能否证明这个方程组有无穷多组解
-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？ 本帖最后由 056254628 于 2011-1-14 21:20 编辑
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。 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 http://bbs.emath.ac.cn/images/common/back.gif
如果t取-7～7，在求解时还会有很多组解，还能证明吗？
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 http://bbs.emath.ac.cn/images/common/back.gif
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。 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？ 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。 6# 056254628
感谢056254628！确实求得了4组解，代码：!using["fcopt","math"];
f(a,b,n,t,y1,y2,y3,y4)=
{
y1=-b*sin(a+6*t)+n-40.4945,
y2=-b*sin(a+7*t)+n-40.5696,
y3=-b*sin(a+8*t)+n-41.0443,
y4=-b*sin(a+9*t)+n-41.4190
};
solve;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 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}} 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},
t=\text{ArcCos}\left[\frac{2249}{4747}\right],
n=\frac{511456557}{12490000}\right\}
页:
[1]