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 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 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！
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