杨学枝的不等式猜想

数学星空

问题转自于http://bbs.cnool.net/cthread-106181906.html

数学星空

我们可以先讨论并证明下面结论

已知\( 5\leq m\lt n \leq 19;m\in N;n\in N;s, t\in R^+\),且\(ms^2+( n-m)t^2=1\), 证明

\[(m-1)\sqrt{1-s^2}+(n-m-1)\sqrt{1-t^2}+2\sqrt{1-st} \geq\sqrt{n(n-1)}\]

hujunhua

2#是令`a_i(i=1,2,\cdots,n)`只有2个不同值的一个结果吧。由于与次序无关，所以这种情况下更一般的结果应该是\[a\sqrt{1-s^2}+2b\sqrt{1-st}+c\sqrt{1-t^2} \geq\sqrt{n(n-1)}\]这里`a+2b+c=n,(a+b)s^2+(b+c)t^2=1`, `a,b,c`皆为非负整数。
因为环排列的缘故，`\sqrt{1-st}`项的系数一定是偶数。

hujunhua

由杨老师的轮换对称式可得到全对称式\[\sum_{i\lt j}{\sqrt{1-a_i a_j}\ge\frac{(n-1)\sqrt{n(n-1)}}{2}}\]全对称式比轮换对称式弱，或许能对 n>1普遍成立呢？

相应地，3#的塌缩式则为\[a(a-1)\sqrt{1-s^2}+2ab\sqrt{1-st}+b(b-1)\sqrt{1-t^2} \ge (n-1)\sqrt{n(n-1)}\]其中`a\cdot s^2+b\cdot t^2=1，a+b=n`, `a,b`为非负整数。

数学星空

关于已知的相关结果，现搜集后贴于下面，供参考：

http://blog.sina.com.cn/s/blog_a9ec2e1b010117en.html

数学星空

hujunhua

用Mathematica10计算搜索了一下，在计算精度设为小数点后50位的情况下，结果如下：

n	全对称不等式		轮换对称不等式
	-log(计算最小值-猜想值)	-log(驻点坐标-平均坐标)	-log(计算最小值-猜想值)	-log(驻点坐标-平均坐标)
3	34	31	34	31
4	32	30	27	27
5	29	28~30	30	29
6	32	30~32	29	29
7	32	30~32	29	28~30
8	30	29~30	27	27~28
9	27	28~29	26	26~28
10	27	28~29	25	25~26
11	28	29~30	25	25~26
12	29	29~31	23	23~24
13	27	28~29	23	23~24
14	27	28	21	22~23
15	26	27~28	19	19~20
16	26	27~28	19	19~20
17	28	29~30	34	33~34
18	25	26~27	28	27~28
19	26	27	23	21~22
20	28	29~31	11	4~6
21	25	27	4	2
22	25	27	4	1~2
23	26	28	4	1~2

平均坐标即`a_i=1/\sqrt n`.
计算结果表明：
当 n=5~19 时，不等式正确性都很高。
当 n >20时，轮换对称不等式不成立，而全对称不等式仍然很有正确性。

数学星空

我们可以看到严文兰老师利用级数展开很巧妙的求出的\(n\)不成立的值（至少可以肯定\(n\geq 20\)不成立）？

由于5#给出的解答过程太简洁，现重新整理一下以便数学爱好者更容易理解：

对于\(\theta=\frac{2\pi}{n},\alpha\in R\)

我们可以得到如下恒等式：

\(\sum_{k=1}^n \sin(k\theta+\alpha)=\cos(\alpha)\frac{\sin(\frac{(n+1)\theta}{2})\sin(\frac{n\theta}{2}}{\sin\frac{\theta}{2}})+\sin(\alpha)(\frac{\sin((\frac{1}{2}+n)\theta)}{2\sin(\frac{\theta}{2})}-\frac{1}{2})=0\)

\(\sum_{k=1}^n \sin(k\theta+\alpha)^2=\frac{1}{2}(\sum_{k=1}^n {(1-\cos(\alpha)\cos(2k\theta)-\sin(\alpha)\sin(2k\theta))})=\frac{n}{2}-\frac{\sin(\theta)\cos(2\alpha)}{4\sin(\theta)}+\frac{\cos(2\alpha)}{4}=\frac{n}{2}\)

\(\sum_{k=1}^n \sin(k\theta+\alpha)\sin((k+1)\theta+\alpha)=\frac{1}{2}(\sum_{k=1}^n {\cos(\theta)-\cos((2k+1)\theta)})=\frac{n\cos(\theta)}{2}\)

\(\sum_{k=1}^n \sin(k\theta+\alpha)^3=0\)

\(\sum_{k=1}^n \sin(k\theta+\alpha)^4=\frac{3n}{8}\)

\(\sum_{k=1}^n \sin(k\theta+\alpha)^5=0\)

\(\sum_{k=1}^n \sin(k\theta+\alpha)^6=\frac{5n}{16}\)

\(\sum_{k=1}^n \sin(k\theta+\alpha)^7=0\)

\(\sum_{k=1}^n \sin(k\theta+\alpha)^8=\frac{35n}{128}\)

\(\sum_{k=1}^n \sin(k\theta+\alpha)^9=0\)

\(\sum_{k=1}^n \sin(k\theta+\alpha)^{10}=\frac{63n}{256}\)

由\(a_i=a+x\sin(i\theta)\)得到

\(\sum_{k=1}^n a_i^2=\sum_{k=1}^n {a^2+2ax\sin(\theta)+x^2\sin(\theta)^2}=na^2+\frac{n}{2}x^2=1\)

\(a=\sqrt{\frac{1}{n}-\frac{x^2}{2}}=\frac{1}{\sqrt{n}}(1-\frac{nx^2}{4}-\frac{n^2x^4}{32}-\frac{n^3x^6}{128}-\frac{5n^4x^8}{2048}-\ldots)\)

\(1－a_i a_{i+1}=1-a^2-ax(\sin(k\theta)+\sin((k+1)\theta))-x^2\sin(k\theta)\sin((k+1)\theta)=1-\frac{1}{n}+\frac{x^2}{2}-\frac{1}{\sqrt{n}}(1-\frac{nx^2}{4}-\frac{n^2x^4}{32}-\frac{n^3x^6}{128}-\frac{5n^4x^8}{2048})x(\sin(k\theta)+\sin((k+1)\theta))-x^2\sin(k\theta)\sin((k+1)\theta)\)

记\(\sin(k\theta)+\sin((k+1)\theta)＝M,\sin(k\theta)\sin((k+1)\theta)=N\)

\(\sqrt{1－a_i a_{i+1}}=\sqrt{1-\frac{1}{n}+\frac{Mx}{\sqrt{n}}+(\frac{1}{2}-N)x^2+\frac{n^2Mx^5}{32\sqrt{n}}+\frac{n^3Mx^7}{128\sqrt{n}}+\frac{5n^4Mx^9}{2048\sqrt{n}}+\ldots}\)

\(=(\sqrt{1-\frac{1}{n}})(1-\frac{-\sqrt{n}M}{2(n-1)}x-\frac{-n(-2n+2+4Nn-4N+M^2)}{8(n-1)^2}x^2-\frac{-n^{3/2}M(2n+4Nn-4N-2n^2+M^2)}{16(n-1)^3}x^3-\frac{n^2(4nM^2+4M^2+24M^2Nn-24M^2N+4n^2-8n+4-16Nn^2+32Nn-16N+16N^2n^2-32N^2n+16N^2-8n^2M^2+5M^4)}{128(n-1)^4}x^4-\frac{n^{5/2}M(2n+4Nn-4N-2n^2+M^2)(12Nn-12N+8+7M^2-10n+2n^2)}{256(n-1)^5}x^5\)

\(-\frac{n^3(8-24n-48N+30M^4+24n^2-8n^3+96N^2+21M^6-64N^3+140M^4Nn+240M^2N^2n^2-480M^2N^2n+144Nn+48n^2M^2N-96n^3M^2N-40n^2M^4-84n^2M^2+48n^3M^2+288N^2n^2+48Nn^3-96N^2n^3+24nM^2+192M^2Nn+12M^2+64N^3n^3-288N^2n-144M^2N+240M^2N^2-140M^4N+192N^3n-192N^3n^2-144Nn^2+10M^4n)}{1024(n-1)^6}x^6\)

\(\frac{n^{7/2}M(2n+4Nn-4N-2n^2+M^2)(33M^4+120M^2Nn+56M^2-120M^2N-4n^2M^2-52nM^2+80N^2n^2-160N^2n-24n^3-96N-80n-128Nn^2+208Nn+68n^2+4n^4+16Nn^3+32+80N^2)}{2048(n-1)^7}x^7-\)

\(\frac{n^4(80-320n-640N+600M^4+80n^4+480n^2-320n^3+1920N^2+3360M^4Nn^2+840M^6-8960M^2N^3-640Nn^4+8960M^2N^3n^3+26880M^2N^3n-26880M^2N^3n^2-2560N^3-5120N^4n+7680N^4n^2-5120N^4n^3-1008M^6n^2+168M^6n+1280N^4n^4+3840n^4M^2N+6720M^4Nn+17280M^2N^2n^2-24960M^2N^2n+2560Nn+1280N^4+2880n^2M^2N-8640n^3M^2N-2280n^2M^4-2400n^2M^2-960n^4M^2+2720n^3M^2+11520N^2n^2+2560Nn^3-7680N^2n^3+480nM^2+10080M^4N^2n^2-20160M^4N^2n+3696M^6Nn+4800M^2Nn+10080M^4N^2-3696M^6N+1920N^2n^4+160M^2+320n^4M^4+10240N^3n^3-2560N^3n^4+429M^8-7680N^2n-2880M^2N+9600M^2N^2-5600M^4N+10240N^3n-15360N^3n^2-4480n^3M^4N-3840Nn^2+400M^4n-3840n^4M^2N^2+960n^3M^4+1920n^3M^2N^2)}{32768(n-1)^8}x^8+\ldots\))

再记\(t=\cos(\frac{2\pi}{n})\)得到

\(\sum_{k=1}^n \sqrt{1－a_i a_{i+1}}＝\sqrt{n(n-1)}-\frac{n^2(-2n+2tn-t+3)x^2}{8(n-1)^2}-\frac{n^3(8t^2n^2-4n^2-32n^2t+8nt^2+76tn-4n-14t-t^2+23)x^4}{256(n-1)^4 }-\frac{(-t^3+72t^3n^2+18t^3n+16n^3t^3+312t^2n^2-11t^2-144n^3t^2+158nt^2+742tn-372n^2t-79t+24tn^3+107+74n-84n^2+8n^3)n^4x^6}{2048(n-1)^6}-\frac{5(2219+2968n-1692t-334t^2+560n^4-1176n^2-1568n^3-60t^3-5t^4+1640t^3n+10560t^3n^2+7272nt^2+160t^4n+1440n^2t^4+1280t^4n^3-6288n^2t+23288tn+1920n^4t^2+1024n^4t-16640n^3t^2-4320tn^3+25800t^2n^2+1920n^3t^3-2048t^3n^4+128t^4n^4)n^5x^8}{262144(n-1)^8}+\ldots\)

由\(t=\cos(\frac{2\pi}{n})\geq 1-\frac{1}{2}(\frac{2\pi}{n})^2\)

可以得到\(x^k(1 \leq k \leq 8)\)系数大于0可以依次得到：

\(k=2\) 时　\(4n\pi^2-2n^2-2\pi^2 \geq 0 \)

得到 \(0.513350523 \leq n \leq 19.22585828\)

\(k=4\)时　\(28n^6-80n^5-32n^4\pi^2-8n^4+184n^3\pi^2-32n^2\pi^4-32n^2\pi^2-32\pi^4n+4\pi^4 \geq 0 \)

得到 \(4.020637469 \leq n\)

\(k=6\)时　\(96n^9+72n^8-432n^7\pi^2-992n^7+936n^6\pi^2-16n^6+384n^5\pi^4+2224n^5\pi^2-208n^4\pi^2-2112n^4\pi^4+128n^3\pi^6-848n^3\pi^4+56n^2\pi^4+576\pi^6n^2+144\pi^6n-8\pi^6 \geq 0 \)

得到 \(0.04671693198 \leq n\)

\(k=8\)时　\(433920n^7\pi^2+69120n^8\pi^4+827520n^8\pi^2+64000n^7\pi^4-1322400n^6\pi^4-61440n^6\pi^6+281600n^5\pi^6-10240n^4\pi^8+652800n^4\pi^6-115200n^2\pi^8-12800\pi^8n-640n^8-7680n^{10}\pi^2-267200n^9\pi^2-102400\pi^8n^3-263040n^5\pi^4-25600n^6\pi^2-3200\pi^6n^2+91200n^3\pi^6+10880n^4\pi^4-176640n^9+400\pi^8-151680n^{10}+96640n^{11}-7920n^{12}\geq 0 \)

得到 \(0.000 \leq n \leq  0.02533870599\)

注：\(x^k(1 \leq k \leq 8)\)奇次项系数为0

数学星空

有谁能解释一下5＃，杨学枝给出的\(f''(x_0)\)特征值及对应的特征向量是如何求得的？

进一步，对于\(n \geq 20\)相对弱一点的极值是什么？

即对于正实数\(a_1,a_2,\ldots,a_n\)满足\(\sum_{k=1}^n a_k^2=1\)

且\(\sqrt{1-a_1a_2}+\sqrt{1-a_2a_3}+\sqrt{1-a_3a_4}+\ldots+\sqrt{1-a_na_1}\geq \lambda_n\sqrt{n(n-1)}\),求 \(\lambda_n\)的最大值？

mathe

多变量函数二阶导数的矩阵指的是其所有二阶偏导数构成的对称矩阵，不如
f(x,y)的二阶导数阵为
\(\begin{bmatrix}\frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x \partial y} \\
\frac{\partial^2 f}{\partial x \partial y} & \frac{\partial^2 f}{\partial y^2}
\end{bmatrix}\)
然后将某一点的值代入，就可以得出在对应点二阶导数对应的矩阵，然后可以分析其特征值和特征向量