GENERAL SOLVINGTRANSCEDENTAL ANDPOLYNOMIALS EQUATIONS-WITHTHE GENERALIZEDTHEOREMOF LAGRANGE N.MantzakourasAbstract:Whileall theapproximate methods ment ioned orothersthatexist,givesome speciIic solutionsoIt hegeneralized transcendental equationsor evenpolynomial, cannot resolvethemcompletely. "Whatweask whenwesolveageneralized transcendentalequation or polyonomi al, is toIindthetotalnumberoIrootsand notseparate sets oI roots in somerandomor speciIiedt his time.Mainlybecause this,toomanycategories transcendentalequations havei nIinitenumber oIsolutions i n thecompl exwhole " TherearesomeparticularequationsorLogarithmic Iunctions Tri gonometric Iunctions which solveparticul ar problemsin Physics, andmostlyneed thegeneralizedsolution.This is nowthe theory G.RLE,todealwi th thehel p oISuper Simple geometric Iunctions, or interlocking withvery sat isIactory answerto all thiscomplex problem.PART I.TheBasic theorem G.RL.E(N.Mantzakouras. 2007) Types offunctions1. DefinitionWedeIine as typeoffunction oneoIthe5generalIormsoIIunctions thatare presented in themathematics thatis theIollowing ones: 1. ExponentialIunct ion.2. Logarithmi c Iunction.3. Tri gonometric Iunction. 4. PowerIunction.5.PowerexponentialIunction. 2. Definiti on1. Primarysimple transcendentalequation is called eachequationoItheIorm( ) ( )0 z q z t o = + = withe t Cthathas root s in C 2. Primarycompositetranscendental equation is calledeach equationoI theIorm( ) ( ) ( )o = ++ = 0 z q z n z t whichhasroots inthe totalentireCMoreover theIactors , 0 nt = which alsotakevalues Irom the set CandingeneraltheIunctions( ) ( ) , q z zareoIdiIIerenttype,with values abovein theC 3.Theoremof existenceandcount of roots of primarycompositeTranscendental equation Eachprimarycomplex transcendental equationoI the Iorm ( ) () () o = ++ =0 z qz n z t has as countoI roots the unionoIindividual Iields oIroots e1 2, L L Ci.e.: ( )==U21. Couni FooisC F L (1) oIthe equations that Iollow: 0 ) ( ) (1= += t : p m : o(o1)0 ) ( ) (2= + = t : q : o(o2)whichcome upwiththe primarycomposite transcendental equation:( ) ( ) ++ =0 q z n z tiIalso provided thatiI:1.The Iunctions ( ) ( ) , qz zare Iunctions oI diIIerent type,oroIadiIIerent Iorm,or oIa diIIerentpoweroI the same type ingeneral. 2.The Iactors , 0 n t = take valuesIromthe total entire C3.The Iields oI1 2, L L rootsoI (o1,o2) equations are solvedaccordingto the theoremoILagrange,andaIterthe primarysimple transcendental functions are solved per case,andbelonginthe total entire CTheCount oIIieldsoI theroots is 2,andconsequentlythe setoI the IieldsoI the roots oI is, 0 ) ( ) ( /\ } 0 ) ( } 0 ) ( 2:1: = ++ = E= = t : p m : q : C: ECL f i o o or==U21L L Case I Let( ) ( ) , qz z and() z and() z be Iunctions oI zanalytic on andinside acontourCsurroundinga point t andletnbe suchthatthe inequality() ( ) < nz z t is satisIiedat allpointszonthe perimeteroIC;letting ( ) q z , =andconsequently()1z q ,= ; thendoinginversionoI theIunctionI take ( )1 q ,= andthenaccordingtothe initial primarycomposite transcendental equation, I have: ( ) ( )++ =0 qz nz t(1)( ) ( ),= = qz nz t(2) IIwhere ( ) ( ) ( )1z q ,= (3) AIterwardsthe equation(2)becomes:( ), = n z tregarded as anequation in , hasonerootin thei nterioroI C; andIurther anyIunctionoI , analyticon theinsideCcanbe expanded as apowerseri es wi th theuseoIavariabl e similart o , by theIormula ( ) ( )( )( ) ( ){ }111!nnnnu tu tnnd u u un du, = (' = + (4)Andwecomeupwith the root( )( )( ) ( ) { }1,111!nnnn u tu tnndz u u un du = ('= + (5) that isalso theIi nalrelation Iorthesol ution oItherootoI theinitialprimarycomposite transcendentalequation( ) ( )++ = 0 qz nz thavi ng countoItheroots,1 zwherethe sum oIthe roots is identiIied withthecountoItheroots oItheprimary simple transcendentalequation ( )qz t = whichalsodetermines t heIield oItheroots oIthe equation( ) ( )= qz nz t thatit is al so1Land concerns onlythis Iorm,that is to say(o1). CaseIIAs previously,weexaminet hesecond case ( ) ( )= 1 tz qzn n(o2) which comeupiI wesolvetheinitialequation( ) ( )++ = 0 qz nz t as Ior ( )z Similarl y theIunctions( ) ( ), qz z and ( ) zarein eIIectwhich areregarded Iunctions oIanalyti c on and i nsideacontour Csurrounding a point t and let nbesuch that theinequalit y( )| |< |\ .1 tqz zn n Thenlet ting ()z , = Doingi nversion oIt he Iunction Itake( ) ( )1 , ,= andIromtheinitialprimarycompl extranscendentalequation( ) ( )++ = 0 q z nz t (1`)Itake( ) ( ),= = 1 tz q zn n(2`) Andi I I take( ) ( )( )1z q ,= (3`)then the equation( ), , = 1 tn n(more thanone iIthe q (z)Iunctionexponential,logarithmic, t rigonometri c,Power Iunction (n~ 1),PowerexponentialIunction)regardedasanequation in, hasone rooti n the int erior oIC;andIurtheranyIunction oI ,analytic on andi nside Ccan be expanded as a powerseries withu t n by the Iormula ( ) ( ) ( ) ( ) { }1111!nnnn u t nu t nndn u u un du, =| | |\ . ('= + (4`)Andwecome upwith the root ( ) ( ) ( ) { }1,2111!nnnnu t nu t nnd nz u u un du=| | |\ . ('= + (5`) that i s the relationIor the solution oIthe rootoIthe ini tialprimarycomposite transcendental equation( ) ( )++ = 0 q z nz t having sumoIroots,2 z sothe count is identiIiedwiththe countoIroots oI the primary simple transcendentalequation( )tzn= whi chdetermines also the Iield oI the roots oI the equation( ) ( )= 1 tz q zn nthat are2L andit also concerns only this Iorm,t hatis to say(o2).The totaloIroots,as simply comes up,will be the unionoItotals oIsol utions thati s represented by the Iields oI the roots1 2, L L i .e. ( )==U21. Couni FooisC F L andbecause the Iunct ions( )q z and ( )z arediIIerentbetween them,accordingly the totals oIsolutions 1 2, L L willbe diIIerentbetweenthem.Ingeneral,however, weacceptthe uni on oIIieldsoItheroots1 2, L L astheIinalIield oIroots oItheequation () ()++ =0 qz nz t which wealso symbolizeas1 2L L L = U1. Generalised Theorem ofexistenceand countofroots, of an accidental transcendentalequation Each accidentaltranscendental equation, oItheIorm ( ) ( )o==+ =10n z n z thasas countoIroots theunionoI i ndi vidualIieldsoItheroots

,i.e.:t heequationsthat Iol low: ( ) ( )= ++ =1 120n n z n z t (o1) ( ) ( )= = ++ = 2 21, 20n n z n z t (o2) ..................................................... ..................................................... ..................................................... ( ) ( )k kk= = ++ =11,0n n z n z t(ok)..................................................... ( ) ( )= ++ =110nn n n z n z t (on)whi chcomeup withthegeneralisedtranscendental equation:( ) ( ) o==+ =10n z n z tiIal so provided thatiI: 1. Functions () zareanalytical Iunctions at all thepoint in theinterioron totalentire Csimultaneously,theyareIunctions oIdiIIerenttype,or oI diIIerentIorm,oroIdiIIerentpower oIthesame typegenerally. 2. TheIactors, 0nt = takevaluesinthetotalentire Cbelong tothesequenceoI n(countoIn Iactors) withatleast2Iactors oI ntobediIIerentoI zero. 3. TheIields oI t herootsnL L L ,... ,2 1oItheno o o ,... ,2 1equations aresolvedaccordingtot hetheoremoI Lagrangeand bel ong in thetotalentire C 4.1heCount of fields of theroots is n, andoonsequently theset of thefields of the rootsof( ) ()k kkk= = ++ = =11,0, 1, 2,...,n n z n z t n is ==U1nL L Case1 ProofLet( ) z and ( ) z and( )z beIunct ions oI zanalyticonandinsideacont ourCsurrounding apoint t and let nbesuch t hat theinequality( ) ( )=< 1211n n z z t nnis satisIi edat allpointszon the perimeter oI C; letting() z , = then doing i nversion oIthe Iunction Itake ( ) ( )11 , ,= and Iromthe generalised transcendentalequation ( ) ( )o==+ =10n z n z t (1)I take ( ) ( ),== = 1 1211/n z n z t nn(2) IIwhere( ) ( )( ), ,== 112n n (3) andalso it is ineIIect( ) ( )11 , ,= then theequation (),== 1211/n n z t nn regardedas anequat ionin , whichhasoneroot in theinterioroIC;(morethanoneiIthe q(z) Iunction exponent ial, logarithmic,trigonometric,PowerIunction(n~ 1), Powerexponential Iunction)and Iurther any Iunction oI,analyticon and inside Ccanbeexpandedas apowerseries withtheuseoIavariable1u t n by theIormula ( ) ( ) ( ) ( ){ }111111!nnnnu t nu t nnnnd u u un du, =| | |\ . ('= + (4) where ( ) ( ) ( ) ( )== 112n u nn u (5)with < n nIorany> = s e 1, , , I n I Zor } ,..., , min2 1 n rm m m m= sothat = s or} ,... , min2 1 n rm m m m = so that k

, then thesolution is represented bythesameIorm(i). B)Forinterval R tet e . s s10butalsogeneralwhere et10 s sincasethatt is complexnumberand when 0 = k ,thenthe solutionsillustrated Iromtheequat ion:)) ( log () 1 () ((111, ,,,iiiiiki Gammam:'cc++ == (i) andin casethat 0 = kthen usingtheIorm) ( ) log( ) (2, , Exp : : : p = = =theLagrangeequationIrom(G.RL.E)transIormedto)) ( ) ( (111) 1 () (( ) ( , ,,,iExp Expiiii GammaimExpk:'cc=++ = (ii)but this speci Iic Iorm translat able to ) ) (1 1) 1 (1) 1 () (( ) ( , ,+

+ =++ =iExpiiii GammaimExpk: because weknow the nthderivative oI) ( ) ( x m Exp m x m Expn= .C)SpeciIicity Iorthe regi onR t e tee . s s1butmore generallye tes s1.Appears a smallanomalyinthe Iorm (i) andas regardsthe complex orrealvalue Ior k 0ini k t+ = t , 2 ) log( .The caseIorCompl ex roots wegetas a solutionoI the equationby the Iorm )) ( log () 1 () ((111, ,,,iiiiiki Gammam:'cc++ = =exceptiI 0 = k.Eventuallythe case k 0ispresentedandthe anomaly inthe approachoIthe inIinite sumin theIorm(ii) 1 1*1( )( ) ( ( 1) ( ))( 1)ii isim: Exp i ExpGammai, , +== + + +but1*/sm m e+=withs~1Because the replaywillbe s t imes and1 ,1> =s :s, wehave to repeat.A verygoodapproximationalsoin thi s specialcase is whenweuse the method approximat e oI NewtonaIterobtaining aninitialroot s:withs 1.2.MAXIMUMTHESURFACE AREA ANDVOLUME OFAHYPERSPHEREN DIM`SInmathematics, ann-sphereis a generalization oIthe surIace oIanordinarysphere toarbitrarydimension.Forany naturalnumbern,an n-sphere oIradiusris deIinedast he setoIpointsin(n 1)-dimensionalEuclideanspacewhichare atdistancer Iroma centralpoint, where the radius rmaybe anypositi verealnumber.It isann-dimensi onal maniIol d in Euclidean (n 1)-space. The n-hypersphere (oItensimply calledthe n-sphere)is a generalizationoIthe circle(call edbygeometersthe 2-sphere)and usualsphere(called by geometers t he 3-sphere) to dimensions n~4.The n-sphere isthereIore deIi ned(again, toa geometer;seebelow) as the setoI n-tuples oI point s ( nx x x ...,, 2 , 1 )suchthat 2 2 2221.... R x x xn= + + + (1)where R isthe radius oIthe hypersphere. Let nJdenote the content oIan n-hypersphere (in the geomet er's sense)oIradius Ri s givenby nR Sdr r S JnnnRn n

= =}10where nS i s thehyper-surIaceareaoIan n-sphereoIunit radius.A unithyperspheremustsatisIy nnn xnx x n rnn S dx e dx dx e dr r e Sn)) 2 / 1 ( ( ) 2 / (21) ( .... ....2 2 2121) .. . ( 10I = I = =} } } } } + + Andto theend ) 2 / ( / ) 2 ( ) 2 / ( / )) 2 / 1 ( ( 22 / 1 1n R n R Sn n n nnI = I I = t(1) ) 2 / 1 ( / ) (2 /n R Jn nn+ I = t(2)Butthegamma Iunctioncan bedeIinedby dr r e mm r 1 2022 ) (}= I Strangely enough,thehyper-surIaceareareaches amaximumand then decreases towards 0 as nincreases.Thepoi ntoImaxi mal hyper-surIaceareasatisIies0 ) 2 / ( / )| 2 / ( |ln ) 2 / ( / ) 2 (02 / 1 2 / 1= I = I = n n R n RdndSn n n n n t t t(3)Where ) ( ) (0x x + = is thedigammaIunct ion.Formaximumvolumethesamethey be calculated 0 )) 2 / 1 ( 2 /( )| 2 / 1 ( |ln ) 2 / 1 ( / ) (02 / 2 /= + I+ = + I = n n R n RdndJn n n n n t t t(4)FromFeng QiandBaiNi-Guoexisttheorem|arXiv:0902.2519v2 |math.CA|19 Jan2011| Forall ) , 0 ( e x ,xe x xxx1) ln( ) (1)21ln( + < < +the constant 56 . 0 =e . Taking advantageoItheprevioustheoremsolved in two levels ie. From(3) wehave2levels : t ln211)2121ln( = +xx(a)and tln211)21ln( = +xe x(b) Both cases, iIresolvedinaccordancewith thet heorem(G.RL.E )Iromby the Iorm.. ) )) 2 / 1 ( 22( 2 () 1 () (( ) 2 / 1 ( 2111,,, ,, cc++==eei Gammame :iiii with) log(t , but 1/ 1+=se m , withs~1 as beIorein1 case.TheinitialvalueIor (a)case is5.59464and Ior(b)caseis 5.48125.Weuse the method approxi mate oINewtonaIter obtaining aninitialroots:withs1 is 7.27218and 7.18109respectively,IinallyaIteraIewiterations.Thisshowsthat ultimat elyweas integerresult theinteger7,Ior maximumhyper-surIacearea. ThereaIterIor thecaseoImaximumvol ume,andbeIoreapplyi ng FromFengQiandBaiNi-Guo Forall) , 0 ( e x, ) log() 121(1)21121ln(( t =+ + +xxandtln1211) 121ln( =+ + +xe x.The resultsin both cases according toequation.. ) )2 ) 2 / 3 ( 22( 2 () 1 () (( ) 2 / 3 ( 2111,,, ,, + cc++= =eei Gammame :iiiiwith) log(t , but1/ 1+=se m wit hs~1 as beIorecase.In two cases endup in theinitialvalues3.59464 and 3.48125.Weusethe methodapproximate oINewtonarrivequickl y i n 5.27218 and5.18109respectively.ThereIorethe integerIorthemaximumvolumehyper-surIaceis the5. 3.THEKEPLER`S EQUATION Thekepler`s equation allows determine therelationoIthet imeangulardisplacementwithinan orbit.Kepler's equationisoI Iundamentalimportanceincelestialmechanics,butcannotbedirectlyinverted interms oI simpleIunctions inordertodeterminewheretheplanetwillbeat agiven time.Let Mbethemeananomaly(aparameterizationoI time) and Etheeccentric anomaly(a paramet erizationoIpolarangle) oI abodyorbiting on an ellipsewitheccentricitye, then. 3) ( ) (21 aT t SinE e E M SinE e E b a f = = =and = p h is angularmomentum, angular Area f =.Eventuallyt heequationoI interestis in Ii nalIormisSinE e E M = andcalculat e theE.TheKepler's equation|14|hasauniquesolution,butisasimpletranscendental equation andso cannotbeinverted andsolveddirectly IorE given anarbitraryM. However,many algori thms have been derivedIor solving theequation as aresult oIitsi mportanceincelestialmechanics.Inessent ially tryingtosolvethe generalequationt Sinx e x = wheree t, are arbitrary i n C moregenerally.According to thetheoryG.RL.Ewehave twobasic cases, = =: : p ) (1(a)and , = = ) ( ) (2: Si n : p(b) whi ch i I thesolveseparately,thetotalsettlementwill resultIromtheunion oIthe2 Iields oItheindividual solutions.TheIirstcasei s this isoIinterest in relationtotheequationKepl er,because1 < e .Fromtheory G.RL.Ewehavethesolution)) ( ' () 1 () ((111, ,,,iiiiiSini Gammae: cc++ ==)) ( () 1 () ((111,,,iiiiiSini Gammae=cc++ = (1) Ior t ,.SincetheexponentsarechangedIromanodd to evenweuset wogeneralexpressi ons Iorthent h derivatives.IIwehaveevenexponentis

and Iorodd exponent is These Iormulas helpgreatlyinIinding the general solutionoIequationKepler,because this is generalize the nth derivative oI) (,iSinassumoIthe two separate cases.SoIrom(1)we can seethe only solutionIor the equationKepler`s with the type (2) The secondcase solution oIthet Sinx e x =according to the theoryG.RL.Ewe canalso Iromthe , = = ) ( ) (2: Sin : p (b)that t k : ArcSin : 2 ) ( + = andalsot ) 1 2 ( ) ( + + = k : ArcSin : .SotheIull solution oIthe equationt Sinx e x = oI the secondIield oIrootsis .) ) 2 ) ( ( )' ( (() 1 () / 1 (( ) 2 ) ( (111iiiiikk ArcSin ArcSini Gammaek ArcSin : t , ,,t , + cc++ + ==(3)Or ) ) ) 1 2 ( ) ( ( )' ( (() 1 () / 1 (( ) ) 1 2 ( ) ( (111iiiiikk ArcSin ArcSini Gammaek ArcSin : t , ,,t , + + cc++ + + ==(4) 211) (,,='ArcSin(5)withe t / , andZ ke.Anexample isthe Jupiter,withdata8622 . 11 / 2 5 t = M with eccentricitye where 04844 . 0 = e , thenIromequation(2)we Iindthevalue oI(x orE) 2.6704 radians. 4.THENEUTRAL DIFFERENTIAL EQUATIONS(D.D.E)Inthispartsolve oItranscendental equations we introduceanother classoI equations dependingonpastand presentvalues butthatinvolve derivatives with delays aswell asIunctionitselI.Such equations historically have beenreIerredasneutraldiIIerentialdiIIerenceequations|15|. The modelnon homogeneous equation is = = =+ += cc +cc

ot1 1 1) ( ) ( ) ( ) ( ) (ii i rmrrrrnkkkkt f v t x w t x a t xxc t xxg(1)Withi r kw a c g , , , isconstants and0 =iw and ) (t fis a continuous Iunction onC.OIcourse any discussionoIspeciIic propertiesoI the characteristic equationwill be muchmore diIIicult sincethisequation transcendental,will be oIthe Iorm:0 ) ( ) ( ) ( ) (21110=++ = = = if vniinffe b e a a ht (2)Where0 ), ( ), ( > f b ai f are polynomials oI degree) ( o + s mand ) (0 ais a polynomial oI degreenalsomusto + s + m n n2 1. The equations (2)also resolved in accordance with the methodG.RL.E andthe generalsolutionis oIas the Iorm

+ =ftf sfe t p t f t x) ( ) ( ) (wherefare the roots oI the equation oI characteristic and fpare polynomials and alsof fs = ingenerally.Asan example we give the D.D.EdiIIerentialequation) ( ) ( ) ( ) ( ) ( t f v t x w t x a r t x C t x + += ' '(3)whichis like anequation) ( has oIcharacteristic 0 ) 1 ( ) ( = = v re w a e C h where 0 , 0 , 0 > > = v r C and w a, constants.5.SOLUTION OF THE EQUATION 0xx m x t + =The solution oI the equation is based mostlyon the solutionoIequationxx : =whichhas solution relyingon the solution oIxx e v=which solvedbeIore. SpeciIicallybecause we know the Iunction ) ( : Wkisproductlog Iunction Z k e,and usingittosolve the equationv e xx= is0 ), ( = = v v W :k.Also Z k e,all the solutions oI the equationxx : =is Ior0 = : .According to this assumptionwe can solve theequation0xx m x t + =withthe help oI the method G.RL.E. Accordingtothe theory G.R.E wehave twobasic cases, = =xx x p ) (1(a) and, = =x x p ) (2(b) whichiIthe solve separately,the totalsettlementwill result Iromthe unionoIthe 2Iields oIthe individualsolutions, e ,.The Iirstcase isoI interestin relationto the equationhas more optionsthan the second,because itcovers alargepartoIthe real andthe complexsolutions.This situationleads to the solutionIorxsuch that itis inthe Iorm |(2 ) | || ProductLog k i Logx et , +=or takingand the otherIorm ( ) ( )( ) ( ) ( )logx W logk,,= Fromtheory G.RL.Ewe have the solution Pr | , 2 log| || Pr | ,2 log| || ' Pr | ,2 log| ||(1( )(11 ( 1)( ) )voductLog h ik oductLog h ik v oductLog h ikkvmvv Gamma ix e e et , t , t ,,+ ++ c= + c= + with theZ k eAnd1 , 0 , 1 = h or more exactly Pr | ,2 log| || Pr | , 2 log| ||(1 Pr | ,2 log| ||)1( )( (11 ( 1)1) )voductLog h ik v oductLog h ikkoductLog h ikvmvv Gamma vx e et , t ,, t ,,++ + + c= +c= + with multiple roots inrelationtok and t ,. Variations presentedincase where, whenwe change the signoIm,t mainly inthe signoIthe complexroots.Even andinanomalyinthe approachoI the inIinite sum we use the transIormationbut 1*/sm m e+=withs~1,a verygoodapproximationalso inthis special case is when we use the methodapproximate oI NewtonaIterobtaininganinitial roots:.Thesecond group oI solutions represents real mainlyroots oI equationwhere , = = x x p ) (2 So we have1 11 11 1( 1/ ) ( 1/ )( ( ) ( ( )( 1) ( 1)v v v vv vv vv vm mxGamma v Gamma v, ,, , , , ,, , = = c c' = += ++ c + c withm t / , ,Ior C t m e ,ingenerally. 6.SOLUTION OF THEEQUATION 0q px m x t + =Anequationseems simple but needsanalysis primarilyonthe distinctionoIm,but alsothe powersspeciIic p,qasto whatlookevery time. Distinguish two maincases:R q p i e , )......The weight methodwould Iollow it takes m ,whichregulates the methodwe will Iollowanytime.Butaccording to the methodG.E.R we have twobasic cases( )1pp x x , = =(a) and ( )2qp x x , = =(b)whose solutiongives the individual acomprehensive solutionoI the equation.For the case under consideration ieq p m > > , 1 transIorms the originalint wo Iormats toassistusi n connectionwith the logic employed bythegeneralrel ationG.RL.E.TheIirsttransIormgiven IromtheIorm0 / ) / 1 ( 0 = = + m t x m x t x m xp q q pwhichisnow in thenormalIormtosolveequation.Firstweneed tosolvetherelationship , =px in C. Following thatwe can get theIorm ( ( ) 2 ) / Log k i qkx e, t +=,, 0, 1, 2, . .. | / 2| k Z k IntegerPart q e = and the count oIroots ismaximum2*IntegerPart|q/2| ingenerality. ThereIore so theIirst IormoIsolutionoItheequation is.. ( ( ) 2 ) / ( ( ) 2 )/( ( ) 2 )/(( )1( 1/ )(11 ( 1)( ) )vLog k i q p Log k i q vkLog k i qeqvmvv Gamma ix e e, t , t, t,,+ ++

c= +c= + Where ( ( ) 2 )/ ( ( ) 2 )/) ( ) / ( ) (Log k i q Log k i qq e e, t , t,,+ +=c,with multiple roots inrelation to kand, 0, 1, 2,... | / 2| k Z k IntegerPart q e = and / t m ,. ButIorthecompletesolutionoIthiscaseandIind theotherrootsoItheequation Iorthispurposeimake thetransIormation1x y=andwehave 0 0p q q px t m x t y m y + + = =andthenwetransIorm in1 0 / 1 / 0p q p p q pm y t y y t m y m += = .In this way weIind awhole otherrootswehaveleIt Iromall the roots.TheIormoIsolutionwillbeas aboveand assumingthethatq p g = wehave..( ( ) 2 )/ ( ( ) 2 )/( ( ) 2 )/(( )1( / )(11 ( 1)( ) )vLog k i g p Log k i g vkLog k i gegvt mvv Gamma iy e e, t , t, t,,+++

c= +c= + andk ky x / 1 = which rootsareinrelation to| 2 / | ,.... 2 , 1 , 0 , g t IntegerPar k Z k = ewith m / 1 ,.The secondcaserelated tom1 hasno procedureIordealingwith the method.Starting Iromtheoriginalequation was originally Ioundon thepandso theIirstt ransIormgiven IromtheIorm 0p qx m x t + =to solvethe relationshi p, =pxi n C,ashelpIultothegeneralequationG.RL.E.So wehave ( ( ) 2 ) / ( ( ) 2 )/( ( ) 2 )/(( )1( )(11 ( 1)( ) )vLog k i p q Log k i p vkLog k i pepvmvv Gamma ix e e, t , t, t,,+ + +

c= +c= + with | 2 / | ,.... 2 , 1 , 0 , p t IntegerPar k Z k = ewith t ,. To settle the issueoIIinding theroots, whererootsariseot herand with m1 then imakethe transIormation 1x y= and wehave 0 0p q q px m x t y m y t++ = =andthenwetransIormin 1/ / 0q q py t y m t+ =witht heprecaseq p > , 1 transIorms the original intwoIormats toassist us inconnectionwith thelogic employedby the general relation G.RL.E.The Iirst transIormgiven Iromthe Iorm 0 (1/ ) / 0p q q px m x t x m x t m + = =whichis now in the normalIorm tosolve equation.First we need to solve the relationship px , = inC.Followingthat wecangetthe Iorm| 2 / | ,.... 2 , 1 , 0 , q t IntegerPar k Z k = e andthe countoI roots is maximum 2*IntegerPart|q/2| ingenerality.The solutionis whenwe analyzethe poweras( ( ) 2 )/ ( ( ) 2 )/( ( ) 2 )/(( )1( 1 / )(11 ( 1)( ) )vLog k i q p Log k i q vkLog k i qeqvmvv Gamma vy e e, t , t, t,,+ + +

c= + c= + andk ky x = whlchrooLs are ln relaLlon Lo | 2 / | ,.... 2 , 1 , 0 , q t IntegerPar k Z k = e wlLh m t / , . 1he remalnlng cases areslmllarLoprevlous wlLh R q p e , . 1hesolechange ls ln relaLlon LoLhenumberofcases ls2 2(( ) / ) Integer a b a +for (+/-xaxes)anda bi: x y i w+= += for any C : w e , 7.2 FAMOUSEQUATIONSOFPHYSICSi)The difractionphenomena due to"capacity"oI the wavesbypass obstacles in theirway, sotobeobservedinregions oIspace behindthe barriers, whichcould be describedas geometric shadow areas . Inessencethe phenomena oIdiIIractionphenomena |17| is contribution, that is due tosuperpositionoIwaves oIthe same Irequencythat coexist at the same point inspace. IIvirusiswherethe intensity atadistancero Iromthe slotat00,ieoppositetotheslit.So Iinally we writetherelat ionshipin the Iorm Themaximumi ntensityappears to correspondt o theext remeIunction sinw/ w.DerivativeoIandequating tozero willtakethe trigonometricequationwtan w asolutionwhich provides thevalues oIwcorresponding tomaxi mumintensity. With theassi stoIa secondoItherelations Wecan then,Ior agivenproblemis know thewavenumberk (orwavelength)and widt h Dtheslit,to calculatetheaddressescorresponding to0 arethegreatest. Consider many tears as a crowdoI2N 1 parallelbetweenthecrackswidth D,thedist anceIromcentert o centerisa andwhich wehavenumbered Irom- N toN.Suchadevicecalledadiffract ion gratingslits.WeacceptthatsuIIicient ly metthecriterionIorFraunhoIerdiIIraction andIind theequationIor the volume. where ThereIringes addresses Iorwhichzero quantitysinu,and thereIoretheintensity oI whichis determinedby theIactor So wemustsolvetherelation w tan w. Whereku/D and ukw,mM.Tryi ng solvingthegeneral IormoItheequationwm`tanwwithm C e , consider2generalIorms oI solut ion, arising IromtheIormCos(w) and( ) ( ) 2 Cos w w ArcCos w k , t = = +, 0, 1, 2, ... k Z k e = so wehave.. 111( )( ( ) 2 ) ( (( ( )' ( | ( ) 2 )| / ( ( ) 2 )) )( 1)i iipiimw ArcCos k ArcCos Sin ArcCos k ArcCosGammai, t , , t , kt,=c= + ++ ++ c and theIorm 111( )( ( ) 2 ) ( (( ( )' ( | ( ) 2 )| / ( ( ) 2 )) )( 1)i iiqiimw ArcCos k ArcCos Sin ArcCos k ArcCosGammai, t , , t , kt,=c= + + + ++ cThen thegeneralsolut ionisq pw w U . ii) The spectraldensityoIblack bodyis given bythe equation accordingto the relationshipoIPlank.The correlated u () By / c v = T = which isextreme iIthe derivative zero.Thus we have the relationship Zeroing the derivativewill have the relationship And iI/ x hc kT = then we getthe equation Findingthe solution oIx we Iind the relationship By / 4.965 b hc k =Is called constant Bin,called displacement law.Then we need tocalculate thegeneral solution oIthe equationby the method G,RLE The Iirst groupoI solutions represents realmainlyroots oI equation where1( ) p x x , = =So we have1111 1( ) ( )( ( | |) ) ( ( )( 1) ( 1)v v vvvv vm mx Exp eGamma v Gammavv ,v, , , , v, = =c'= + = + + c + witht ,,Ior , mt C e. In this case t5 andm-5,we calculate the x4.9651142317442763037 the nearest 20ignored. Because apply relation ( )( )rxw r xwrrxw r xwre w exe w ex c=cc=c ThesecondgroupoIsolutions representscomplex roots oIequationwhere 2( ) log( ) 2xp x e x i , , kt = = = +ButthisdoesnotreIerto realsolutionsand notthephysicalEvolequations Iorthis andomitted.