IVA Reading: Chapter 02, Section 06 Exercises

IVA読書会 chap02-sect06 宿題

 

No.5

lex順序でS多項式を求める計算問題。

RECAP
$$S(f, g) = \frac{x^\gamma}{{\scriptstyle \text{LT}}(f)}\cdot f - \frac{x^\gamma}{{\scriptstyle \text{LT}}(g)}\cdot g$$
where $$x^\gamma = \text{LCM}({\scriptstyle \text{LM}}(f), {\scriptstyle \text{LM}}(g))$$

計算には PARI/GP を利用。
ただ Leading Term を返す関数が見つからなかったので、LM, LC を手で計算して与えた。

? spol(f, g, lm_f, lm_g, lc_f, lc_g) = lcm(lm_f, lm_g)/(lm_f * lc_f) * f - lcm(lm_f, lm_g)/(lm_g * lc_g) * g
%11 = (f,g,lm_f,lm_g,lc_f,lc_g)->lcm(lm_f,lm_g)/(lm_f*lc_f)*f-lcm(lm_f,lm_g)/(lm_g*lc_g)*g
? spol(4*x^2*z - 7*y^2, x*y*z^2 + 3*x*z^4, x^2*z, x*y*z^2, 4, 1)
%12 = -3*z^4*x^2 - 7/4*z*y^3
? spol(x^4*y - z^2, 3*x*z^2 - y, x^4*y, x*z^2, 1, 3)
%18 = 1/3*y^2*x^3 - z^4
? spol(x^7*y^2*z + 2*i*x*y*z, 2*x^7*y^2*z + 4, x^7*y^2*z, x^7*y^2*z, 1, 2)
%17 = 2*i*z*y*x - 2
? spol(x*y + z^3, z^2 - 3*z, x*y, z^2, 1, 1)
%15 = 3*z*y*x + z^5

a.
$ \begin {eqnarray*} S(4x^2z - 7y^2, xyz^2 + 3xz^4) &=& \frac{x^2yz^2}{4x^2z}\cdot (4x^2z - 7y^2) - \frac{x^2yz^2}{xyz^2}\cdot (xyz^2 + 3xz^4)\\ &=& \frac{1}{4}yz \cdot (4x^2z - 7y^2) - x \cdot (xyz^2 + 3xz^4) \\ &=& x^2yz^2 - \frac{7}{4}y^3z - x^2yz^2 - 3x^2z^4 \\ &=& -3x^2z^4-\frac{7}{4}y^3z \end {eqnarray*}$

以下、途中式は省略。

b.
$ S(x^4y - z^2, 3xz^2 - y) = \frac{1}{3}x^3y^2 - z^4$

c.
$ S(x^7y^2z + 2ixyz, 2x^7y^2z + 4) = 2ixyz - 2$

d.
$ S(xy+z^3, z^2-3z) = 3xyz + z^5$

 

No.11




左辺を変形

$$\begin {eqnarray*} S(x^\alpha f, x^\beta g) &=& \frac{\text{LCM}({\scriptstyle \text{LM}}(x^\alpha f), {\scriptstyle \text{LM}}(x^\beta g))}{{\scriptstyle \text{LT}}(x^\alpha f)}\cdot x^\alpha f - \frac{\text{LCM}({\scriptstyle \text{LM}}(x^\alpha f), {\scriptstyle \text{LM}}(x^\beta g))}{{\scriptstyle \text{LT}}(x^\beta g)}\cdot x^\beta g \\ &=& \frac{\text{LCM}(x^\alpha {\scriptstyle \text{LM}}(f), x^\beta {\scriptstyle \text{LM}}(g))}{x^\alpha {\scriptstyle \text{LT}}(f)}\cdot x^\alpha f - \frac{\text{LCM}(x^\alpha {\scriptstyle \text{LM}}(f), x^\beta {\scriptstyle \text{LM}}(g))}{x^\beta {\scriptstyle \text{LT}}(g)}\cdot x^\beta g \\ &=& \frac{\text{LCM}(x^\alpha {\scriptstyle \text{LM}}(f), x^\beta {\scriptstyle \text{LM}}(g))}{{\scriptstyle \text{LT}}(f)}\cdot f - \frac{\text{LCM}(x^\alpha {\scriptstyle \text{LM}}(f), x^\beta {\scriptstyle \text{LM}}(g))}{{\scriptstyle \text{LT}}(g)}\cdot g \end {eqnarray*} $$

右辺を変形

$$\begin {eqnarray*} x^\gamma S(f,g) &=& x^\gamma \cdot \Bigl(\frac{\text{LCM}({\scriptstyle \text{LM}}(f), {\scriptstyle \text{LM}}(g))}{{\scriptstyle \text{LT}}(f)}\cdot f - \frac{\text{LCM}({\scriptstyle \text{LM}}(f), {\scriptstyle \text{LM}}(g))}{{\scriptstyle \text{LT}}(g)}\cdot g\Bigr) \\ &=& \frac{\text{LCM}(x^\alpha {\scriptstyle \text{LM}}(f), x^\beta {\scriptstyle \text{LM}}(g))}{\text{LCM}({\scriptstyle \text{LM}}(f), {\scriptstyle \text{LM}}(g))} \cdot \Bigl(\frac{\text{LCM}({\scriptstyle \text{LM}}(f), {\scriptstyle \text{LM}}(g))}{{\scriptstyle \text{LT}}(f)}\cdot f - \frac{\text{LCM}({\scriptstyle \text{LM}}(f), {\scriptstyle \text{LM}}(g))}{{\scriptstyle \text{LT}}(g)}\cdot g\Bigr) \\ &=& \text{LCM}(x^\alpha {\scriptstyle \text{LM}}(f), x^\beta {\scriptstyle \text{LM}}(g)) \cdot \Bigl(\frac{1}{{\scriptstyle \text{LT}}(f)}\cdot f - \frac{1}{{\scriptstyle \text{LT}}(g)}\cdot g\Bigr) \\ &=& \frac{\text{LCM}(x^\alpha {\scriptstyle \text{LM}}(f), x^\beta {\scriptstyle \text{LM}}(g))}{{\scriptstyle \text{LT}}(f)}\cdot f - \frac{\text{LCM}(x^\alpha {\scriptstyle \text{LM}}(f), x^\beta {\scriptstyle \text{LM}}(g))}{{\scriptstyle \text{LT}}(g)}\cdot g \end {eqnarray*} $$

したがって、$S(x^\alpha f, x^\beta g) = x^\gamma S(f,g)$

□

 

また、
$\text{LCM}(x^\alpha {\scriptstyle \text{LM}}(f), x^\beta {\scriptstyle \text{LM}}(g)) = \text{LCM}({\scriptstyle \text{LM}}(x^\alpha f), {\scriptstyle \text{LM}}(x^\beta g))$
${\scriptstyle \text{LM}}(f) \mid {\scriptstyle \text{LM}}(x^\alpha f)$
${\scriptstyle \text{LM}}(g) \mid {\scriptstyle \text{LM}}(x^\alpha g)$
より、
$\text{LCM}({\scriptstyle \text{LM}}(f), {\scriptstyle \text{LM}}(g)) \mid \text{LCM}(x^\alpha {\scriptstyle \text{LM}}(f), x^\beta {\scriptstyle \text{LM}}(g))$

$x^\alpha$, $x^\beta$, ${\scriptstyle \text{LM}}(f)$, ${\scriptstyle \text{LM}}(g)$ はいずれも単項式であるから、$x^\gamma$ もまた単項式である。