## ~kb/open-notes

c696bf6d5c2aa59d6f3bb4010e3cd99b47bb10e9 — kb 7 months ago
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==

---