我们设\(x_k=\frac{a_k}{a_{k+1}},k=1..n\),则有极值条件：
\(x_{k+2}^2=x_{k+1}^2+x_{k+1}-\frac{x_{k}^2}{x_{k+1}}\)
对于n=7,我们得到：
\(x_1^2=x_7^2+x_7-\frac{x_6^2}{x_7}\)
\(x_2^2=x_1^2+x_1-\frac{x_7^2}{x_1}\)
\(x_3^2=x_2^2+x_2-\frac{x_1^2}{x_2}\)
\(x_4^2=x_3^2+x_3-\frac{x_2^2}{x_3}\)
\(x_5^2=x_4^2+x_4-\frac{x_3^2}{x_4}\)
\(x_6^2=x_5^2+x_5-\frac{x_4^2}{x_5}\)
\(x_7^2=x_6^2+x_6-\frac{x_5^2}{x_6}\)
通过艰辛的计算，我们消元得到(超出想象的结果,\(x_1,x_2,x_3,x_4,x_5,x_6,x_7\)为下列关于y方程的根，详细表达式见附件）：
\(y^{1575}+237y^{1574}+27960y^{1573}+2189167y^{1572}+127967055y^{1571}+
5956542111y^{1570}+229966468039y^{1569}+7573740288747y^{1568}+217195916251734y^
{1567}+5509224934586649y^{1566}+125136090704063355y^{1565}+2570697121556978829y^
{1564}+48156540002089434273y^{1563}+828275961122268316399y^{1562}+
13156417844524062956717y^{1561}+193959627276535346931317y^{1560}+
2665471049415845840780533y^{1559}+34274013692711101754535718y^{1558}+
413737859400601836305780338y^{1557}+4702496720097649823655021586y^{1556}+
50455171343554283441379325067y^{1555}+512229855678807802192283264868y^{1554}+
4930683484539424975189646038889y^{1553}+45085742320062585790190905818592y^{1552}
+392269284750521293203307247108736y^{1551}+3252297692356540986585666729214139y
^{1550}+25729825589613508859427216137618490y^{1549}+
194464748622873082362576842998104622y^{1548}+
1405597526808002206155756147100493505y^{1547}+
9725246507446609876489454720237816927y^{1546}+\dots+-71793810348771483375391108000228145
y^{30}-13100909023550720373698929424205502y^{29}-\
2296874872643888083204839838615397y^{28}-386643414798137816051162164327259y^{27}
-62440236022275419435274445188562y^{26}-9664237519210817455694811960715y^{25}-\
1431896440799115559001206901440y^{24}-202819370853057431834671945578y^{23}-\
27421542096528982240583950780y^{22}-3532672638900873004315266906y^{21}-\
432808771922657726356622019y^{20}-50317756233030382350028192y^{19}-\
5537563785067957159898811y^{18}-575310020991268693223646y^{17}-\
56252156607650435518263y^{16}-5158569399568432942730y^{15}-441947025365891770693
y^{14}-35214265373062071162y^{13}-2596193691265762792y^{12}-176045275408316310y^
{11}-10902389172376620y^{10}-611481429273170y^9-30746301739012y^8-1368654303590
y^7-53085298369y^6-1757100738y^5-48242847y^4-1054508y^3-17199y^2-186y-1=0\)
解得：
{-2.5313754361572083274, -2.2897435314085089836, -2.1834962340662334071, -2.1253313509433058988, -1.7875749231432755320, -1.7423863064272187694,
-1.6770641724445321307, -1.6243773891112491293, -1.5415473045208114337, -1.3881929471272967637, -1.3328098201143265808, -1.3146415925398947908,
-1.3077838701985277477, -1.3059632128632167189, -1.2151831983216832552, -1.2100157148300979496, -1.2064913022022663453, -1.2034626937323212323,
-1.1498770384577404686, -1.1377798045432585345, -1.0312842256127762898, -1.0308402673719049890, -1.0221997159959176423, -0.92331967132561932280,
-0.90883720222180845989, -0.78607477684269818152, -0.78412461246543316888, -0.78397910048394251469, -0.77526882129506576722, -0.76910744508814982268,
-0.69212265623708541717, -0.66663655058268905479, -0.64591036580872210630, -0.37970446355950123768, -0.36499631361226621680, -0.29905968201527480038,
-0.22359585813136146356, -0.18249015231659363020, -0.17387273230980682958, -0.10301647889050310476, 0.79863335015334249027, 0.81831197137120640244,
0.84713240968364325923, 0.90020230214287103567, 0.90667579762331956742, 0.91222229382793599056, 0.93953457502418666744, 0.94450390184391910245,
0.97923337970653848310, 0.98724364494215151574, 0.99197152733571484758, 1.0151757673093954822, 1.0201960406524940163, 1.0407750733403786538,
1.0497215743716858584, 1.0644601871440451014, 1.0889696744406762472, 1.0889944527757833722, 1.0995090477768181806, 1.1099664528720066446,
1.1188511257163850983, 1.1327891624940687097, 1.1413799553375473493, 1.1442564456368120404, 1.2049578377713863665, 1.2200423816708315112,
1.2298948000619613306, 1.2311306286792130033, 1.2833612147705397827, 1.3423373691686473528, 1.3870442957496284705, 1.3954604220135717331,
1.5511049214726247284, 1.6117716462358246318, 1.6202221715093102880, 1.7227089659927969040, 1.9699437824343871243}
可以对比wayne给出的结果：(说明结果是准确的）
\(x_1 =\frac{a_1}{a_2}=-1.206491302, x_2 = \frac{a_2}{a_3}=1.109966453, x_3 =\frac{a_3}{a_4}=1.015175767,
x_4 =\frac{a_4}{a_5}=0.9122222938, x_5 =\frac{a_5}{a_6}=-0.7839791009, x_6 =\frac{a_6}{a_7}=0.9445039018, x_7 =\frac{a_7}{a_1}=1.088969674\)
对于奇数n,不妨设为\(N=2n+1\),极值记为\(S(N)\)
我们根据wayne 给出的结果知当n足够大时，极值时\(x_n^2=(\frac{a_n}{a_{n+1}})^2 \to 1,S(N) \to N\)
又因为\(\sum_{k=1}^{2n+1} \cos(k\theta+t)=0\),其中\(\theta=\frac{2\pi}{2n+1}\)
因此可分析最简单的情形:令\(x_k=\frac{\cos(k\theta)}{\cos((k+1)\theta)}\),\(k=0...2n+1\)
渐近分析结果：
\(x_k^2=1+\frac{(k + 1)^2\pi^2 - k^2\pi^2}{n^2}+\frac{-(k + 1)^2\pi^2 + k^2\pi^2}{n^3}+\frac{\frac{3k^4\pi^4}{4} +\frac{5\pi^4k^3}{3}+ \frac{5\pi^4k^2}{2}+\frac{5\pi^4k}{3}+ \frac{5\pi^4}{12}+\frac{3\pi^2k}{2}+\frac{3\pi^2}{4}+ \frac{(k + 1)^4\pi^4}{4 }-k^2\pi^4(k + 1)^2}{n^4}+\dots\)
\(S(2n+1)=2n+1+4\pi^2+\frac{8\pi^4}{3}+\frac{16\pi^4+4\pi^2}{n}+\frac{32\pi^4-2\pi^2}{n^2}+\frac{79\pi^4+9\pi^2}{3n^3}+\frac{28\pi^4+9\pi^2}{4n^4}+\dots\)
\(S(N)=N(1+\frac{12\pi^2+8\pi^4}{3N}+\frac{32\pi^4+8\pi^2}{N^2}+\frac{160\pi^4}{N^3}+\dots)\)
因此我们可以试着构造：\(\theta=\frac{2\pi}{n}\)
\(a_k=u_1\cos(k\theta)+u_2\cos((k+1)\theta)+u_3\cos((k+2)\theta)+\dots+u_{n}\cos((n+k-1)\theta)\) .........................(1)
或者
\(a_k=\cos(k\theta+t_1)+\cos((k+1)\theta+t_2)+\cos((k+2)\theta+t_3)+\dots+\cos((n+k-1)\theta+t_n)\).........................(2)
显然（1），(2)都满足
\(a_1+a_2+\dots+a_n=0\)
\(a_k=a_{n+k}\)
理论上将上面一般式代入极值条件：
\(x_{k+2}^2=x_{k+1}^2+x_{k+1}-\frac{x_{k}^2}{x_{k+1}}\)
\(x_k=\frac{a_k}{a_{k+1}},k=1..n\)
是可以得到渐近表达式的，但涉及到三角函数和展开太复杂了，不知@mathe 能否给出更简洁的计算方案
我们可以设：
\(x_k=\frac{a_k}{a_{k+1}}\)
\(x_k=1+\frac{a_0}{n}+\frac{b_0+b_1 k}{n^2}+\frac{c_0+c_1 k+c_2 k^2}{n^3}+\frac{d_0+d_1 k +d_2 k^2+d_3 k^3}{n^4}+\frac{m_0+m_1 k +m_2 k^2+m_3 k^3+m_4 k^4}{n^5}+\frac{n_0+n_1 k +n_2 k^2+n_3 k^3+n_4 k^4+n_5 k^5}{n^6}\)
代入极值条件：
\(x_{k+2}^2=x_{k+1}^2+x_{k+1}-\frac{x_k}{x_{k+1}}\)
利用渐近分析方法得到：
\(\frac{t_{10}}{n^3}+\frac{t_{20}+t_{21}k}{n^4}+\frac{t_{30}+t_{31}k+t_{32}k^2}{n^5}+\frac{t_{40}+t_{41}k+t_{42}k^2+t_{43}k^3}{n^6}+\frac{t_{50}+t_{51}k+t_{52}k^2+t_{53}k^3+t_{54}k^4}{n^7}+\frac{t_{60}+t_{61}k+t_{62}k^2+t_{63}k^3+t_{64}k^4+t_{65}k^5}{n^8}+\dots=0\)
由于上式对\(k=1..n\) 都成立，因此有所有\(k^i(i=0...5)\)的系数项均为0，与k无关
得到21个方程，谁有兴趣算一下数值解
若能求出上面21个未知数，就能得到极值：
S(n)=n + 2*a0 + b1 + (2*c2)/3 + d3/2 + (2*m4)/5 + n5/3 + (d3 + 1/2*m3 + 1/2*a0*d3 + 1/2*b1*c2 + a0^2 + 2*b0 + c1 + 1/3*b1^2 + 2/3*d2 + 2/5*b1*d3 + a0*b1 + 2/3*a0*c2 + m4 + 1/5*c2^2 + 2/5*n4 + n5 + c2 + b1 + 2/5*a0*m4)/n + (1/2*c1*c2 + 2/3*b1*c1 + 2/3*b0*c2 + 1/3*c2 + 5/6*n5 + 1/2*c2^2 + 1/2*n3 + 1/2*d3 + a0*b1 + c1 + 2*c0 + 1/2*a0*m3 + d2 + m3 + 2/3*m2 + 2/3*m4 + a0*m4 + b1*d3 + b1*c2 + b0*b1 + a0*d3 + a0*c2 + a0*c1 + d1 + n4 + 1/2*b0*d3 + 2/3*a0*d2 + 1/2*b1*d2 + 2*a0*b0 + 1/2*b1^2)/n^2 + (m1 + 2/3*n4 + 1/3*c2^2 + 1/2*a0*d3 + 1/2*b1*c2 + m2 + 1/6*b1^2 + 1/3*c1^2 + 2/3*n2 + a0*m3 + c1*c2 + b0*d3 + b1*d2 + a0*d1 + b0*c1 + b1*c0 + a0*d2 + b0*c2 + b1*c1 + b0*b1 + a0*c1 + d1 + n3 + b0^2 + 2*a0*c0 + 2/3*b1*d1 + 2/3*b1*d3 + 2*d0 + 1/2*m3 + 1/3*d2 + 1/3*a0*c2 + 2/3*b0*d2 + 2/3*c0*c2 + 2/3*a0*m2 + 2/3*a0*m4)/n^3 + (m1 + 1/2*c1*c2 + 1/2*n3 + 1/2*a0*m3 + 2*m0 + c0*c1 + b1*d0 + b0*d1 + a0*m2 + a0*m1 + c0*c2 + b1*d1 + b0*d2 + a0*d1 + b0*c1 + b1*c0 + n1 + 1/3*m2 - 1/15*m4 + 1/2*c1^2 + 1/2*b0*d3 + 1/2*b1*d2 + 1/3*b0*c2 + 1/3*b1*c1 + 2*a0*d0 + 2*b0*c0 - 1/6*n5 + 1/3*a0*d2 + n2)/n^4 + (1/3*n2 + 1/3*b1*d1 + 1/3*c0*c2 + c0^2 + 2*n0 + n1 - 1/15*a0*m4 - 1/15*b1*d3 + 2*a0*m0 + 2*b0*d0 + 1/3*b0*d2 + a0*m1 + b0*d1 + b1*d0 + c0*c1 + 1/3*a0*m2 - 1/30*c2^2 - 1/15*n4 + 1/6*c1^2)/n^5