - 7 :Notice the-minus-sign! -==-+-:—-- -= _ oe .. mn a. ee . Investigation indetailshowathat@-\ andgl@-\_arerelatedikesot - @ yore et . : YQ. —Bron dooqedoS a ek ©@oOSeprseyonbunDLondeapondindacageexpe.unonty.duatchinn ~—-—MakNgprcksista."nianiraleee. 2(setQOframe. -=Yoske\seWrare.cmouy lar , @SaksyapmgKbe omamamper eee TS 2TAeat Ryn SE SwTR L 7 a - -Dy,EEteNeaandduactbe. coeeeee 4 .eon NERETO joornAsalshade,Se-ag,BDVawaydown feMarshes "Bk keREE Gos, Jdeake.se . ae- ae SL pe een Senandeloy eA conto _e aaSiteGRIME et “ogMoats. ree sucw |Re] a : - - cee — -E-- ~~ PH---Gonun, dose88e=oL ~Se@. -ke -. -. .eco SMTR LL. : od .“ _--feel ay swe |LL. a rs 7Gina One8onealUasqadhe yg an - ea ee ~~ RA sir CR | wee 7RPP eo. Seon(xsi® ty Le. Qumk &SecsuuieS ome, 7SLIT. . -K»- mpQnCRQ+ix).—H--- Re) LL. a REI STE) SHEE a Scrceee>)neeee)Ln we eo |séxR rn+goattory|=SWRA CTeek TF =akitssmallest, _laa «a\aexh. . “ROL iee-i8)) .@DSaxLo _ Soe Ny eR Ce eee ne we ee Qe eee pee eeNe eee Seog ae JS RIS VACA SSK TEE - ae nS Se Lee eR aTOe baa ne LeLe meSraGta/9) asweLITT.a -— ee Se ne | . a ee pe-Se en ees -SOA MSS JLE od aQu BreVic.2S.cowon,Jyrand.Qwok.C=a=\30..aSs.DawovRagyO.2eeeeee -:a- S©.suppose theabove vertical.integral doescolifeSge... Thenitis-fairly clearthat ~~ — —the-upper half.vertical-integral running upthe.imaginary axiswillconverge; --—=since.we are.assuming our.integraid -hasexpo.deacy-in.all directions exceptreal. - ‘This inturn would imply. (assuming no.singulerities inupper half. plane-first --=quadrant} thatthe-horizontal_upper -contour-integral_going-off to.theright--- . amst beconverging. So what.weget isthat: .you can.drop great circle given expodecayinany.direction exceptreal_pravided thattherealintegral converges. Itneed only bepower convergent, ofcourse. This explains whyyouareonlyallowed toopenthecontour in.our.1ittle.example herewhenRe(p)S.0. _Actually, - what really seems.to count.is that.the integrand must approach zero asyou gooffinto -sheveeldirection, Then.thelittlecritical integration xegment willnotcuaseany trouble. ee. @roy omddy, omsison. Sree ke. - - a. Tt Rg ie - ere @j\~.@ Te,wateakakontudiklswot GC.MoSiM\>0, ~- =HORA OM.ee ee ee -- - see eee eee : cee Seen eet -OhyPrickm: S8.A= Grohe -A.M@>0, CawYeucom. “ode Daant. -| -- :‘ a hak &-- -R - Qa 2, QS x eS hay Kt Y.kesvete =eanwangna BhV8SaUae.Bansae oa,a.prookhak o. 5 akeGeew) - ==Sfae Senger dou.else, \Q\4AT: ~Roe : e-Ma.fodQosawnrrawkk NowRusa cocarming Qesaree. \- @a5 NYouaduiecnsereDWyearamvY=Se).- owed Roanweo_sqasde oe =VE)=g(@)Lda a(S)ancdylic. be i Be ne _©SKS=oun sndiesabnanotoe,Oran VOFOMon 8 =8OsMHS+GOMEe- ~Fa hel, SOR SNE32=\80-t9\ =(eto) oe oe“Ease Fo _--23). =2-5). —AO) e -- - a ~o™ dehWuggnSemana’. oplptityyeesamoraggathBlatahe . ge Sigus SOVSO.NanA@DsMys TOR -- @G-8) 2a8WO)a._ae)= [SsSes}. Neo SaaS —ald)nae agitVeda aRGe —6|Taga DLOaanataliccoonanSpaedERE), |-|Waren.trainavanar Sanction =9@)=£@).sites <=--f-cnodayhis. conanS\_onsogt, (9) elNoweoe_acongle. (8-81) —-Ldariwatin Say. Sa,AAMAA FO,Dreioe ~--|ttomda gawk - — get westion: Suppose someone. gives you.a real_function,defined, onsome porti function?” -~- —" “ ~ ~ eaip N®=OeaudYorn wee ee 2.|RF tyeLe oo, —o SK: © eee eee Hee -a RET SaSake - - ~Bt K-% L or Ifyou can ignore the great circle and the possible hardness of~the~endpoints ofthe Antegrval, then weseem tohave aformula tellingusthecandidate analytic function which corresponds toany h(x), namely: . . - Rae oY ane - - .. “nee |. . oo. 2.Examples h@=SK~)K x4-l are 4 al~« Ras’S ex =1Gd.Sxidx. 7art x card Taney @crust presstoR -)- -+e t. - =e) FCey ) Wake Mel | peve LL Se) "GY" xe +pled pace! - --- + . — ~ _ek « ~— — — -aoe aaSS.FGotive e) Le RyBawey: ae - Ay -Meee= ea ey)7 - ant Page = CA)Fayei-asSE)ex) ) WeeSf@-®)5 . pe oe . te . - atPacer =BEE) ROME BYR) 0 A A Ye - S@=ayGCayrteay Baaaare(>=) L. @oo. CapSoweetpee)! wet aH ' Woonadn FE(4-av8). 2wd atheiesoe e -- - a wbe-8., tose. bance =k! atsgeath eek. 2Le Ra sak aaab=egat - a - e-b =) base =o wbteea se 2LL ae \ eo eee - wee Fe ROTC ngPOLAT ASoy cee : - PAP CY TOYFQ). ~— se a FOr Gaysha Beare Gs) FO,aiate e). reg 0 . pam woe ee i meee Poesy=GAPES), “canaries rons) =aaa oe yee co SoPRGrasrays) =-teewy neds)FLECaa tg) +EKCs) e -- _ - -ATA = 43. .ave Se eee ag dQ= 05" AL |ae6) . - x.. =Bea =x a a . Brere, E70seoakSewnyoumnabte.. Quweg a - a a ~ A Mes HAS %)aut = Summarize this example: -- =e - ° « _— Gd=OGe%) x _ Coy Xe 4> FR)= te) tee EQanjates ZF)4Ee)tei “ . 7~Usher =tO FO,-43 455) - $0,this was anexample inwhich Istarted with arather arbitrary_discontinuity .and. \worked backwards toconstruct theanalytic function whosediscontinuity thiswap!_ At one_time Ithought thatthisexample wasimpossible. Ithought. therewasna.__eLe angiytic function whichhadthissimple discontinnity starting at_x\x,.1 But nowIhave foundsuchafunction, Notanelementary function, of.course. . "Furtéher exploration ofthisexample: whatdoestheanalytic function f(z)doat “the branch poimt? _ __ _ —————-— x weeeeeaeAFQSdoled Se)en(Aeo re . : foe tee Be ee a fe (asb=she9 ee - >——The-standerd-test-fails. —- == --¢—— -—— -Iconjecture-the-foHowings—as youapproaeh the-branch-point-of-f(z);-f(z)-diverges ~ -—-—- -in-some-way- log-or powers—f{z)}-ts-not-finites However —the-endcap-intepration “aroun —— -——the-brenchpeint- vanishes, Iguesss-Moreover;-as we-know;~the discontinuity -sutidenly ~~~~~ -—dreps-to-zero-frem-a-finite-value- as-you-pass~over~the ranch “point. - ~ ~ ae ‘— —dTf-£{2)}-had-e-finite- value -atthe branch~point;-then-F dontsee-how-the disc + eould drop suddenly to-zero: -That-would not bepossible.- ~~ ©--———-- —- ~~ ~—-—— -—- Example- of-g-branchpoitn- which-hasthese~propertiest --++s@ S@\aBer 2 WG)=eyAg . oo~—e ~bySB8@re=OsaMRAONaf!ee ——— -@\diA).=.Foci,causal |heope-we tly teseve, aa Some facts about branch points. a 1.First, considér the logarithmic exetiple. We‘hayer @ es oe ee RP eeRie ae dias82)=(WRAP), —GR-E) HZ, adap4,&. - - = oh YRI@= KRYao[Maric] =AWRBR Or | - wR % : wee oe (a) asyou approach the branch point from different directions, you get different complex ribibér's, eSpéially different imaginary’ part =angle ofapproach. At ~fhebranchpoint,f(z)is"logarithmically infinite’: — (b) the discontinuity across the cut isaconstnat. Atthe end ofthe cut, it suddenly dropsto2éro. wee (c) the circle integration around the branchpoint, depends onR,but. ascircle closes down onto the branchpoint, integral vanihses because the contour gets "shortér" faster than thé function diverges. ~ 2.Second, consider the simple power example: a i N CeCe - Saez asRe® »Sey= R| . aT AA mitre) tede . acegaye LST) =aisuoraR = wr ay nal. AWA inno . en : Saedah= RSgeROEZR“tllea‘co Rk ~~ ue ae. ee = awk = > . =BR sure rs Que: bare . -. Loar Bart bk Mado Ee6hao. Gee? a Q-1 ae aee - GaN a= 0 Load: la=~\ . (a)again, asyouapproach thebranch point; thedirection youchoose controls the phsgeofthef(z),sodirections donotallgivethesamething. . 'b)the cut disc ingeneral depends onR. Atthe branch point the-dise does: 1)vanishes smoothly atbranchpoint - 23,5) disc blows upatbranchpoint (c)the circle integration depends onR.Ascloses onbranchpoint you find: 1,2) circle “integration vanishes - - 3)cirlceintegration blowsup. a . (a) the actual f(z) blows upatbranchpoint incase 2,3 but not 1. eC 4) hach(t) 2 huaso ° ° ° \ . =)<(ea go o ~ _o a Gad-\ o -° “ 4. Due Dao onak fe) w 7 wesf -2- Coninenits: 1.Even ifthedise blows up,thecircle integration maystill vanish. (2.2) [email protected](z)blowsup,thecircleintegration maystillvanish. ~3«Thecircle integration certainly vanibhes iff(z)=.Qatbranchpoint. _ 4,Ifthe Aiscontiuity drops"from a-constant tozero atbranchpointy youshould =~ ~suspect alogarithmic branchpoint. See —--- Comments: Basically, there are not many kind ofbranchpoints you.can make with . elementary functions. You can make Iplne and higher chains ofsame, oryou can combine powers wit logs. 9 7 - : -Isuspect, however, .thetthereeremuchfancier branchpoints possible in.functions which are integral representations ofelementary functions, ie, special functions ‘and higher~functions. For example, F(a,bjcjz) hasabranch points atz=1 where sometimes the function isaconstant atthe branch point. None.of my.examples above could dothis. . Conments:. Inparticle physics; the scattering amplitude issupposed tohave thresholds Mhich are “soft” but the functions aresupposed also tereggeize, ;. - . . . ae a |Rulefordiscontinuityofaproductmanddividendoftwofunctionswithcuts. 4 4 Quaid: am)[O\-Le-ies] +bG[aGeriey~200] =farid) brie)—UnrlebO-e)P =Forby(xt)—FLYx-e}| +ACE —bate _ ; b Se ACL) =alkdao'itsak) Ab+b(aboeits ott)Aa e4x8§@)=aeyve). Oude“ At=ak)ab+b&)da Suagget cheredrone Deaalllany* oo Nev ofdb+LKOR =ale)[BOesied-boa) 4bE)[ocxsa(eeeel) =aGesie\bletie)— LGrie)afere) Se, Conclusion: Suppose you are given fact that f(z) =a(z)b(z) where aand bhave cuts inz.You want toknow the disc ofthe function f. Ifyou agree todraw one cut pulled above and the other cut pulled below, then the resultant formula is unambiguos, itdoes not matter which one you pull upand which one you pull down. |Omardinssbe=2Que: (Yeswhoboe) ‘ =a-+—se= TA e AenvUs)ira) =ameeie) a(x)e(¥-) g oF=a[Ede] ~Abaye)ee)ext) co) e(e-&) =te |serioAw—a)“| e@\ ee) Fanen) ontonze, ‘ = 2s La = =a abeboat BOY=aia~tay Revencres) 3 Tyee [Leorie)00=earl, iS) Se SO) \ \ .=—-_- extte)Do~ooae| =|4%eo)&(xt1)|t ° Ghee «utBraun(abnt\\s @