~kb/open-notes

c696bf6d5c2aa59d6f3bb4010e3cd99b47bb10e9 — kb 7 months ago aea9626
CM1015 midterm WIP
1 files changed, 82 insertions(+), 30 deletions(-)

M notes/study/CM1015/CM1015-midterm.md
M notes/study/CM1015/CM1015-midterm.md => notes/study/CM1015/CM1015-midterm.md +82 -30
@@ 93,8 93,39 @@ $$

## Question 1 (d)

==TODO need a good notation for showing binary operations==
### i.

$$
1110001_{2}-110010_{2} = 111111_{2}
$$

$$
\begin{array}{lllllllllllllllllllllllllllllllll}
  & 1  & 1  &  1 &  1 & 1  & 0  &  0 & \it{borrow}  &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &  \\ %carry digits
  & 1  & 1  & 1  & 0  & 0  & 0  & 1  &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &  \\	%1st number
- & 0  &  1 &  1 & 0  & 0  & 1  & 0  &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &  \\ %2nd number
\hline
  & 0  & 1  & 1  &  1 & 1  & 1  & 1  &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &  \\ %result
\end{array}
$$

## ii.

$$
11011_{2}\times 111_{2} = 11011_{2} + 110110_{2} + 1101100_{2} = 10111101_{2}
$$
$$
\begin{array}{lllllllllllllllllllllllllllllllll}
  & 1 &  1 &  1 & 1  &  1 & 1 & 0  & 0& \it{carry}   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &  \\ %carry digits
  &   &   &  & 1  & 1  & 0  & 1  & 1  &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &  \\	%1st number
+ &   &   & 1  & 1  & 0  & 1  & 1  & 0  &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &  \\ %2nd number
+ &   &  1 &  1 & 0  & 1  & 1  & 0  &  0 &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &  \\ %3rd number
\hline
  & 1  & 0  & 1  &  1 & 1  & 1  & 0  & 1  &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &   &  \\ %result
\end{array}
$$

This may then be converted to hexadecimal by first segmenting $1011\ 1101$, noting the representation $8+2+1 \quad 8+4+1$ and lastly representing as base 16 characters $BD_{16}$.

## Question 1 (e)



@@ 138,17 169,31 @@ To express $25.25_{10}$ in binary we may segment the integer and fractional comp

To manually calculate this conversion the integer component may first be repeated dived by the new base:

| Step | 
$$
\begin{align*}
\frac{25}{2} &= \\
5\enclose{longdiv}{25}\\
\frac{25}{2} &= 12 R1\\
\\
\frac{12}{2} &= 6 R0\\
\\
\frac{6}{2} &= 3R0\\
\\
\frac{3}{2} &= 1 R 1\\
\\
\frac{1}{2} &= 0 R 1
\end{align*}
$$
==TODO notation for showing long div==
https://opentextbc.ca/pressbooks/chapter/how-do-i-write-long-division/
https://learnsharewithdp.wordpress.com/2020/05/19/how-to-write-long-division-in-latex/

For the fractional component we may then formed from subsequent multiplication by the new base
$$
\begin{align*}
0.25 \cdot 2 &= 0.5 C 0\\
\\
0.5 \cdot 2 &= 1 C 0\\
\\
1 \cdot 2 &= 2 C 1
\end{align*}
$$
Reforming this provides $25.25_{10} = 11001.001_{2}$

---
---


@@ 310,40 355,37 @@ $$
\begin{align*}
\sum\limits_{n=1}^{\infty} \frac{n}{2^{n}}
&= \frac{1}{2^{1}} + \frac{2}{2^{2}} + \frac{3}{2^{3}} + \frac{4}{2^{4}} + \cdots\\
&= \frac{1}{2} + \frac{2}{4}+ \frac{3}{8} + \frac{4}{16} + \cdots\\
\\
&= \frac{1}{2} + \frac{4}{8}+ \frac{3}{8} + \frac{2}{8} + \cdots
&= \frac{1}{2} + \frac{2}{4}+ \frac{3}{8} + \frac{4}{16} + \cdots\\\\

&= 2
\end{align*}
 $$
we may reform as
$$
\sum\limits_{n=1}^{\infty} n \cdot 2^{-n}
$$
==TODO return to this - split into first and rest?==

this will converge.


## Question 2 (f)

For a ball dropped from an initial height of 50cm reaching $\frac{1}{3}$ the previous height for each subsequent bounce we may define the progression of heights as follows
For a ball dropped from an initial height of 50cm reaching $\frac{1}{3}$ the previous height for each subsequent bounce.

We may define the initial drop as $d = 50$.

The first bound will then reach a height of $50 \cdot \frac{1}{3}$. Noting there will be both an upward and downward motion the gives $b_{1} = 50 \cdot \frac{2}{3}$.

Continuing $b_{2} = 50 \cdot \frac{2}{3} \cdot \frac{2}{3}$.

We may then noted that the progression for each bounce
$$
b_{n} = 50 \cdot \left( \frac{1}{3} \right)^n
b_{n} = 50 \cdot \left( \frac{2}{3} \right)^n
$$
Noting that each bounce includes both an upwards and downwards motion we may then define the distance traveling for any bounce
To model the complete movement it then becomes possible to express
$$
d_{n} = 100 \cdot \left( \frac{1}{3} \right)^n
50 \sum_{n=1}^{\infty} \left( \frac{2}{3} \right)^n
$$
Combining both the initial fall and subsequent bounced we may then state the total movement


, we may then state
Giving the total distance as
$$
\begin{align*}
50 + \sum\limits_{n=1}^{\infty} 2 \cdot 50 \cdot \left( \frac{1}{3} \right)^n\\
\\
100 \cdot \left( \frac{1}{3} \right)^n
\end{align*}
\frac{50}{1-{\frac{2}{3}}} = 150\text{cm}
$$
==TODO and all this.. study with a 2yo screaming is completely fucking impossible==


## Question 2 (g)



@@ 351,6 393,16 @@ $$

## Question 2 (h)

$$
\{a_0,9,a_2,a_3,0.125,\ldots\}
$$
$$
a_1=9=a_{0}\cdot r^1
$$
$$
a_{4}= 0.125 = a_{0} \cdot r^{4}
$$

==TODO==

---