Sa se determine limitele sirurilor \( (a_{n}),(b_{n}),(c_{n}),(d_{n}) \) definite astfel
\( \left( \begin{array}{cc} {a_n} & {b_n} \\ {c_n} & {d_n} \end{array} \right)= \left( \begin{array}{cc} {1} & {\frac{\alpha}{n}} \\ {-\frac{\alpha}{n}} & {1} \end{array} \right)^n (a \in R). \)
Limite de siruri
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- Ciprian Oprisa
- Pitagora
- Posts: 55
- Joined: Tue Feb 19, 2008 8:01 pm
- Location: Lyon sau Cluj sau Baia de Cris
Se poate arata usor ca a inmulti doua matrici de forma \( \left(
\begin{array}{cc}
a & b \\
-b & a
\end{array} \right) \) este echivalent cu a inmulti doua numere complexe de forma \( a+bi \).
Deci calculam \( \lim\limits_{n\rightarrow\infty}(1+\frac{i\alpha}{n})^n=\lim\limits_{n\rightarrow\infty}(1+\frac{i\alpha}{n})^{\frac{n}{i\alpha}i\alpha}=e^{i\alpha}=\cos\alpha+i\sin\alpha \)
Obtinem \( \lim\limits_{n\rightarrow\infty}\left(
\begin{array}{cc}
1 & \frac{\alpha}{n} \\
-\frac{\alpha}{n} & 1
\end{array} \right)^n=
\left(
\begin{array}{cc}
\cos\alpha & sin\alpha \\
-sin\alpha & cos\alpha
\end{array} \right) \)
\begin{array}{cc}
a & b \\
-b & a
\end{array} \right) \) este echivalent cu a inmulti doua numere complexe de forma \( a+bi \).
Deci calculam \( \lim\limits_{n\rightarrow\infty}(1+\frac{i\alpha}{n})^n=\lim\limits_{n\rightarrow\infty}(1+\frac{i\alpha}{n})^{\frac{n}{i\alpha}i\alpha}=e^{i\alpha}=\cos\alpha+i\sin\alpha \)
Obtinem \( \lim\limits_{n\rightarrow\infty}\left(
\begin{array}{cc}
1 & \frac{\alpha}{n} \\
-\frac{\alpha}{n} & 1
\end{array} \right)^n=
\left(
\begin{array}{cc}
\cos\alpha & sin\alpha \\
-sin\alpha & cos\alpha
\end{array} \right) \)
Un lucru este ceea ce este, nu ceea ce pare a fi.
- Ciprian Oprisa
- Pitagora
- Posts: 55
- Joined: Tue Feb 19, 2008 8:01 pm
- Location: Lyon sau Cluj sau Baia de Cris
Ok, am scos factor comun r, am aratat k \( r^n\rightarrow1 \), am calculat si limitele din acei arccos si arcsin, mi-au dat \( \alpha \), dupa ce am substituit \( \frac{1}{n} \) cu x si am aplicat l'Hopital, dar n-am de gand sa scriu toate astea in latex k mi-e lene 
Un lucru este ceea ce este, nu ceea ce pare a fi.