level 2
左地GG6
楼主
各位大佬求问一下,我最近在学习mathmatic,我作一个for循环,,我把我认为该赋值的参数都赋值了,但是还是提示为for提供的参数过少,,程序如下aa = {};
For[Subscript[\[CapitalEpsilon], f] = 0.10;
Subscript[\[CapitalEpsilon], f] <= 0.33;
Subscript[\[CapitalEpsilon], f] += 0.01, \[Tau] = 0.1;
Subscript[\[Mu], c] = 0.3; e = 1.6*10^-19; \[CurlyEpsilon] = 2.25;
Subscript[\[CurlyEpsilon], h] = 2.25;
a = 100; \[HBar] = 4.135*10^-15; \[Lambda] = 25;
Subscript[E, 0] = 5*10^6; c = 3*10^8;
m = Sqrt[\[CurlyEpsilon]/Subscript[\[CurlyEpsilon], h]]; \[Alpha] =
Sqrt[Subscript[\[Mu], 0]/(
Subscript[\[CurlyEpsilon], 0] Subscript[\[CurlyEpsilon], h])];
x = k*a; k = Subscript[k, 0] Sqrt[Subscript[\[CurlyEpsilon], h]];
Subscript[\[ScriptCapitalV], f] =
2 Subscript[\[Pi]\[CapitalEpsilon], f]/\[HBar]k; \[Omega] = (
2 \[Pi]c)/\[Lambda];
Subscript[\[Sigma], 0] =
ie^2 Subscript[\[Mu], c]/[\[Pi]\[HBar]^2 (\[Omega] + i/\[Tau])];
Subscript[\[Sigma],
3] = -i9e^4 Subscript[\[ScriptCapitalV],
f]^2/(8 Subscript[\[Pi]\[Mu], c] \[HBar]^2 \[Omega]^3);
Subscript[\[Sigma], g] =
Subscript[\[Sigma], 0] +
Subscript[\[Sigma],
3] \[LeftBracketingBar]Subscript[E, t]\[RightBracketingBar]^2;
Subscript[
\!\(\*OverscriptBox[\(F\), \(~\)]\), n] = (Subscript[H, n] (x)*
\!\(\*SubsuperscriptBox[\(J\), \(n\), \(\[Prime]\)]\) (x) - \!\(
\*SubsuperscriptBox[\(H\), \(n\), \(\[Prime]\)] \((x)\)*
\*SubscriptBox[\(J\), \(n\)] \((x)\)\))/(Subscript[H, n] (x)*
\!\(\*SubsuperscriptBox[\(J\), \(n\), \(\[Prime]\)]\) (mx) - m*
\!\(\*SubsuperscriptBox[\(H\), \(n\), \(\[Prime]\)]\) (x)*Subscript[J,
n] (mx) - i*Subscript[
\!\(\*OverscriptBox[\(\[Sigma]\), \(~\)]\), g]*\[Alpha]*
\!\(\*SubsuperscriptBox[\(H\), \(n\), \(\[Prime]\)]\) (x)*
\!\(\*SubsuperscriptBox[\(J\), \(n\), \(\[Prime]\)]\) (mx));
Subscript[k, 1] = Subscript[k, 0] Sqrt[\[CurlyEpsilon]];
J = BesselJ[0, 1];
Subscript[H, n] ==
HankelH1[0, 1]; \[LeftBracketingBar]Subscript[E,
t]\[RightBracketingBar]^2 = \[LeftBracketingBar]Subscript[E,
0]\[RightBracketingBar]^2 \[LeftBracketingBar]\!\(
\*UnderoverscriptBox[\(\[Sum]\), \(n = \(-0\)\), \(2\)]\(
\*SuperscriptBox[\(i\), \(n + 1\)]
\*SubscriptBox[
OverscriptBox[\(F\), \(~\)], \(n\)]
\*SubsuperscriptBox[\(J\), \(n\), \(\[Prime]\)] \((
\*SubscriptBox[\(k\), \(1\)] a)\)
\*SubscriptBox[
OverscriptBox[\(e\), \(^\)], \(\[CurlyPhi]\)]
\*SuperscriptBox[\(e\), \(in\[CurlyPhi]\)]\)\)\[RightBracketingBar];
\!\(\*OverscriptBox[\(B\), \(~\)]\) = (
2 Subscript[\[CurlyEpsilon], h])/(\[CurlyEpsilon] +
Subscript[\[CurlyEpsilon], h] + \[Gamma]); \[Gamma] = Subscript[
i\[Sigma], g]/Subscript[\[Omega]a\[CurlyEpsilon],
0]; \[LeftBracketingBar]Subscript[E,
t]\[RightBracketingBar]^2 = \[LeftBracketingBar]
\!\(\*OverscriptBox[\(B\), \(~\)]\)\[RightBracketingBar]^2 \
\[LeftBracketingBar]Subscript[E, 0]\[RightBracketingBar]^2;
AppendTo[aa, {\[LeftBracketingBar]Subscript[E,
0]\[RightBracketingBar]^2, \[LeftBracketingBar]Subscript[E,
t]\[RightBracketingBar]^2}]]
ListPlot[aa, PlotJoined -> True]
Export["E:\data\1.dat", aa, "Table"]由于刚使用mathmatic时间不足两个月,还请指教和帮助,多谢大佬了

2020年05月11日 09点05分
1
For[Subscript[\[CapitalEpsilon], f] = 0.10;
Subscript[\[CapitalEpsilon], f] <= 0.33;
Subscript[\[CapitalEpsilon], f] += 0.01, \[Tau] = 0.1;
Subscript[\[Mu], c] = 0.3; e = 1.6*10^-19; \[CurlyEpsilon] = 2.25;
Subscript[\[CurlyEpsilon], h] = 2.25;
a = 100; \[HBar] = 4.135*10^-15; \[Lambda] = 25;
Subscript[E, 0] = 5*10^6; c = 3*10^8;
m = Sqrt[\[CurlyEpsilon]/Subscript[\[CurlyEpsilon], h]]; \[Alpha] =
Sqrt[Subscript[\[Mu], 0]/(
Subscript[\[CurlyEpsilon], 0] Subscript[\[CurlyEpsilon], h])];
x = k*a; k = Subscript[k, 0] Sqrt[Subscript[\[CurlyEpsilon], h]];
Subscript[\[ScriptCapitalV], f] =
2 Subscript[\[Pi]\[CapitalEpsilon], f]/\[HBar]k; \[Omega] = (
2 \[Pi]c)/\[Lambda];
Subscript[\[Sigma], 0] =
ie^2 Subscript[\[Mu], c]/[\[Pi]\[HBar]^2 (\[Omega] + i/\[Tau])];
Subscript[\[Sigma],
3] = -i9e^4 Subscript[\[ScriptCapitalV],
f]^2/(8 Subscript[\[Pi]\[Mu], c] \[HBar]^2 \[Omega]^3);
Subscript[\[Sigma], g] =
Subscript[\[Sigma], 0] +
Subscript[\[Sigma],
3] \[LeftBracketingBar]Subscript[E, t]\[RightBracketingBar]^2;
Subscript[
\!\(\*OverscriptBox[\(F\), \(~\)]\), n] = (Subscript[H, n] (x)*
\!\(\*SubsuperscriptBox[\(J\), \(n\), \(\[Prime]\)]\) (x) - \!\(
\*SubsuperscriptBox[\(H\), \(n\), \(\[Prime]\)] \((x)\)*
\*SubscriptBox[\(J\), \(n\)] \((x)\)\))/(Subscript[H, n] (x)*
\!\(\*SubsuperscriptBox[\(J\), \(n\), \(\[Prime]\)]\) (mx) - m*
\!\(\*SubsuperscriptBox[\(H\), \(n\), \(\[Prime]\)]\) (x)*Subscript[J,
n] (mx) - i*Subscript[
\!\(\*OverscriptBox[\(\[Sigma]\), \(~\)]\), g]*\[Alpha]*
\!\(\*SubsuperscriptBox[\(H\), \(n\), \(\[Prime]\)]\) (x)*
\!\(\*SubsuperscriptBox[\(J\), \(n\), \(\[Prime]\)]\) (mx));
Subscript[k, 1] = Subscript[k, 0] Sqrt[\[CurlyEpsilon]];
J = BesselJ[0, 1];
Subscript[H, n] ==
HankelH1[0, 1]; \[LeftBracketingBar]Subscript[E,
t]\[RightBracketingBar]^2 = \[LeftBracketingBar]Subscript[E,
0]\[RightBracketingBar]^2 \[LeftBracketingBar]\!\(
\*UnderoverscriptBox[\(\[Sum]\), \(n = \(-0\)\), \(2\)]\(
\*SuperscriptBox[\(i\), \(n + 1\)]
\*SubscriptBox[
OverscriptBox[\(F\), \(~\)], \(n\)]
\*SubsuperscriptBox[\(J\), \(n\), \(\[Prime]\)] \((
\*SubscriptBox[\(k\), \(1\)] a)\)
\*SubscriptBox[
OverscriptBox[\(e\), \(^\)], \(\[CurlyPhi]\)]
\*SuperscriptBox[\(e\), \(in\[CurlyPhi]\)]\)\)\[RightBracketingBar];
\!\(\*OverscriptBox[\(B\), \(~\)]\) = (
2 Subscript[\[CurlyEpsilon], h])/(\[CurlyEpsilon] +
Subscript[\[CurlyEpsilon], h] + \[Gamma]); \[Gamma] = Subscript[
i\[Sigma], g]/Subscript[\[Omega]a\[CurlyEpsilon],
0]; \[LeftBracketingBar]Subscript[E,
t]\[RightBracketingBar]^2 = \[LeftBracketingBar]
\!\(\*OverscriptBox[\(B\), \(~\)]\)\[RightBracketingBar]^2 \
\[LeftBracketingBar]Subscript[E, 0]\[RightBracketingBar]^2;
AppendTo[aa, {\[LeftBracketingBar]Subscript[E,
0]\[RightBracketingBar]^2, \[LeftBracketingBar]Subscript[E,
t]\[RightBracketingBar]^2}]]
ListPlot[aa, PlotJoined -> True]
Export["E:\data\1.dat", aa, "Table"]由于刚使用mathmatic时间不足两个月,还请指教和帮助,多谢大佬了
