Entire function - Wikipedia

In complex analysis, an entire function, also called an integral function, is a complex-valued function that is holomorphic on the whole complex plane. Entirefunction FromWikipedia,thefreeencyclopedia Jumptonavigation Jumptosearch Functionthatisholomorphiconthewholecomplexplane Incomplexanalysis,anentirefunction,alsocalledanintegralfunction,isacomplex-valuedfunctionthatisholomorphiconthewholecomplexplane.Typicalexamplesofentirefunctionsarepolynomialsandtheexponentialfunction,andanyfinitesums,productsandcompositionsofthese,suchasthetrigonometricfunctionssineandcosineandtheirhyperboliccounterpartssinhandcosh,aswellasderivativesandintegralsofentirefunctionssuchastheerrorfunction.Ifanentirefunctionf(z)hasarootatw,thenf(z)/(z−w),takingthelimitvalueatw,isanentirefunction.Ontheotherhand,thenaturallogarithm,thereciprocalfunction,andthesquarerootareallnotentirefunctions,norcantheybecontinuedanalyticallytoanentirefunction. Atranscendentalentirefunctionisanentirefunctionthatisnotapolynomial. Contents 1Properties 2Growth 3Orderandtype 3.1Examples 3.1.1Orderρ 3.1.2Order0 3.1.3Order1/4 3.1.4Order1/3 3.1.5Order1/2 3.1.6Order1 3.1.7Order3/2 3.1.8Order2 3.1.9Orderinfinity 4Genus 5Otherexamples 6Seealso 7Notes 8References Properties[edit] Everyentirefunctionf(z)canberepresentedasapowerseries f ( z ) = ∑ n = 0 ∞ a n z n {\displaystylef(z)=\sum_{n=0}^{\infty}a_{n}z^{n}} thatconvergeseverywhereinthecomplexplane,henceuniformlyoncompactsets.Theradiusofconvergenceisinfinite,whichimpliesthat lim n → ∞ | a n | 1 n = 0 {\displaystyle\lim_{n\to\infty}|a_{n}|^{\frac{1}{n}}=0} or lim n → ∞ ln ⁡ | a n | n = − ∞ . {\displaystyle\lim_{n\to\infty}{\frac{\ln|a_{n}|}{n}}=-\infty.} Anypowerseriessatisfyingthiscriterionwillrepresentanentirefunction. If(andonlyif)thecoefficientsofthepowerseriesareallrealthenthefunctionevidentlytakesrealvaluesforrealarguments,andthevalueofthefunctionatthecomplexconjugateofzwillbethecomplexconjugateofthevalueatz.Suchfunctionsaresometimescalledself-conjugate(theconjugatefunction, F ∗ ( z ) {\displaystyleF^{*}(z)} ,beinggivenby F ¯ ( z ¯ ) {\displaystyle{\bar{F}}({\bar{z}})} ).[1] Iftherealpartofanentirefunctionisknowninaneighborhoodofapointthenboththerealandimaginarypartsareknownforthewholecomplexplane,uptoanimaginaryconstant.Forinstance,iftherealpartisknowninaneighborhoodofzero,thenwecanfindthecoefficientsforn>0fromthefollowingderivativeswithrespecttoarealvariabler: Re ⁡ a n = 1 n ! d n d r n Re ⁡ f ( r ) at  r = 0 Im ⁡ a n = 1 n ! d n d r n Re ⁡ f ( r e − i π 2 n ) at  r = 0 {\displaystyle{\begin{aligned}\operatorname{Re}a_{n}&={\frac{1}{n!}}{\frac{d^{n}}{dr^{n}}}\operatorname{Re}f(r)&&{\text{at}}r=0\\\operatorname{Im}a_{n}&={\frac{1}{n!}}{\frac{d^{n}}{dr^{n}}}\operatorname{Re}f\left(re^{-{\frac{i\pi}{2n}}}\right)&&{\text{at}}r=0\end{aligned}}} (Likewise,iftheimaginarypartisknowninaneighborhoodthenthefunctionisdetermineduptoarealconstant.)Infact,iftherealpartisknownjustonanarcofacircle,thenthefunctionisdetermineduptoanimaginaryconstant.(Forinstance,iftherealpartisknownonpartoftheunitcircle,thenitisknownonthewholeunitcirclebyanalyticextension,andthenthecoefficientsoftheinfiniteseriesaredeterminedfromthecoefficientsoftheFourierseriesfortherealpartontheunitcircle.)Notehoweverthatanentirefunctionisnotdeterminedbyitsrealpartonallcurves.Inparticular,iftherealpartisgivenonanycurveinthecomplexplanewheretherealpartofsomeotherentirefunctioniszero,thenanymultipleofthatfunctioncanbeaddedtothefunctionwearetryingtodetermine.Forexample,ifthecurvewheretherealpartisknownistherealline,thenwecanadditimesanyself-conjugatefunction.Ifthecurveformsaloop,thenitisdeterminedbytherealpartofthefunctionontheloopsincetheonlyfunctionswhoserealpartiszeroonthecurvearethosethatareeverywhereequaltosomeimaginarynumber. TheWeierstrassfactorizationtheoremassertsthatanyentirefunctioncanberepresentedbyaproductinvolvingitszeroes(or"roots"). Theentirefunctionsonthecomplexplaneformanintegraldomain(infactaPrüferdomain).Theyalsoformacommutativeunitalassociativealgebraoverthecomplexnumbers. Liouville'stheoremstatesthatanyboundedentirefunctionmustbeconstant.Liouville'stheoremmaybeusedtoelegantlyprovethefundamentaltheoremofalgebra. AsaconsequenceofLiouville'stheorem,anyfunctionthatisentireonthewholeRiemannsphere(complexplaneandthepointatinfinity)isconstant.Thusanynon-constantentirefunctionmusthaveasingularityatthecomplexpointatinfinity,eitherapoleforapolynomialoranessentialsingularityforatranscendentalentirefunction.Specifically,bytheCasorati–Weierstrasstheorem,foranytranscendentalentirefunctionfandanycomplexwthereisasequence ( z m ) m ∈ N {\displaystyle(z_{m})_{m\in\mathbb{N}}} suchthat lim m → ∞ | z m | = ∞ , and lim m → ∞ f ( z m ) = w . {\displaystyle\lim_{m\to\infty}|z_{m}|=\infty,\qquad{\text{and}}\qquad\lim_{m\to\infty}f(z_{m})=w.} Picard'slittletheoremisamuchstrongerresult:anynon-constantentirefunctiontakesoneverycomplexnumberasvalue,possiblywithasingleexception.Whenanexceptionexists,itiscalledalacunaryvalueofthefunction.Thepossibilityofalacunaryvalueisillustratedbytheexponentialfunction,whichnevertakesonthevalue0.Onecantakeasuitablebranchofthelogarithmofanentirefunctionthatneverhits0,sothatthiswillalsobeanentirefunction(accordingtotheWeierstrassfactorizationtheorem).Thelogarithmhitseverycomplexnumberexceptpossiblyonenumber,whichimpliesthatthefirstfunctionwillhitanyvalueotherthan0aninfinitenumberoftimes.Similarly,anon-constant,entirefunctionthatdoesnothitaparticularvaluewillhiteveryothervalueaninfinitenumberoftimes. Liouville'stheoremisaspecialcaseofthefollowingstatement: Theorem — AssumeM,Rarepositiveconstantsandnisanon-negativeinteger.Anentirefunctionfsatisfyingtheinequality | f ( z ) | ≤ M | z | n {\displaystyle|f(z)|\leqM|z|^{n}} forallzwith | z | ≥ R , {\displaystyle|z|\geqR,} isnecessarilyapolynomial,ofdegreeatmostn.[2]Similarly,anentirefunctionfsatisfyingtheinequality M | z | n ≤ | f ( z ) | {\displaystyleM|z|^{n}\leq|f(z)|} forallzwith | z | ≥ R , {\displaystyle|z|\geqR,} isnecessarilyapolynomial,ofdegreeatleastn. Growth[edit] Entirefunctionsmaygrowasfastasanyincreasingfunction:foranyincreasingfunctiong:[0,∞)→[0,∞)thereexistsanentirefunctionfsuchthatf(x)>g(|x|)forallrealx.Suchafunctionfmaybeeasilyfoundoftheform: f ( z ) = c + ∑ k = 1 ∞ ( z k ) n k {\displaystylef(z)=c+\sum_{k=1}^{\infty}\left({\frac{z}{k}}\right)^{n_{k}}} foraconstantcandastrictlyincreasingsequenceofpositiveintegersnk.Anysuchsequencedefinesanentirefunctionf(z),andifthepowersarechosenappropriatelywemaysatisfytheinequalityf(x)>g(|x|)forallrealx.(Forinstance,itcertainlyholdsifonechoosesc :=g(2)and,foranyinteger k ≥ 1 {\displaystylek\geq1} onechoosesanevenexponent n k {\displaystylen_{k}} suchthat ( k + 1 k ) n k ≥ g ( k + 2 ) {\displaystyle\left({\frac{k+1}{k}}\right)^{n_{k}}\geqg(k+2)} ). Orderandtype[edit] Theorder(atinfinity)ofanentirefunction f ( z ) {\displaystylef(z)} isdefinedusingthelimitsuperioras: ρ = lim sup r → ∞ ln ⁡ ( ln ⁡ ‖ f ‖ ∞ , B r ) ln ⁡ r , {\displaystyle\rho=\limsup_{r\to\infty}{\frac{\ln\left(\ln\|f\|_{\infty,B_{r}}\right)}{\lnr}},} whereBristhediskofradiusrand ‖ f ‖ ∞ , B r {\displaystyle\|f\|_{\infty,B_{r}}} denotesthesupremumnormof f ( z ) {\displaystylef(z)} onBr.Theorderisanon-negativerealnumberorinfinity(exceptwhen f ( z ) = 0 {\displaystylef(z)=0} forallz).Inotherwords,theorderof f ( z ) {\displaystylef(z)} istheinfimumofallmsuchthat: f ( z ) = O ( exp ⁡ ( | z | m ) ) , as  z → ∞ . {\displaystylef(z)=O\left(\exp\left(|z|^{m}\right)\right),\quad{\text{as}}z\to\infty.} Theexampleof f ( z ) = exp ⁡ ( 2 z 2 ) {\displaystylef(z)=\exp(2z^{2})} showsthatthisdoesnotmeanf(z)=O(exp(|z|m))if f ( z ) {\displaystylef(z)} isoforderm. If 0 < ρ < ∞ , {\displaystyle0