| − | 其中<math>f\left(x^{*}, r_{c}\right)=0</math>,且因为<math>x^*</math>是[[不动点]],且由相切情况可得<math>\left. \frac {\partial f}{\partial x}\right|_{\left(x^{*} ; r_{c}\right)}=0</math>,再将<math>\left. \frac {\partial f}{\partial r}\right|_{\left(x ; r_{c}\right)}</math>视为常数<math>a</math>,将<math>\left. \frac {1}{2} \frac{\partial^{2} f}{\partial x^{2}}\right|_{\left(x ; r_{c}\right)}</math>视为常数<math>b</math>。忽略后面的高阶项,即可得到<math>\dot{x} = a(r-r_c)+b(x-x^*)^2</math> | + | 其中<math>f\left(x^{*}, r_{c}\right)=0</math>,且因为<math>x^*</math>是[[不动点]],由相切情况可得<math>\left. \frac {\partial f}{\partial x}\right|_{\left(x^{*} ; r_{c}\right)}=0</math>,再将<math>\left. \frac {\partial f}{\partial r}\right|_{\left(x ; r_{c}\right)}</math>视为常数<math>a</math>,将<math>\left. \frac {1}{2} \frac{\partial^{2} f}{\partial x^{2}}\right|_{\left(x ; r_{c}\right)}</math>视为常数<math>b</math>。忽略后面的高阶项,即可得到<math>\dot{x} = a(r-r_c)+b(x-x^*)^2</math> |