能否证明这个方程组有无穷多组解

forcal

-b*sin(a+6*t)+n-40.4945=0 -b*sin(a+7*t)+n-40.5696=0 -b*sin(a+8*t)+n-41.0443=0 -b*sin(a+9*t)+n-41.4190=0 在求解这个方程组时，感觉解非常多，能否证明这个方程组有无穷多组解？

056254628

设 $a1,b1,n1 ,t1$是方程组的一组解， 那么，$a1,b1,n1,t1+2*k*pi$都是方程组的解 ，k属于任何整数。

forcal

设 $a1,b1,n1 ,t1$是方程组的一组解， 那么，$a1,b1,n1,t1+2*k*pi$都是方程组的解 ，k属于任何整数。 056254628 发表于 2011-1-14 21:19

如果t取-7～7，在求解时还会有很多组解，还能证明吗？

northwolves

如果t取-7～7，在求解时还会有很多组解，还能证明吗？ forcal 发表于 2011-1-14 21:48

设$ a1,b1,n1,t1$是方程组的一组解， 那么，$a1+2kπ,b1,n1,t1$都是方程组的解 ，k属于任何整数。

forcal

谢谢楼上两位！最基本的东西都忘了！ 将a和t都限定为0～2pi（6.28），求一下还有几个解，应该是有限个解了吧？ 我只能数值求解，不知理论推导能推出几个？

056254628

设u=a+7.5t -b*sin(u-1.5t)+n=40.4945 式1 -b*sin(u-0.5t)+n=40.5696 式2 -b*sin(u+0.5t)+n=41.0443 式3 -b*sin(u+1.5t)+n=41.4190 式4 式2-式1得 -2b*cos(u-t)*sin(0.5t)=0.0751 式5 式3-式2得 -2b*cos(u)*sin(0.5t)=0.4747 式6 式4-式3得 -2b*cos(u+t)*sin(0.5t)=0.3747 式7 式5+式7得 -2b*sin(0.5t)*(cos(u+t)+cos(u-t))=0.4498 即-2b*sin(0.5t)*2*cos(u)*cos(t)=0.4498 式8 式6代入式8得 0.4747*2*cos(t)=0.4498 所以cos(t)=0.4498/(0.4747*2)=0.47377290920581419844112070781546 若0<=t<2*pi，那么t有2解， t1=1.0772262149940490008272798975116 t2=2*pi-1.0772262149940490008272798975116=5.2059590921855374760980068690474 tan(t)=±1.8587957121697277908778293159514 式7/式5得 cos(u+t)*0.0751 =cos(u-t)*0.3747 即 0.0751cos(u)*cos(t)-0.0751sin(u)*sin*(t)=0.3747cos(u)*cos(t)+0.3747sin(u)*sin*(t) -0.4498*sin*(u)*sin(t)=0.2996cos(u)*cos(t) tan(u)=-0.2996/0.4498/tan(t) =-0.66607381058248110271231658514896/tan(t)=±0.35833620995660091634903055212031 b= -0.4498/(4*sin(0.5t)*cos(u)*cos(t)) n=(40.5696+41.0443)/2 +b*sin(u)*cos(0.5t) --------------------------------------------------------------------- a、t都在0和2pi以内应有有4组解，验证一下是否都对就可以了。

forcal

6# 056254628 感谢056254628！确实求得了4组解，代码：

1. !using["fcopt","math"];
2. f(a,b,n,t,y1,y2,y3,y4)=
3. {
4. y1=-b*sin(a+6*t)+n-40.4945,
5. y2=-b*sin(a+7*t)+n-40.5696,
6. y3=-b*sin(a+8*t)+n-41.0443,
7. y4=-b*sin(a+9*t)+n-41.4190
8. };
9. solve[HFor("f"), optrange,0,2*pi,-1e50,1e50,-1e50,1e50,0,2*pi];

复制代码

算法有待改进，需反复求解，得4组解： 1.001499450029047 0.4915300827061889 40.94928398718976 1.077226214994026 5.024295867788081e-015 4.143092103618644 -0.4915300827061838 40.94928398718975 1.077226214994054 5.024295867788081e-015 2.140093205431478 0.4915300826222994 40.94928398713199 5.205959091908552 6.146226145805337e-011 5.281685857150924 -0.4915300827061913 40.94928398718974 5.205959092185513 1.280949133595751e-014

wayne

7# forcal Mathematica可以给出任意精度准确的解!!! 四组解， 保留50位精度,{a,b,t,n} 分别是:

{{-2.1400932035609147586637969667258776810633067249051, -0.49153008270618919640083077724753090756036174814013, 1.0772262149940490008272798975116339887809997209734, 40.949283987189751801441152922337870296236989591673}, \ {-1.0014994500288784797988464165536252031338626744700, \ -0.49153008270618919640083077724753090756036174814013, \ -1.0772262149940490008272798975116339887809997209734, 40.949283987189751801441152922337870296236989591673}, \ {2.1400932035609147586637969667258776810633067249051, 0.49153008270618919640083077724753090756036174814013, \ -1.0772262149940490008272798975116339887809997209734, 40.949283987189751801441152922337870296236989591673}, \ {1.0014994500288784797988464165536252031338626744700, 0.49153008270618919640083077724753090756036174814013, 1.0772262149940490008272798975116339887809997209734, 40.949283987189751801441152922337870296236989591673}}

wayne

8# wayne 准确的解： $a=\left\{-\text{ArcCos}\left[-\frac{126738209717915290245033631483 \sqrt{\frac{1249}{23402724241}}}{54316855764315064809674563}\right]$, b=-\frac{4747 \sqrt{\frac{23402724241}{3498}}}{24980000}[/TeX], t=\text{ArcCos}\left[\frac{2249}{4747}\right][/TeX], $n=\frac{511456557}{12490000}\right\}$