level 2
尚随明007
楼主
头大,帮忙看看呗。
$RecursionLimit::reclim2: 在 Subscript[\[Rho], min] 计算过程中超过 1024 的递归深度.
begin
M=1000;
pts:={};
Subscript[\[Rho], step]=(Subscript[\[Rho], max]-Subscript[\[Rho], min])/M;
for i:=0 to M
\[Rho]:=Subscript[\[Rho], min]+i*Subscript[\[Rho], step];
c(q)=-1;
J(\[Rho],s):=\!\(
\*SubsuperscriptBox[\(\[Integral]\), \(0\), \(\[Rho]\)]\(
\*FractionBox[\(1\),
SqrtBox[\(F \((\[Rho])\) - F \((t)\)\)]] \[DifferentialD]t\)\)+\!\(
\*SubsuperscriptBox[\(\[Integral]\), \(s\), \(\[Rho]\)]\(
\*FractionBox[\(1\),
SqrtBox[\(F \((\[Rho])\) - F \((t)\)\)]] \[DifferentialD]t\)\)-(c(q)*s)/Sqrt[[F(\[Rho])-F(s)]]
q=FindRoot[J(\[Rho],s),{s,0}];
F(\[Rho])=\!\(
\*SubsuperscriptBox[\(\[Integral]\), \(0\), \(\[Rho]\)]\(Exp[u] \[DifferentialD]u\)\);
F(q)=\!\(
\*SubsuperscriptBox[\(\[Integral]\), \(0\), \(q\)]\(Exp[u] \[DifferentialD]u\)\);
\[Lambda]=(c(q)*q)^2/(2 [F(\[Rho])-F(q)]);
pts:=AppendTo[pts,{\[Lambda],\[Rho]}]
end
ListPlot[pts]
end
2018年10月23日 02点10分
1
$RecursionLimit::reclim2: 在 Subscript[\[Rho], min] 计算过程中超过 1024 的递归深度.
begin
M=1000;
pts:={};
Subscript[\[Rho], step]=(Subscript[\[Rho], max]-Subscript[\[Rho], min])/M;
for i:=0 to M
\[Rho]:=Subscript[\[Rho], min]+i*Subscript[\[Rho], step];
c(q)=-1;
J(\[Rho],s):=\!\(
\*SubsuperscriptBox[\(\[Integral]\), \(0\), \(\[Rho]\)]\(
\*FractionBox[\(1\),
SqrtBox[\(F \((\[Rho])\) - F \((t)\)\)]] \[DifferentialD]t\)\)+\!\(
\*SubsuperscriptBox[\(\[Integral]\), \(s\), \(\[Rho]\)]\(
\*FractionBox[\(1\),
SqrtBox[\(F \((\[Rho])\) - F \((t)\)\)]] \[DifferentialD]t\)\)-(c(q)*s)/Sqrt[[F(\[Rho])-F(s)]]
q=FindRoot[J(\[Rho],s),{s,0}];
F(\[Rho])=\!\(
\*SubsuperscriptBox[\(\[Integral]\), \(0\), \(\[Rho]\)]\(Exp[u] \[DifferentialD]u\)\);
F(q)=\!\(
\*SubsuperscriptBox[\(\[Integral]\), \(0\), \(q\)]\(Exp[u] \[DifferentialD]u\)\);
\[Lambda]=(c(q)*q)^2/(2 [F(\[Rho])-F(q)]);
pts:=AppendTo[pts,{\[Lambda],\[Rho]}]
end
ListPlot[pts]
end
