@@ 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==
---