如何創建具有浮點系數的多項式環Julia

我想創建一個具有這樣的浮點系數的多項式環。我可以用整數創建，但是浮點數不起作用。

using Oscar

S, (a,b,c,d) = PolynomialRing(QQ,["a","b","c","d"])
RR = AbstractAlgebra.RealField
s1 = S( 8*a - RR(0.51234)*a*(1 RR(1/2)*a RR(1/3)*b RR(1/4)*c) - 8)
s2 = S( 8*b - RR(0.51234)*b*(1 RR(2/3)*a RR(2/4)*b RR(2/5)*c) - 8)
s3 = S( 8*c - RR(0.51234)*c*(1 RR(3/4)*a RR(3/5)*b RR(3/6)*c) - 8)
s4 = S( 8*d - RR(0.51234)*d*(1 RR(4/5)*a RR(4/6)*b RR(4/7)*c) - 8)

它給了我這個錯誤。我怎樣才能創建這樣的多項式。

ERROR: LoadError: MethodError: no method matching (::FmpqMPolyRing)(::BigFloat)
Closest candidates are:
  (::FmpqMPolyRing)() at ~/.julia/packages/Nemo/5CDLD/src/flint/fmpq_mpoly.jl:1063
  (::AbstractAlgebra.Ring)(::Singular.n_RingElem{Singular.RingElemWrapper{S, T}}) where {S, T} at ~/.julia/packages/Singular/uG7uo/src/number/n_unknown.jl:358
  (::AbstractAlgebra.Ring)(::Union{Singular.n_FieldElem{T}, Singular.n_RingElem{T}} where T) at ~/.julia/packages/Oscar/iRpOQ/src/Rings/mpoly.jl:736
  ...
Stacktrace:
 [1] *(x::BigFloat, y::fmpq_mpoly)
   @ AbstractAlgebra ~/.julia/packages/AbstractAlgebra/mQIYL/src/Rings.jl:84
 [2] top-level scope
   @ /mnt/c/Users/yusuf/Desktop/7.Semester/bitirme/Repo_Resultant_System/resultant-system/chandra4.jl:7
in expression starting at /mnt/c/Users/yusuf/Desktop/7.Semester/bitirme/Repo_Resultant_System/resultant-system/chandra4.jl:7

uj5u.com熱心網友回復：

雖然我以前沒有使用過此特定（從外觀上看，相當復雜）包Oscar.jl的經驗，但決議此錯誤訊息告訴我，您嘗試呼叫的函式被BigFloat作為輸入提供，但根本沒有方法對于那種型別。

起初這有點令人驚訝，因為BigFloat您的輸入中沒有 s，但經過一番調查，似乎罪魁禍首如下

julia> RR = AbstractAlgebra.RealField
Floats

julia> RR(1/3)
0.333333333333333314829616256247390992939472198486328125

julia> typeof(ans)
BigFloat

但是，將這些輸入更改BigFloat為更標準Float64并不能解決問題；S也沒有辦法。但是，它確實具有Rationals諸如1//3. 因此，一個簡單的明顯解決方法是撰寫

using Oscar

S, (a,b,c,d) = PolynomialRing(QQ,["a","b","c","d"])
RR = AbstractAlgebra.Rational # Note the change here!
s1 = S( 8*a - RR(0.51234)*a*(1 RR(1/2)*a RR(1/3)*b RR(1/4)*c) - 8)
s2 = S( 8*b - RR(0.51234)*b*(1 RR(2/3)*a RR(2/4)*b RR(2/5)*c) - 8)
s3 = S( 8*c - RR(0.51234)*c*(1 RR(3/4)*a RR(3/5)*b RR(3/6)*c) - 8)
s4 = S( 8*d - RR(0.51234)*d*(1 RR(4/5)*a RR(4/6)*b RR(4/7)*c) - 8)

它運行沒有錯誤。

或者更簡潔一點，從一開始就直接將系數作為有理數輸入：

S, (a,b,c,d) = PolynomialRing(QQ,["a","b","c","d"])
RR = AbstractAlgebra.Rational
s1 = S( 8*a - RR(51234//100000)*a*(1 RR(1//2)*a RR(1//3)*b RR(1//4)*c) - 8)
s2 = S( 8*b - RR(51234//100000)*b*(1 RR(2//3)*a RR(2//4)*b RR(2//5)*c) - 8)
s3 = S( 8*c - RR(51234//100000)*c*(1 RR(3//4)*a RR(3//5)*b RR(3//6)*c) - 8)
s4 = S( 8*d - RR(51234//100000)*d*(1 RR(4//5)*a RR(4//6)*b RR(4//7)*c) - 8)

產生

julia> s1 = S( 8*a - RR(51234//100000)*a*(1 RR(1//2)*a RR(1//3)*b RR(1//4)*c) - 8)
-25617//100000*a^2 - 8539//50000*a*b - 25617//200000*a*c   374383//50000*a - 8

julia> s2 = S( 8*b - RR(51234//100000)*b*(1 RR(2//3)*a RR(2//4)*b RR(2//5)*c) - 8)
-8539//25000*a*b - 25617//100000*b^2 - 25617//125000*b*c   374383//50000*b - 8

julia> s3 = S( 8*c - RR(51234//100000)*c*(1 RR(3//4)*a RR(3//5)*b RR(3//6)*c) - 8)
-76851//200000*a*c - 76851//250000*b*c - 25617//100000*c^2   374383//50000*c - 8

julia> s4 = S( 8*d - RR(51234//100000)*d*(1 RR(4//5)*a RR(4//6)*b RR(4//7)*c) - 8)
-25617//62500*a*d - 8539//25000*b*d - 25617//87500*c*d   374383//50000*d - 8

在后一種情況下，RR包裝器似乎不是必需的，因為它不會改變輸入的型別，但我想它不會受到傷害。

標籤：

上一篇：如何檢查X是否是60的任何整數冪的除數？

下一篇：f(x)的n次導數的數值實作？