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
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=？ 本帖最后由 chyanog 于 2019-12-5 22:22 编辑
{y == a y + 1/2 (-a^2 + b) y + 1/6 (a^3 - 3 a b + 2 c) y, y == a, y == b, y == c};
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}
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}] math_humanbeing 发表于 2019-12-5 22:16
x^5+y^5+z^5答案是不是
-1
NO math_humanbeing 发表于 2019-12-5 22:40
1/6
NO,6 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.
chyanog 发表于 2019-12-5 22:13
Clear["Global`*"];(*Clear all variables*)
aa=Solve;
bb=x^5+y^5+z^5/.aa;
cc=FullSimplify
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} 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 发表于 2019-12-6 09:01
结果是
{11, 11, 11, 11, 11, 11}
Clear["Global`*"];(*Clear all variables*)
(*求解出方程的根*)
(*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*)
aa=NSolve;
bb={};(*用来保存求解结果的变量*)
Do;(*用数值来得到近似值*)
bb=Append,(*把结果搞到一起去*)
{k,5,20}];
\[
\begin{array}{cc}
5 & 11 \\
6 & \frac{223}{12} \\
7 & \frac{268}{9} \\
8 & \frac{3473}{72} \\
9 & \frac{1415}{18} \\
10 & \frac{55099}{432} \\
11 & \frac{22361}{108} \\
12 & \frac{290587}{864} \\
13 & \frac{353897}{648} \\
14 & \frac{1532459}{1728} \\
15 & \frac{1866475}{1296} \\
16 & \frac{72743411}{31104} \\
17 & \frac{29531999}{7776} \\
18 & \frac{383658403}{62208} \\
19 & \frac{467269063}{46656} \\
20 & \frac{6070407673}{373248} \\
\end{array}
\]
{{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}} 本帖最后由 王守恩 于 2019-12-6 17:28 编辑
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, ................ \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*}
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