代数方程的虚根判定问题

dingjifen

mathe 发表于 2020-1-20 08:10
f=1+x/2!+x^2/4!+x^3/6!
polroots(f)
%11 = [-2.4646042998750414383960323257279253689 + 0.E-38*I, -13 ...

有啥法子来证明——当n为所有正整数时，1楼两个方程有无虚根呢？

zeroieme

dingjifen 发表于 2020-1-20 20:03
看不明白是啥意思？  能否具体说明一下？

两系列方程从n=1算到n=20

dingjifen

zeroieme 发表于 2020-1-20 17:37
{{n=1,{{x->-2.}}},{n=2,{{x->-9.4641},{x->-2.5359}}},{n=3,{{x->-2.4646},{x->-13.7677+10.1285 I},{ ...

是不是可以这样说，通过计算表明——当方程次数n大于2时，1楼两个方程始终有虚根？

dingjifen

zeroieme 发表于 2020-1-20 17:37
{{n=1,{{x->-2.}}},{n=2,{{x->-9.4641},{x->-2.5359}}},{n=3,{{x->-2.4646},{x->-13.7677+10.1285 I},{ ...

理论上可以证明——当方程次数n相当大时，1楼两个方程无虚根。

wayne

（1）1+x/2!+x²/4!+x³/6!+……+xⁿ/(2n)!=0
（2）1+x/3!+x²/5!+x³/7!+……+xⁿ/(2n+1)!=0

高斯代数基本定理: 复域内一元n次方程有n个复根.  算复根没啥子劲,我们推高一下楼主的问题的难度.

容易用软件验证,两个方程 大概有$n/4$个实根,${3n}/4$个虚根.  比如$n=1000$的时候,第一个方程有$236$个实根,第二个方程有$234$个实根.
更一般的结论是, 当n为偶数的时候, 第一个方程的实根个数 比第二个方程的实根个数 多两个.  当n为奇数的时候, 第一个方程的实根个数 跟 第二个方程的实根个数 相等.

这两个猜想 有人能证明或者正否吗?
下面给的数据是n=1-500的时候,实根的个数, 按8取模分成8组,发现,规律非常明显. 基本上都是线性递增,方差为2,偶尔会在一两处重复.

1. {{1,1},{9,3},{17,5},{25,7},{33,9},{41,11},{49,13},{57,15},{65,17},{73,19},{81,19},{89,21},{97,23},{105,25},{113,27},{121,29},{129,31},{137,33},{145,35},{153,37},{161,39},{169,41},{177,43},{185,45},{193,47},{201,49},{209,49},{217,51},{225,53},{233,55},{241,57},{249,59},{257,61},{265,63},{273,65},{281,67},{289,69},{297,71},{305,73},{313,75},{321,77},{329,79},{337,79},{345,81},{353,83},{361,85},{369,87},{377,89},{385,91},{393,93},{401,95},{409,97},{417,99},{425,101},{433,103},{441,105},{449,107},{457,109},{465,109},{473,111},{481,113},{489,115}}
   
2. {{2,2},{10,4},{18,6},{26,6},{34,8},{42,10},{50,12},{58,14},{66,16},{74,18},{82,20},{90,22},{98,24},{106,26},{114,28},{122,30},{130,32},{138,34},{146,36},{154,38},{162,38},{170,40},{178,42},{186,44},{194,46},{202,48},{210,50},{218,52},{226,54},{234,56},{242,58},{250,60},{258,62},{266,64},{274,66},{282,68},{290,68},{298,70},{306,72},{314,74},{322,76},{330,78},{338,80},{346,82},{354,84},{362,86},{370,88},{378,90},{386,92},{394,94},{402,96},{410,98},{418,98},{426,100},{434,102},{442,104},{450,106},{458,108},{466,110},{474,112},{482,114},{490,116}}
   
3. {{3,1},{11,3},{19,5},{27,7},{35,9},{43,11},{51,13},{59,15},{67,17},{75,19},{83,21},{91,23},{99,25},{107,25},{115,27},{123,29},{131,31},{139,33},{147,35},{155,37},{163,39},{171,41},{179,43},{187,45},{195,47},{203,49},{211,51},{219,53},{227,55},{235,57},{243,57},{251,59},{259,61},{267,63},{275,65},{283,67},{291,69},{299,71},{307,73},{315,75},{323,77},{331,79},{339,81},{347,83},{355,85},{363,87},{371,87},{379,89},{387,91},{395,93},{403,95},{411,97},{419,99},{427,101},{435,103},{443,105},{451,107},{459,109},{467,111},{475,113},{483,115},{491,117}}
   
4. {{4,2},{12,4},{20,6},{28,8},{36,10},{44,12},{52,14},{60,14},{68,16},{76,18},{84,20},{92,22},{100,24},{108,26},{116,28},{124,30},{132,32},{140,34},{148,36},{156,38},{164,40},{172,42},{180,44},{188,44},{196,46},{204,48},{212,50},{220,52},{228,54},{236,56},{244,58},{252,60},{260,62},{268,64},{276,66},{284,68},{292,70},{300,72},{308,74},{316,76},{324,76},{332,78},{340,80},{348,82},{356,84},{364,86},{372,88},{380,90},{388,92},{396,94},{404,96},{412,98},{420,100},{428,102},{436,104},{444,106},{452,106},{460,108},{468,110},{476,112},{484,114},{492,116}}
   
5. {{5,1},{13,3},{21,5},{29,7},{37,9},{45,11},{53,13},{61,15},{69,17},{77,19},{85,21},{93,23},{101,25},{109,27},{117,29},{125,31},{133,33},{141,33},{149,35},{157,37},{165,39},{173,41},{181,43},{189,45},{197,47},{205,49},{213,51},{221,53},{229,55},{237,57},{245,59},{253,61},{261,63},{269,63},{277,65},{285,67},{293,69},{301,71},{309,73},{317,75},{325,77},{333,79},{341,81},{349,83},{357,85},{365,87},{373,89},{381,91},{389,93},{397,93},{405,95},{413,97},{421,99},{429,101},{437,103},{445,105},{453,107},{461,109},{469,111},{477,113},{485,115},{493,117}}
   
6. {{6,2},{14,4},{22,6},{30,8},{38,10},{46,12},{54,14},{62,16},{70,18},{78,20},{86,22},{94,22},{102,24},{110,26},{118,28},{126,30},{134,32},{142,34},{150,36},{158,38},{166,40},{174,42},{182,44},{190,46},{198,48},{206,50},{214,52},{222,52},{230,54},{238,56},{246,58},{254,60},{262,62},{270,64},{278,66},{286,68},{294,70},{302,72},{310,74},{318,76},{326,78},{334,80},{342,82},{350,82},{358,84},{366,86},{374,88},{382,90},{390,92},{398,94},{406,96},{414,98},{422,100},{430,102},{438,104},{446,106},{454,108},{462,110},{470,112},{478,112},{486,114},{494,116}}
   
7. {{7,3},{15,5},{23,7},{31,9},{39,11},{47,11},{55,13},{63,15},{71,17},{79,19},{87,21},{95,23},{103,25},{111,27},{119,29},{127,31},{135,33},{143,35},{151,37},{159,39},{167,41},{175,41},{183,43},{191,45},{199,47},{207,49},{215,51},{223,53},{231,55},{239,57},{247,59},{255,61},{263,63},{271,65},{279,67},{287,69},{295,71},{303,71},{311,73},{319,75},{327,77},{335,79},{343,81},{351,83},{359,85},{367,87},{375,89},{383,91},{391,93},{399,95},{407,97},{415,99},{423,101},{431,101},{439,103},{447,105},{455,107},{463,109},{471,111},{479,113},{487,115},{495,117}}
   
8. {{8,2},{16,4},{24,6},{32,8},{40,10},{48,12},{56,14},{64,16},{72,18},{80,20},{88,22},{96,24},{104,26},{112,28},{120,30},{128,30},{136,32},{144,34},{152,36},{160,38},{168,40},{176,42},{184,44},{192,46},{200,48},{208,50},{216,52},{224,54},{232,56},{240,58},{248,60},{256,60},{264,62},{272,64},{280,66},{288,68},{296,70},{304,72},{312,74},{320,76},{328,78},{336,80},{344,82},{352,84},{360,86},{368,88},{376,90},{384,90},{392,92},{400,94},{408,96},{416,98},{424,100},{432,102},{440,104},{448,106},{456,108},{464,110},{472,112},{480,114},{488,116},{496,118}}

复制代码

mathe

wayne的修改还不错。dingfen需要注意一下。虽然说论坛中提供了大家自由交流的机会，但是大家在发布内容时还是应该尽量避免犯一些低级的错误，至少像本题中题目应该提前做一些数值验证。如果对数学软件不熟悉，可以趁机熟悉一下。

mathe

我们可以考虑函数$f_{2n}(x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}-....+(-1)^n\frac{x^{2n}}{(2n)!}$
类似定义$f_{2n+1}(x)=\frac{x}{1!}-\frac{x^3}{3!}+\frac{x^5}{5!}-....+(-1)^n\frac{x^{2n+1}}{(2n+1)!}$
以偶数情况为例子，我们查看
$cos(x)-f_{2n}(x)=\sum_{h=n+1}^{\infty}(-1)^{h}\frac{x^{2h}}{(2h)!}$
由Stirling公式，在$(2h)!>(\frac{2h}{e})^{2h}$, 所以在$|x|<\frac{2n+2}{e}$,
上面级数中每项绝对值小于1，而且逐项绝对值单调递减趋向0，由于级数符号是交错的，我们可以得出
$|cos(x)-f_{2n}(x)|<|\frac{x^{2n+2}}{(2n+2)!}|<1$ (其中$|x|<\frac{2n+2}{e}$）
有因为$cos(h\pi)=(-1)^h$,, 所以我们知道在$|h\pi|<\frac{2n+2}{e}|$时， $f_{2n}(h\pi)$和$cos(h\pi)$同号
，所以$f_{2n}(x)$在$|x|<\frac{2n+2}{e}$的范围内实数根不少于$cos(x)$,折算成wayne题目里面实数根数目约等于$\frac{2n}{e\pi}$

wayne

厉害厉害,  $1000 *2/{\pi e} = 234.199$ ,跟数程序计算出来的 $236$ 非常接近

dingjifen

kastin 发表于 2020-1-19 23:03
详见杨翠红、朱思铭、梁肇军的论文《多项式代数方程根的完全分类及其应用》。

理论上可以证明——当方程次数n相当大时，1楼两个方程无虚根。

mathe

随着n增大，虚根数目是不减的，所以必然虚根数目只会越来越多，而实根的比例接近$\frac2{\pi e}$才是合理的，
只是所有的虚根绝对值越来越大而已，而绝对值较小的都会逐步转化为实根。