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结式的幂简洁型

本帖最后由 青青子衿 于 2019-11-6 10:10 编辑

\begin{align*}
\operatorname{Reslt}\left(\left.
\begin{matrix}
u^2+\alpha\,\!u+\beta\\
u^3+\gamma\,\!u+\delta\\
\end{matrix}\right|\,u\right)
&=\frac{1}{4}
(\alpha^3-3\alpha\beta+\alpha\gamma-2\delta)^2-\frac{1}{4}(\alpha^2-4 \beta) (\alpha^2-\beta+\gamma)^2\\
\operatorname{Reslt}\left(\left.
\begin{matrix}  
u^2+\alpha\,\!u+\beta\\  
u^2+\gamma\,\!u+\delta\\  
\end{matrix}\right|\,u\right)
&=\frac{1}{4}
(\alpha^2-2\beta-\alpha\gamma+2\delta)^2-\frac{1}{4}(\alpha^2-4 \beta) (\alpha-\gamma)^2  
\end{align*}

\begin{align*}
&&g_0+2g_1p+g_2p^2\phantom{+}\phantom{+g_3p^3}&&\longrightarrow&&\dfrac{g_0}{g_2}+\dfrac{2g_1}{g_2}p+\phantom{\dfrac{1g_3}{g_3}}p^2\,\,\phantom{+}\phantom{p^3}\\
&&h_0+3h_1p+3h_2p^2+h_3p^3&&\longrightarrow&&\dfrac{h_0}{h_3}+\dfrac{3h_1}{h_3}p+\dfrac{3h_2}{h_3}p^2+p^3
\end{align*}

\begin{align*}
\dfrac{g_0}{g_2}+\dfrac{2g_1}{g_2}p+p^2\,
\xrightarrow[]{\quad\,p\,=\,q-\frac{\Large{h}_{\small2}}{\Large{h}_{\small3}}\quad}
\left(\dfrac{g_0}{g_2}-\dfrac{2g_1h_2}{g_2h_3}+\dfrac{{h_2}^2}{{h_3}^2}\right)+2\left(\dfrac{g_1}{g_2}-\dfrac{h_2}{h_3}\right)q+q^2
\end{align*}

\begin{align*}
\dfrac{h_0}{h_3}+\dfrac{3h_1}{h_3}p+\dfrac{3h_2}{h_3}p^2+p^3\qquad\qquad\\
\\
\overset{|}{\overset{|}{\downarrow}}\overset{{\quad\,p\,=\,q-\frac{\Large{h}_{\small2}}{\Large{h}_{\small3}}\quad}}{}\qquad\qquad\qquad\\
\\
\left(\dfrac{h_0}{h_3}-\dfrac{3h_1h_2}{{h_3}^2}+\dfrac{2{h_2}^3}{{h_3}^3}\right)+3\left(\dfrac{h_1}{h_3}-\dfrac{{h_2}^2}{{h_3}^2}\right)q+q^3
\end{align*}
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本帖最后由 青青子衿 于 2019-11-4 10:41 编辑

回复 1# 青青子衿
幂简洁形式只是相对的,其实也没有多简洁……
\begin{align*}
&\left(3g_0g_1g_2h_3-3g_0{g_2}^2h_2-4{g_1}^3h_3+6{g_1}^2g_2h_2-3g_1{g_2}^2h_1+{g_2}^3h_0\right)^2\\
=\,&\left({g_1}^2-g_0g_2\right)\left(g_0g_2h_3-4{g_1}^2h_3+6g_1g_2h_2-3{g_2}^2h_1\right)^2
\end{align*}
\begin{align*}
\left(\dfrac{3g_0g_1g_2h_3-3g_0{g_2}^2h_2-4{g_1}^3h_3+6{g_1}^2g_2h_2-3g_1{g_2}^2h_1+{g_2}^3h_0}{g_0g_2h_3-4{g_1}^2h_3+6g_1g_2h_2-3{g_2}^2h_1}\right)^2={g_1}^2-g_0g_2\geqslant0
\end{align*}
...
\begin{gather*}
\begin{vmatrix}   
f_{xx}&2f_{xy}&f_{yy}&&\\   
&f_{xx}&2f_{xy}&f_{yy}&\\   
&&f_{xx}&2f_{xy}&f_{yy}\\   
f_{xxx}&3f_{xxy}&3f_{xyy}&f_{yyy}&\\   
&f_{xxx}&3f_{xxy}&3f_{xyy}&f_{yyy}\\
\end{vmatrix}\\
\\
=\dfrac{1}{{f_{yy}}^3}\left(\begin{split}
3f_{xx}f_{xy}f_{yy}\cdot\,\!f_{yyy}-3f_{xx}{f_{yy}}^2\cdot\,\!f_{xyy}-4{f_{xy}}^3\cdot\,\!f_{yyy}
+6{f_{xy}}^2f_{yy}\cdot\,\!f_{xyy}-3f_{xy}{f_{yy}}^2\cdot\,\!f_{xxy}+{f_{yy}}^3\cdot\,\!f_{xxx}
\end{split}\right)^2\\
-\dfrac{{f_{xy}}^2-f_{xx}f_{yy}}{{f_{yy}}^3}\left(f_{xx}f_{yy}\cdot\,\!f_{yyy}-4{f_{xy}}^2\cdot\,\!f_{yyy}+6f_{xy}f_{yy}\cdot\,\!f_{xyy}-3{f_{yy}}^2\cdot\,\!f_{xxy}\right)^2
\end{gather*}

\begin{gather*}
\operatorname{Reslt}\left(\left.
\begin{matrix}  
f_{xx}+2f_{xy}p+f_{yy}p^2\\  
f_{xxx}+3f_{xxy}p+3f_{xyy}p^2+f_{yyy}p^3\\  
\end{matrix}\right|\,p\right)
=
\begin{vmatrix}   
f_{xx}&2f_{xy}&f_{yy}&&\\   
&f_{xx}&2f_{xy}&f_{yy}&\\   
&&f_{xx}&2f_{xy}&f_{yy}\\   
f_{xxx}&3f_{xxy}&3f_{xyy}&f_{yyy}&\\   
&f_{xxx}&3f_{xxy}&3f_{xyy}&f_{yyy}\\
\end{vmatrix}\equiv0\\
\\
\Downarrow\\
\left(\begin{split}
3f_{xx}f_{xy}f_{yy}\cdot\,\!f_{yyy}-3f_{xx}{f_{yy}}^2\cdot\,\!f_{xyy}-4{f_{xy}}^3\cdot\,\!f_{yyy}
+6{f_{xy}}^2f_{yy}\cdot\,\!f_{xyy}-3f_{xy}{f_{yy}}^2\cdot\,\!f_{xxy}+{f_{yy}}^3\cdot\,\!f_{xxx}
\end{split}\right)^2\\
\equiv\left({f_{xy}}^2-f_{xx}f_{yy}\right)\left(f_{xx}f_{yy}\cdot\,\!f_{yyy}-4{f_{xy}}^2\cdot\,\!f_{yyy}+6f_{xy}f_{yy}\cdot\,\!f_{xyy}-3{f_{yy}}^2\cdot\,\!f_{xxy}\right)^2
\end{gather*}

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本帖最后由 青青子衿 于 2019-11-11 10:33 编辑

回复 2# 青青子衿
\begin{gather*}
\operatorname{Reslt}\left(\left.
\begin{matrix}  
u^2+\alpha\,\!u+\beta\\  
u^3+\gamma\,\!u+\delta\\  
\end{matrix}\right|\,u\right)\\
\\
=\left(\alpha^2-\beta+\gamma\right)^2\beta-(\alpha\beta+\delta)\left(\alpha^3-2\alpha\beta+\alpha\gamma-\delta\right)\\
=(\alpha \beta+\delta)^2-\left(\alpha^2-\beta+\gamma\right) \left(\alpha\delta+\beta^2-\beta\gamma\right)
\end{gather*}
\begin{gather*}
\operatorname{Reslt}\left(\left.  
\begin{matrix}   
u^2+\alpha\,\!u+\beta\\   
u^2+\gamma\,\!u+\delta\\   
\end{matrix}\right|\,u\right)\\
\\
=(\beta -\delta )^2-(\gamma -\alpha ) (\alpha  \delta -\beta  \gamma )
\end{gather*}

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