dlpg070
发表于 2020-10-7 19:59:26
本帖最后由 dlpg070 于 2020-10-7 20:05 编辑
wayne 发表于 2020-10-7 17:31
谢谢你的代码
我找到加快的办法了.和你的思路相同
我的代码基本没改,只是NMaximize 去掉N -> Maximize
速度比较,位数<10000000 我们的速度基本一样,
但 10000000,100000000 我的突然变得特别慢,能指出原因吗?
我只计算到1千万位!没来得及比较数值是否相同
wayne dlpg
1000 0.0000779358 n=30.0000211678
10000 0.0000763322 n=40.0000202056
100000 0.0000737664 n=50.0000372039
1000000 0.0000628618 n=60.000149136
10000000 0.284114 n=7739.19
1000000003.96726 n=8没计算
dlpg070
发表于 2020-10-8 15:03:13
dlpg070 发表于 2020-10-7 19:59
谢谢你的代码
我找到加快的办法了.和你的思路相同
我的代码基本没改,只是NMaximize 去掉N -> Maximiz ...
1亿位的最后1000位比较: (最后1000位完全相同)
0258435688720118463303852633386720970982705135884310234348367547806974864950128341089836156441898182397216247290024318822536369575611957266956828066077684094406532148408875750176446859750636674815485336704365392769131678590057636783527488071530274061288344716742323555915915315151527061509777040395662428944037435515663651654148663742973728703646649699976151855093045446145522913825220022772794769403607676753148345991639851858073210426976379864872558008585704389852256853811145509578048515157248057180201958131409832586557737475328743461718654068424040627042556979851788733457679904926722152864633719375417139355195934667773942424731131055254171241233704166047680178087406531836273794128405054745491651466550041384390838998405690836550099571210200576746503703679501319431755122190248943109739729271256964728679754967232136693907831212228744144859546638807782553303799843761761606218397671793624146901739917433749683276183666934360958158134979396469095079602777452445890909321462774499298588167196671 Waynelast
025843568872011846330385263338672097098270513588431023434836754780697486495012834108983615644189818239721624729002431882253636957561195726695682806607768409440653214840887575017644685975063667481548533670436539276913167859005763678352748807153027406128834471674232355591591531515152706150977704039566242894403743551566365165414866374297372870364664969997615185509304544614552291382522002277279476940360767675314834599163985185807321042697637986487255800858570438985225685381114550957804851515724805718020195813140983258655773747532874346171865406842404062704255697985178873345767990492672215286463371937541713935519593466777394242473113105525417124123370416604768017808740653183627379412840505474549165146655004138439083899840569083655009957121020057674650370367950131943175512219024894310973972927125696472867975496723213669390783121222874414485954663880778255330379984376176160621839767179362414690173991743374968327618366693436095815813497939646909507960277745244589090932146277449929858816719667176627974354632002798335617 dlpg 多27位
多出27位纯属意外,有"赶"的想法,没有"超"的愿望,
这个节日虽没吃月饼,但收获良多,过得挺痛快,心中有"数"就好!
aimisiyou 的猜想坚定了我大小园方案
mathe的求解公式开创了公式解的先河,内切圆半径为1的设定,影响了其后所有公式
wayne的公式另辟蹊径,1亿位的精度令人叹为观止,鞭策我跃马扬鞭,奋蹄急追
王守恩的公式简单易懂漂亮,突飞猛进,
我也依样画葫芦,搞了个自以为不算太丑的公式和代码
经过大家的讨论,k的最大值计算获得出乎意料的圆满结果
王守恩的扩展题:三角形内放2个园,可能更复杂,更具挑战性,望有意者另立主题
茶馆很开心,开心来茶馆
王守恩
发表于 2020-10-12 13:34:37
本帖最后由 王守恩 于 2020-10-12 13:36 编辑
问题(1):三角形内放1个圆,圆与三角形面积比=k,求k最大值
NMaximize\(\D\bigg[\frac{1+\frac{3a(1-a^0)}{1-a}}{\cot^2(\theta)\tan(2\theta)},\ \frac{\pi}{2} >\theta> 0, \ \ \theta\ \bigg]\)
问题(2):三角形内放2个圆,圆与三角形面积比=k,求k最大值
NMaximize\(\D\bigg[\frac{1+\frac{b(1-b)}{1-b}+\frac{2c(1-c^0)}{1-c}}{\cot^2(\theta)\tan(2\theta)},\ \frac{\pi}{2} >\theta> 0, \ \ \theta\ \bigg]\)
问题(3):三角形内放3个圆,圆与三角形面积比=k,求k最大值
NMaximize\(\D\bigg[\frac{1+\frac{2c(1-c)}{1-c}+\frac{b(1-b^0)}{1-b}}{\cot^2(\theta)\tan(2\theta)},\ \frac{\pi}{2} >\theta> 0, \ \ \theta\ \bigg]\)
问题(4):三角形内放4个圆,圆与三角形面积比=k,求k最大值
NMaximize\(\D\bigg[\frac{1+\frac{3a(1-a)}{1-a}}{\cot^2(\theta)\tan(2\theta)},\ \frac{\pi}{2} >\theta> 0, \ \ \theta\ \bigg]\)
问题(5):三角形内放5个圆,圆与三角形面积比=k,求k最大值
NMaximize\(\D\bigg[\frac{1+\frac{b(1-b^2)}{1-b}+\frac{2c(1-c)}{1-c}}{\cot^2(\theta)\tan(2\theta)},\ \frac{\pi}{2} >\theta> 0, \ \ \theta\ \bigg]\)
问题(6):三角形内放6个圆,圆与三角形面积比=k,求k最大值
NMaximize\(\D\bigg[\frac{1+\frac{2c(1-c^2)}{1-c}+\frac{b(1-b)}{1-b}}{\cot^2(\theta)\tan(2\theta)},\ \frac{\pi}{2} >\theta> 0, \ \ \theta\ \bigg]\)
问题(7):三角形内放7个圆,圆与三角形面积比=k,求k最大值
NMaximize\(\D\bigg[\frac{1+\frac{3a(1-a^2)}{1-a}}{\cot^2(\theta)\tan(2\theta)},\ \frac{\pi}{2} >\theta> 0, \ \ \theta\ \bigg]\)
问题(8):三角形内放8个圆,圆与三角形面积比=k,求k最大值
NMaximize\(\D\bigg[\frac{1+\frac{b(1-b^3)}{1-b}+\frac{2c(1-c^2)}{1-c}}{\cot^2(\theta)\tan(2\theta)},\ \frac{\pi}{2} >\theta> 0, \ \ \theta\ \bigg]\)
问题(9):三角形内放9个圆,圆与三角形面积比=k,求k最大值
NMaximize\(\D\bigg[\frac{1+\frac{2c(1-c^3)}{1-c}+\frac{b(1-b^2)}{1-b}}{\cot^2(\theta)\tan(2\theta)},\ \frac{\pi}{2} >\theta> 0, \ \ \theta\ \bigg]\)
记\(a=\big(\frac{1-\sin(\pi/2)}{1+\sin(\pi/2)}\big)^2,\ b=\big(\frac{1-\cos(2\theta)}{1+\cos(2\theta)}\big)^2,\ c=\big(\frac{1-\sin(\theta)}{1+\sin(\theta)}\big)^2\)
dlpg070
发表于 2020-10-15 10:52:51
本帖最后由 dlpg070 于 2020-10-15 12:18 编辑
王守恩 发表于 2020-10-12 13:34
问题(1):三角形内放1个圆,圆与三角形面积比=k,求k最大值
NMaximize\(\D\bigg[\frac{1+\frac{3a(1-a^0)}{1 ...
王守恩的公式和代码去手误后
王守恩的扩展9题:
计算结果: 10项
ok问题1底角=60. k=0.6046
ok问题2底角=68.2244 k=0.697033
ok问题3底角=50.8823 k=0.762235
ok问题4底角=60. k=0.806133
? 问题5底角=64.4386 k=0.816406
? 问题6底角=55.6499 k=0.823511
? 问题7底角=60. k=0.828526
? 问题8底角=61.6506 k=0.829522
? 问题9底角=58.4847 k=0.830333
? 问题10 底角=60. k=0.831014 (模仿放10园)
?问题100 底角=60. k=0.831325 (模仿放100园)
王守恩
发表于 2020-10-19 11:49:37
本帖最后由 王守恩 于 2020-10-19 11:57 编辑
dlpg070 发表于 2020-10-15 10:52
王守恩的公式和代码去手误后
王守恩的扩展9题:
计算结果: ...
茶馆很开心,开心来茶馆。允许我拓展一下:
1,主帖是从面积到面积,其实也可以面积到周长,周长到面积,周长到周长。
2,主帖是从三角形到圆,其实也可以是四角形到圆,五角形到圆,.........
3,有这样一串数:
\(\frac{\pi}{3\sqrt{3}},\frac{4\pi}{9\sqrt{3}},\frac{37\pi}{81\sqrt{3}},\frac{334\pi}{729\sqrt{3}},\frac{3007\pi}{6561\sqrt{3}},\frac{27064\pi}{59049\sqrt{3}},\frac{243577\pi}{531441\sqrt{3}},...\)
极限是\(\frac{11\pi}{24\sqrt{3}}=0.8313247.....\)
看分子:1,4,37,334,3007,27064,243577,2192194,......,规律怎么找?
dlpg070
发表于 2020-10-19 19:08:28
王守恩 发表于 2020-10-19 11:49
茶馆很开心,开心来茶馆。允许我拓展一下:
1,主帖是从面积到面积,其实也可以面积到周长,周长到面积, ...
分子数列:20项,对吗?
{1,4,37,334,3007,27064,243577,2192194,19729747,177567724,1598109517,
14382985654,129446870887,1165021837984,10485196541857,94366768876714,
849300919890427,7643708279013844,68793374511124597,619140370600121374}
dlpg070
发表于 2020-10-19 21:56:49
dlpg070 发表于 2020-10-19 19:08
分子数列:20项,对吗?
{1,4,37,334,3007,27064,243577,2192194,19729747,177567724,1598109517,
14382 ...
已经验证前34项
{1,4,37,334,3007,27064,243577,2192194,19729747,177567724,1598109517,14382985654,129446870887,1165021837984,10485196541857,94366768876714,849300919890427,7643708279013844,68793374511124597,619140370600121374,5572263335401092367,50150370018609831304,451353330167488481737,4062179971507396335634,36559619743566567020707,329036577692099103186364,2961329199228891928677277,26651962793060027358095494,239867665137540246222859447,2158808986237862216005735024,19429280876140759944051615217,174863527885266839496464536954,1573771750967401555468180832587,14163945758706613999213627493284}
dlpg070
发表于 2020-10-20 08:39:52
本帖最后由 dlpg070 于 2020-10-20 11:17 编辑
dlpg070 发表于 2020-10-19 21:56
已经验证前34项
{1,4,37,334,3007,27064,243577,2192194,19729747,177567724,1598109517,14382985654, ...
那串数的通项公式
an := 1/8 3^(-(3/2) - 2 n) (-3 + 11 9^n) \; n>=0
分子的通项 an*Sqrt/Pi求分子即可
(* 通项和极限 *)
Clear["Global`*"];
n0=30
an:=1/8 3^(-(3/2)-2 n) (-3+11 9^n) \;
tab=Table,
{n,0,33}];
Grid]]},{i,1,Length}],Alignment->Left]
listnew=Table]*Sqrt/Pi ],{i,1,34}];
Print["分子数列:34项"]
Flatten
list1new=Table]*Sqrt/Pi ],{i,1,34}];
Print["分母数列:34项"]
Flatten
lim=Limit,n->\];
Print["极限 limit=",lim," = ",N];
Print["=== 通项 end ==="]
计算结果:
1 \/(3 Sqrt)
2 (4 \)/(9 Sqrt)
3 (37 \)/(81 Sqrt)
4 (334 \)/(729 Sqrt)
5 (3007 \)/(6561 Sqrt)
6 (27064 \)/(59049 Sqrt)
7 (243577 \)/(531441 Sqrt)
8 (2192194 \)/(4782969 Sqrt)
9 (19729747 \)/(43046721 Sqrt)
10 (177567724 \)/(387420489 Sqrt)
11 (1598109517 \)/(3486784401 Sqrt)
12 (14382985654 \)/(31381059609 Sqrt)
13 (129446870887 \)/(282429536481 Sqrt)
14 (1165021837984 \)/(2541865828329 Sqrt)
15 (10485196541857 \)/(22876792454961 Sqrt)
16 (94366768876714 \)/(205891132094649 Sqrt)
17 (849300919890427 \)/(1853020188851841 Sqrt)
18 (7643708279013844 \)/(16677181699666569 Sqrt)
19 (68793374511124597 \)/(150094635296999121 Sqrt)
20 (619140370600121374 \)/(1350851717672992089 Sqrt)
21 (5572263335401092367 \)/(12157665459056928801 Sqrt)
22 (50150370018609831304 \)/(109418989131512359209 Sqrt)
23 (451353330167488481737 \)/(984770902183611232881 Sqrt)
24 (4062179971507396335634 \)/(8862938119652501095929 Sqrt)
25 (36559619743566567020707 \)/(79766443076872509863361 Sqrt)
26 (329036577692099103186364 \)/(717897987691852588770249 Sqrt)
27 (2961329199228891928677277 \)/(6461081889226673298932241 Sqrt)
28 (26651962793060027358095494 \)/(58149737003040059690390169 Sqrt)
29 (239867665137540246222859447 \)/(523347633027360537213511521 Sqrt)
30 (2158808986237862216005735024 \)/(4710128697246244834921603689 Sqrt)
31 (19429280876140759944051615217 \)/(42391158275216203514294433201 Sqrt)
32 (174863527885266839496464536954 \)/(381520424476945831628649898809 Sqrt)
33 (1573771750967401555468180832587 \)/(3433683820292512484657849089281 Sqrt)
34 (14163945758706613999213627493284 \)/(30903154382632612361920641803529 Sqrt)
分子数列:
{1,4,37,334,3007,27064,243577,2192194,19729747,177567724,1598109517,14382985654,129446870887,1165021837984,10485196541857,94366768876714,849300919890427,7643708279013844,68793374511124597,619140370600121374,5572263335401092367,50150370018609831304,451353330167488481737,4062179971507396335634,36559619743566567020707,329036577692099103186364,2961329199228891928677277,26651962793060027358095494,239867665137540246222859447,2158808986237862216005735024,19429280876140759944051615217,174863527885266839496464536954,1573771750967401555468180832587,14163945758706613999213627493284}
极限 limit=(11 \)/(24 Sqrt) = 0.831324708607349848188952534753
王守恩
发表于 2020-10-25 08:28:38
本帖最后由 王守恩 于 2020-10-25 08:30 编辑
dlpg070 发表于 2020-10-20 08:39
那串数的通项公式
an := 1/8 3^(-(3/2) - 2 n) (-3 + 11 9^n) \; n>=0
LinearRecurrence[{10, -9}, {4, 37}, 19] 有趣的数列
{4, 37, 334, 3007, 27064, 243577, 2192194, 19729747, 177567724, 1598109517,
14382985654, 129446870887, 1165021837984, 10485196541857, 94366768876714,
849300919890427, 7643708279013844, 68793374511124597, 619140370600121374,}
dlpg070
发表于 2020-10-26 11:09:01
王守恩 发表于 2020-10-25 08:28
LinearRecurrence[{10, -9}, {4, 37}, 19] 有趣的数列
{4, 37, 334, 3007, 27064, 243577, 219219 ...
也来个1行代码,简单的:
In:= Table,{n,0,33}]
Out= {1,4,37,334,3007,27064,243577,2192194,19729747,177567724,1598109517,14382985654,129446870887,1165021837984,10485196541857,94366768876714,849300919890427,7643708279013844,68793374511124597,619140370600121374,5572263335401092367,50150370018609831304,451353330167488481737,4062179971507396335634,36559619743566567020707,329036577692099103186364,2961329199228891928677277,26651962793060027358095494,239867665137540246222859447,2158808986237862216005735024,19429280876140759944051615217,174863527885266839496464536954,1573771750967401555468180832587,14163945758706613999213627493284}