三數之和 - 拼圖
By Kelly
at 2019-12-23T11:20
at 2019-12-23T11:20
Table of Contents
※ 引述《Ryow (哈扣)》之銘言:
: : 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 = ?
: : 延伸問題:x^n+y^n+z^n = ?
: : 提示:請別強行去解x,y,z,會少許多解題樂趣!
: 三個未知數 三個方程式
: 先求出它們的關係吧
: x+y+z=1
: xy+yz+zx=?
: xyz=?
: (x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)
: =>1=2+2(xy+yz+zx)
: =>xy+yz+zx=-(1/2)
: x^3+y^3+z^3-3xyz=(x+y+z)[x^2+y^2+z^2-(xy+yz+zx)]
: =>3-3xyz=1[2+(1/2)]
: =>xyz=1/6
: 不會化簡x^5+y^5+z^5 只好拜託估狗 找到了這個
: (x+y+z)^5-x^5-y^5-z^5=5(y+z)(x+y)(z+x)(x^2+y^2+z^2+xy+yz+zx)
: 又x+y+z=1
: 所以
: (x+y+z)^5-x^5-y^5-z^5=5(1-x)(1-y)(1-z)(x^2+y^2+z^2+xy+yz+zx)
: =5[1-(x+y+z)+(xy+yz+zx)-xyz](x^2+y^2+z^2+xy+yz+zx)
: =>1-(x^5+y^5+z^5)=5[1-1-(1/2)-(1/6)][2-(1/2)]
: =>x^5+y^5+z^5=6
: 順便算了一下x^4+y^4+z^4=25/6
: 可能還要多算幾項才能找到x^n+y^n+z^n的一般式吧 我放棄 O_Q
其實只差一點點 XD
(x^n + y^n + z^n) * (x + y + z) = F(n) * F(1)
= (x^{n+1} + y^{n+1} + z^{n+1}) + xy^n + xz^n + yx^n + yz^n + zx^n + zy^n
= F(n+1) + xy(y^{n-1} + x^{n-1}) + yz(y^{n-1}+z^{n-1}) + xz(x^{n-1}+z^{n-1})
= F(n+1) + (xy + yz + xz)(x^{n-1} + y^{n-1} + z^{n-1}) - (xyz^{n-1} + xy^{n-1}z + x^{n-1}yz)
= F(n+1) + (xy + yz + xz)F(n-1) - xyz F(n-2)
代入 xy+yz+zx = -1/2, xyz = 1/6, F(1) = 1
F(n) = F(n+1) - F(n-1)/2 - F(n-2)/6
整理一下:
F(n+1) = F(n) + F(n-1)/2 + F(n-2)/6
--
: : 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 = ?
: : 延伸問題:x^n+y^n+z^n = ?
: : 提示:請別強行去解x,y,z,會少許多解題樂趣!
: 三個未知數 三個方程式
: 先求出它們的關係吧
: x+y+z=1
: xy+yz+zx=?
: xyz=?
: (x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)
: =>1=2+2(xy+yz+zx)
: =>xy+yz+zx=-(1/2)
: x^3+y^3+z^3-3xyz=(x+y+z)[x^2+y^2+z^2-(xy+yz+zx)]
: =>3-3xyz=1[2+(1/2)]
: =>xyz=1/6
: 不會化簡x^5+y^5+z^5 只好拜託估狗 找到了這個
: (x+y+z)^5-x^5-y^5-z^5=5(y+z)(x+y)(z+x)(x^2+y^2+z^2+xy+yz+zx)
: 又x+y+z=1
: 所以
: (x+y+z)^5-x^5-y^5-z^5=5(1-x)(1-y)(1-z)(x^2+y^2+z^2+xy+yz+zx)
: =5[1-(x+y+z)+(xy+yz+zx)-xyz](x^2+y^2+z^2+xy+yz+zx)
: =>1-(x^5+y^5+z^5)=5[1-1-(1/2)-(1/6)][2-(1/2)]
: =>x^5+y^5+z^5=6
: 順便算了一下x^4+y^4+z^4=25/6
: 可能還要多算幾項才能找到x^n+y^n+z^n的一般式吧 我放棄 O_Q
其實只差一點點 XD
(x^n + y^n + z^n) * (x + y + z) = F(n) * F(1)
= (x^{n+1} + y^{n+1} + z^{n+1}) + xy^n + xz^n + yx^n + yz^n + zx^n + zy^n
= F(n+1) + xy(y^{n-1} + x^{n-1}) + yz(y^{n-1}+z^{n-1}) + xz(x^{n-1}+z^{n-1})
= F(n+1) + (xy + yz + xz)(x^{n-1} + y^{n-1} + z^{n-1}) - (xyz^{n-1} + xy^{n-1}z + x^{n-1}yz)
= F(n+1) + (xy + yz + xz)F(n-1) - xyz F(n-2)
代入 xy+yz+zx = -1/2, xyz = 1/6, F(1) = 1
F(n) = F(n+1) - F(n-1)/2 - F(n-2)/6
整理一下:
F(n+1) = F(n) + F(n-1)/2 + F(n-2)/6
--
Tags:
拼圖
All Comments
By Charlotte
at 2019-12-28T09:12
at 2019-12-28T09:12
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