结式的幂简洁型

青青子衿

\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{{\>\,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*}

青青子衿

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*}

青青子衿

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*}

青青子衿

回复 3# 青青子衿

青青子衿

青青子衿 发表于 2020-6-15 22:14
\begin{gather*}
\operatorname{Reslt}\left(\left.
\begin{matrix}  
u^2+\alpha\,\!u+\beta\\  
u^3+\gamma\,\!u+\delta\\  
\end{matrix}\right|\,u\right)\\
\\
\end{gather*}

\begin{align*}
\left(a f g-b e g+d e^2\right)^2-\left(a f^2-a e g-b e f+c e^2\right) \left(a g^2-c e g+d e f\right)\\
\left(a g^2-c e g+d e f\right)^2-\left(a f g-b e g+d e^2\right) \left(b g^2-c f g-d e g+d f^2\right)
\end{align*}

1. GroebnerBasis[{a + b*p + c*p^2 + d*p^3, e + f*p + g*p^2}, {p}]
2. GroebnerBasis[#, p][[1]] & /@ {{d e f - c e g +
3.      a g^2 + (d f^2 - d e g - c f g + b g^2) p,
4.     d e^2 - b e g + a f g + (d e f - c e g + a g^2) p},
5.    {d e f - c e g + a g^2 + (d f^2 - d e g - c f g + b g^2) p,
6.     c e^2 - b e f + a f^2 - a e g + (d e^2 - b e g + a f g) p},
7.    {d e^2 - b e g + a f g + (d e f - c e g + a g^2) p,
8.     c e^2 - b e f + a f^2 -
9.      a e g + (d e^2 - b e g + a f g) p}} // Factor
10. (d e^2 - b e g + a f g)^2 - (d e f - c e g + a g^2) (c e^2 - b e f +
11.      a f^2 - a e g) // Factor
12. (d e f - c e g + a g^2) (d e^2 - b e g + a f g) - (d f^2 - d e g -
13.      c f g + b g^2) (c e^2 - b e f + a f^2 - a e g) // Factor
14. (d e f - c e g + a g^2)^2 - (d f^2 - d e g - c f g + b g^2) (d e^2 -
15.      b e g + a f g) // Factor

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青青子衿

\begin{align*}
\Delta_3&=\dfrac{1}{48}\begin{vmatrix}
\dfrac{\partial}{\partial\,\!x^2}&\dfrac{\partial^2}{\partial\,\!x\partial\,\!y}\\
\dfrac{\partial}{\partial\,\!x\partial\,\!y}&\dfrac{\partial}{\partial\,\!y^2}\\
\end{vmatrix}^2\Big(a_0 x^3+a_1 x^2 y+a_2 x y^2+a_3 y^3\Big)\\
\\
&=\dfrac{1}{12}\begin{vmatrix}
2 \left(3 a_0 a_2-a_1^2\right) & 9 a_0 a_3-a_1 a_2 \\
9 a_0 a_3-a_1 a_2 & 2 \left(3 a_1 a_3-a_2^2\right) \\
\end{vmatrix}\\
\\
&=\dfrac{1}{a_0}
\begin{vmatrix}
a_0 & a_1 & a_2 & a_3 & 0 \\
0 & a_0 & a_1 & a_2 & a_3 \\
3 a_0 & 2 a_1 & a_2 & 0 & 0 \\
0 & 3 a_0 & 2 a_1 & a_2 & 0 \\
0 & 0 & 3 a_0 & 2 a_1 & a_2 \\
\end{vmatrix}
\end{align*}

\begin{gather*}
\big(\det\!\operatorname{Hess}(F)\big)^3+
\big(\det\!\operatorname{Jac}[F,\det\!\operatorname{Hess}(F)]\big)^2+9\det\!\operatorname{Hess}(\det\!\operatorname{Hess}(F))\cdot\,\!F^2=0\\
F=a_0x^3+a_1x^2y+a_2xy^2+a_3y^3
\end{gather*}

1. (ResourceFunction["HessianDeterminant"][
2.     a*x^3 + b*x^2*y + c*x*y^2 + d*y^3, {x,
3.      y}])^3 + (ResourceFunction[
4.       "JacobianDeterminant"][{a*x^3 + b*x^2*y + c*x*y^2 + d*y^3,
5.       ResourceFunction["HessianDeterminant"][
6.        a*x^3 + b*x^2*y + c*x*y^2 + d*y^3, {x, y}]}, {x, y}])^2 +
7.   9 ResourceFunction["HessianDeterminant"][
8.     ResourceFunction["HessianDeterminant"][
9.      a*x^3 + b*x^2*y + c*x*y^2 + d*y^3, {x, y}], {x,
10.      y}] (a*x^3 + b*x^2*y + c*x*y^2 + d*y^3)^2 // Factor

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青青子衿

\begin{align*}
\begin{vmatrix}
\begin{vmatrix}
4 a & b \\
3 b & 2 c \\
\end{vmatrix}
&
\begin{vmatrix}
4 a & b \\
2 c & 3 d \\
\end{vmatrix}
&
\begin{vmatrix}
4 a & b \\
d & 4 e \\
\end{vmatrix} \\
\begin{vmatrix}
4 a & b \\
2 c & 3 d \\
\end{vmatrix}
&
\begin{vmatrix}
4 a & b \\
d & 4 e \\
\end{vmatrix}+\begin{vmatrix}
3 b & 2 c \\
2 c & 3 d \\
\end{vmatrix}
&
\begin{vmatrix}
3 b & 2 c \\
d & 4 e \\
\end{vmatrix}
\\
\begin{vmatrix}
4 a & b \\
d & 4 e \\
\end{vmatrix}
&
\begin{vmatrix}
3 b & 2 c \\
d & 4 e \\
\end{vmatrix}
&
\begin{vmatrix}
2 c & 3 d \\
d & 4 e \\
\end{vmatrix}
\end{vmatrix}
\end{align*}

\begin{align*}
\begin{vmatrix}
3 b^2-8 a c & b c-6 a d & b d-16 a e \\
b c-6 a d & -4 a e-2 b d+c^2 & c d-6 b e \\
b d-16 a e & c d-6 b e & 3 d^2-8 c e \\
\end{vmatrix}=4\operatorname{Discr}[a x^4+b x^3+c x^2+d x+e]
\end{align*}

1. ResourceFunction["BezoutMatrix"][4 a*x^3 + 3 b*x^2 + 2 c*x + d,
2.   b*x^3 + 2 c*x^2 + 3 d*x + 4 e, x] // MatrixForm
3. ResourceFunction["BezoutMatrix"][4 a*x^3 + 3 b*x^2 + 2 c*x + d,
4.    b*x^3 + 2 c*x^2 + 3 d*x + 4 e, x] // Det // Factor
5. Discriminant[a*x^4 + b*x^3 + c*x^2 + d*x + e, x]
6. ( {
7.     {Det[({{a, e},{b, f}} )], a, e, 0},
8.     {Det[({{a, e},{c, g}} )], b, f, e},
9.     {Det[( {{a, e},{d, 0}} )], c, g, f},
10.     {0, d, 0, g}
11. } ) // Det // Factor
12. ( {
13.     {Det[({{a, e},{b, f}} )],
14.      Det[({{a, e},{c, g}} )], e},
15.     {Det[({{a, e},{c, g}} )],
16.      Det[({{a, e},{d, 0}} )]
17. + Det[({{b, f},{c, g}} )], f},
18.     {Det[({{a, e},{d, 0}})],
19.      Det[({{b, f},{d, 0}} )], g}
20.    } ) // Det // Factor
21. GroebnerBasis[{e*x^2 + f*x + g, a*x^3 + b*x^2 + c*x + d}, {x}][[1]]

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\begin{align*}
\begin{vmatrix}
\begin{vmatrix}
a & e \\
b & f
\end{vmatrix}
& a & e & 0 \\
\begin{vmatrix}
a & e \\
c & g \\
\end{vmatrix}
& b & f & e \\
\begin{vmatrix}
a & e \\
d & 0
\end{vmatrix}
& c & g & f \\
0 & d & 0 & g \\
\end{vmatrix}=
\begin{vmatrix}
\begin{vmatrix}
a & e \\
b & f \\
\end{vmatrix}
&
\begin{vmatrix}
a & e \\
c & g \\
\end{vmatrix}
& e \\
\begin{vmatrix}
a & e \\
c & g \\
\end{vmatrix}
&
\begin{vmatrix}
a & e \\
d & 0 \\
\end{vmatrix}
+\begin{vmatrix}
b & f \\
c & g \\
\end{vmatrix} & f \\
\begin{vmatrix}
a & e \\
d & 0 \\
\end{vmatrix}
&
\begin{vmatrix}
b & f \\
d & 0 \\
\end{vmatrix} & g \\
\end{vmatrix}
\end{align*}