假设$\displaystyle x_0=1,x_n=x_{n-1}+\cos x_{n-1}(n=1,2,\cdots )$,证明:当$x\rightarrow \infty $时, $\displaystyle x_n-\frac{\pi }{2}=o\left(\frac{1}{n^n}\right)$.
证.先证$1\leq x_n<\frac\pi /2$,得到$x_n-x_{n-1}>0$,由单调有界定理可知$x_n$极限存在且$\lim_{n\to\infty}=\frac\pi2$.下面用归纳法证明$\lim_{n\rightarrow\infty}n^n\left(x_n-\frac{\pi}{2}\right)=0$.假设
$$\lim_{n\rightarrow\infty}n^n\left(x_n-\frac{\pi}{2}\right)=0.$$
我们有
\begin{align*}\lim_{n\rightarrow\infty}\left(n+1\right)^{n+1}\left(x_{n+1}-\frac{\pi}{2}\right)&=\lim_{n\rightarrow\infty}\left(n+1\right)^{n+1}\left(x_n+\cos x_n-\frac{\pi}{2}\right)\\&=\lim_{n\rightarrow\infty}\left(n+1\right)^{n+1}\left(x_n+\sin\left(\frac{\pi}{2}-x_n\right)-\frac{\pi}{2}\right)\\&=\lim_{n\rightarrow\infty}\left(n+1\right)^{n+1}\left(x_n+\left(\frac{\pi}{2}-x_n\right)-\frac{1}{6}\left(\frac{\pi}{2}-x_n\right)^3-\frac{\pi}{2}\right)\\&=-\frac{1}{6}\lim_{n\rightarrow\infty}\left(n+1\right)^{n+1}\left(\frac{\pi}{2}-x_n\right)^3\\&=-\frac{1}{6}\lim_{n\rightarrow\infty}\frac{\left(n+1\right)^{n+1}}{n^{3n}}\left[n^n\left(x_n-\frac{\pi}{2}\right)\right]^3=0.\end{align*}
TangSong：令$y_n=\frac\pi2-x_n$,得到$y_n=y_{n-1}-\sin y_{n-1}$.可以证明$$\lim_{n\rightarrow\infty}\frac{y_{n+1}}{y_{n}^{3}}=\frac{1}{6}.$$因此当$n>N$时,我们有$$\frac{y_{n+1}}{y_{n}^{3}}<\frac{1}{2}.$$因此$$0<y_n<\frac{1}{2}y_{n-1}^{3}<\left(\frac{1}{2}\right)^{1+3}y_{n-2}^{3^2}<\cdots <\left(\frac{1}{2}\right)^{1+3+\cdots +3^{n-N-2}}y_{N+1}^{3^{n-N-1}},$$
即$$0<y_n<\left(\frac{1}{2}\right)^{\left(3^{n-N-1}-1\right)/2}y_{N+1}^{3^{n-N-1}}.$$
设$$\left({\begin{array}{*{20}{c}}{{x_{3n}}}\\{{x_{3n + 1}}}\\{{x_{3n + 2}}}\end{array}}\right)=\left({\begin{array}{*{20}{c}}3&{ - 2}&1\\4&{ - 1}&0\\4&{ - 3}&2\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{x_{3n - 3}}}\\{{x_{3n - 2}}}\\{{x_{3n - 1}}}\end{array}} \right).$$给定初值$a_0=5,a_1=7,a_2=8$,求$x_n$的通项.
解.我们先求矩阵的Jordan标准型,得到$$M = \left( {\begin{array}{*{20}{c}}3&{ - 2}&1\\4&{ - 1}&0\\4&{ - 3}&2\end{array}} \right) = SJ{S^{ - 1}} = \left( {\begin{array}{*{20}{c}}1&{1/2}&3\\2&0&4\\2&0&5\end{array}} \right)\left( {\begin{array}{*{20}{c}}1&1&0\\0&1&0\\0&0&2\end{array}} \right)\left( {\begin{array}{*{20}{c}}0&{5/2}&{ - 2}\\2&1&{ - 2}\\0&{ - 1}&1\end{array}} \right).$$因此
\begin{align*}\left( {\begin{array}{*{20}{c}}{{x_{3n - 3}}}\\{{x_{3n - 2}}}\\{{x_{3n - 1}}}\end{array}} \right) &= {\left( {\begin{array}{*{20}{c}}3&{ - 2}&1\\4&{ - 1}&0\\4&{ - 3}&2\end{array}} \right)^{n - 1}}\left( {\begin{array}{*{20}{c}}{{x_0}}\\{{x_1}}\\{{x_2}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1&{1/2}&3\\2&0&4\\2&0&5\end{array}} \right){\left( {\begin{array}{*{20}{c}}1&1&0\\0&1&0\\0&0&2\end{array}} \right)^{n - 1}}\left( {\begin{array}{*{20}{c}}0&{5/2}&{ - 2}\\2&1&{ - 2}\\0&{ - 1}&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}5\\7\\8\end{array}} \right)\\&= \left( {\begin{array}{*{20}{c}}1&{1/2}&3\\2&0&4\\2&0&5\end{array}} \right)\left( {\begin{array}{*{20}{c}}1&{n - 1}&0\\0&1&0\\0&0&{{2^{n - 1}}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}0&{5/2}&{ - 2}\\2&1&{ - 2}\\0&{ - 1}&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}5\\7\\8\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{n + 1 + 3 \times {2^{n - 1}}}\\{2n + 1 + {2^{n + 1}}}\\{2n + 1 + 5 \times {2^{n - 1}}}\end{array}} \right).\end{align*}
空间中四点$O,A,B,C$使得\[\angle AOB=\frac{\pi}{2},\angle BOC=\frac{\pi}{3},\angle COA=\frac{\pi}{4}\]
设$AOB$决定的平面为$\pi_1$,$BOC$决定的平面为$\pi_2$,求$\pi_1,\pi_2$二面角.求出二面角的余弦值即可.
解.
设线性空间$V$, $\delta$是线性映射,其中向量$\beta$是$\delta$以特征值$\lambda$的特征向量.求证:对任意不全为零的$k_i(1\leq i \leq n)$,都存一组$V$的基使得$\beta$可以表示为该组基以$k_i$的线性组合.
证明存在矩阵$A$使得, $Aa_1=b_1,Aa_2=b_2$,其中$a_1\neq a_2,b_1\neq b_2$且$|A|=1$.
2022年8月17日 19:01
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