作业帮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
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/10 20:56:36
\)),\*TemplateBox[{](/uploads/image/z/2479525-61-5.jpg?t=mathematica%E5%A6%82%E4%BD%95%E5%AE%9A%E6%9C%89%E4%B8%80%E4%B8%AA%E5%88%86%E6%AE%B5%E5%87%BD%E6%95%B0%3FWp+%3D+3%2A2%2APi%3BWs+%3D+12%2A2%2APi%3BRp+%3D+0.1%3BRs+%3D+60%3B%5C%28TraditionalForm%5C%60Cn+%3D+Piecewise%5B%7B%7Bcos%28Njie%5C+%5C%28%5C%2ASuperscriptBox%5B%5C%28cos%5C%29%2C%5C%28-1%5C%29%5D%28x%29%5C%29%29%2C%5C%2ATemplateBox%5B%7B%22x%22%7D%2C%22Abs%22%5D+%3C+1%7D%2C%7BCosh%5BNjie%2AArcCosh%5Bx%5D%5D+%7B%5C%2ATemplateBox%5B%7B%22x%22%7D%2C%22Ab)
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
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，
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
Cn=Piecewise[{{Cos[Njie ArcCos[x]],Abs[x]1}}]