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楼主: 数学星空

[讨论] 关于五次代数方程可根式求解问题

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 楼主| 发表于 2014-12-11 22:30:28 | 显示全部楼层
1.设\(\alpha\)是方程\(\frac{x^{11}-1}{x-1}=0\)最小幅角的复根.

   \(32x^5+3349456x^4-5941616812296x^3-585145514845851080x^2+147013447513276833423286x+15377302441624829616294559439\)

   \(=(\alpha^8+\alpha^3-x)(\alpha^9+\alpha^2-x)(\alpha^9+\alpha^8+\alpha^7+\alpha^6+\alpha^5+\alpha^4+\alpha^3+\alpha^2+x+1)(\alpha^7+\alpha^4-x)(\alpha^6+\alpha^5-x)\)


2.设\(\alpha\)是方程\(\frac{x^{31}-1}{x-1}=0\)最小幅角的复根
   \(x^5+x^4-12x^3-21x^2+x+5\)

   \(=(\alpha^{29}+\alpha^{28}+\alpha^{27}+\alpha^{24}+\alpha^{23}+\alpha^{22}+\alpha^{21}+\alpha^{20}+\alpha^{19}+\alpha^{18}+\alpha^{17}+\alpha^{16}+\alpha^{15}+\alpha^{14}+\alpha^{13}+\alpha^{12}+\alpha^{11}+\alpha^{10}+\alpha^9+\alpha^8+\alpha^7+\alpha^4+\alpha^3+\alpha^2+x+1)(\alpha^{29}+\alpha^{21}+\alpha^{19}+\alpha^{12}+\alpha^{10}+\alpha^2-x)(\alpha^{27}+\alpha^{24}+\alpha^{20}+\alpha^{11}+\alpha^7+\alpha^4-x)(\alpha^{28}+\alpha^{18}+\alpha^{16}+\alpha^{15}+\alpha^{13}+\alpha^3-x)(\alpha^{23}+\alpha^{22}+\alpha^{17}+\alpha^{14}+\alpha^9+\alpha^8-x)\)

3.设\(\beta\)是方程\(\frac{x^{41}-1}{x-1}=0\)最小幅角的复根
   \(y^5+y^4-16y^3+5y^2+21y-9\)

   \(=(\beta^{39}+\beta^{37}+\beta^{36}+\beta^{35}+\beta^{34}+\beta^{33}+\beta^{31}+\beta^{30}+\beta^{29}+\beta^{28}+\beta^{26}+\beta^{25}+\beta^{24}+\beta^{23}+\beta^{22}+\beta^{21}+\beta^{20}+\beta^{19}+\beta^{18}+\beta^{17}+\beta^{16}+\beta^{15}+\beta^{13}+\beta^{12}+\beta^{11}+\beta^{10}+\beta^8+\beta^7+\beta^6+\beta^5+\beta^4+\beta^2+y+1)(\beta^{33}+\beta^{31}+\beta^{30}+\beta^{24}+\beta^{17}+\beta^{11}+\beta^{10}+\beta^8-y)(\beta^{34}+\beta^{25}+\beta^{22}+\beta^{21}+\beta^{20}+\beta^{19}+\beta^{16}+\beta^7-y)(\beta^{37}+\beta^{36}+\beta^{29}+\beta^{26}+\beta^{15}+\beta^{12}+\beta^5+\beta^4-y)(\beta^{39}+\beta^{35}+\beta^{28}+\beta^{23}+\beta^{18}+\beta^{13}+\beta^6+\beta^2-y)\)

4.设\(\beta\)是方程\(\frac{x^{61}-1}{x-1}=0\)最小幅角的复根
   \(y^5+y^4-24y^3-17y^2+41y-13\)

   \(=(\beta^{59}+\beta^{58}+\beta^{42}+\beta^{39}+\beta^{35}+\beta^{33}+\beta^{28}+\beta^{26}+\beta^{22}+\beta^{19}+\beta^3+\beta^2-y)(\beta^{57}+\beta^{56}+\beta^{55}+\beta^{52}+\beta^{44}+\beta^{38}+\beta^{23}+\beta^{17}+\beta^9+\beta^6+\beta^5+\beta^4-y)(\beta^{53}+\beta^{51}+\beta^{49}+\beta^{46}+\beta^{43}+\beta^{34}+\beta^{27}+\beta^{18}+\beta^{15}+\beta^{12}+\beta^{10}+\beta^8-y)(\beta^{59}+\beta^{58}+\beta^{57}+\beta^{56}+\beta^{55}+\beta^{54}+\beta^{53}+\beta^{52}+\beta^{51}+\beta^{49}+\beta^{46}+\beta^{45}+\beta^{44}+\beta^{43}+\beta^{42}+\beta^{41}+\beta^{39}+\beta^{38}+\beta^{37}+\beta^{36}+\beta^{35}+\beta^{34}+\beta^{33}+\beta^{31}+\beta^{30}+\beta^{28}+\beta^{27}+\beta^{26}+\beta^{25}+\beta^{24}+\beta^{23}+\beta^{22}+\beta^{20}+\beta^{19}+\beta^{18}+\beta^{17}+\beta^{16}+\beta^{15}+\beta^{12}+\beta^{10}+\beta^9+\beta^8+\beta^7+\beta^6+\beta^5+\beta^4+\beta^3+\beta^2+y+1)(\beta^{54}+\beta^{45}+\beta^{41}+\beta^{37}+\beta^{36}+\beta^{31}+\beta^{30}+\beta^{25}+\beta^{24}+\beta^{20}+\beta^{16}+\beta^7-y)\)



5.设\(\beta\)是方程\(\frac{x^{71}-1}{x-1}=0\)最小幅角的复根

   \(y^5+y^4-28y^3+37y^2+25y+1\)

   \(=(\beta^{67}+\beta^{65}+\beta^{62}+\beta^{57}+\beta^{50}+\beta^{49}+\beta^{38}+\beta^{33}+\beta^{22}+\beta^{21}+\beta^{14}+\beta^9+\beta^6+\beta^4-y)(\beta^{69}+\beta^{68}+\beta^{67}+\beta^{66}+\beta^{65}+\beta^{64}+\beta^{63}+\beta^{62}+\beta^{61}+\beta^{60}+\beta^{59}+\beta^{58}+\beta^{57}+\beta^{56}+\beta^{55}+\beta^{54}+\beta^{53}+\beta^{52}+\beta^{50}+\beta^{49}+\beta^{47}+\beta^{46}+\beta^{44}+\beta^{43}+\beta^{42}+\beta^{40}+\beta^{38}+\beta^{36}+\beta^{35}+\beta^{33}+\beta^{31}+\beta^{29}+\beta^{28}+\beta^{27}+\beta^{25}+\beta^{24}+\beta^{22}+\beta^{21}+\beta^{19}+\beta^{18}+\beta^{17}+\beta^{16}+\beta^{15}+\beta^{14}+\beta^{13}+\beta^{12}+\beta^{11}+\beta^{10}+\beta^9+\beta^8+\beta^7+\beta^6+\beta^5+\beta^4+\beta^3+\beta^2+y+1)(\beta^{69}+\beta^{68}+\beta^{64}+\beta^{60}+\beta^{52}+\beta^{46}+\beta^{40}+\beta^{31}+\beta^{25}+\beta^{19}+\beta^{11}+\beta^7+\beta^3+\beta^2-y)(\beta^{66}+\beta^{63}+\beta^{59}+\beta^{53}+\beta^{44}+\beta^{43}+\beta^{42}+\beta^{29}+\beta^{28}+\beta^{27}+\beta^{18}+\beta^{12}+\beta^8+\beta^5-y)(\beta^{61}+\beta^{58}+\beta^{56}+\beta^{55}+\beta^{54}+\beta^{47}+\beta^{36}+\beta^{35}+\beta^{24}+\beta^{17}+\beta^{16}+\beta^{15}+\beta^{13}+\beta^{10}-y)\)

毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
 楼主| 发表于 2014-12-11 22:36:40 | 显示全部楼层
关于26楼的分解结果,我现在有几个问题需要大家讨论:

1.关于整系数的五次方程若可以利用三角函数表达分解,且根记为\(\frac{x^k-1}{x-1}=0\),那么\(k\)可以取哪些整数?

2.如何利用单位根构造整系数的五次方程?

3.如何判断一个整系数的五次方程可以用单位根分解,其单位根如何求得?
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
 楼主| 发表于 2014-12-14 00:55:45 | 显示全部楼层
通过计算发现:\(k=11,31,41,61,71,101,..\).时可以构造整系数的五次方程?

对于\(k=101\)我还没有构造出整系数的五次方程,谁有兴趣构造一个看看?
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
发表于 2014-12-14 09:04:42 来自手机 | 显示全部楼层
不要依赖于数学工具。你说它是复数就是复数,说它实数就是实数。需要注意,复数范围冪运算是多值函数。而且你这种写法不是很规范

点评

多值函数,数学工具只给出其中一个值,所以结果就应工具而异了  发表于 2014-12-15 17:12
数学工具可以检验猜测的正确性哈..我修改了26#的描述  发表于 2014-12-14 11:58
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
 楼主| 发表于 2014-12-14 12:20:51 | 显示全部楼层
通过几天的努力终于找到了\(k=101,131,151\)的整系数五次方程:

\(k=101时, y^5+y^4-40y^3+93y^2-21y-17\)

\(k=131时, y^5+y^4-52y^3-89y^2+109y+193\)

\(k=151时, y^5+y^4-60y^3-12y^2+784y+128\)

三角函数域内分解.png

有兴趣的可以按照图片中的命令验算一下.
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
 楼主| 发表于 2014-12-14 12:39:10 | 显示全部楼层
寻找k值的思路如下:

1.先试着分解\(\frac{x^k-1}{x-1}\),\(k=10m+1\),若不能分解,则转到下一步

2.寻找 \(y^i \mod k (i=1..k,y=1..k)\)使其解集数目\(=2m\) 的\(y\)值,找到后,取最小的\(y\)值,及其解集\(A\)

3.作差集\(\{0...{k-1}\}-A\)得到集合\(B\)

4.则有\(y=-\sum_{j\in B}^{} \beta^j\)

5.寻找最小五次多项式使\(f(y)=0\) 即可

6.使用maple软件验算,\(f(y)\)能否在\(Root(\frac{x^k-1}{x-1})\)三角函数域内分解.
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
 楼主| 发表于 2014-12-14 16:18:24 | 显示全部楼层
对于一般的整系数\(p\)次方程寻找\(k\)的问题:我们可以按照楼上的方案得到找到对应的分圆域内的分解实例.

只需要\(k=2pm+1\),且\(p,k\)均为素数.其它的求解步骤是一致的.

例如对于\(p=7\)有

我们可设\(\beta\)为\(\frac{x^{29}-1}{x-1}=0\)最小幅角的复根.

\(y^7+y^6-12y^5-7y^4+28y^3+14y^2-9y+1\)

\(=(\beta^{18}+\beta^{16}+\beta^{13}+\beta^{11}-y)(\beta^{25}+\beta^{19}+\beta^{10}+\beta^{4}-y)(\beta^{26}+\beta^{22}+\beta^{7}+\beta^{3}-y)(\beta^{27}+\beta^{24}+\beta^{5}+\beta^{2}-y)(\beta^{23}+\beta^{15}+\beta^{14}+\beta^{6}-y)(\beta^{27}+\beta^{26}+\beta^{25}+\beta^{24}+\beta^{23}+\beta^{22}+\beta^{21}+\beta^{20}+\beta^{19}+\beta^{18}+\beta^{16}+\beta^{15}+\beta^{14}+\beta^{13}+\beta^{11}+\beta^{10}+\beta^9+\beta^8+\beta^7+\beta^6+\beta^5+\beta^4+\beta^3+\beta^2+y+1)(\beta^{21}+\beta^{20}+\beta^9+\beta^8-y)\)

可以得到一般的可利用\(k\)次分圆域进行分解的\(p\)次整系数有如下形式:

\(y^p+y^{p-1}-m(p-1)y^{p-2}+...=0 \)

上面的问题还有另一种描述:

记\(A_0\)={\(e^{\frac{2n\pi I}{k}},n=0..2pm\)},则需要寻找一种分类方式满足下列要求:

1.对于集合\(A_0\)需要除掉\(2m\)个元素后得到集合\(B_0\)

2.然后对\(B_0\)除去1元素后,分成\(p-1\)组子集,使每组有\(2m\)个元素,并记每个子集为\(B_i ( i=1..p-1)\)

3.分别记\(B_j (j=0..p-1)\)中各个元素之和为\(s_j (j=0..p-1)\)

4.使其\(s_j(j=0..p-1)\)为\(p\)次整系数方程的根.



毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
发表于 2014-12-15 10:23:46 来自手机 | 显示全部楼层
如果你那个k是素数,很容易证明p模10余1
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
发表于 2014-12-15 10:28:37 来自手机 | 显示全部楼层
另外由于Q[x^k-1]是Q的k-1次阶扩张而且是可解群,那么子群都是正规的
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
 楼主| 发表于 2014-12-15 19:47:58 | 显示全部楼层
不知mathe能否算出能在分圆域内(\(k=2mp+1\))分解的一般五次方程的具体系数?

通过计算,我们可以得到

\(x^5+a_4 x^4+a_3 x^3+a_2 x^2+a_1 x+a_0=0\)

\(a_4=1\)

\(a_3=-m(p-1)\)

\(a_2=?\)

\(a_1=?\)

\(a_0=?\)

毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
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