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mathematica如何定有一个分段函数?Wp = 3*2*Pi;Ws = 12*2*Pi;Rp = 0.1;Rs = 60;\(TraditionalForm\`Cn = Piecewise[{{cos(Njie\ \(\*SuperscriptBox[\(cos\),\(-1\)](x)\)),\*TemplateBox[{"x"},"Abs"] < 1},{Cosh[Njie*ArcCosh[x]] {\*TemplateBox[{"x"},"Ab
题目详情
mathematica如何定有一个分段函数?
Wp = 3*2*Pi;
Ws = 12*2*Pi;
Rp = 0.1;
Rs = 60;
\(TraditionalForm\`Cn = Piecewise[{{cos(Njie\ \(
\*SuperscriptBox[\(cos\),\(-1\)](x)\)),\*
TemplateBox[{"x"},
"Abs"] < 1},{Cosh[Njie*ArcCosh[x]] {\*
TemplateBox[{"x"},
"Abs"] > 1}}}]\);
自己搞明白了，格式打错了
Cn=Piecewise[{{cos(Njie cos^-1(x)),\[LeftBracketingBar]x\[RightBracketingBar]<1},{Cosh[Njie*ArcCosh[x]],\[LeftBracketingBar]x\[RightBracketingBar]>1}}]
就对了，
Wp = 3*2*Pi;
Ws = 12*2*Pi;
Rp = 0.1;
Rs = 60;
\(TraditionalForm\`Cn = Piecewise[{{cos(Njie\ \(
\*SuperscriptBox[\(cos\),\(-1\)](x)\)),\*
TemplateBox[{"x"},
"Abs"] < 1},{Cosh[Njie*ArcCosh[x]] {\*
TemplateBox[{"x"},
"Abs"] > 1}}}]\);
自己搞明白了，格式打错了
Cn=Piecewise[{{cos(Njie cos^-1(x)),\[LeftBracketingBar]x\[RightBracketingBar]<1},{Cosh[Njie*ArcCosh[x]],\[LeftBracketingBar]x\[RightBracketingBar]>1}}]
就对了，
▼优质解答
答案和解析
Cn=Piecewise[{{Cos[Njie ArcCos[x]],Abs[x]1}}]
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