多项式的余式小题

青青子衿

多项式\(\,f(x)=-1+x^5+x^{20}+x^{2019}\,\)除以多项式\(\,g(x)=x^4+x^3+x+1\,\)所得余式为\(\,r(x)\,\)，求\(\,r(0)\,\)=?

1. PolynomialRemainder[-1 + x^5 + x^20 + x^2019, x^4 + x^3 + x + 1, x]

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1. f[x_]:=-1+x^5+x^20+x^2019
2. Df[x_]=D[f[x],x]
3. r[x_]:=Subscript[a,0]+Subscript[a,1]x+Subscript[a,2]x^2+Subscript[a,3]x^3
4. Dr[x_]=D[r[x],x]
5. f[-1]
6. Df[-1]
7. f[E^((I/3)*Pi)]//FullSimplify
8. f[E^(-(I/3)*Pi)]//FullSimplify
9. r[-1]
10. Dr[-1]
11. ...
12. Solve[Subscript[a,0]-Subscript[a,1]+Subscript[a,2]-Subscript[a,3]==-2&&
13.   Subscript[a,1]-2 Subscript[a,2]+3 Subscript[a,3]==2004&&
14.   Subscript[a,0]+E^((I/3)*Pi) Subscript[a,1]-E^(-(I/3)*Pi) Subscript[a,2]-Subscript[a,3]==-2&&
15.   Subscript[a,0]+E^(-(I/3)*Pi) Subscript[a,1]-E^((I/3)*Pi) Subscript[a,2]-Subscript[a,3]==-2,
16.   Subscript[a,#]&/@{0,1,2,3}]

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https://www.quora.com/unanswered/What-will-be-the-remainder-if-1-x-5-x-20-x-2019-is-divided-by-x-4-x-3-x-1

tommywong

$f(x)\equiv -1+(-1)x^2+(-1)^6 x^2+(-1)^{673}\equiv -2\pmod{x^3+1}$

$r(x)=k(x^3+1)-2$

$f’(-1)=5-20+2019=2004$

$r(x)=668(x^3+1)-2$

爪机专用

回复 2# tommywong

看了半天才反应过来，奶思！果然是多项式高手，学习鸟

色k

写个笨方法，虽然麻烦，总算是靠自己撸的，计算量也不算大……

首先因式分解除式 `g(x)=(x+1)^2(x^2-x+1)`，由此可得
\[\frac3{g(x)}=\frac1{x+1}+\frac1{(x+1)^2}+\frac{1-x}{x^2-x+1},\]进一步分解最后一项，记 `p`, `q` 为 `x^2-x+1` 的两根，则 `p^3=q^3=-1`，且
\[\frac{1-x}{x^2-x+1}=\frac{1-p}{p-q}\cdot\frac1{x-p}-\frac{1-q}{p-q}\cdot\frac1{x-q},\]于是利用二项式定理，有
\begin{align*}
\frac{3f(x)}{g(x)}={}&\frac{x^{2019}+x^{20}+x^5-1}{x+1}+\frac{x^{2019}+x^{20}+x^5-1}{(x+1)^2}\\
&+\frac{1-p}{p-q}\cdot\frac{x^{2019}+x^{20}+x^5-1}{x-p}-\frac{1-q}{p-q}\cdot\frac{x^{2019}+x^{20}+x^5-1}{x-q}\\
={}&T(x)+\frac{(-1)^{2019}+(-1)^{20}+(-1)^5-1}{x+1}\\
&+\frac{(-1)^{2019}+2019(-1)^{2018}(x+1)+(-1)^{20}+20(-1)^{19}(x+1)+(-1)^5+5(-1)^4(x+1)-1}{(x+1)^2}\\
&+\frac{1-p}{p-q}\cdot\frac{p^{2019}+p^{20}+p^5-1}{x-p}-\frac{1-q}{p-q}\cdot\frac{q^{2019}+q^{20}+q^5-1}{x-q}\\
={}&T(x)+\frac{-2}{x+1}+\frac{2004x+2002}{(x+1)^2}+\frac{1-p}{p-q}\cdot\frac{-2}{x-p}-\frac{1-q}{p-q}\cdot\frac{-2}{x-q}\\
={}&T(x)+\frac{-2}{x+1}+\frac{2004x+2002}{(x+1)^2}+(-2)\cdot\frac{1-x}{x^2-x+1}\\
={}&T(x)+\frac{6(334x^3+333)}{(x^2-x+1)(1+x)^2},
\end{align*}其中 `T(x)` 为多项式，所以 `r(x)=2(334x^3+333)`。

青青子衿

\begin{align*}
f(x)&=-1+x^5+x^{20}+x^{2019}\\
f'(x)&=5x^4+20x^{19}+2019x^{2018}
\end{align*}
\begin{align*}
g(x)&=1+x+x^{3}+x^{4}\\
&=(1+x)^2(1-x+x^2)
\end{align*}
\begin{align*}
f(x)&=g(x)q(x)+r(x)\\
&=(1+x)^2(1-x+x^2)q(x)+r(x)\\
&=(1+x)^2(\small\frac{1-\sqrt{3}\,i}{2}-x)(\small\frac{1+\sqrt{3}\,i}{2}-x)q(x)+r(x)\\
&=(1+x)^2(e^{\frac{\pi}{3}i}-x)(e^{{\small{-}}\frac{\pi}{3}i}-x)q(x)+(a_0+a_1x+a_2x^2+a_3x^3)\\
\end{align*}
\begin{align*}
r(x)&=a_0+a_1x+a_2x^2+a_3x^3\\
r'(x)&=a_1+2a_2x+3a_3x^2
\end{align*}

\begin{align*}
g(-1)&=0&g(e^{\frac{\pi}{3}i})&=0&f(-1)&=-2&f(e^{\frac{\pi}{3}i})&=-2&\\
g'(-1)&=0&g(e^{{\small{-}}\frac{\pi}{3}i})&=0&f'(-1)&=2004&f(e^{{\small{-}}\frac{\pi}{3}i})&=-2\\
\end{align*}
\begin{align*}
\begin{cases}
\begin{split}
r(-1)&=f(-1)\\
r'(-1)&=f'(-1)\\
r(e^{\frac{\pi}{3}i})&=f(e^{\frac{\pi}{3}i})\\
r(e^{{\small{-}}\frac{\pi}{3}i})&=f(e^{{\small{-}}\frac{\pi}{3}i})
\end{split}
\end{cases}
\Rightarrow
\begin{cases}
\begin{split}
a_0-a_1+a_2-a_3&=-2\\
a_1-2a_2+3a_3&=2004\\
a_0+e^{\frac{\pi}{3}i}a_1-e^{{\small{-}}\frac{\pi}{3}i}a_2-a_3&=-2\\
a_0+e^{{\small{-}}\frac{\pi}{3}i}a_1-e^{\frac{\pi}{3}i}a_2-a_3&=-2
\end{split}
\end{cases}
\end{align*}
\begin{cases}
\begin{split}
a_0&&&&=666\\
&a_1&&&=0\\
&&a_2&&=0\\
&&&a_3&=668
\end{split}
\end{cases}