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 ... )

Stephen Hawking RIP

3 days ago

## No comments:

## Post a Comment