7.2 Puiseux Expansions

Transcription

1 7. Puiseux Expansions Given an algebraic funcion on an algebraic curve, we wish o compue is principle pars by locaing is poles and compuing series expansions here. Since he powers of y form a Cx basis for he curve s funcion field PROOF, our primary goal is o compue series expansions for y a arbirary poins on he curve. Wih such expansions in hand, i is sraighforward o consruc expansions for any algebraic funcion. A any poin where he discriminan is non-zero, a series expansion for y exiss as a power series in xα, which is a sraighforward applicaion of he Implici Funcion Theorem. Theorem 7.. Implici Funcion Theorem [Ru??] Theorem 9.8 is a real version of he heorem. [Gu05] Lecure 7 sars wih a complex version of he heorem. See hps://mah.sackexchange.com/quesions/ The wo-dimensional complex analyic version: Le f be an analyic mapping of an open se E C ino C, such ha fx, y = 0 and 0, hen an analyic funcion gx exiss such ha fx, gx = 0. df dy For ramificaion poins, we ll use a subsiuion of he form x = r α, where r is he ramificaion index, and consruc a power series, no in x α, bu in = x α /r, a Puiseux series. As a funcion of, boh x and y are analyic in an open neighborhood of = 0. x is analyic as a funcion of because of he I.F.T. applied o x = r α. y is analyic as a funcion of x because of he I.F.T. applied o he defining equaion of he curve. Composiion of analyic funcions are analyic, showing ha y is analyic as funcion of. These argumens hold in an open neighborhood of = 0. Coninuiy of he roos shows ha y is coninuous a = 0. Then use exisence of he Lauren series Silverman. and coninuiy of he roos HOW? o esablish analyiciy a he ramificaion poin. Singular poins will admi muliple Puiseux series, each one corresponding o a single cycle. The mos sraighforward way o compue Puiseux series is o use Newon polygons o deermine ramificaion, hen seup a rial series wih he correc ramificaion and subsiue i ino he curve s defining equaion. Le s assume ha we re expanding around he poin 0, 0, as his simplifies he analysis wih no loss of generaliy. Consider facoring he defining polynomial of he algebraic curve:

2 p n y n p n y n p 0 = y r y r y r n How migh we do his, if he polynomial is irreducible? We need o exend o a larger field where he polynomial s roos exis. The analysis above shows ha Puiseux series form a suiable exension. For each roo r i, define is order as he lowes power of ha appears in is Puiseux expansion, divided by is ramificaion index. Muliplying facors ogeher adds heir orders, so p 0 s order will be he sum of all of he r i s orders. Now le s consider increasing i by one. How does p 0 s order change? p is formed by adding ogeher all producs of n roos, so p s order will be lower han p 0 s order by he larges of r i s orders, unless here are muliple r i s wih he same order. In his case, cancellaion beween hese muliple erms could resul in p having a larger order han oherwise expeced. If here are j r i s wih he same larges order, increasing i by j will lower p i s order by j imes ha larges order. The Newon polygon is formed by ploing he orders of he p i coefficiens, wih i varying along he horizonal axis and he order ploed verically. The easies way o do his is o plo he powers of he monomials ha appear in he equaion, and consruc he polygon s lower convex hull. Thus, a segmen on he lower convex hull of he Newon polygon will correspond o as many soluions as he widh of he line segmen, each wih order equal o he change in heigh divided by he widh, i.e, he negaive slope of he line segmen. The denominaor of he slope will be he ramificaion index, and he numeraor of he slope will be he lowes exponen expeced in he expansion of y. Consider a Puiseux series corresponding o a single line segmen of he Newon polygon. Leing α be he x exponen and β be he y exponen, so he monomials in f have he form x α y β, hen he equaion of he line segmen is rα sβ = p, where r, s, and p are inegers and r and s are relaively prime. Making he subsiuion x = r and y = s u, we obain: fx, y = A αβ x α y β = A αβ rα sβ u β = p A αβ rαsβp u β }{{} g,u A leas wo of he rαsβp exponens will be zero hose monomials corresponding o he endpoins of he line segmen on he Newon polygon; all of he remaining exponens will be posiive. This means ha if we expand u in a power series in :

3 hen any power of u will have he form: u = u 0 u u u u β = U 0 u 0 U u 0, u U u 0, u, u U u 0, u, u, u and g, u will also have he form: g, u = G 0 u 0 G u 0, u G u 0, u, u G u 0, u, u, u In order for g, u = 0 a = 0, G 0 u 0 mus be zero, and since G 0 u 0 is a polynomial in u 0, his gives us a finie number of values for u 0 ha can solve our equaion. Now, by seing g, u = 0, can we obain u as a funcion of? The Implici Funcion Theorem saes ha we can, if δg δu o expand around. is no zero a he poin we wish δg δu 0, u 0 = δ δu G 0u 0 In shor, he roos of G 0 u 0 give us he saring values for our series expansion, and if he roos are simple, hen he Implici Funcion Theorem guaranees ha we ll have a unique series expansion for u as a funcion of. If any of he roos are no simple, hen we can repea his procedure for g, u. I can be shown [Bl47] 5 ha his procedure always erminaes.

4 puiseuxf,x,y,x0,y0,deg,[rafunc], adoped from implici_aylor in he Maxima disribuion, compues he Puiseux series of a funcion f in variables x and y, cenered around place x0, y0, o degree deg. An opional sevenh argumen rafunc specifies a raional funcion in x and y o be expanded; he defaul is y. rafunc may also be a differenial. Eiher x0 or y0, or boh, may be inf o reques expansions a infiniy. y0 may be false o reques expansions a all poins of he curve lying over x0. Limiaions. puiseux consrucs he Puiseux series coefficiens using Maxima s buil-in solve rouine, which is limied o solving polynomials using radicals. Therefore, puiseux can fail on algebraic curves of fifh degree and higher. Also, puiseux does no ye implemen he recursion case described above when δg is zero. δu puiseux_fracexp conrols wheher puiseux s resuls should be reurned as fracional powers in x, or using an auxilary variable. newon_polygon graphs he Newon polygon corresponding o an algebraic curve. Several auxilary funcions are also used; hese are inernal o he namespace puiseux: puiseux4 compues a Puiseux series corresponding o a single place on he algebraic curve; i is called muliple imes by puiseux if y0 is false. puiseux compues a Puiseux series corresponding o a single line segmen of he Newon polygon. puiseux compues a Puiseux series of y corresponding o a single line segmen of he Newon polygon. disribue_denominaorfrac, var, lim disribues he numeraor of a fracion over is denominaor, limiing he powers of var o be no higher han lim. recener_curvef,x,y,x0,y0 changes variables on an algebraic curve so ha a given poin x0,y0 is moved o 0,0. monomial_powersf,x,y,x0,y0 reurns a lis of he powers ha appear in he monomials of a recenered curve. lower_convex_poinslis akes a monomial_powers lis as inpu and reurns he subse corresponding o he lower convex hull of he Newon polygon. lower_convex_hulllis, x is used by newon_polygon o graph he lower convex hull.

5 %i4 in_namespacepuiseux$ %i5 recener_curvef,x,y,x0,y0 := block[xcoeff, newf:0], if x0 = inf hen f:subsx=/x, f, x0:0, if y0 = inf hen f:subsy=/y, f, y0:0, f : numrasimpf, f : expandsubs[y=yy0, x=xx0], f, for xpow: 0 hru hipowf, x do xcoeff : coefff, x, xpow, for ypow: 0 hru hipowxcoeff, y do newf : newf radcanrasimpcoeffxcoeff, y, ypow, newf $ %i /* why doesn' his produce cdos? */ aylorsinx,x,0,; %o x x %i7 monomial_powersf,x,y,x0,y0 := block [resul: [], xcoeff], f : recener_curvef,x,y,x0,y0, for xpow: 0 hru hipowf, x do xcoeff : coefff, x, xpow, if xcoeff # 0 hen for ypow: 0 hru hipowxcoeff, y do if coeffxcoeff, y, ypow # 0 hen resul: endcons[ypow,xpow], resul, resul $

6 %i8 /* lis is a lis of pairs, like [[0,], [0,], [,0 */ /* his funcion reurns he poins on he lower convex hull */ lower_convex_poinslis := block [resul, nx, slopes, lowes_slope], $ /* sep - sor he lis firs by increasing x values, */ /* hen by increasing y values */ lis: sorlis, lambda[u,v], u[] < v[] or u[] = v[] and u[] < v[], /* sep - discard all poins direcly above oher poins */ lis: lreducelambda[u,v], if lasu[] = v[] hen U else endconsv,u, lis, [lis[, /* sep - sar wih he poin on he y-axis, */ /* hen roae around he lower convex hull, */ /* adding poins as we go */ resul: [poplis], while lenghlis > 0 do nx: lasresul, slopes: maplambda[u], nx - u[] / nx - u[], lis, lowes_slope: sorslopes[], while lenghslopes > 0 and slopes[] > lowes_slope do popslopes, poplis, while lenghslopes > 0 and slopes[] = lowes_slope do popslopes, resul:endconspoplis, resul, resul %i9 /* his funcion graphs a line around he lower convex hull */ lower_convex_hullinlis, x := block[lis], lis: sorinlis, lambda[u,v], u[] < v[], while lenghlis >= and lis[][] < x do poplis, if lenghlis >= hen lis[][] x - lis[][] * lis[][] - lis[][] / lis[][] - lis[][] else %nan $

7 %i0 newon_polygonf, x, y, x0, y0 := block [newon_poins, convex_hull_poins, maxx, maxy, maxxy, filename], $ newon_poins : monomial_powersf,x,y,x0,y0, convex_hull_poins : lower_convex_poinsnewon_poins, maxx: lreducemax, mapfirs, convex_hull_poins, maxy: lreducemax, mapsecond, convex_hull_poins, maxxy: maxmaxx, maxy, filename: nex_pdf_filename, plod[[discree, newon_poins], 'lower_convex_hullconvex_hull_poins, l], [l,0,maxxy], [y,0,maxxy], [syle, poins, [lines, 5, [color, red, blue], [poin_ype, bulle], [legend, false], [xlabel, false], [xics, ], [yics, ], [ylabel, false], [pdf_file, filename], embed_laex_graphicfilename

8 %i puiseuxf,x,y,x0,y0,deg,r,s,p := block[n:maxdeg,s, xexpansion, yexpansion, res, allans, ans, eqns, deqn], if x0 # inf hen xexpansion: x0 ^r else xexpansion: /^r, yexpansion: y0 sum^i*a[i],i,s,n, res: subs[x=xexpansion, y=yexpansion], f/^p, res: expandnumradcanrasimpres, eqns: creae_liscoeffres,,i,i,0,n-s, /* eqns[] should be a polynomial in a[s] */ /* We can only handle is simple roos, so compue */ /* is derivaive o check ha */ deqn: diffeqns[], a[s], allans: sorradcansolveeqns[],a[s], /* Maxima's solve can' handle arbirary high powered */ /* polynomials. Check o see if i worked. */ if apply"", mulipliciies # hipoweqns[], a[s] hen error"nyi: solve didn' work in puiseux", /* If r >, hen we expec each soluion o be duplicaed */ /* r imes, differing only by r'h roos of uniy. These */ /* differen soluions all correspond o he same branch, */ /* so we'll remove all bu one of each. */ /* We exclude zero as a soluion, since zero corresponds */ /* o a higher degree soluion han he ones we're seeking. */ ans: [], for a in allans do if rhsa # 0 and no memberrasimpa^r, rasimpans^r hen if rasimpsubsa[s]=a, deqn = 0 hen error"nyi: muliple roo in puiseux" else ans: endconsa, ans, /* For each soluion o eqns[], solve he remaining eqns. */ /* IFT guaranees unique soluions o all eqns. */ ans: mapmakelis, ans, maplambda[oneans], block[eqns:subsoneans, eqns], for i: hru n-s do block[s:radcansolveeqns[i],a[si, if no lisps or lenghs # hen error"puiseux: inernal error", oneans: endconss[], oneans, eqns: subsoneans, eqns, subsoneans, yexpansion, ans $

9 %i /* puiseux_fracexp conrols wheher series are reurned wih * fracional exponens, or using an auxilary variable. */ puiseux_fracexp : false $ %i disribue_denominaorfrac, var, lim := block [resul: 0, n: numexpandfrac, d: denomexpandfrac], for i: lopown, var hru minhipown, var, lim hipowd, var do resul : resul radcanrecformrasimpcoeffn, var, i/d*var^i, resul $ %i4 puiseuxf,x,y,x0,y0,deg,r,s,p,b := block[xsubs, subs, differenial:, adjusmen:, maxpow, maxypow, rf:0, yserieslis, resul], if x0 # inf hen xsubs: x = x0 ^r, subs: ^r = x - x0 else xsubs: x = / ^r, subs: ^r = /x, /* deg is he highes -degree we wan in our final resul */ maxpow: deg, if no freeofdelx, b hen b: rasimpb/delx, if no freeofdelx, b hen error"puiseux: requesed differenial can' be normalized", differenial: del, if x0 # inf hen /* x= ^r -> dx = r ^r- d */ /* so we don' need as many erms o ge o maxpow */ maxpow : maxpow - r-, adjusmen : r * ^r- else /* x= ^-r -> dx = -r ^-r- d */ /* so we need more erms o ge o maxpow */ maxpow : maxpow r, adjusmen : -r * ^-r-, /* Expand each of he coefficiens of he requesed */ /* funcion as a Taylor or Lauren series in x, */ /* leaving y alone. We'll laer subsiue a Puiseux */ /* series for y. */ maxypow : 0, for i: 0 hru hipowb, y do block[icomponen], /* s is he lowes -degree expeced in he expansion of y */ /* if y0 is non-zero, hen use mins,0 insead of s */ /* herefore, maxpow-s*i is he highes -degree required in rf's expansion */ /* ceilingmaxpow-s*i/r is he highes x-degree required */ icomponen: rasimpsubsxsubs, aylorrasimpcoeffb, y, i, x, x0, ceilingmaxpow-mins,0*i/r * adjusmen, /* maxpowrf = minpowicomponen maxpowy * i */ if i # 0 hen maxypow: ceilingdeg - lopowicomponen,/i, rf: rf icomponen * y^i, /* expand y o he requesed degree */ yserieslis: puiseuxf,x,y,x0,y0,maxypow,r,s,p, maplambda[yseries], resul: rasimpsubsy=yseries, rf, resul: disribue_denominaorresul,, deg, if puiseux_fracexp hen subs=x-x0^/r, resul * differenial else [resul * differenial, subs], yserieslis $

10 %i5 puiseux4f,x,y,x0,y0,deg,rf := block[resul: [], lasslope : 0, sign, mp, pairs], if y0 # inf hen sign: - else /* If we're expanding y a infiniy, don' recener y, */ /* bu use upward sloping segmens on he Newon polygon */, y0 : 0, sign: /* Make a lis of all line segmens on he Newon polygon */ mp: lower_convex_poinsmonomial_powersf,x,y,x0,y0, if sign = - and mp[] = [0,0] hen errorconca"puiseux: coordinaes do no solve equaion of curve", if sign = and lasmp[] = 0 hen errorconca"puiseux: coordinaes do no solve equaion of curve", pairs: makelis[mp[i], mp[i, i, lenghmp-, for pair in pairs do block[dela, g, r, s, p], dela: pair[]-pair[], if dela[] * sign > 0 hen g: gcddela[],dela[], r: dela[]/g, s: -dela[]/g, p: pair[][]*r pair[][]*s, if s/r # lasslope hen resul: appendresul, puiseuxf,x,y,x0,y0,deg,r,s,p,rf, lasslope : s/r, resul $

11 %i puiseuxf,x,y,x0,y0,deg,[rafunc] := block[resul: [], rf], /* If he caller requesed a specific raional funcion o */ /* be expaned, use modulo o normalize i. Defaul is y. */ if lenghrafunc > 0 hen rf: maxima modulorafunc[], f, y else rf: y, /* If y0 is false, compue he possible soluions for y0. */ if y0 = false hen block[ys, f0], if x0 # inf hen f0 : subsx=x0, f else f0 : subsx=0, numrasimpsubsx=/x, f, ys : uniqueradcanrecformsolvef0, y, if apply"", mulipliciies # hipowf0, y hen error"nyi: solve didn' work in puiseux", for yi in ys do resul: appendresul, puiseux4f,x,y,x0,rhsyi,deg,rf, /* If subsiuing x0 ino f reduced hipowy, hen */ /* some of our soluions are a infiniy. */ if hipowf,y # hipowf0,y hen resul: appendresul, puiseux4f,x,y,x0,inf,deg,rf else resul: puiseux4f,x,y,x0,y0,deg,rf, resul $ %i7 %i8 %i9 exporpuiseux, puiseux_fracexp, newon_polygon$ in_namespacemaxima$ imporpuiseux$

12 Example 7.4. Consruc Puiseux expansions of y a he muliple poins of he curve y = x We normalize he defining polynomial by wriing i as y x = 0. Where does i have muliple poins? We compue he discriminan: %i4 discriminany^ x^ -, y; %o4 4 x x The muliple poins of y = x lie a he roos of he discriminan, which are x = ±. In boh cases, y = 0 is he only soluion, so he curve has muliple poins a x, y = ±, 0. The parial derivaive of he defining polynomial wih respec o x is x, which is non-zero, so neiher of hese muliple poins are singular; we ll ge ramificaion insead. The analysis is almos he same in boh cases, so I ll jus do, 0. Firs, consrucion of he Newon polygon requires recasing he curve s polynomial ino a form cenered abou he poin being analyzed, i.e, y x x = 0. Nex, we consruc he Newon polygon by ploing he monomial powers, puing he y exponens on he horizonal axis and he x exponens on he verical: 0 0 The only segmen on he Newon polygon s lower convex hull has slope / and widh, elling us ha wo of our roos he widh of he segmen will require a single Puiseux series wih ramificaion index he denominaor of he slope: x =

13 We know ha y can be expressed as a power series in wih iniial exponen he numeraor of he slope: y = a a a Now, subsiuing hese expressions for x and y ino he curve s defining equaion y x = 0 and seing all coefficiens of o zero, we find: %i45 rasimpsubs [x=^, y=a[]*a[]*^a[]*^], y^x^-; %o45 a a a 5 a a a 4 a a a a = 0 a a = 0 a a a = 0 The firs equaion ells us ha a = ± i, he second equaion ells us ha a = 0 and he hird equaion ells us ha a = ± i, so 4 x = ; y = ± [ ] i 4 i I would seem ha we have wo differen series o chose from. This is no really he case, as hey differ by only a 80 roaion in he -plane, as can been seen by subsiuing =, which ransforms one of he y-series ino he oher, while leaving he x-series unchanged. Le s confirm his resul wih Maxima: %i4 puiseuxy^ x^ -, x, y,, 0, ; %o4 [[ i i, = x Now, le s analyze he poin a infiniy. We move infiniy o a finie poin 0 wih he subsiuion x = u, hen combine all of our erms over a common denominaor and discard he denominaor. Our curve becomes: y u u = 0

14 0 0 The Newon polygon s lower convex hull has a single line segmen, slope, lengh, elling us ha we ll have wo separae poles, each wih ramificaion index. Thus, u can be used direcly as a uniformizing variable, and we posulae an expansion for y in he form: y = a u a 0 a u a u a u Plugging his ino y u u and seing all he resuling coefficiens o zero, we conclude: %i48 rasimpsubs [y=a[-]*/ua[0]a[]*ua[]*u^a[]*u^], y^*u^-u^; %o48 a u 8 a a u 7 a a a u a 0 a a a u 5 a a a 0 a a u 4 a a a 0 a u a a a 0 u a a 0 u a a = ±i; a 0 = 0; a = i; a = 0; a = 8 i y = ±i u iu 8 iu This ime, here is no ramificaion, since u, and no a power of u, is x. We acually have wo disinc series ha will yield wo differen values of y for each value of u. Geomerically, we have wo shees ha approach each oher and ouch a a singular poin where he curve is no locally Euclidean.

15 %i49 puiseuxy^ x^ -, x, y, inf, inf,, y; Example 7.5. %o49 [[ i 8 i i, = ], [ i x 8 i i, = x [Bl47] 8 Compue expansions a all muliple poins of y x y x = 0 We begin by compuing he discriminan of he equaion, which gives us he locaions of he muliple poins. %i50 discriminany^ x^*y x, y; %o50 x 4 x 7 7 The muliple poins lie over he roos of his equaion: x = 0 and he seven roos of 4x 7 7 = 0. Infiniy also needs o be examined. We begin wih x = 0: %i5 puiseuxy^ x^*y x, x, y, 0, false, 0; %o5 [[ 8, = x This resul shows ha we have a single cycle a x, y = 0, 0 wih hree shees. Now, le s look a a specimen roo of 4x 7 7 = 0: %i5 g: -7/4^/7; %o %i5 puiseuxy^ x^*y x, x, y, g, false, ;

16 %o5 [[ 7 i , = x ], [ , = x 7 7 We have one shee of wo cycles a g, /g and an ordinary poin a g, /g. Finally, le s look a wha happens when x goes o infiniy: %i54 puiseuxy^ x^*y x, x, y, inf, false, 0; 4 7 %o54 [[ 9, = ] [ 4, x i, = x Here we have an ordinary poin a, 0 and a single cycle of wo shees a,. Example 7.. Find he principal pars of y on he curve y = x The principal par of an algebraic funcion is he par of is series expansion wih negaive exponens. Theorem 7. saes ha an algebraic funcion is compleely deermined, up o adding a consan, by is principal pars. The firs sep is o locae he funcion s poles, which in his case is simply he places where he denominaor is zero, and ha s jus x = ±. Now, if we use puiseux, we can jus reques a series runcaed a he erm: %i55 puiseuxy^ x^ -, x, y,, 0, -, /y; %o55 [[ i, = x Example 7.7. Find he principal pars of x y dx on he curve y = x Differenial forms are no funcions, and have differen series expansions, due o he presence of he differenial, which mus be adjused a ramificaion poins. Le s expand x y a x = : %i5 puiseuxy^ x^ -, x, y,, 0, 5, x/y;

17 %o5 [[ 7 i i i 5 i, = x Now x =, so dx = d. Thus, muliplying x by dx and changing our y variable o will muliply all of he erms in our expansion by : %i57 puiseuxy^ x^ -, x, y,, 0,, x/y*delx; %o57 [[ 7 i 5 i 4 9 i Even hough x has a pole a x =, x dx does no! y y Is behavior a infiniy also requires analysis. i d, = x %i58 puiseuxy^ x^ -, x, y, inf, inf,, x/y; %o58 [[ i, = ] [, i, = x x x has no poles a infiniy, and approaches he limiing values ±i as x and y approach infiniy. The differenial x dx, on he oher hand, requires us o muliply y y by dx, and since x =, dx = d. %i59 puiseuxy^ x^ -, x, y, inf, inf, -, x/y * delx; %o59 [[ i d, = ] [ i d, x, = x In shor, while x has poles only a ±, 0, x dx has poles only a infiniy. y y Example 7.8. [Bl47] 8. Compue expansions a all muliple poins of he exercises in To faciliae his example, le s define an auxiliary funcion o perform he analysis:

18 %i0 analyze_muliple_poinsf, [args] := block [discriminan: discriminanf, y, deg], displaydiscriminan, deg : if lenghargs > 0 hen args[] else, for r in solvediscriminan, x do disppuiseuxf, x, y, rhsr, false, deg, disppuiseuxf, x, y, inf, false, 5 $ y y x = 0 %i analyze_muliple_poinsy^ - *y *x$ discriminan = 08 x x [[ ] 9, = x, [ 8 4 ] ] 9, = x [ [ 8 4 ] [ 9, = x i, 9, = x [[ 5 5, = x y y x %i analyze_muliple_poinsy^ *y - x$ discriminan = 7 x 4

19 [[ i 8 i, = x ] [ i i, 4 ] ] 9 i, = x i [ [ i 4 ] [ 9 i, = x i, 8 i i, = x i [[ 5, = x y y x %i analyze_muliple_poinsy^ - *y^ - x$ discriminan = 7 x x 4 [[ 4 ] 9, = x 4, [ 8, = x 4 [[ 8 i ], = x, [ 4 9, = x [[ , = x y 4 4y x This curve produces coordinaes ha aren recener ed properly.

20 puiseux: coordinaes do no solve equaion of curve %i4 errcachanalyze_muliple_poinsy^4-4*y - x$ discriminan = 5 x x x 9 puiseux: coordinaes do no solve equaion of curve y 4 xy %i5 analyze_muliple_poinsy^4*-*x*y^$ discriminan = 409 x x [[ ] i i, = x, [ i i, = x [[ ] [, = x,, = x [[ 5 8, = ] [ 5, x 8, = x y y x %i analyze_muliple_poinsy^-*y^x^, $ discriminan = 7 x x x

21 i i 7 i, = x, i i, = x 75 i i 5 59 i i i i i i 5 i 5 4 7, = x i, 4 59 i i 4 5 i 7 8 i, = x i [[ ] [ 8, = x 4, , = x

22 i i 7 i, = x, i 4 5 i 7 8 i, = x 75 i i 5 97 i 59 5 i 7 i i i 7 i, = x, 9 i 4 5 i 7 8 i, = x i i 5 97 i i i [[ ] [ 8, = x, 5 i i, = x 4 97 i

23 [[ ] [ ] 8, = x, 8, = x, [ 9, = x [ [ 4, = ] [ i 4, x ] [ i i, = i 4, x i i, = x y y x x %i7 analyze_muliple_poinsy^-*y*x^*-x^$ discriminan = 08 x x x 4 x , = x, 4 7, = x 9

24 5 9, = x i , 4, = x 5 i i 4, = x , 4 7 i 9, = x 5 i i 4, = x , 4 7 i 9, = x

25 [ [8 ] [ 4 9 9, = x, 5 ] [ 4 9, 5, = x, = x [ [8 ] [ 4 9 9, = x, 5 ] [ 4 9, 5, = x, = x [[ 4 4 4, = x y y x x %i8 analyze_muliple_poinsy^-*y*x^*-x^$ discriminan = 08 x x x x x

26 i 4 i 4 i i i, = x, 9 5 i 5, = x 7 i 4 4 i i i i, = x,, = x 9 5 i 5 7

27 4, i, = x 0 7, = x 4 i 5, = x 5 4, = x, 0 7 i i 4 4 i, = x i 9 5 i 5, i i, = x 7

28 i 4 i 4, = x 9 5 i 5, i 7, = x [[ i, = x ] [ i i i, ] [ i i, = x, i i, = x [[ i, = x ] [ i i, i, = x ] [ i i, i i, = x

29 [ [, = x ] [, ] [, = x,, = x [ ] , = i 5, x 4 i 4 4 i i 5 4 i, = i 5 4 i 4 4, x 5 i i i, = x 4 xx y 4 4x x y 4/7x 4 %i9 analyze_muliple_poins*x*x-*y^4-4*x-*x-*y^ 4/7*x-^4$ discriminan = 77 x x 4 x x x i 5 i 8 8 9, = x i 5, 8 i 8 9, = x 9 i i 8, 8 7 i, = x 9 i i 8, i 9 8, = x 7 9

30 [ 5 4 i i ], = x, 5 i i i i 4, = x, 5888, = x [ [ 4 ], = x, i 4 [ 5 i 8 5 i, = x i i ] [, = x, 5 i ], [ ] ], = x [[ i i 4 i, 4 = x i i i i i i , i i i i , i i i i , i i

31 i 5 [[ 5 i 4 7 i i 7 i 0 4 i, = x i ], 5 i i i 5 [ 5 i 4 48 i y 5 x y /5 5 x x, = x This funcion produces complicaed fifh degree equaions ha Maxima can solve. NYI: solve didn work in puiseux %i70 errcachanalyze_muliple_poins y^5 x^-*y^4-4^4/5^5*x^*x^-$ discriminan = x 4 x x 4 x 4 x x x 4 x x NYI: solve didn work in puiseux y axy x This funcion conains an exra variable and is, in fac, a family of algebraic curves. %i7 analyze_muliple_poinsy^ - *a*x*y x^$ discriminan = 7 x x 4 a i aan 0, a π aan 0, a π a i sin cos i a 4 i 4 a, = x, 7 a i 4 i 4 a i a, = x

32 i aan 0, a π aan 0, a π a i sin cos i a 4 i 4 a, = x, 7 a i 4 i 4 a i a, = x [[ aan 0, a π aan 0, a π a i sin cos ] [ a, = x 4 a, 7 a 5 4 a, = x 4 a [[ ] [ a, = x, aan 0, a i sin aan 0, a a cos, = x [ [4 a 5 a4 a 9 a, = ] [ i a 4, x ] [ i a i a, = i a 4, x i a i a, = x y xy x %i7 analyze_muliple_poinsy^ - x*y - x^$

33 discriminan = x 7 x 4 [[ 5 i 9, = x 4 ], [ , = x 4 7 [ [, = x ] [,, = x [[ , = x y x y x %i7 analyze_muliple_poinsy^ - *x^*y *x$ discriminan = 08 x x x x [ [ i 4 4 ] [ 9 i, = x i, 7 9 i i, = x i [[ 7 i 9 i, = x ] i i, [ 4 4 ] ] 9 i, = x i

34 [[ 7 9 i ] [, = x, 4 4 9, = x [[, = x [[ 4 4 ] [ 7 9, = x, 9, = x [ [8 5 8, = x ] [, = x, ] [, , = x y xy x %i74 analyze_muliple_poinsy^ - *x*y *x^$ discriminan = 08 x x [[ 8 8 ] [, = x,, = x [[ ] [ 5 9, = x, 9 i, = x [[ , = x

35 y y x %i75 analyze_muliple_poinsy^ - *y x^$ discriminan = 7 x x i i, = x sin 7 π i, = x i i cos 7 π, i [ 87 i 9 ] 7 5 i i, = x i sin 7 π i cos 7 π i, [ 87 i 9 7, = x ] i

36 i, = x, 5 i 5 i 9 7, = x i i, = x sin π i, = x i i cos π, i [ 87 i 9 ] 7 5 i i sin π i cos π, = x i i, [ 87 i 9 7, = x ] i

37 [ i, = x 5 i 5 π i sin π cos ], 9 7, = x 87 i 9 7, = x i 5 i 5, [ i, = x i i ] 87 i 9 7, = x i 5 i 5, i, = x [ i i ] [[ 9 7 ] [, = x, 5, = x

38 87 i 9 7, = x i 5 i 5, i, = x [ i i ] 87 i 9 7, = x i 5 i 5, i, = x [ i i ] [[ 9 7 ] [, = x 5, i, = x [ [, = ] [ ] [ i,, = i,, = x x x