楼主: tulipsliu
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[学科前沿] 金融学前沿研究R代码实现 [推广有奖]

21
tulipsliu 在职认证  发表于 2020-12-8 14:58:58
$$\frac{8}{{\mathrm{tan}\left(\frac{x}{2}\right)}^2 +9}+\frac{1}{4\,\mathrm{cos}\left(x\right)+5}$$

22
tulipsliu 在职认证  发表于 2020-12-8 15:00:42
$$
\begin{array}{l}
\left(\begin{array}{c}
\frac{3\,\sigma_2 -17\,{{\left(\sigma_5 -\sigma_6 +\frac{15}{68}\right)}}^2 -\sigma_4 +\sigma_1 +\frac{11}{34}}{\sigma_6 +2\,\sigma_2 -\sigma_5 -\frac{219}{68}}\\
-\frac{\sigma_4 +17\,{{\left(\sigma_6 +\sigma_5 -\frac{15}{68}\right)}}^2 +3\,\sigma_3 +\sigma_1 -\frac{11}{34}}{\sigma_6 -2\,\sigma_3 +\sigma_5 -\frac{219}{68}}
\end{array}\right)\\
\mathrm{}\\
\textrm{where}\\
\mathrm{}\\
\;\;\sigma_1 =\frac{\sigma_7 }{{\sigma_9 }^{1/6} \,{\sigma_8 }^{1/4} }\\
\mathrm{}\\
\;\;\sigma_2 ={{\left(\sigma_5 -\sigma_6 +\frac{15}{68}\right)}}^3 \\
\mathrm{}\\
\;\;\sigma_3 ={{\left(\sigma_6 +\sigma_5 -\frac{15}{68}\right)}}^3 \\
\mathrm{}\\
\;\;\sigma_4 =\frac{\sqrt{\sigma_8 }}{{\sigma_9 }^{1/6} }\\
\mathrm{}\\
\;\;\sigma_5 =\frac{\sigma_7 }{6\,{\sigma_9 }^{1/6} \,{\sigma_8 }^{1/4} }\\
\mathrm{}\\
\;\;\sigma_6 =\frac{\sqrt{\sigma_8 }}{6\,{\sigma_9 }^{1/6} }\\
\mathrm{}\\
\;\;\sigma_7 =\sqrt{\frac{337491\,\sqrt{6}\,\sqrt{\frac{3\,\sqrt{3}\,\sqrt{178939632355}}{9826}+\frac{2198209}{9826}}}{39304}+\frac{2841\,{\sigma_9 }^{1/3} \,\sqrt{\sigma_8 }}{578}-9\,{\sigma_9 }^{2/3} \,\sqrt{\sigma_8 }-\frac{361\,\sqrt{\sigma_8 }}{289}}\\
\mathrm{}\\
\;\;\sigma_8 =\frac{2841\,{\sigma_9 }^{1/3} }{1156}+9\,{\sigma_9 }^{2/3} +\frac{361}{289}\\
\mathrm{}\\
\;\;\sigma_9 =\frac{\sqrt{3}\,\sqrt{178939632355}}{176868}+\frac{2198209}{530604}
\end{array}
$$

23
tulipsliu 在职认证  发表于 2020-12-8 15:02:08
$$
\begin{array}{l}
\left(\begin{array}{cc}
\frac{6\,{{\left(3\,{x_1 }^2 -1\right)}}^2 -36\,x_1 \,{\left(-{x_1 }^3 +x_1 +x_2 \right)}+2}{\sigma_2 }-\frac{{\sigma_3 }^2 }{{\sigma_2 }^2 } & \sigma_1 \\
\sigma_1  & \frac{6}{\sigma_2 }-\frac{{{\left(-6\,{x_1 }^3 +6\,x_1 +6\,x_2 \right)}}^2 }{{\sigma_2 }^2 }
\end{array}\right)\\
\mathrm{}\\
\textrm{where}\\
\mathrm{}\\
\;\;\sigma_1 =\frac{{\left(-6\,{x_1 }^3 +6\,x_1 +6\,x_2 \right)}\,\sigma_3 }{{\sigma_2 }^2 }-\frac{18\,{x_1 }^2 -6}{\sigma_2 }\\
\mathrm{}\\
\;\;\sigma_2 ={{\left(x_1 -\frac{4}{3}\right)}}^2 +3\,{{\left(-{x_1 }^3 +x_1 +x_2 \right)}}^2 +1\\
\mathrm{}\\
\;\;\sigma_3 =6\,{\left(3\,{x_1 }^2 -1\right)}\,{\left(-{x_1 }^3 +x_1 +x_2 \right)}-2\,x_1 +\frac{8}{3}
\end{array}
$$

24
tulipsliu 在职认证  发表于 2020-12-8 15:03:50
$$
\begin{array}{l}
\left(\begin{array}{c}
\sigma_1 -\frac{b}{3\,\sigma_1 }\\
\frac{b}{6\,\sigma_1 }-\frac{\sigma_1 }{2}-\frac{\sqrt{3}\,{\left(\frac{b}{3\,\sigma_1 }+\sigma_1 \right)}\,\mathrm{i}}{2}\\
\frac{b}{6\,\sigma_1 }-\frac{\sigma_1 }{2}+\frac{\sqrt{3}\,{\left(\frac{b}{3\,\sigma_1 }+\sigma_1 \right)}\,\mathrm{i}}{2}
\end{array}\right)\\
\mathrm{}\\
\textrm{where}\\
\mathrm{}\\
\;\;\sigma_1 ={{\left(\sqrt{\frac{b^3 }{27}+\frac{c^2 }{4}}-\frac{c}{2}\right)}}^{1/3}
\end{array}
$$
$$
\begin{array}{l}
\left(\begin{array}{c}
\sigma_1 -\frac{b}{3\,\sigma_1 }\\
\frac{b}{6\,\sigma_1 }-\frac{\sigma_1 }{2}-\frac{\sqrt{3}\,{\left(\frac{b}{3\,\sigma_1 }+\sigma_1 \right)}\,\mathrm{i}}{2}\\
\frac{b}{6\,\sigma_1 }-\frac{\sigma_1 }{2}+\frac{\sqrt{3}\,{\left(\frac{b}{3\,\sigma_1 }+\sigma_1 \right)}\,\mathrm{i}}{2}
\end{array}\right)\\
\mathrm{}\\
\textrm{where}\\
\mathrm{}\\
\;\;\sigma_1 ={{\left(\sqrt{\frac{b^3 }{27}+\frac{c^2 }{4}}-\frac{c}{2}\right)}}^{1/3}
\end{array}
$$

25
tulipsliu 在职认证  发表于 2020-12-8 15:05:02
$$
{{\left(\sqrt{\frac{b^3 }{27}+\frac{c^2 }{4}}-\frac{c}{2}\right)}}^{1/3} -\frac{b}{3\,{{\left(\sqrt{\frac{b^3 }{27}+\frac{c^2 }{4}}-\frac{c}{2}\right)}}^{1/3} }
$$

26
tulipsliu 在职认证  发表于 2020-12-8 15:11:57
$$
C_r = v_b - v_a / u_a - u_b
$$
$$
\frac{\partial^2 }{\partial t^2 }\;H\left(t\right)=g
$$
$$
\frac{g\,t^2 }{2}+v_0 \,t+h_0
$$
$$
-\frac{981\,t^2 }{200}+15\,t+10
$$
$$
\frac{dh}{dt} = v
$$

27
tulipsliu 在职认证  发表于 2020-12-8 15:12:25
$$
\frac{h(t+\Delta t) - h(t)}{\Delta t} = v(t)
$$

28
tulipsliu 在职认证  发表于 2020-12-8 15:13:09
$$
\left(\begin{array}{c}
m\,\frac{\partial^2 }{\partial t^2 }\;x\left(t\right)=\frac{F\left(t\right)\,x\left(t\right)}{r}\\
m\,\frac{\partial^2 }{\partial t^2 }\;y\left(t\right)=\frac{F\left(t\right)\,y\left(t\right)}{r}-g\,m\\
{x\left(t\right)}^2 +{y\left(t\right)}^2 =r^2
\end{array}\right)
$$

29
tulipsliu 在职认证  发表于 2020-12-8 15:13:39
$$
\left(\begin{array}{c}
m\,\frac{\partial }{\partial t}\;{\textrm{Dxt}}_1 \left(t\right)-\frac{F\left(t\right)\,x\left(t\right)}{r}\\
g\,m+m\,\frac{\partial }{\partial t}\;{\textrm{Dyt}}_1 \left(t\right)-\frac{F\left(t\right)\,y\left(t\right)}{r}\\
-r^2 +{x\left(t\right)}^2 +{y\left(t\right)}^2 \\
{\textrm{Dxt}}_1 \left(t\right)-\frac{\partial }{\partial t}\;x\left(t\right)\\
{\textrm{Dyt}}_1 \left(t\right)-\frac{\partial }{\partial t}\;y\left(t\right)
\end{array}\right)
$$

30
tulipsliu 在职认证  发表于 2020-12-8 15:15:14
$$
\left(\begin{array}{c}
x\left(t\right)\\
y\left(t\right)\\
F\left(t\right)\\
{\textrm{Dxt}}_1 \left(t\right)\\
{\textrm{Dyt}}_1 \left(t\right)
\end{array}\right)
$$
$$
\left(\begin{array}{c}
m\,\frac{\partial }{\partial t}\;{\textrm{Dxt}}_1 \left(t\right)-\frac{F\left(t\right)\,x\left(t\right)}{r}\\
g\,m+m\,\frac{\partial }{\partial t}\;{\textrm{Dyt}}_1 \left(t\right)-\frac{F\left(t\right)\,y\left(t\right)}{r}\\
-r^2 +{x\left(t\right)}^2 +{y\left(t\right)}^2 \\
{\textrm{Dxt}}_1 \left(t\right)-\frac{\partial }{\partial t}\;x\left(t\right)\\
{\textrm{Dyt}}_1 \left(t\right)-\frac{\partial }{\partial t}\;y\left(t\right)
\end{array}\right)

$$
$$
\left(\begin{array}{c}
-\frac{F\left(t\right)\,x\left(t\right)-m\,r\,{\textrm{Dxt1t}}_1 \left(t\right)}{r}\\
\frac{g\,m\,r-F\left(t\right)\,y\left(t\right)+m\,r\,{\textrm{Dyt1t}}_1 \left(t\right)}{r}\\
-r^2 +{x\left(t\right)}^2 +{y\left(t\right)}^2 \\
2\,{\textrm{Dxt}}_1 \left(t\right)\,x\left(t\right)+2\,{\textrm{Dyt}}_1 \left(t\right)\,y\left(t\right)\\
2\,{{\textrm{Dxt}}_1 \left(t\right)}^2 +2\,{{\textrm{Dyt}}_1 \left(t\right)}^2 +2\,{\textrm{Dxt1t}}_1 \left(t\right)\,x\left(t\right)+2\,{\textrm{Dyt1t}}_1 \left(t\right)\,y\left(t\right)\\
{\textrm{Dyt1t}}_1 \left(t\right)-\frac{\partial }{\partial t}\;{\textrm{Dyt}}_1 \left(t\right)\\
{\textrm{Dyt}}_1 \left(t\right)-\frac{\partial }{\partial t}\;y\left(t\right)
\end{array}\right)

$$

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