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[函数] 多项式恒等式的规律

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其妙 Posted at 2013-10-18 18:58:15 |Read mode
多项式.gif

上述多项式恒等式有如下规律:
1、左边是关于(x+1)的多项式,右边是关于(x)的多项式,
2、如果左边次幂的系数为正,则右边的相应次幂的系数与左边的一样;如果左边次幂的系数为负,则右边的相应次幂的系数取为左边的相反数!

是否能给出此种类型的更高次的恒等式或一般情况???
妙不可言,不明其妙,不着一字,各释其妙!

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kuing Posted at 2013-10-18 19:02:54
你第二句话就不能简略一下……直接说右边系数为左边对应项系数的绝对值不就好了

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kuing Posted at 2013-10-18 23:36:27
还有,你给出的 f(x) 还有一个规律,就是左边系数都是正负交替的,有没有这一要求?

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kuing Posted at 2013-10-19 00:08:28
还有,你给出的 f(x) 还有一个规律,就是左边系数都是正负交替的,有没有这一要求? ...
kuing 发表于 2013-10-18 23:36

看来应该没有这要求,否则就没什么难度了。
刚才找了一个,左边系数不正负交替的:$(x+1)^4-2(x+1)^3+(x+1)^2+4(x+1)-2=x^4+2x^3+x^2+4x+2$

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 Author| 其妙 Posted at 2013-10-19 12:38:19
一般的n次多项式呢?
妙不可言,不明其妙,不着一字,各释其妙!

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isee Posted at 2013-10-21 20:52:10
还真美,至少一眼猜不出来

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tommywong Posted at 2013-10-26 19:15:50
Last edited by tommywong at 2013-10-28 13:57:00解线性方程即可

$\displaystyle a(x+1)^3-\frac{3a}{2}(x+1)^2+b(x+1)-(-\frac{a}{4}+\frac{b}{2})=ax^3+\frac{3a}{2}x^2+bx+(-\frac{a}{4}+\frac{b}{2})$

$
\begin{pmatrix}
-1-1 & 1\\0 & 1-1
\end{pmatrix}
\begin{pmatrix}
c_0\\c_1
\end{pmatrix}
=
\begin{pmatrix}
0\\0
\end{pmatrix}
$

$
\begin{pmatrix}
1-1 & -1 & 1\\0 & -1-1 & 2\\0 & 0 & 1-1
\end{pmatrix}
\begin{pmatrix}
c_0\\c_1\\c_2
\end{pmatrix}
=
\begin{pmatrix}
0\\0\\0
\end{pmatrix}
$

$
\begin{pmatrix}
-1-1 & 1 & -1 & 1\\0 & 1-1 & -2 & 3\\0 & 0 & -1-1 & 3\\0 & 0 & 0 & 1-1
\end{pmatrix}
\begin{pmatrix}
c_0\\c_1\\c_2\\c_3
\end{pmatrix}
=
\begin{pmatrix}
0\\0\\0\\0
\end{pmatrix}
$
现充已死,エロ当立。
维基用户页:https://zh.wikipedia.org/wiki/User:Tttfffkkk
Notable algebra methods:https://artofproblemsolving.com/community/c728438
《方幂和及其推广和式》 数学学习与研究2016.

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 Author| 其妙 Posted at 2013-10-27 15:56:26
回复 7# tommywong
妙不可言,不明其妙,不着一字,各释其妙!

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爪机专用 Posted at 2013-10-28 03:01:48
回复 8# 其妙

你看懂了?

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tommywong Posted at 2013-10-28 08:58:36
Last edited by tommywong at 2013-10-28 13:39:00以下再推广到$a_0 (x+k)^0+...+a_n (x+k)^n$的展开式。

$\displaystyle\sum_{i=1}^{n+1}a_{i-1}(x+k)^{i-1}= \sum_{i=1}^{n+1}x^{i-1}\sum_{j=1}^{n-i+1}C_{j-1}^{i-1}k^{j-i}a_{j-1}= \sum_{i=1}^{n+1}b_{i-1}x^{i-1}$

$
\begin{pmatrix}
b_0\\b_1\\b_2\\b_3
\end{pmatrix}
=
\begin{pmatrix}
C_0^0 k^0 & C_1^0 k^1 & C_2^0 k^2 & C_3^0 k^3\\
0 & C_1^1 k^{1-1} & C_2^1 k^{2-1} & C_3^1 k^{3-1}\\
0 & 0 & C_2^2 k^{2-2} & C_3^2 k^{3-2}\\
0 & 0 & 0 & C_3^3 k^{3-3}
\end{pmatrix}
\begin{pmatrix}
a_0\\a_1\\a_2\\a_3
\end{pmatrix}
$

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tommywong Posted at 2013-10-28 10:03:19
Last edited by tommywong at 2013-10-28 14:08:00多项式的矩阵化真奇妙。

$(x+2)(2x-5)=2x^2-x-10$

$
\begin{pmatrix}
c_0\\c_1\\c_2
\end{pmatrix}
=
\begin{pmatrix}
2 & 0\\
1 & 2\\
0 & 1
\end{pmatrix}
\begin{pmatrix}
-5\\2
\end{pmatrix}
=
\begin{pmatrix}
-10\\-1\\2
\end{pmatrix}
$

$(x+2)^3+2(x+2)^2+3(x+2)+1=x^3+8x^2+23x+23$

$
\begin{pmatrix}
b_0\\b_1\\b_2\\b_3
\end{pmatrix}
=
\begin{pmatrix}
1 & 2 & 4 & 8\\
0 & 1 & 4 & 12\\
0 & 0 & 1 & 6\\
0 & 0 & 0 & 1
\end{pmatrix}
\begin{pmatrix}
1\\3\\2\\1
\end{pmatrix}
=
\begin{pmatrix}
23\\23\\8\\1
\end{pmatrix}
$
现充已死,エロ当立。
维基用户页:https://zh.wikipedia.org/wiki/User:Tttfffkkk
Notable algebra methods:https://artofproblemsolving.com/community/c728438
《方幂和及其推广和式》 数学学习与研究2016.

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isee Posted at 2013-10-28 10:38:27
Last edited by isee at 2013-10-28 10:49:00这像是LaTex,直接丢代码上来多好

矩阵这块全部遗忘了~看不懂了~

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青青子衿 Posted at 2014-6-21 16:57:15
这种东西可以用综合除法、二项式定理或泰勒公式解决
\[f(x)=x^4+3x^2+4
=(x-1+1)^4+3(x-1+1)^2+4\]
(用二项式定理展开得,记得系数1、4、6、4、1)
\[=(x-1)^4+4(x-1)^3+6(x-1)^2+4(x-1)+1+3(x-1)^2+6(x-1)+3+3\]
\[=(x-1)^4+7(x-1)^3+6(x-1)^2+10(x-1)+7\]

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 Author| 其妙 Posted at 2014-6-21 18:25:51
回复 13# 青青子衿
干嘛?

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青青子衿 Posted at 2014-6-21 20:48:38
回复 14# 其妙
回复  青青子衿
干嘛?
其妙 发表于 2014-6-21 18:25

只是(ˇˍˇ) 想~说
将多项式$f(x)$表成$x-c$的方幂和
设$f(x)=b_n(x-c)^n+b_{n-1}(x-c)^{n-1}+\cdots+b_1(x-c)+b_0$
问题:如何求系数$b_n, b_{n-1},\cdots,b_1,b_0$?
因$f(x)=[b_n(x-c)^{n-1}+b_{n-1}(x-c)^{n-2}+\cdots+b_1](x-c)+b_0$
则$b_0$是$f(x)$被$x-c$除所得的余数;方括号内为商式,记作$q_1(x)$,又因
   $q_1(x)=b_n(x-c)^{n-1}+b_{n-1}(x-c)^{n-2}+\cdots+b_1
   =[b_n(x-c)^{n-2}+b_{n-1}(x-c)^{n-3}+\cdots+b_2](x-c)+b_1$
则$b_1$是$q_1(x)$被$x-c$除所得的余数;方括号内为商式,记作$q_2(x)$,如此继续,所得的余数即为$b_n, b_{n-1},\cdots,b_1,b_0$
将$f(x)=x^4- 3x^2+3$表成$x-1$的方幂和
\[\begin{gathered}
  \left. {\underline {\, 
 {\begin{array}{*{20}{r}}
  1&0&{ - 3}&0&3 \\ 
  {}&1&1&{ - 2}&{ - 2} 
\end{array}} \,}}\! \right| \begin{array}{*{20}{c}}
  1 \\ 
  {} 
\end{array} \\
  \left. {\underline {\, 
 {\begin{array}{*{20}{r}}
  1&1&{ - 2}&{ - 2}&{1\left( { = {b_0}} \right)} \\ 
  {}&1&2&0&{} 
\end{array}} \,}}\! \right| \begin{array}{*{20}{c}}
  {} \\ 
  {} 
\end{array} \\
  \left. {\underline {\, 
 {\begin{array}{*{20}{r}}
  1&2&0&{ - 2\left( { = {b_1}} \right)}&{} \\ 
  {}&1&3&{}&{} 
\end{array}} \,}}\! \right| \begin{array}{*{20}{c}}
  {} \\ 
  {} 
\end{array} \\
  \left. {\underline {\, 
 {\begin{array}{*{20}{r}}
  1&3&{3\left( { = {b_2}} \right)}&{}&{} \\ 
  {}&1&{}&{}&{} 
\end{array}} \,}}\! \right| \begin{array}{*{20}{c}}
  {} \\ 
  {} 
\end{array} \\
  \begin{array}{*{20}{c}}
  {1\left( { = {b_4}} \right)}&{4\left( { = {b_3}} \right)}&{}&{} 
\end{array} \\ 
\end{gathered} \]
$f(x)=(x-1)^4+4(x-1)^3+3(x-1)^2-2(x-1)+1$

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 Author| 其妙 Posted at 2014-6-21 23:14:35
回复 15# 青青子衿
哦,我以为你在解决楼主的猜想呢

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爪机专用 Posted at 2014-6-21 23:42:18
看上去是机器码。。。

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 Author| 其妙 Posted at 2014-6-22 16:34:21
回复 17# 爪机专用
,软件做的?

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