103和平高中

如題

12.
已知\( f(x) \)是一個三次實係數多項式，且\( x_1,x_2,x_3,x_4,x_5 \)為等差數列。試證：\( C_0^4 f(x_1)-C_1^4 f(x_2)+C_2^4 f(x_3)-C_3^4 f(x_4)+C_4^4 f(x_5)=0 \)

10.
給定二階方陣\( \displaystyle A=\Bigg[\; \matrix{2 & 7 \cr 1 & -4} \Bigg]\ \)，若\( \displaystyle A \Bigg[\; \matrix{x_{n-1} \cr y_{n-1}} \Bigg]\; = \Bigg[\; \matrix{x_n \cr y_n} \Bigg]\; \)，\( n \in N \)，\( n \ge 2 \)求\( \displaystyle \lim_{n \to \infty}\frac{x_n}{y_n} \)

9.
w為\( x^2+x+1=0 \)的根\( \langle\; a_n \rangle\;=\langle\; 1+w^n+w^{2n} \rangle\; \)，若\( \displaystyle S_m=\sum_{k=0}^m a_k C_k^m \)，\( m=3k \)，\( k \in N \)
(1)\( S_m=3C_0^m+3C_3^m+\ldots+3C_6^m+3C_m^m \)。
(2)求\( S_{138} \)的首位數為何？

[ 本帖最後由 bugmens 於 2014-10-19 10:09 PM 編輯 ]

第九題第二小題是不是要這樣做.
令\(f(x)=3(1+x)^{138}=a_0+a_1x+....a_{138}x^{138}\), 則所求為\(a_0+a_3+a_6+\cdots+a_{138}\).
令\(\omega_1,\omega_2\)為\(x^2+x+1=0\)之兩根, 則所求為\(\frac{f(1)+f(\omega_1)+f(\omega_2)}{3}\). 其中
\(f(1)=3(1+1)^{138}=3\cdot2^{138}\),
\(f(\omega_1)=3(1+\omega_1)^{138}=3(-\omega_1^2)^{138}=3\)
\(f(\omega_2)=3(1+\omega_2^2)^{138}=3(-\omega_2^2)^{138}=3\)
故\(S_{138}=\frac{3\cdot2^{138}+3+3}{3}=2^{138}+2\). 計算log值之首位數時, 2可以忽略, 得\(\log 2^{138}\approx 0.3010\times138=41.538\),
由\(\log3=0.4771,\log4=0.602\)知首位數為3.(謝謝Sandy指正)

[ 本帖最後由 David 於 2014-5-3 09:45 PM 編輯 ]

第九題第一小題: 以3的倍數分類討論\(a_n\).
1. 若\(n=3k\), 則\(a_n=a_{3k}=1+\omega^{3k}+\omega^{6k}=1+1+1=3\)
2. 若\(n=3k+1\), 則\(a_{3k+1}=1+\omega^{3k+1}+\omega^{6k+2}=1+\omega+\omega^2=0\)
3, 若\(n=3k+2\), 則\(a_{3k+2}=1+\omega^{3k+2}+\omega^{3k+4}=1+\omega^2+\omega=0\)
故只剩\(a_{3k}=3\)

請大家指教

[ 本帖最後由 David 於 2014-5-3 07:04 PM 編輯 ]

第十二題, 仿瑋大對巴貝奇定理的講解, 不知道可以不可以:
令\(g_1(x)=f(x+d)-f(x)\), 則\(g_1(x)\)為二次,
令\(g_2(x)=g_1(x+d)-g_1(x)\), 則\(g_2(x)\)為一次,
令\(g_3(x)=g_2(x+d)-g_2(x)\), 則\(g_3(x)\)為0次.
故\(g_3(x+d)-g_3(x)=0\).
依序代入\(g_3,g_2,g_1\)定義, 得
$$
\begin{aligned}
0=&g_3(x+d)-g_3(x)\\
=&[g_2(x+2d)-g_2(x+d)]-[g_2(x+d)-g_2(x)]\\
=&g_2(x+2d)-2g_2(x+d)+g_2(x)\\
=&[g_1(x+3d)-g_1(x+2d)]-2[g_1(x+2d)-g_1(x+d)]+[g_1(x+d)-g_1(x)]\\
=&g_1(x+3d)-3g_1(x+2d)+3g_1(x+d)-g_1(x)\\
=&[f(x+4d)-f(x+3d)]-3[f(x+3d)-f(x+2d)]+3[f(x+2d)-f(x+d)]-[f(x+d)-f(x)]\\
=&f(x+4d)-4(x+3d)+6f(x+2d)-4f(x+d)+f(x)
\end{aligned}
$$

[ 本帖最後由 David 於 2014-5-11 08:56 PM 編輯 ]

題目是問首位數，不是首數喔！

啊! 謝謝指正!

依稀記得有ㄧ題是
f為ㄧ實係數多項式函數，f(6-t)=f(t+3)，t為實數，
f(x)=0有40個相異實根，求40個根之和。

當下有想到圖形有對稱性，但還是忘了怎麼繼續下去....

老梗題
f(x) = f(9 - x)
所求 = 9 * (40/2) = 180

下面圖檔的做法哪裡有誤呢?我認為是增根。還有圖檔的作法，利用了小技巧。因為根本不知道是否收斂?



10、 給定二階方陣，\(A = \left[ {\begin{array}{*{20}{c}}2&7\\1&{ - 4}\end{array}} \right]\)，若\(A\left[ {\begin{array}{*{20}{c}}
{{x_{n - 1}}}\\{{y_{n - 1}}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{{x_n}}\\{{y_n}}\end{array}} \right]\)，\(n \in N\)，\(n \ge 2\)，求\(\mathop {\lim }\limits_{n \to \infty } \frac{{{x_n}}}{{{y_n}}} = \)

解 (1)
\(A = \left[ {\begin{array}{*{20}{c}}2&7\\1&{ - 4}\end{array}} \right]\),
\(|A - \lambda I| = \left| {\begin{array}{*{20}{c}}{2 - \lambda }&7\\1&{ - 4 - \lambda }\end{array}} \right| = {\lambda ^2} + 2\lambda  - 15 = \left( {\lambda  + 5} \right)\left( {\lambda  - 3} \right) = 0 \Rightarrow \lambda  =  - 5\;,\;\lambda  = 3\)

(2)  \(\lambda  =  - 5\)
\(\left[ {\begin{array}{*{20}{c}}
7&7\\
1&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
x\\
y
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
0\\
0
\end{array}} \right] \Rightarrow \;\;{v_1} = \left( {1, - 1} \right)\)

(3)\(\lambda  = 3\)
\(\left[ {\begin{array}{*{20}{c}}
{ - 1}&7\\
1&{ - 7}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
x\\
y
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
0\\
0
\end{array}} \right] \Rightarrow \;\;{v_2} = \left( {7,1} \right)\)

(4) \(AP=PD\)  ~~~   \(A = PD{P^{ - 1}}\)
(5)

\[\begin{array}{l}
P = \left[ {\begin{array}{*{20}{c}}
1&7\\
{ - 1}&1
\end{array}} \right]\;\;,\;\;{P^{ - 1}} = \frac{1}{8}\left[ {\begin{array}{*{20}{c}}
1&{ - 7}\\
1&1
\end{array}} \right]\;\;\;,\;\;\\
A = \frac{1}{8}\left[ {\begin{array}{*{20}{c}}
1&7\\
{ - 1}&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
{ - 5}&0\\
0&3
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
1&{ - 7}\\
1&1
\end{array}} \right]\\
\left[ {\begin{array}{*{20}{c}}
{{x_n}}\\
{{y_n}}
\end{array}} \right] = {A^{n - 1}}\left[ {\begin{array}{*{20}{c}}
{{x_1}}\\
{{y_1}}
\end{array}} \right]\\
\;\;\;\;\;\;\; = {\left( {\frac{1}{8}} \right)^{n - 1}}\left[ {\begin{array}{*{20}{c}}
1&7\\
{ - 1}&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
{{{\left( { - 5} \right)}^{n - 1}}}&0\\
0&{{3^{n - 1}}}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
1&{ - 7}\\
1&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
{{x_1}}\\
{{y_1}}
\end{array}} \right]\\
\;\;\;\;\;\;\; = {\left( {\frac{1}{8}} \right)^{n - 1}}\left[ {\begin{array}{*{20}{c}}
{{{\left( { - 5} \right)}^{n - 1}}\left( {{x_1} - 7{y_1}} \right) + 7{{\left( 3 \right)}^{n - 1}}\left( {{x_1} + {y_1}} \right)}\\
{ - {{\left( { - 5} \right)}^{n - 1}}\left( {{x_1} - 7{y_1}} \right) + {{\left( 3 \right)}^{n - 1}}\left( {{x_1} + {y_1}} \right)}
\end{array}} \right]\\
\\
\\
\frac{{{x_n}}}{{{y_n}}} = \frac{{\frac{{{{\left( { - 5} \right)}^{n - 1}}}}{{{{\left( { - 5} \right)}^{n - 1}}}}\left( {{x_1} - 7{y_1}} \right) + 7\frac{{{{\left( 3 \right)}^{n - 1}}}}{{{{\left( { - 5} \right)}^{n - 1}}}}\left( {{x_1} + {y_1}} \right)}}{{ - \frac{{{{\left( { - 5} \right)}^{n - 1}}}}{{{{\left( { - 5} \right)}^{n - 1}}}}\left( {{x_1} - 7{y_1}} \right) + \frac{{{{\left( 3 \right)}^{n - 1}}}}{{{{\left( { - 5} \right)}^{n - 1}}}}\left( {{x_1} + {y_1}} \right)}}\\
\;\;\;\; = \frac{{\left( {{x_1} - 7{y_1}} \right) + 7\frac{{{{\left( 3 \right)}^{n - 1}}}}{{{{\left( { - 5} \right)}^{n - 1}}}}\left( {{x_1} + {y_1}} \right)}}{{ - 1\left( {{x_1} - 7{y_1}} \right) + \frac{{{{\left( 3 \right)}^{n - 1}}}}{{{{\left( { - 5} \right)}^{n - 1}}}}\left( {{x_1} + {y_1}} \right)}}
\end{array}\]

\({x_1},{y_1}\)為定值， \({\left( {\frac{{ - 3}}{5}} \right)^{n - 1}}\)逼近於 0，當n逼近無限大

所以 當 n逼近於無限大
\(\frac{{{x_n}}}{{{y_n}}}\)逼近於 \(-1\)



\( \displaystyle \Bigg[\; \matrix{2 & 7 \cr 1 & -4} \Bigg]\; \Bigg[\; \matrix{x_{n-1} \cr y_{n-1}} \Bigg]\;=\Bigg[\; \matrix{x_n \cr y_n} \Bigg]\; \)

\( \displaystyle \Rightarrow \cases{x_n=2x_{n-1}+7y_{n-1} \cr y_n=x_{n-1}-4y_{n-1}} \)

\( \displaystyle \frac{x_n}{y_n}=\frac{2x_{n-1}+7y_{n-1}}{x_{n-1}-4y_{n-1}} \)(同除以\( y_{n-1} \))\( \displaystyle =\frac{2 (\frac{x_{n-1}}{y_{n-1}})+7}{(\frac{x_{n-1}}{y_{n-1}})-4} \)(取極限)

令\( \displaystyle \lim_{n \to \infty}\frac{x_n}{y_n}=\lim_{n \to \infty}\frac{x_{n-1}}{y_{n-1}}=t \)

\( \displaystyle t=\frac{2t+7}{t-4} \Rightarrow t=7,-1 \)

[ 本帖最後由 bugmens 於 2014-5-22 05:04 AM 編輯 ]

10. 因為未給初始值，造成答案不唯一

當 \( \begin{bmatrix}x_{1}\\
y_{1}
\end{bmatrix}=c\begin{bmatrix}7\\
1
\end{bmatrix} , c\neq0 \), \( \frac{x_{n}}{y_{n}} \) 恆為 7。

當 \( \begin{bmatrix}x_{1}\\
y_{1}
\end{bmatrix} \) 滿足 \( x_{1}-7y_{1}\neq0 \)，時 \( \lim_{n\to\infty}\frac{x_{n}}{y_{n}}=-1 \)