The MST209 exam has a different format than MS221 and M208 have. In 2006, for example, the format was as follows:
Part A. 15 multiple-choice questions, 2 marks each = 30 points. ( 1 hour )
Part B. 8 questions, 5 marks each = 40 points. ( 1 hour 15 min )
Part C. 3 out of 7, 15 marks each = 45 points. ( 45 min )
Yes. Maximum score is 115. Scores above 100 are set to 100.
The exam looks doable. And again questions on eigenvalues and eigenvectors. That would be three in a row: MS221, M208 and MST209. Considering the fact that one can prove Binet's formula for the Fibonacci numbers with them it's worthwhile having it firm under your math-belt.
There is also an analytical function for the Fibonacci numbers which rounded, gives an exact Fibonacci number if the input variable is an integer. Here is the related math.
Let $F_n = F_{n-1} + F_{n-2}, F_0=0, F_1=1$, show that $Fa_n=\frac{1}{\sqrt{5}}e^{n \cdot \log{\phi}}$, where $\phi$ is the Golden Ratio or $\frac{1+\sqrt{5}}{2}$. ( Round $Fa_n$ to get $F_n$. )
If we define the elements $F_{n}$ and $F_{n+1}$ as the vector $s_n= \left(
\begin{array}{c}
F_{n+1}\\
F_{n}
\end{array}
\right)$
then $F_n$ simply becomes
$F_n= \left(
\begin{array}{cc}
1 & 1\\
1 & 0
\end{array}
\right)^n
\cdot s_{0}$.
We can calculate the power of a matrix by diagonalizing the matrix. And this is where eigenvalues and vectors come in. If $\lambda_1, \lambda_2$ are eigenvectors with respective eigenvectors $E= \left( e_1, e_2 \right)$ we get $F_n= E^{-1}
\cdot
\left(
\begin{array}{cc}
\lambda_1^n & 0\\
0 & \lambda_2^n
\end{array}
\right)
\cdot
E
\cdot s_{0}$
The eigenvectors are the roots of the characteristic equation $\left|
\begin{array}{cc}
1-\lambda & 1\\
1 & -\lambda
\end{array} \right| = 0$ and are thus $\frac{1}{2} + \frac{1+\sqrt{5}}{2}$ and $\frac{1}{2} - \frac{1+\sqrt{5}}{2}$.
( TO BE CONTINUED ... )
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