已知 x+y+z=1 ，x^2+y^2+z^2=2 ，x^3+y^3+z^3=5 ，求 x^5+y^5+z^5

王守恩

mathematica 发表于 2019-12-6 09:20
\[
\begin{array}{cc}
5 & 11 \\

     ”爬楼梯“问题
\(x\ +\ y\ +\ z\ =1\)
\(x^2+y^2+z^2=2\)
\(x^3+y^3+z^3=3\)
............
\(x^n+y^n+z^n=a(n)\)
LinearRecurrence[{ 1, 1/2 , 1/6 }, { 1, 2, 3 }, n]

     ”爬楼梯“问题
\(x\ +\ y\ +\ z\ =1\)
\(x^2+y^2+z^2=2\)
\(x^3+y^3+z^3=5\)
............
\(x^n+y^n+z^n=a(n)\)
LinearRecurrence[{ 1, 1/2 , 5/6 }, { 1, 2, 5 }, n]

     ”爬楼梯“问题
\(x\ +\ y\ +\ z\ =a\)
\(x^2+y^2+z^2=b\)
\(x^3+y^3+z^3=c\)
............
\(x^n+y^n+z^n=a(n)\)
LinearRecurrence[{a, (b-a^2)/2 , (2c-3ab+a^3)/6}, {a, b, c}, n]

     ”爬楼梯“问题
\(x\ +\ y\ +\ z\ =1\)
\(x^2+y^2+z^2=3\)
\(x^3+y^3+z^3=4\)
............
\(x^n+y^n+z^n=a(n)\)
LinearRecurrence[{1, 1, 0}, {1, 3, 4}, n]
a(n)=1, 3, 4, 7, 11, 18, 29, 47, 76, 123, 199, 322, 521, 843,
1364, 2207, 3571, 5778, 9349, 15127, 24476, 39603, 64079,
103682, 167761, 271443, 439204, 710647, 1149851, 1860498,
3010349, 4870847, 7881196, 12752043, 20633239, ................

葡萄糖

\begin{align*}
6\left(u^{\color{red}5}+v^{\color{red}5}+w^{\color{red}5}\right)=&\phantom{+1}\left(u+v+w\right)^{\color{red}5}\\
&-{\color{red}5}\left(u+v+w\right)^3\left(u^2+v^2+w^2\right)\\
&+{\color{red}5}\left(u+v+w\right)^2\left(u^3+v^3+w^3\right)\\
&+ {\color{red}5}\left(u^2+v^2+w^2\right)\left(u^3+v^3+w^3\right)
\end{align*}

wayne

这个问题 论坛讨论过很多次. 轮换对称多项式有一个牛顿恒等式.
设$e_1=x+y+z,e_2=xy+yz+zx, e_3=xyz$,  $p_k = x^k+y^k+z^k$,那么 $p_1 =e_1 =1, p_2=e_1^2-2e_2 =2 ,p_3=e_1^3-3e_1e_2+3e_3 =3$ ,解得$e_1=1,e_2=-1/2,e_3=1/6$
所以,$p_5=e_1^5-5 e_2 e_1^3+5 e_3 e_1^2+5 e_2^2 e_1-5 e_2 e_3 =6$

更进一步的,对于一般情况,$ke_{k}=\sum _{i=1}^{k}(-1)^{i-1}p_{i}e_{k-i}$,  得知存在 恒等式: $x^{k+1} + y^{k+1} + z^{k+1} = (x+y+z)(x^k + y^k + z^k) - (x y + y z + z x) (x^{k-1} + y^{k-1} + z^{k-1}) + x y z (x^{k-2} + y^{k-2} + z^{k-2}) $ ,所以最后一问也得证了.

参考链接:https://en.wikipedia.org/wiki/Ring_of_symmetric_functions

classstruct

视为三阶线性递推数列an,a(n+3)=(x+y+z)a(n+2)-(xy+yz+zx)a(n+1)+xyza(n),得a(n+3)=a(n+2)+1/2a(n+1)+1/6a(n).
故a4=25/6,a5=6,103/72

mathematica

classstruct 发表于 2019-12-6 14:13
视为三阶线性递推数列an,a(n+3)=(x+y+z)a(n+2)-(xy+yz+zx)a(n+1)+xyza(n),得a(n+3)=a(n+2)+1/2a(n+1)+1/6a( ...

原来是这个恒等式，我现在彻底明白了！