\hESSKJr SSKJr SSKJrJr SSKJr SSK J r SSK J r SSK Jr SSKJrJrJr SS KJr SS KJr SS KJrJr SS KJr SS KJr SSKJr SSKJ r J!r! SSK"J#r# Sr$SSjr%Sr&Sr'Sr(SS.Sjr)g))Add) factor_terms) expand_log_mexpand)Pow)S)ordered)Dummy)LambertWexplog)root)roots)Polyfactor) separatevars)collect)powsimp)solve_invert)uniqc"URVs1sHo!UR;dMUiM nn[U5HMnSU- nX#;dMXC;dMUR5S[R LaUnUR U5 MO U$s snf)aprocess the generators of ``poly``, returning the set of generators that have ``symbol``. If there are two generators that are inverses of each other, prefer the one that has no denominator. Examples ======== >>> from sympy.solvers.bivariate import _filtered_gens >>> from sympy import Poly, exp >>> from sympy.abc import x >>> _filtered_gens(Poly(x + 1/x + exp(x)), x) {x, exp(x)} )gens free_symbolslistas_numer_denomrOneremove)polysymbolgrags O/var/www/auris/envauris/lib/python3.13/site-packages/sympy/solvers/bivariate.py_filtered_gensr%s{$yy =y!ann$s B B Ncj^URT5Vs/sHVo2(aKUR(aX#R;d+UR(aM<URU5(dMTUPMX nn[ U5S:XaUS$U(a [ [ [U55U4SjS9$gs snf)aReturns the term in lhs which contains the most of the func-type things e.g. log(log(x)) wins over log(x) if both terms appear. ``func`` can be a function (exp, log, etc...) or any other SymPy object, like Pow. If ``X`` is not ``None``, then the function returns the term composed with the most ``func`` having the specified variable. Examples ======== >>> from sympy.solvers.bivariate import _mostfunc >>> from sympy import exp >>> from sympy.abc import x, y >>> _mostfunc(exp(x) + exp(exp(x) + 2), exp) exp(exp(x) + 2) >>> _mostfunc(exp(x) + exp(exp(y) + 2), exp) exp(exp(y) + 2) >>> _mostfunc(exp(x) + exp(exp(y) + 2), exp, x) exp(x) >>> _mostfunc(x, exp, x) is None True >>> _mostfunc(exp(x) + exp(x*y), exp, x) exp(x) rrc&>URT5$N)count)xfuncs r$_mostfunc..Ps  )keyN)atoms is_Symbolrhaslenmaxrr )lhsr+Xtmpftermss ` r$ _mostfuncr9/s6!YYt_)_cQ --- KKGGAJ_F) 6{aay 4(.EFF )s;B0B0,B0cp[UR55nURU5up#UR(a(UR(a[ X15upEnX$-X%-U4$UR(dSnX#pdOUn[ U5RUSS9upFUR5(aU*nU*nXEU4$)aReturn ``a, b, X`` assuming ``arg`` can be written as ``a*X + b`` where ``X`` is a symbol-dependent factor and ``a`` and ``b`` are independent of ``symbol``. Examples ======== >>> from sympy.solvers.bivariate import _linab >>> from sympy.abc import x, y >>> from sympy import exp, S >>> _linab(S(2), x) (2, 0, 1) >>> _linab(2*x, x) (2, 0, x) >>> _linab(y + y*x + 2*x, x) (y + 2, y, x) >>> _linab(3 + 2*exp(x), x) (2, 3, exp(x)) rFas_Add)rexpandas_independentis_Mulis_Add_linabrcould_extract_minus_sign)argr!inddepabr*s r$rArATs( szz| $C!!&)HC zzcjj%auceQ :: 1 C //u/E!!## B B 7Nr.cn^^[[U55n[U[U5nU(d/$UR US5n[ U*[5(a[X- R X"R S5nUR Sn[ U[5(d/$U*R S*nX- nXR;a/$[X15upEn[X- U5nURU5nUbXR;a/$UR Sn [X5upn X:wa/$[S5m[U T- U5n SS/n/nX-X-- U- U - R5unnUR5unn[UU- 5n[S5n[!UU-U- U5R#5Vs/sH nXHU -- U-PM nnUH\nUHSn[%UU5nU(aUR&(dM)U *U - X- U--mUR)UU4SjU 55 MU M^ U$s snf)z Given an expression assumed to be in the form ``F(X, a..f) = a*log(b*X + c) + d*X + f = 0`` where X = g(x) and x = g^-1(X), return the Lambert solution, ``x = g^-1(-c/b + (a/d)*W(d/(a*b)*exp(c*d/a/b)*exp(-f/a)))``. rrhstc3F># UHoRTT5v M g7fr()subs).0xurIus r$ _lambert..s92wwq#s!)rrr9r rM isinstanceargsrrAras_coefficientr rr as_coeff_Mulr rkeysr is_realextend)eqr*mainlogotherdfX2logtermrFlogargrGcX1xusolnslambert_real_branchessolnumdenperKrTrCkwrIrPs @@r$_lambertrmys! *R. !BC#G  GGGQ E5&#j  w Q 8,,q/'3''I&q!! """ eHA"bj'*Gw'AyA' \\!_Ff HA" x  e ABFAG G CQS! A --/HC    FAs CG A c A$QTAXq1668 98!AsGAI8D 9&Aa A"Q$!#q.C JJ99 9 ' J :s8H2c* ^^^U4SjnURTSS9upEU*nTVs/sHNnUR[[4;d/UR(dM0TURR ;dMLUPMP nnU(d [ 5eUR(dUR(Gad[S 0TRD6mURU4SjU4Sj5nUR(aURT5(aURTS5n XY- n Xi- n U R(ddU (a]U R[R[R 5(d*[#[U 5[U 5- 5n U"U TT5$OkUR(aZU(aS[#[U5SS9n[U5nURT5(aUR(aXV- n U"U TT5$UR%TT05n['[)USS95n[5n [+X]- T5upUR%X05n/nU(Gd-[-U[T5nU(GaUR(a(US:wa"[/[U5[U5- T5nOUR(aURUS5nU(aUR(dUR1[25Vs/sHnTUR ;dMUPM sn(aUU(d[U5[UU- 5- nO[UU- 5[UU- 5- n[/[#U5T5nO[/XV- T5nU(d[-U[T5nU(a[5UU5nUR(a1US:wa+[/[#[U5[U5- 5T5nOUR(a|URUS5nUU- nUU- nUR75(aUR75(a US -nUS -n[U5[U5- n[/[#U5T5nU(d[-U[2T5nU(aTURR ;a[5UU5nUR(a1US:wa+[/[#[U5[U5- 5T5nOYUR(aHURUS5nUU- nUU- n[U5[U5- n[/[#U5T5nU(d[ S U-5e[9[;U55$s snfs snf) apReturn solution to ``f`` if it is a Lambert-type expression else raise NotImplementedError. For ``f(X, a..f) = a*log(b*X + c) + d*X - f = 0`` the solution for ``X`` is ``X = -c/b + (a/d)*W(d/(a*b)*exp(c*d/a/b)*exp(f/a))``. There are a variety of forms for `f(X, a..f)` as enumerated below: 1a1) if B**B = R for R not in [0, 1] (since those cases would already be solved before getting here) then log of both sides gives log(B) + log(log(B)) = log(log(R)) and X = log(B), a = 1, b = 1, c = 0, d = 1, f = log(log(R)) 1a2) if B*(b*log(B) + c)**a = R then log of both sides gives log(B) + a*log(b*log(B) + c) = log(R) and X = log(B), d=1, f=log(R) 1b) if a*log(b*B + c) + d*B = R and X = B, f = R 2a) if (b*B + c)*exp(d*B + g) = R then log of both sides gives log(b*B + c) + d*B + g = log(R) and X = B, a = 1, f = log(R) - g 2b) if g*exp(d*B + h) - b*B = c then the log form is log(g) + d*B + h - log(b*B + c) = 0 and X = B, a = -1, f = -h - log(g) 3) if d*p**(a*B + g) - b*B = c then the log form is log(d) + (a*B + g)*log(p) - log(b*B + c) = 0 and X = B, a = -1, d = a*log(p), f = -log(d) - g*log(p) c>SVs/sHo0RXU-05PM snupE[XBT5nXT:waUR[XRT55 [[ U55$s snf)aReturn the unique solutions of equations derived from ``expr`` by replacing ``t`` with ``+/- symbol``. Parameters ========== expr : Expr The expression which includes a dummy variable t to be replaced with +symbol and -symbol. symbol : Symbol The symbol for which a solution is being sought. Returns ======= List of unique solution of the two equations generated by replacing ``t`` with positive and negative ``symbol``. Notes ===== If ``expr = 2*log(t) + x/2` then solutions for ``2*log(x) + x/2 = 0`` and ``2*log(-x) + x/2 = 0`` are returned by this function. Though this may seem counter-intuitive, one must note that the ``expr`` being solved here has been derived from a different expression. For an expression like ``eq = x**2*g(x) = 1``, if we take the log of both sides we obtain ``log(x**2) + log(g(x)) = 0``. If x is positive then this simplifies to ``2*log(x) + log(g(x)) = 0``; the Lambert-solving routines will return solutions for this, but we must also consider the solutions for ``2*log(-x) + log(g(x))`` since those must also be a solution of ``eq`` which has the same value when the ``x`` in ``x**2`` is negated. If `g(x)` does not have even powers of symbol then we do not want to replace the ``x`` there with ``-x``. So the role of the ``t`` in the expression received by this function is to mark where ``+/-x`` should be inserted before obtaining the Lambert solutions. )rJr)xreplace_solve_lambertrYrr)exprrKr!sgnnlhsplhssolsrs r$_solve_even_degree_expr/_solve_lambert.._solve_even_degree_exprshV7>?6=sMM1&j/ *g? dD1 < KKtT: ; DJ?sA)Tr;c>UR=(a, URT:H=(a URR$r()is_Powbaser is_even)ir!s r$r, _solve_lambert..(s(?QVVv-?!%%--?r.c">TUR-$r()r )r}rKs r$r,r~*s155r.r)force)deeprJz:%s does not appear to have a solution in terms of LambertW)rK)r>r+r r rzrNotImplementedErrorr@r?r assumptions0replacer2rMrComplexInfinityNaNrrprrrr9rmr0rrrBrr )r^r!rrwnrhsr5rIr7lamcheckt_indept_term_rhsrZrr}solnr[r\diffmainexpmaintermmainpowrKs `` @r$rqrqspD4 l   5ID %C#BtHHc * &#''*>*> >tHB !## zzSZZZ  -,, -kk @  ::#''!**hhq!nG]F=D==TJJq00!%%88F c$i 78.r1f== ZZCSXT2Cc(CwwqzzcjjY.r1f==llAv;' &4( )C A SWf %FA **aX C D Cf- zzcQhC3s8 3V<!,',{{3'737'7!S%5%55'737"5zC ,<<"3;/#cEk2BB#Jt$4f=D$CIv6D Cf- #w'CzzcQh 3s8c#h+> ?H!,;Ek55770022NH2IC8}s3x/ 4 0&9 Cf- v!9!99#w'CzzcQh 3s8c#h+> ?H!,;Ek8}s3x/ 4 0&9 !# "##$% %  EB@37s/V V 2V ?VVTfirstc ^^[SSS9nU(a[UTT5nUR5n[5n[5n[[UR TUTU05Xg5XgSS9nU(a2UTUT0n USR U 5USR U 5US4$g UnUR5n[ R"UR55n /n U HLn [U R TUT- 55n U Rn TU ;dTU ;a O#U RU 5 MN TT-[ U 6U4$UU4S jn/n URT5nURT5U:Xa_[URTU-5U5n [URTU-5U5nU"UTUUT-- U - 5n U b U T-UT--X4$/n URT5nURT5U:Xa[S5Hun[URTU-TU--5U5n [URTU-5U5nU"UTUUT-- U - T- 5n U bU T-T-UT--X4s $TTsmmMw g g ) aoGiven an expression, f, 3 tests will be done to see what type of composite bivariate it might be, options for u(x, y) are:: x*y x+y x*y+x x*y+y If it matches one of these types, ``u(x, y)``, ``P(u)`` and dummy variable ``u`` will be returned. Solving ``P(u)`` for ``u`` and equating the solutions to ``u(x, y)`` and then solving for ``x`` or ``y`` is equivalent to solving the original expression for ``x`` or ``y``. If ``x`` and ``y`` represent two functions in the same variable, e.g. ``x = g(t)`` and ``y = h(t)``, then if ``u(x, y) - p`` can be solved for ``t`` then these represent the solutions to ``P(u) = 0`` when ``p`` are the solutions of ``P(u) = 0``. Only positive values of ``u`` are considered. Examples ======== >>> from sympy import solve >>> from sympy.solvers.bivariate import bivariate_type >>> from sympy.abc import x, y >>> eq = (x**2 - 3).subs(x, x + y) >>> bivariate_type(eq, x, y) (x + y, _u**2 - 3, _u) >>> uxy, pu, u = _ >>> usol = solve(pu, u); usol [sqrt(3)] >>> [solve(uxy - s) for s in solve(pu, u)] [[{x: -y + sqrt(3)}]] >>> all(eq.subs(s).equals(0) for sol in _ for s in sol) True rPT)positiveFrrrNcp>[URX55nURnTU;dTU;aS$U$r()rrMr)r^vrbnewfreer*ys r$okbivariate_type..oks7qvva|$T Q$Yt8S8r.)r ras_exprbivariate_typerMrpr make_argsrrappenddegreercoeff_monomialrange)r^r*rrrPri_x_yrvrepsrTrrFrrr]rGitrys `` r$rrs{N cD!A AqM IIK W W DB2!7@"PU V 2q>Da5>>$'A)=r!uD D A A == %D C  QVVAqs^ $~~ 9T  1 sCIq  9 C  Axx{a !!!Q$' + !!!Q$' +AAaC{# ?Q319c$ $ C  Axx{a!HDQ%%ad1a4i0!4AQ%%ad+Q/AQA!GQ;q=)Cs1uqs{C**aDAq r.r()*sympy.core.addrsympy.core.exprtoolsrsympy.core.functionrrsympy.core.powerrsympy.core.singletonrsympy.core.sortingr sympy.core.symbolr &sympy.functions.elementary.exponentialr r r (sympy.functions.elementary.miscellaneousrsympy.polys.polyrootsrsympy.polys.polytoolsrrsympy.simplify.simplifyrsympy.simplify.radsimprrsympy.solvers.solversrrsympy.utilities.iterablesrr%r9rArmrqrr.r$rsb-4 "&#GG9'.0*+0*8"J"JEP]@&*\r.