求sin(cosA)的极值

王守恩

zeroieme 发表于 2018-1-18 10:47
知识渊博不敢接受，这只是微积分入门课程，离本论坛各位大佬的水平还有几光年。

还有更短的。

\(\D\lim_{\theta\to 0}\ \ \ \ \frac{2\sin 1\theta}{\sin 2\theta}=1\)

\(\D\lim_{\theta\to 0}\ \ \ \ \frac{3\sin 2\theta}{\sin 3\theta}=2\)

\(\D\lim_{\theta\to 0}\ \ \ \ \frac{4\sin 3\theta}{\sin 4\theta}=3\)

\(\D\lim_{\theta\to 0}\ \ \ \ \frac{5\sin 4\theta}{\sin 5\theta}=4\)

\(\D\lim_{\theta\to 0}\ \ \ \ \frac{6\sin 5\theta}{\sin 6\theta}=5\)

\(\D\lim_{\theta\to 0}\ \ \ \ \frac{7\sin 6\theta}{\sin 7\theta}=6\)

\(\D\lim_{\theta\to 0}\ \ \ \ \frac{8\sin 7\theta}{\sin 8\theta}=7\)

\(\D\lim_{\theta\to 0}\ \ \ \ \frac{9\sin 8\theta}{\sin 9\theta}=8\)

\(\D\lim_{\theta\to 0}\ \ \ \ \frac{10\sin 9\theta}{\sin 10\theta}=9\)

\(\D\lim_{\theta\to 0}\ \ \ \ \frac{11\sin 10\theta}{\sin 11\theta}=10\)

\(\D\lim_{\theta\to 0}\ \ \ \ \frac{12\sin 11\theta}{\sin12\theta}=11\)

\(\D\lim_{\theta\to 0}\ \ \ \ \frac{13\sin 12\theta}{\sin 13\theta}=12\)

\(\D\lim_{\theta\to 0}\ \ \ \ \frac{14\sin 13\theta}{\sin 14\theta}=13\)

\(\D\lim_{\theta\to 0}\ \ \ \ \frac{15\sin 14\theta}{\sin 15\theta}=14\)

\(\D\lim_{\theta\to 0}\ \ \ \ \frac{16\sin 15\theta}{\sin 16\theta}=15\)

\(\D\lim_{\theta\to 0}\ \ \ \ \frac{17\sin 16\theta}{\sin 17\theta}=16\)

\(\D\lim_{\theta\to 0}\ \ \ \ \frac{18\sin 17\theta}{\sin 18\theta}=17\)

\(\D\lim_{\theta\to 0}\ \ \ \ \frac{19\sin 18\theta}{\sin 19\theta}=18\)

..............

王守恩

zeroieme 发表于 2018-1-18 10:47
知识渊博不敢接受，这只是微积分入门课程，离本论坛各位大佬的水平还有几光年。

同样更短的。

\(\D\lim_{\theta\to 0}\ \ \ \ \frac{1cos 3\theta}{\cos 2\theta}=1\)

\(\D\lim_{\theta\to 0}\ \ \ \ \frac{2cos 4\theta}{\cos 3\theta}=2\)

\(\D\lim_{\theta\to 0}\ \ \ \ \frac{3cos 5\theta}{\cos 4\theta}=3\)

\(\D\lim_{\theta\to 0}\ \ \ \ \frac{4cos 6\theta}{\cos 5\theta}=4\)

\(\D\lim_{\theta\to 0}\ \ \ \ \frac{5cos 7\theta}{\cos 6\theta}=5\)

\(\D\lim_{\theta\to 0}\ \ \ \ \frac{6cos 8\theta}{\cos 7\theta}=6\)

\(\D\lim_{\theta\to 0}\ \ \ \ \frac{7cos 9\theta}{\cos 8\theta}=7\)

\(\D\lim_{\theta\to 0}\ \ \ \ \frac{8cos 10\theta}{\cos 9\theta}=8\)

\(\D\lim_{\theta\to 0}\ \ \ \ \frac{9cos 11\theta}{\cos 10\theta}=9\)

\(\D\lim_{\theta\to 0}\ \ \ \ \frac{10cos 12\theta}{\cos 11\theta}=10\)

\(\D\lim_{\theta\to 0}\ \ \ \ \frac{11cos 13\theta}{\cos 12\theta}=11\)

\(\D\lim_{\theta\to 0}\ \ \ \ \frac{12cos 14\theta}{\cos 13\theta}=12\)

\(\D\lim_{\theta\to 0}\ \ \ \ \frac{13cos 15\theta}{\cos 14\theta}=13\)

\(\D\lim_{\theta\to 0}\ \ \ \ \frac{14cos 16\theta}{\cos 15\theta}=14\)

\(\D\lim_{\theta\to 0}\ \ \ \ \frac{15cos 17\theta}{\cos 16\theta}=15\)

\(\D\lim_{\theta\to 0}\ \ \ \ \frac{16cos 18\theta}{\cos 17\theta}=16\)

\(\D\lim_{\theta\to 0}\ \ \ \ \frac{17cos 19\theta}{\cos 18\theta}=17\)

\(\D\lim_{\theta\to 0}\ \ \ \ \frac{18cos 20\theta}{\cos 19\theta}=18\)

............

王守恩

zeroieme 发表于 2018-1-18 10:47
知识渊博不敢接受，这只是微积分入门课程，离本论坛各位大佬的水平还有几光年。

出道难一点的题。

\(\D\lim_{\theta\to 0}\ \ \ \ \frac{\tan3\theta\tan4\theta\tan5\theta\tan6\theta\tan7\theta\tan8\theta\tan9\theta-\sin3\theta\sin4\theta\sin5\theta\sin6\theta\sin7\theta\sin8\theta\sin9\theta}{\tan\theta\tan2\theta\tan3\theta\tan4\theta\tan5\theta\tan6\theta\tan7\theta-\sin\theta\sin2\theta\sin3\theta\sin4\theta\sin5\theta\sin6\theta\sin7\theta}=72\)

zeroieme

王守恩 发表于 2018-1-19 19:52
出道难一点的题。

\(\D\lim_{\theta\to 0}\ \ \ \ \frac{\tan3\theta\tan4\theta\tan5\theta\tan6\the ...

如果你真心喜爱数学，就应当进一步自学更高级的数学课程和尝试使用数学软件，这里有很多热心的朋友会帮助你。
你摆弄的这些数字游戏在用软件解决只是毫秒等级的问题，用人手算也不是难题。

王守恩

zeroieme 发表于 2018-1-19 22:14
如果你真心喜爱数学，就应当进一步自学更高级的数学课程和尝试使用数学软件，这里有很多热心的朋友会帮 ...

用4个  "2"  能构成所有的自然数。
ln[ln(2)÷ln(根号2)]÷2ln(根号2)=1
ln[ln(2)÷ln(根号,根号2)]÷2ln(根号2)=2
ln[ln(2)÷ln(根号,根号,根号2)]÷2ln(根号2)=3
ln[ln(2)÷ln(根号,根号,根号,根号2)]÷2ln(根号2)=4
ln[ln(2)÷ln(根号,根号,根号,根号,根号2)]÷2ln(根号2)=5
ln[ln(2)÷ln(根号,根号,根号,根号,根号,根号2)]÷2ln(根号2)=6
ln[ln(2)÷ln(根号,根号,根号,根号,根号,根号,根号2)]÷2ln(根号2)=7
ln[ln(2)÷ln(根号,根号,根号,根号,根号,根号,根号,根号2)]÷2ln(根号2)=8
ln[ln(2)÷ln(根号,根号,根号,根号,根号,根号,根号,根号,根号2)]÷2ln(根号2)=9
............................

用4个\(\ \ \D\theta\ \ \ \)也能构成所有的自然数。

\(\D\lim_{\theta\to0}\ \ \frac{\sin \theta-\sin(\sin \theta)}{\tan \theta-\sin(\tan \theta)}=1\)

\(\D\lim_{\theta\to0}\ \ \frac{\sin(\tan \theta)-\sin \theta}{\tan \theta-\sin(\tan \theta)}=2\)

\(\D\lim_{\theta\to0}\ \ \frac{\tan \theta-\sin \theta}{\tan \theta-\sin(\tan \theta)}=3\)

\(\D\lim_{\theta\to0}\ \ \frac{\tan \theta-\sin(\sin \theta)}{\tan \theta-\sin(\tan \theta)}=4\)

\(\D\lim_{\theta\to0}\ \ \frac{\tan(\tan \theta)-\sin \theta}{\tan \theta-\sin(\tan \theta)}=5\)

\(\D\lim_{\theta\to0}\ \ \frac{\tan(\tan \theta)-\sin(\sin \theta)}{\tan \theta-\sin(\tan \theta)}=6\)

\(\D\lim_{\theta\to0}\ \ \frac{\tan(\tan(\tan \theta))-\sin \theta}{\tan \theta-\sin(\tan \theta)}=7\)

.................

求助：来个提示，给出更简洁的表达式。

王守恩

王守恩 发表于 2018-1-24 15:40
用4个  "2"  能构成所有的自然数。
ln[ln(2)÷ln(根号2)]÷2ln(根号2)=1
ln[ln(2)÷ln(根号,根号2)] ...

求助：在此先谢谢大家！

\(\D\lim_{\theta\to 0}\ \ \ \ \frac{e^{\theta}×\cos \theta}{e^{\theta}-\cos \theta}=\frac{1}{\theta}\)

王守恩

王守恩 发表于 2018-1-25 10:29
求助：在此先谢谢大家！

\(\D\lim_{\theta\to 0}\ \ \ \ \frac{e^{\theta}×\cos \theta}{e^{\theta ...

请大家指点，这样证明可以吗？

\(\D\lim_{\theta\to 0}\ \ \ \frac{e^{\theta}×\cos \theta}{e^{\theta}-\cos \theta}\)

\(\D=\frac{(1+\theta+\frac{\theta^2}{2}+0(\theta^2))×(1-\frac{\theta^2}{2}+0(\theta^2))}{(1+\theta+\frac{\theta^2}{2}+0(\theta^2))-(1-\frac{\theta^2}{2}+0(\theta^2))}\)

\(\D=\frac{1+\theta+\frac{\theta^2}{2}-\frac{\theta^2}{2}+0(\theta^2)}{\theta+\theta^2+0(\theta^2)}\)

\(\D=\frac{(1+\theta)+0(\theta^2)}{\theta(1+\theta)+0(\theta^2)}\)

\(\D=\frac{1+0(\theta^2)}{\theta+0(\theta^2)}\)

\(\D=\frac{1}{\theta}\)

王守恩

王守恩 发表于 2018-1-26 11:06
请大家指点，这样证明可以吗？

\(\D\lim_{\theta\to 0}\ \ \ \frac{e^{\theta}×\cos \theta}{e^{\the ...

数学并不神秘，用这样的方法也能构造出所有自然数。



          \(\D\frac{e^{\frac{1}{2}}×\cos(\frac{1}{2})}{e^{\frac{1}{2}}-\cos(\frac{1}{2})}=2\)

          \(\D\frac{e^{\frac{1}{3}}×\cos(\frac{1}{3})}{e^{\frac{1}{3}}-\cos(\frac{1}{3})}=3\)

          \(\D\frac{e^{\frac{1}{4}}×\cos(\frac{1}{4})}{e^{\frac{1}{4}}-\cos(\frac{1}{4})}=4\)

          \(\D\frac{e^{\frac{1}{5}}×\cos(\frac{1}{5})}{e^{\frac{1}{5}}-\cos(\frac{1}{5})}=5\)

          \(\D\frac{e^{\frac{1}{6}}×\cos(\frac{1}{6})}{e^{\frac{1}{6}}-\cos(\frac{1}{6})}=6\)

          \(\D\frac{e^{\frac{1}{7}}×\cos(\frac{1}{7})}{e^{\frac{1}{7}}-\cos(\frac{1}{7})}=7\)

          \(\D\frac{e^{\frac{1}{8}}×\cos(\frac{1}{8})}{e^{\frac{1}{8}}-\cos(\frac{1}{8})}=8\)

          \(\D\frac{e^{\frac{1}{9}}×\cos(\frac{1}{9})}{e^{\frac{1}{9}}-\cos(\frac{1}{9})}=9\)

王守恩

zeroieme 发表于 2018-1-12 20:02
极大值Sin(1)

感谢论坛的帮助！还是想问个幼稚的问题。设：

\(A=\frac{k^0}{0!}+\frac{k^2}{2!}+\frac{k^4}{4!}+\frac{k^6}{6!}+\frac{k^8}{8!}+......\)

\(B=\frac{k^1}{1!}+\frac{k^3}{3!}+\frac{k^5}{5!}+\frac{k^7}{7!}+\frac{k^9}{9!}+......\)

问：

\(\frac{A}{B}= 1 时，k=?\)

王守恩

王守恩 发表于 2018-1-29 10:37
感谢论坛的帮助！还是想问个幼稚的问题。设：

\(A=\frac{k^0}{0!}+\frac{k^2}{2!}+\frac{k^4}{4!}+\fr ...

感谢高手们的帮助！这样说可以吗？

\(\D\frac{\frac{(\frac{1}{1})^0}{0!}+\frac{(\frac{1}{1})^2}{2!}+\frac{(\frac{1}{1})^4}{4!}+\frac{(\frac{1}{1})^6}{6!}+\frac{(\frac{1}{1})^8}{8!}+......}{\frac{(\frac{1}{1})^1}{1!}+\frac{(\frac{1}{1})^3}{3!}+\frac{(\frac{1}{1})^5}{5!}+\frac{(\frac{1}{1})^7}{7!}+\frac{(\frac{1}{1})^9}{9!}+......}\implies1\)

\(\D\frac{\frac{(\frac{1}{2})^0}{0!}+\frac{(\frac{1}{2})^2}{2!}+\frac{(\frac{1}{2})^4}{4!}+\frac{(\frac{1}{2})^6}{6!}+\frac{(\frac{1}{2})^8}{8!}+......}{\frac{(\frac{1}{2})^1}{1!}+\frac{(\frac{1}{2})^3}{3!}+\frac{(\frac{1}{2})^5}{5!}+\frac{(\frac{1}{2})^7}{7!}+\frac{(\frac{1}{2})^9}{9!}+......}\implies2\)

\(\D\frac{\frac{(\frac{1}{3})^0}{0!}+\frac{(\frac{1}{3})^2}{2!}+\frac{(\frac{1}{3})^4}{4!}+\frac{(\frac{1}{3})^6}{6!}+\frac{(\frac{1}{3})^8}{8!}+......}{\frac{(\frac{1}{3})^1}{1!}+\frac{(\frac{1}{3})^3}{3!}+\frac{(\frac{1}{3})^5}{5!}+\frac{(\frac{1}{3})^7}{7!}+\frac{(\frac{1}{3})^9}{9!}+......}\implies3\)

\(\D\frac{\frac{(\frac{1}{4})^0}{0!}+\frac{(\frac{1}{4})^2}{2!}+\frac{(\frac{1}{4})^4}{4!}+\frac{(\frac{1}{4})^6}{6!}+\frac{(\frac{1}{4})^8}{8!}+......}{\frac{(\frac{1}{4})^1}{1!}+\frac{(\frac{1}{4})^3}{3!}+\frac{(\frac{1}{4})^5}{5!}+\frac{(\frac{1}{4})^7}{7!}+\frac{(\frac{1}{4})^9}{9!}+......}\implies4\)

\(\D\frac{\frac{(\frac{1}{5})^0}{0!}+\frac{(\frac{1}{5})^2}{2!}+\frac{(\frac{1}{5})^4}{4!}+\frac{(\frac{1}{5})^6}{6!}+\frac{(\frac{1}{5})^8}{8!}+......}{\frac{(\frac{1}{5})^1}{1!}+\frac{(\frac{1}{5})^3}{3!}+\frac{(\frac{1}{5})^5}{5!}+\frac{(\frac{1}{5})^7}{7!}+\frac{(\frac{1}{5})^9}{9!}+......}\implies5\)

\(\D\frac{\frac{(\frac{1}{6})^0}{0!}+\frac{(\frac{1}{6})^2}{2!}+\frac{(\frac{1}{6})^4}{4!}+\frac{(\frac{1}{6})^6}{6!}+\frac{(\frac{1}{6})^8}{8!}+......}{\frac{(\frac{1}{6})^1}{1!}+\frac{(\frac{1}{6})^3}{3!}+\frac{(\frac{1}{6})^5}{5!}+\frac{(\frac{1}{6})^7}{7!}+\frac{(\frac{1}{6})^9}{9!}+......}\implies6\)

\(\D\frac{\frac{(\frac{1}{7})^0}{0!}+\frac{(\frac{1}{7})^2}{2!}+\frac{(\frac{1}{7})^4}{4!}+\frac{(\frac{1}{7})^6}{6!}+\frac{(\frac{1}{7})^8}{8!}+......}{\frac{(\frac{1}{7})^1}{1!}+\frac{(\frac{1}{7})^3}{3!}+\frac{(\frac{1}{7})^5}{5!}+\frac{(\frac{1}{7})^7}{7!}+\frac{(\frac{1}{7})^9}{9!}+......}\implies7\)

\(\D\frac{\frac{(\frac{1}{8})^0}{0!}+\frac{(\frac{1}{8})^2}{2!}+\frac{(\frac{1}{8})^4}{4!}+\frac{(\frac{1}{8})^6}{6!}+\frac{(\frac{1}{8})^8}{8!}+......}{\frac{(\frac{1}{8})^1}{1!}+\frac{(\frac{1}{8})^3}{3!}+\frac{(\frac{1}{8})^5}{5!}+\frac{(\frac{1}{8})^7}{7!}+\frac{(\frac{1}{8})^9}{9!}+......}\implies8\)

\(\D\frac{\frac{(\frac{1}{9})^0}{0!}+\frac{(\frac{1}{9})^2}{2!}+\frac{(\frac{1}{9})^4}{4!}+\frac{(\frac{1}{9})^6}{6!}+\frac{(\frac{1}{9})^8}{8!}+......}{\frac{(\frac{1}{9})^1}{1!}+\frac{(\frac{1}{9})^3}{3!}+\frac{(\frac{1}{9})^5}{5!}+\frac{(\frac{1}{9})^7}{7!}+\frac{(\frac{1}{9})^9}{9!}+......}\implies9\)