数学研发论坛

 找回密码
 欢迎注册
查看: 508|回复: 45

[原创] 解方程 z^(1/z) =1

[复制链接]
发表于 2019-9-1 12:54:21 | 显示全部楼层 |阅读模式

马上注册,结交更多好友,享用更多功能,让你轻松玩转社区。

您需要 登录 才可以下载或查看,没有帐号?欢迎注册

x
求方程 z^(1/z) =1 在复域的所有解.
ps: 这个图形挺好看的,^_^
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
发表于 2019-9-1 13:51:59 | 显示全部楼层
$z^{1/z}$在附属范围是多值函数,不大好定义
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 2019-9-1 14:26:07 | 显示全部楼层
本帖最后由 .·.·. 于 2019-9-1 15:14 编辑

(前面那个版本推错了,对复数不熟,出了BUG)
z^(1/z) =1
|z|^(1/z) =1
|z|^(1/z)=1
设1/z=c1+c2i
|z|^(c1+c2i)=1
于是|z|=1或1/z=c2i==>z=-i/c2是纯虚数

第一种情况是,我们想求所有角度theta使得z=e^(i*theta)且e^(i*theta*(e^(-i*theta)))=1
于是i*theta*(e^(-i*theta))=2k*pi,这里k为整数
e^(-i*theta)==2k*pi/(i*theta)
两边取模,1=2k*pi/|theta|,于是必须有theta=2k*pi
带回原式验证知,此时全部 z^(1/z) =1的解只有z=1

第二种情况,(c i)^(i/c)=1(这里c是实数)
c^(i/c) exp(i*pi/2)^(i/c)=1
c^(i/c) exp(-pi/2/c)=1
exp(-pi/2/c)=(1/c)^(i/c)
记d=1/c
exp(-d*pi/2)=d^(d*i)=exp(i*d*log(d))
右边的模长为1,带到左边推得d=0,这是不可能的,故这种情况无解

这样算没错吧

点评

请后世写小说的如果看到这个帖子,一定不要把这层楼当正确答案  发表于 2019-9-1 22:50
……找到问题了……我错误地以为a^b=abs(a)^b,而事实上复数操作比我相信的更见鬼一点  发表于 2019-9-1 22:50
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 2019-9-1 15:12:00 | 显示全部楼层
虚部和实部的函数图像如下,目前来看,不止一个解:
[BTW: 这个图案有没有觉得像极了海里面的水母,小章鱼]
  1. n=1;f[x_] := x^(1/x) - 1;
  2. ContourPlot[{Re[f[x+I y]]==0,Im[f[x+I y]]==0},{x,-n,n},{y,-n,n},PlotPoints->200,Mesh->{Range[0,0],Range[0,0]},MeshFunctions->{Re[#2]&,Im[#2]&},MeshStyle->{Directive[Thickness[.005],Red],Directive[Dashed,Thickness[.005],Blue]},RegionFunction->Function[{z,f},Abs[f]>=0],BoundaryStyle->None,Frame->False,Axes->True,ContourStyle->{Directive[Thickness[.003],Red],Directive[Dashed,Thickness[.003],Blue]}]
复制代码


1.png

点评

蓝虚线跟红实线的交点  发表于 2019-9-1 15:47
第二个解在哪儿?我用模长分析的结果是,只有一个解啊  发表于 2019-9-1 15:46
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 2019-9-1 15:51:39 | 显示全部楼层
wayne 发表于 2019-9-1 15:12
虚部和实部的函数图像如下,目前来看,不止一个解:

In[23]:= Reduce[z^(1/z) == 1, z]

Out[23]= C[1] \[Element]
  Integers && ((C[1] != 0 &&
     z != 0 && (z == E^-ProductLog[-1, -2 I \[Pi] C[1]] ||
       z == E^-ProductLog[-2 I \[Pi] C[1]] ||
       z == E^-ProductLog[1, -2 I \[Pi] C[1]])) || z == 1)

这是Mathematica给出的结果
然而,比如在z == E^-ProductLog[-1, -2 I \[Pi] C[1]]这个式子里代入C[1]=2
In[35]:= N[(E^-ProductLog[-1, -4 I \[Pi]])^E^
ProductLog[-1, -4 I \[Pi]]]

Out[35]= 0.260873 + 0.0925388 I
返回的并不是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
 楼主| 发表于 2019-9-1 16:02:26 | 显示全部楼层

  1. FindRoot[z^(1/z) == 1, {z, (1 + I)/5}, WorkingPrecision -> 100]
复制代码

通过画图工具的坐标拾取功能,再FindRoot,找到六个解:
  1. 0.1516470112665713905204173940029219628603867983585655060601360985617741306714619614502703747846509247+ 0.2133322736921120994941810069972356770928609128256466476372760720088570456846193738202853589477296211 I
  2. 0.08312880120865399879299978743561608756414380292422761341187530554767721069797155315764691853415162260+0.1431324538050240937502404575907663153892540094778484931233639177536372818300824213506663342399762397 I
  3. 0.05776985230069788122293630341355646840840247949476850567608998641847403033040680622784801894348854500+0.1104633391928621218014073556110027085448090321087229766434737063634039112108053003441557567889831104 I
  4. 0.04443934318760827606648693959667064039954135522939569937499701762809257342509361230976000371516208575+0.09108451548053340629264489442496095472233834332804291306306783147449326808594883822671850522112124331 I
  5. 0.03618538111116619893637488994096977762484811231546247313066631295790555257465145227253662420381644352+0.07807489389384545245815743468522807423391343758123928345098282637393023612578429907923640281081826444 I
  6. 0.03055850548490778885678880428835085635257933446605025858667273594417493787926796932240772978735310191+0.06865607446817378158522934164752129377418930719993746053413700492706881097974643797403693742529210560 I
复制代码
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 2019-9-1 16:22:30 | 显示全部楼层
解是图中绿色和黄色曲线的交点,且都在蓝色曲线上。

图象.jpg

点评

没看懂图像,跟我的不太一样呢  发表于 2019-9-1 17:30
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 2019-9-1 16:45:04 | 显示全部楼层
经过半手工半软件化简,证实,虚部和实部的交点 在分别在 实曲线\[(x^2+y^2)^{\frac{x}{2 (x^2+y^2)}}  - e^ {-\frac{y (4k\pi\left(x^2+y^2\right)+y \log \left(x^2+y^2\right))}{2x(x^2+y^2)}} =0,   k\in Z\]
1.png

幅角满足关系: \[Arg[x+yi] = \frac{4k\pi\left(x^2+y^2\right)+y \log \left(x^2+y^2\right)}{2 x}\]

模满足关系 \[\frac{ln r}{r} = 2k\pi sin\theta,    x=r cos\theta,y= r sin\theta\]
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 2019-9-1 18:09:06 | 显示全部楼层
z可以写成$\exp(\ln(z)+2m\pi i)$, 所以$z^{1/z}$可以写成$\exp({\ln(z)+2m\pi i}/z)$,要求其为1就是存在整数n使得${\ln(z)+2m\pi i}/z=2n\pi i$,即$\ln(z)=2n\pi i z -2m\pi i $
或者可以写成$\ln(r)+i\theta=-2n\pi r \sin(\theta) +2n\pi r \cos(\theta) i -2m\pi i$
所以${\ln(r)}/r = -2n\pi\sin(\theta), \theta = 2n\pi r\cos(\theta)-2m\pi$

点评

由于$\sin(t)$很小,t接近$\pi$的整数倍,所以要求$2nr$也接近整数,也就是对于较大的r,需要接近半整数  发表于 2019-9-2 08:17
还可以改为$\ln(r)=-2n\pi r\sin(t), t=2n\pi r\cos(t)$,只是t的绝对值可以任意大,可以看出除了r很小的情况,通常$sin(t)$的绝对值可以很小(取n很大时,于是$\cos(t)$接近1, t接近$2n\pi r$  发表于 2019-9-2 08:15
我们重新定义变量$t=\theta+ 2m\pi$,那么 $\ln(r) = -2n\pi r\sin(t), t = 2n\pi r\cos(t)$, ^_^  发表于 2019-9-2 08:12
应该不可以。只是数学软件在计算时,通常只会选择m=0,也就是辐角看成$[0,2\pi)$ (也有可能选择$[-\pi,\pi)$?)  发表于 2019-9-2 08:02
这个$m$可以直接取0吧,反正我们可以把$\theta$放在$[0,2\pi]$里  发表于 2019-9-2 07:51
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 2019-9-2 08:54:37 | 显示全部楼层
现在完美了, 每设定一个$n$值,得到一个解,跟前面的计算完全一致.
目标就是解方程组:
\[\log (r)=-2 \pi  n r \sin (\theta ),\theta =2 \pi  n r \cos (\theta )\]

  1. Table[{n,r Cos[\[Theta]]+I r Sin[\[Theta]]/.FindRoot[{Log[r]==-2n r Pi Sin[\[Theta]],\[Theta]==2n Pi r Cos[\[Theta]]},{r,1},{\[Theta],Pi/2},WorkingPrecision->100]},{n,20}]//Column
复制代码

  1. {1,0.1516470112665713905204173940029219628603867983585655060601360985617741306714619614502703747846509259+0.2133322736921120994941810069972356770928609128256466476372760720088570456846193738202853589477296203 I}
  2. {2,0.0831288012086539987929997874356160875641438029242276134118753055476772106979715531576469185341516226+0.1431324538050240937502404575907663153892540094778484931233639177536372818300824213506663342399762397 I}
  3. {3,0.0577698523006978812229363034135564684084024794947685056760899864184740303304068062278480189434885450+0.1104633391928621218014073556110027085448090321087229766434737063634039112108053003441557567889831104 I}
  4. {4,0.0444393431876082760664869395966706403995413552293956993749970176280925734250936123097600037151620857+0.0910845154805334062926448944249609547223383433280429130630678314744932680859488382267185052211212433 I}
  5. {5,0.0361853811111661989363748899409697776248481123154624731306663129579055525746514522725366242038164435+0.0780748938938454524581574346852280742339134375812392834509828263739302361257842990792364028108182644 I}
  6. {6,0.0305585054849077888567888042883508563525793344660502585866727359441749378792679693224077297873531045+0.0686560744681737815852293416475212937741893071999374605341370049270688109797464379740369374252921071 I}
  7. {7,0.02647041318110019626665462300810380149767341211900563150700914893095530272581314803371779359341696469+0.06147968252981000247476281928533015879344716925026374349569502160595755489375061195925342675989166292 I}
  8. {8,0.02336251439753375179028089363324510927248222955685674749547083153471979435470956519676568248911072416+0.05580622297037512818946664593616474723260749566426199324060280759438573921266671862243661036835451075 I}
  9. {9,0.02091812461544603162932543792536068410613923594268748863988176499620393206326661547527167020999531045+0.05119373038932076774684364240787050040744249960437433290846230211001325251811068034051561142793544638 I}
  10. {10,0.01894409649680947551024167505894875663628721836392610640222944476737200177799882996298379846003643641+0.04736061729203994669532458194836462357289263408286344761948107997659432087274375791543396869472400122 I}
  11. {11,0.01731583255728843294625096270920639761733696016174611377386098419172704266085043746017749417586010314+0.04411836524065593888748495721627977916576350520311557146536421941376076763019787246726001205462896359 I}
  12. {12,0.01594929975431846154867497681605190987570146687438586572682620366907235759741695433275501555641833653+0.04133567063208957623358240488745714864543066584377843220788120993459429431268560351767127778329340263 I}
  13. {13,0.01478573488851878402436623656418000753854578361309075428433047209300402003245884807551944960041518811+0.03891807283923533967659339619764285073122919147945469107360904995126236124022888227934063279901737995 I}
  14. {14,0.01378279300004241966789466962324865423890865665979104165869985863807974371698565714465542516402703965+0.03679575870906313760136004598010113635674424834936616745075900613182479488689486697754964671923461809 I}
  15. {15,0.01290917732352009440934885880089908524556068324420916946307719491637464795297163915656900194851089429+0.03491593957964537545482289302567518018703578907189182946900934812506953995815956608339461260912974904 I}
  16. {16,0.01214125011013250887365467008997554211631499352349513227531919412257738006287179824461039281348775439+0.03323791004101481444922052799363501095638323419805668995404290907399423114357233267524983375906928453 I}
  17. {17,0.01146082047850419356898502198247625253946373439041235737034793695309816122860420249249172818769303754+0.03172974404803113988702275909249241679077343246301331527835459050071773468019398013313067026886137962 I}
  18. {18,0.01085365806810113203509796081369725276089854269752654597191099310068666202263261096336570393686134097+0.03036602577062283035976027475394245629515016655072636337909820694134747061254847526925022816602503543 I}
  19. {19,0.01030846882825902618067780108839888751161551334341270124161236262183514245155215577942427692682574843+0.02912625418148743525885871145809873892376118555880663532041719018746916091865636911868104293701070368 I}
  20. {20,0.00981617341562437429805816257355258523298474885581608610270743533090887987993188867541403091385869450+0.02799369795293926677443080587875184221188204847863068738846482064872628837749963551228331434756062680 I}

复制代码
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
您需要登录后才可以回帖 登录 | 欢迎注册

本版积分规则

小黑屋|手机版|数学研发网 ( 苏ICP备07505100号 )

GMT+8, 2019-9-23 00:32 , Processed in 0.066363 second(s), 20 queries .

Powered by Discuz! X3.4

© 2001-2017 Comsenz Inc.

快速回复 返回顶部 返回列表