The MST209 exam has a different format than MS221 and M208 have. In 2006, for example, the format was as follows:
Part A. 15 multiplechoice 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 mathbelt.
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 $F_n = F_{n1} + F_{n2}, F_0=0, F_1=1$, 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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