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

markfang2050

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

补充内容 (2019-12-6 14:45):
已知 x+y+z=1 ，x^2+y^2+z^2=2 ，x^3+y^3+z^3=3 ，求 x^5+y^5+z^5=？

chyanog

1. {y[n] == a y[n - 1] + 1/2 (-a^2 + b) y[n - 2] + 1/6 (a^3 - 3 a b + 2 c) y[n - 3], y[1] == a, y[2] == b, y[3] == c};
   
2. 
   
3. LinearRecurrence[{a, 1/2 (-a^2 + b), 1/6 (a^3 - 3 a b + 2 c)}, {a, b,  c}, 10] /. {a -> 1, b -> 2, c -> 3}

复制代码

补充内容 (2019-12-6 10:22):
Solve[{x+y+z==a,x^2+y^2+z^2==b,x^3+y^3+z^3==c,x y+y z+z x==p,x y z==q},{p,q},{x,y,z}]

markfang2050

math_humanbeing 发表于 2019-12-5 22:16
x^5+y^5+z^5答案是不是
-1

NO

markfang2050

math_humanbeing 发表于 2019-12-5 22:40
1/6

NO,6

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.

mathematica

chyanog 发表于 2019-12-5 22:13

1. Clear["Global`*"];(*Clear all variables*)
   
2. aa=Solve[x+y+z==1&&x^2+y^2+z^2==2&&x^3+y^3+z^3==5,{x,y,z}];
   
3. bb=x^5+y^5+z^5/.aa;
   
4. cc=FullSimplify[bb]
   

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结果是
{11, 11, 11, 11, 11, 11}

6次方是
{223/12, 223/12, 223/12, 223/12, 223/12, 223/12}

王守恩

这串数也可以。
\(x\ +\ y\ +\ z\ =1=a(1)\)
\(x^2+y^2+z^2=3=a(2)\)
\(x^3+y^3+z^3=4=a(3)\)
............
\(x^n+y^n+z^n=a(n)\)
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, ................

mathematica

mathematica 发表于 2019-12-6 09:01
结果是
{11, 11, 11, 11, 11, 11}

1. Clear["Global`*"];(*Clear all variables*)
   
2. (*求解出方程的根*)
   
3. (*解析解求解太累人，太累机器了，所以用数值解，虽然数值，但是求解很快*)
   
4. aa=NSolve[x+y+z==1&&x^2+y^2+z^2==2&&x^3+y^3+z^3==5,{x,y,z},200];
   
5. bb={};(*用来保存求解结果的变量*)
   
6. Do[cc=Union@RootApproximant[x^k+y^k+z^k/.aa];(*用数值来得到近似值*)
   
7.    bb=Append[bb,Flatten@{k,cc}],(*把结果搞到一起去*)
   
8.    {k,5,20}];
   
9. Print[bb]
   

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\[
\begin{array}{cc}
5 & 11 \\
6 & \frac{223}{12} \\
7 & \frac{268}{9} \\
8 & \frac{3473}{72} \\
9 & \frac{1415}{18} \\
10 & \frac{55099}{432} \\
11 & \frac{22361}{108} \\
12 & \frac{290587}{864} \\
13 & \frac{353897}{648} \\
14 & \frac{1532459}{1728} \\
15 & \frac{1866475}{1296} \\
16 & \frac{72743411}{31104} \\
17 & \frac{29531999}{7776} \\
18 & \frac{383658403}{62208} \\
19 & \frac{467269063}{46656} \\
20 & \frac{6070407673}{373248} \\
\end{array}
\]

{{5, 11}, {6, 223/12}, {7, 268/9}, {8, 3473/72}, {9, 1415/18}, {10,
  55099/432}, {11, 22361/108}, {12, 290587/864}, {13, 353897/
  648}, {14, 1532459/1728}, {15, 1866475/1296}, {16, 72743411/
  31104}, {17, 29531999/7776}, {18, 383658403/62208}, {19, 467269063/
  46656}, {20, 6070407673/373248}}

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*}