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[求助] 如何求下面的函数的极值?

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发表于 2019-11-8 12:55:49 | 显示全部楼层 |阅读模式

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求下面这个表达式的极值\[
y=\frac{\sin (a+x) \sin (b+x)}{\sin (c+x) \sin (d+x)}
\]令导数=0易得驻点条件\[
\frac{\sin (a+x) \sin (b+x)}{\sin (c+x) \sin (d+x)}=\frac{\sin(2x+a+b)}{\sin(2x+c+d)}
\]怎么解这个驻点方程呢?

题目背景:土木专业,求岩土边坡上的破裂角
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
发表于 2019-11-10 20:03:04 | 显示全部楼层
K>> syms a b c d x y
y=sin(x+a)*sin(x+b)/(sin(x+c)*sin(x+d))
solve(diff(y))

有解的情况下,返回
$
ans =
     -log(-((exp(a*2i)*exp(b*2i) - exp(c*2i)*exp(d*2i) + ((exp(a*2i) - exp(c*2i))*(exp(a*2i) - exp(d*2i))*(exp(b*2i) - exp(c*2i))*(exp(b*2i) - exp(d*2i)))^(1/2))/(exp(a*2i)*exp(b*2i)*exp(c*2i) + exp(a*2i)*exp(b*2i)*exp(d*2i) - exp(a*2i)*exp(c*2i)*exp(d*2i) - exp(b*2i)*exp(c*2i)*exp(d*2i)))^(1/2))*1i
-log(-(-(- exp(a*2i)*exp(b*2i) + exp(c*2i)*exp(d*2i) + ((exp(a*2i) - exp(c*2i))*(exp(a*2i) - exp(d*2i))*(exp(b*2i) - exp(c*2i))*(exp(b*2i) - exp(d*2i)))^(1/2))/(exp(a*2i)*exp(b*2i)*exp(c*2i) + exp(a*2i)*exp(b*2i)*exp(d*2i) - exp(a*2i)*exp(c*2i)*exp(d*2i) - exp(b*2i)*exp(c*2i)*exp(d*2i)))^(1/2))*1i
     -log(((exp(a*2i)*exp(b*2i) - exp(c*2i)*exp(d*2i) + ((exp(a*2i) - exp(c*2i))*(exp(a*2i) - exp(d*2i))*(exp(b*2i) - exp(c*2i))*(exp(b*2i) - exp(d*2i)))^(1/2))/(exp(a*2i)*exp(b*2i)*exp(c*2i) + exp(a*2i)*exp(b*2i)*exp(d*2i) - exp(a*2i)*exp(c*2i)*exp(d*2i) - exp(b*2i)*exp(c*2i)*exp(d*2i)))^(1/2))*1i
  -log((-(- exp(a*2i)*exp(b*2i) + exp(c*2i)*exp(d*2i) + ((exp(a*2i) - exp(c*2i))*(exp(a*2i) - exp(d*2i))*(exp(b*2i) - exp(c*2i))*(exp(b*2i) - exp(d*2i)))^(1/2))/(exp(a*2i)*exp(b*2i)*exp(c*2i) + exp(a*2i)*exp(b*2i)*exp(d*2i) - exp(a*2i)*exp(c*2i)*exp(d*2i) - exp(b*2i)*exp(c*2i)*exp(d*2i)))^(1/2))*1i$
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
发表于 2019-11-11 10:08:24 | 显示全部楼层
本帖最后由 chyanog 于 2019-11-11 11:06 编辑

\(x\to \arctan \left(-\tan (a)-\tan (b)+\tan (c)+\tan (d),\ \tan (a) \tan (b)-\tan (c) \tan (d)+\sqrt{(\tan (a)-\tan (c)) (\tan (b)-\tan (c)) (\tan (a)-\tan (d)) (\tan (b)-\tan (d))}\right)\)
\(x\to \arctan \left(\tan (a)+\tan (b)-\tan (c)-\tan (d),\ -\tan (a) \tan (b)+\tan (c) \tan (d)+\sqrt{(\tan (a)-\tan (c)) (\tan (b)-\tan (c)) (\tan (a)-\tan (d)) (\tan (b)-\tan (d))}\right)\)
Mathematica检验代码
  1. Sin[c+x] Sin[d+x] Sin[a+b+2 x]-Sin[a+x] Sin[b+x] Sin[c+d+2 x] /.{
  2.    {x->ArcTan[-Tan[a]-Tan[b]+Tan[c]+Tan[d], Tan[a] Tan[b]-Tan[c] Tan[d]+Sqrt[(Tan[a]-Tan[c]) (Tan[b]-Tan[c]) (Tan[a]-Tan[d]) (Tan[b]-Tan[d])]]},
  3.    {x->ArcTan[Tan[a]+Tan[b]-Tan[c]-Tan[d], -Tan[a] Tan[b]+Tan[c] Tan[d]+Sqrt[(Tan[a]-Tan[c]) (Tan[b]-Tan[c]) (Tan[a]-Tan[d]) (Tan[b]-Tan[d])]]}
  4.   } // TrigExpand // Simplify
复制代码

输出
{0, 0}

点评

我只想给你点一万个赞!  发表于 2019-11-11 10:38
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
 楼主| 发表于 2019-11-11 10:36:38 | 显示全部楼层
chyanog 发表于 2019-11-11 10:08
\(x\to \arctan \left(-\tan (a)-\tan (b)+\tan (c)+\tan (d), \tan (a) \tan (b)-\tan (c) \tan (d)+\sqrt ...

牛人,你这个怎么算出来的,
你这个ArcTan[-2,-1]是180度+atan0.5吗?

点评

就是解方程,但是需要自己引导化简,不能全靠软件  发表于 2019-11-11 13:16
https://zh.wikipedia.org/zh-hans/Atan2 Mathematica中两个参数的ArcTan相当于很多语言里的atan2,但两个参数的顺序是反的  发表于 2019-11-11 11:00
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
 楼主| 发表于 2019-11-11 11:07:10 | 显示全部楼层
chyanog 发表于 2019-11-11 10:08
\(x\to \arctan \left(-\tan (a)-\tan (b)+\tan (c)+\tan (d),\ \tan (a) \tan (b)-\tan (c) \tan (d)+\sqr ...
  1. Sin[c+x] Sin[d+x] Sin[a+b+2 x]-Sin[a+x] Sin[b+x] Sin[c+d+2 x]/.{{x->ArcTan[-Tan[a]-Tan[b]+Tan[c]+Tan[d],Tan[a] Tan[b]-Tan[c] Tan[d]-Sqrt[(Tan[a]-Tan[c]) (Tan[b]-Tan[c]) (Tan[a]-Tan[d]) (Tan[b]-Tan[d])]]},
  2. {x->ArcTan[Tan[a]+Tan[b]-Tan[c]-Tan[d],-Tan[a] Tan[b]+Tan[c] Tan[d]-Sqrt[(Tan[a]-Tan[c]) (Tan[b]-Tan[c]) (Tan[a]-Tan[d]) (Tan[b]-Tan[d])]]}}//TrigExpand//Simplify
复制代码


如我所猜,这两个也是解!
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
 楼主| 发表于 2019-11-11 11:14:38 | 显示全部楼层
mathematica 发表于 2019-11-11 11:07
如我所猜,这两个也是解!
  1. Clear["Global`*"];(*Clear all variables*)
  2. y1=Sin[x+1]*Sin[x+2]
  3. y2=Sin[x+3]*Sin[x+4]
  4. yx=D[y1,x]*y2-y1*D[y2,x]
  5. Reduce[yx==0,{x}]//FullSimplify
复制代码

输出结果
\[c_1\in \mathbb{Z}\land \left(x=2 \left(\tan ^{-1}\left(\text{Root}\left[\text{$\#$1}^4 (\sin (1) \sin (2) \sin (3) \cos (4)-\sin (1) \sin (4) \sin (3) \cos (2)-\sin (2) \sin (4) \sin (3) \cos (1)+\sin (1) \sin (2) \sin (4) \cos (3))+\text{$\#$1}^3 (4 \sin (3) \sin (4) \cos (1) \cos (2)-4 \sin (1) \sin (2) \cos (3) \cos (4))+\text{$\#$1}^2 (4 \sin (1) \cos (2) \cos (3) \cos (4)-2 \sin (2) \sin (3) \sin (1) \cos (4)-2 \sin (2) \sin (4) \sin (1) \cos (3)+2 \sin (3) \sin (4) \sin (1) \cos (2)+4 \sin (2) \cos (1) \cos (3) \cos (4)-4 \sin (3) \cos (1) \cos (2) \cos (4)-4 \sin (4) \cos (1) \cos (2) \cos (3)+2 \sin (2) \sin (3) \sin (4) \cos (1))+\text{$\#$1} (4 \sin (1) \sin (2) \cos (3) \cos (4)-4 \sin (3) \sin (4) \cos (1) \cos (2))+\sin (1) \sin (2) \sin (3) \cos (4)+\sin (1) \sin (2) \sin (4) \cos (3)-\sin (1) \sin (3) \sin (4) \cos (2)-\sin (2) \sin (3) \sin (4) \cos (1)\&,1\right]\right)+\pi  c_1\right)\lor x=2 \left(\tan ^{-1}\left(\text{Root}\left[\text{$\#$1}^4 (\sin (1) \sin (2) \sin (3) \cos (4)-\sin (1) \sin (4) \sin (3) \cos (2)-\sin (2) \sin (4) \sin (3) \cos (1)+\sin (1) \sin (2) \sin (4) \cos (3))+\text{$\#$1}^3 (4 \sin (3) \sin (4) \cos (1) \cos (2)-4 \sin (1) \sin (2) \cos (3) \cos (4))+\text{$\#$1}^2 (4 \sin (1) \cos (2) \cos (3) \cos (4)-2 \sin (2) \sin (3) \sin (1) \cos (4)-2 \sin (2) \sin (4) \sin (1) \cos (3)+2 \sin (3) \sin (4) \sin (1) \cos (2)+4 \sin (2) \cos (1) \cos (3) \cos (4)-4 \sin (3) \cos (1) \cos (2) \cos (4)-4 \sin (4) \cos (1) \cos (2) \cos (3)+2 \sin (2) \sin (3) \sin (4) \cos (1))+\text{$\#$1} (4 \sin (1) \sin (2) \cos (3) \cos (4)-4 \sin (3) \sin (4) \cos (1) \cos (2))+\sin (1) \sin (2) \sin (3) \cos (4)+\sin (1) \sin (2) \sin (4) \cos (3)-\sin (1) \sin (3) \sin (4) \cos (2)-\sin (2) \sin (3) \sin (4) \cos (1)\&,2\right]\right)+\pi  c_1\right)\lor x=2 \left(\tan ^{-1}\left(\text{Root}\left[\text{$\#$1}^4 (\sin (1) \sin (2) \sin (3) \cos (4)-\sin (1) \sin (4) \sin (3) \cos (2)-\sin (2) \sin (4) \sin (3) \cos (1)+\sin (1) \sin (2) \sin (4) \cos (3))+\text{$\#$1}^3 (4 \sin (3) \sin (4) \cos (1) \cos (2)-4 \sin (1) \sin (2) \cos (3) \cos (4))+\text{$\#$1}^2 (4 \sin (1) \cos (2) \cos (3) \cos (4)-2 \sin (2) \sin (3) \sin (1) \cos (4)-2 \sin (2) \sin (4) \sin (1) \cos (3)+2 \sin (3) \sin (4) \sin (1) \cos (2)+4 \sin (2) \cos (1) \cos (3) \cos (4)-4 \sin (3) \cos (1) \cos (2) \cos (4)-4 \sin (4) \cos (1) \cos (2) \cos (3)+2 \sin (2) \sin (3) \sin (4) \cos (1))+\text{$\#$1} (4 \sin (1) \sin (2) \cos (3) \cos (4)-4 \sin (3) \sin (4) \cos (1) \cos (2))+\sin (1) \sin (2) \sin (3) \cos (4)+\sin (1) \sin (2) \sin (4) \cos (3)-\sin (1) \sin (3) \sin (4) \cos (2)-\sin (2) \sin (3) \sin (4) \cos (1)\&,3\right]\right)+\pi  c_1\right)\lor x=2 \left(\tan ^{-1}\left(\text{Root}\left[\text{$\#$1}^4 (\sin (1) \sin (2) \sin (3) \cos (4)-\sin (1) \sin (4) \sin (3) \cos (2)-\sin (2) \sin (4) \sin (3) \cos (1)+\sin (1) \sin (2) \sin (4) \cos (3))+\text{$\#$1}^3 (4 \sin (3) \sin (4) \cos (1) \cos (2)-4 \sin (1) \sin (2) \cos (3) \cos (4))+\text{$\#$1}^2 (4 \sin (1) \cos (2) \cos (3) \cos (4)-2 \sin (2) \sin (3) \sin (1) \cos (4)-2 \sin (2) \sin (4) \sin (1) \cos (3)+2 \sin (3) \sin (4) \sin (1) \cos (2)+4 \sin (2) \cos (1) \cos (3) \cos (4)-4 \sin (3) \cos (1) \cos (2) \cos (4)-4 \sin (4) \cos (1) \cos (2) \cos (3)+2 \sin (2) \sin (3) \sin (4) \cos (1))+\text{$\#$1} (4 \sin (1) \sin (2) \cos (3) \cos (4)-4 \sin (3) \sin (4) \cos (1) \cos (2))+\sin (1) \sin (2) \sin (3) \cos (4)+\sin (1) \sin (2) \sin (4) \cos (3)-\sin (1) \sin (3) \sin (4) \cos (2)-\sin (2) \sin (3) \sin (4) \cos (1)\&,4\right]\right)+\pi  c_1\right)\right)\]

点评

这些解都可以在我上面那两个解的基础上加上k π (k∈Z) 得到  发表于 2019-11-11 11:59
周期为π  发表于 2019-11-11 11:55
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
发表于 2019-11-11 13:45:39 | 显示全部楼层
chyanog 发表于 2019-11-11 10:08
\(x\to \arctan \left(-\tan (a)-\tan (b)+\tan (c)+\tan (d),\ \tan (a) \tan (b)-\tan (c) \tan (d)+\sqr ...

这个解怎么算出来的,代码?
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
发表于 2019-11-11 14:00:17 | 显示全部楼层
markfang2050 发表于 2019-11-11 13:45
这个解怎么算出来的,代码?
  1. Solve[D[(Sin[x + a] Sin[x + b])/(Sin[x + c] Sin[x + d]), x] == 0 /. x -> ArcTan[t] // TrigExpand, t]
  2. % /. Sin[x_] :> Cos[x] HoldForm@Tan[x] // Simplify
  3. % /. k_ Sqrt[x_] :> Sqrt[k^2 x] // Simplify // ReleaseHold
复制代码

点评

不敢当,论坛里比我聪明的人多了  发表于 2019-11-11 18:18
我发现你很聪明,比我聪明多了,软件就是没人聪明!人懂得变通办法,人会研究会思考,但是软件永远都是机器不懂得如何思考!  发表于 2019-11-11 16:08
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
发表于 2019-11-11 14:04:44 | 显示全部楼层
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
 楼主| 发表于 2019-11-11 16:05:39 | 显示全部楼层

x -> ArcTan[t]
你这么一替换,就漏解了,
因为ArcTan[t]的范围是从-90度到90度,
而x的范围没这个限制
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
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