You are invited to give your feedback on this module here.
Final CA results below. Unless you excelled, you are identified by the last five digits of your student number.
CA is your total continuous assessment marks out of 30.
PP is your passing percentage: what you need on the final paper to pass.
70P is the percentage you need on the final paper for a first (70%).
AW is the number of attendance warnings: for future reference you can see the fairly strong relationship between attendance and assessment performance.
I will have the assessments with me in today and tomorrow’s tutorials and will send on the marking scheme in a few minutes.
S/N |
Test 1 |
Test 2 |
CA |
PP |
70P |
AW |
O Leary Garvey |
96.2 |
100 |
29.4 |
15.1 |
58.0 |
0 |
McGrath |
91 |
98.1 |
28.4 |
16.6 |
59.5 |
0 |
Xie |
94.9 |
86.5 |
27.2 |
18.3 |
61.1 |
0 |
Cashman |
87.2 |
88.5 |
26.4 |
19.5 |
62.4 |
1 |
62916 |
94.9 |
78.8 |
26.1 |
19.9 |
62.8 |
0 |
63379 |
82.1 |
90.4 |
25.9 |
20.2 |
63.0 |
0 |
64423 |
82.1 |
88.5 |
25.6 |
20.6 |
63.4 |
0 |
62647 |
73.7 |
96.2 |
25.5 |
20.7 |
63.6 |
0 |
62469 |
79.4 |
89.4 |
25.3 |
21.0 |
63.8 |
1 |
62740 |
91.7 |
73.1 |
24.7 |
21.8 |
64.7 |
0 |
64717 |
82.1 |
82.7 |
24.7 |
21.8 |
64.7 |
0 |
63549 |
70.5 |
86.5 |
23.6 |
23.5 |
66.4 |
0 |
64314 |
83.3 |
73.1 |
23.5 |
23.6 |
66.5 |
3 |
64860 |
76.9 |
78.8 |
23.4 |
23.8 |
66.6 |
0 |
63657 |
70.5 |
78.8 |
22.4 |
25.2 |
68.0 |
1 |
64078 |
72.4 |
76 |
22.3 |
25.3 |
68.2 |
0 |
61981 |
81.4 |
65.4 |
22.0 |
25.7 |
68.5 |
2 |
64100 |
65.4 |
78.8 |
21.6 |
26.2 |
69.1 |
0 |
62970 |
66.7 |
76.9 |
21.5 |
26.4 |
69.2 |
0 |
69532 |
67.3 |
71.2 |
20.8 |
27.5 |
70.3 |
0 |
64370 |
66.7 |
69.2 |
20.4 |
28.0 |
70.9 |
3 |
63904 |
65.4 |
67.3 |
19.9 |
28.7 |
71.6 |
2 |
62523 |
60.3 |
71.2 |
19.7 |
29.0 |
71.8 |
2 |
63815 |
50 |
80.8 |
19.6 |
29.1 |
72.0 |
3 |
63193 |
78.2 |
51.9 |
19.5 |
29.3 |
72.1 |
2 |
63651 |
73.7 |
53.8 |
19.1 |
29.8 |
72.7 |
1 |
62360 |
55.1 |
70.2 |
18.8 |
30.3 |
73.2 |
0 |
63606 |
61.5 |
57.7 |
17.9 |
31.6 |
74.5 |
0 |
64052 |
46.2 |
72.1 |
17.7 |
31.8 |
74.7 |
0 |
62502 |
36.5 |
52.9 |
13.4 |
38.0 |
80.8 |
1 |
64320 |
17.9 |
69.2 |
13.1 |
38.5 |
81.3 |
0 |
64259 |
77 |
0 |
11.6 |
40.6 |
83.5 |
3 |
64281 |
20.5 |
55.8 |
11.4 |
40.8 |
83.7 |
2 |
64273 |
30.1 |
44.2 |
11.1 |
41.2 |
84.1 |
0 |
61818 |
55.8 |
0 |
8.4 |
45.2 |
88.0 |
3 |
63677 |
48.1 |
0 |
7.2 |
46.8 |
89.7 |
3 |
64956 |
26.3 |
0 |
3.9 |
51.5 |
94.4 |
3 |
62151 |
23.7 |
0 |
3.6 |
52.1 |
94.9 |
0 |
63740 |
74.4 |
I |
I |
I |
I |
I |
62812 |
64.1 |
WD |
WD |
WD |
WD |
WD |
In Week 12 we finished Recursion between Monday and Tuesday. Friday was an extra tutorial.
In Week 13, we will have five tutorials (normal rooms and times) of which you are invited to up to four (your own tutorial slot plus the up to three of the lecture slots).
Some students need to do extra work outside tutorials. Please feel free to ask me questions about the exercises via email or even better on this webpage.
Anyone who is missing notes is to email me.
Please see the Student Resources tab on the top of this page for information on the Academic Learning Centre, etc. There are some excellent notes on Blackboard for MATH6055.
You are invited to give your feedback on this module here.
Results not too far away… watch this space.
We looked at centroids and centres of gravity.
There is an exam paper at the back of your notes — I will go through this on the board in the lecture times (in the usual venues):
We will also have tutorial time in the tutorial slots. You can come to as many tutorials as you like.
Please feel free to ask me questions about the exercises via email or even better on this webpage.
Please see the Student Resources tab on the top of this page for information on the Academic Learning Centre, etc.
You are invited to give your feedback on this module here.
Results have been emailed to you. You have a chance to see your work this Friday in tutorial. Some comments here.
I postponed Monday’s lecture. We nearly finished off Chapter 4 by looking at Error Analysis. I had to email ye on the last two rounding error examples.
We will go through last year’s exam on the board and then I will answer your questions if there are any. If there are none I will help one-to-one. Usual class times and locations.
We will also have tutorials on Friday 8 December in the usual times and venues.
Please feel free to ask me questions about the exercises via email or even better on this webpage.
Please see the Student Resources tab on the top of this page for information on the Academic Learning Centre, etc.
I told this story in class on Friday. I wasn’t sure if it was true but it appears that it is.
‘Texas‘ is a poker game where a number of players sit around a table. Two cards are dealt to each player. There after follows a round of betting, the reveal of three more cards (the flop), more betting, another card (the turn), another round of betting, another card (the river), and another round of betting:
What we are interested in is what happens after all this, before another hand is dealt?
The deck is shuffled.
A shuffle is required to mix up the deck. Here we used three terms: deck, shuffle, mixed up. These can all be given a precise mathematical realisation (see the introduction here for more). Mixed up means ‘close to random’. Here let me introduce a mathematical realisation of random:
If one is handed a deck of cards, face down, and if each possible order of
the cards is equally possible then the deck is considered random.
Note there are possible orders that a deck can be in so when a deck is random the probability that a deck is in a specific order is
.
One popular method of shuffling cards is the riffle shuffle. In a remarkable 1992 paper by Bayer & Diaconis, with a really cool name: Trailing the Dovetail Shuffle to Its Lair, it is shown that seven riffle shuffles are necessary and sufficient to get a deck close to random:
Here we see , distance to random, plotted against , number of shuffles. After five shuffles the deck is still far from random, but then there is a fairly abrupt convergence to random. After seven shuffles the distance to random is less than .
So the idea is after, say, ten shuffles (or, equivalently, about ten rounds of hands), the deck is mixed up or close to random: each of the 52! orders are approximately likely.
This says something quite remarkable about the uniqueness of card games. Imagine each of the approximately 8 billion people on Earth own a deck of cards. Suppose we all spent 8 hours a day for 50 years shuffling our cards… at one shuffle per minute that is only about deck orders…
This is a tiny, tiny, fraction of the 52! possible orders! Only approximately
th
of all decks have ever been seen… so the probability of two orders ever have been repeated is tiny. When you play a deck of cards, the likelihood is that no other deck of cards have ever been in that order!
Now when we move online we need an algorithm to shuffle the deck. One algorithm works as follows.
Number all the different orders from to .
Start with a random number. This can be found by turning on a computer and having the computer record the number of milliseconds, since midnight (or from some other time).
The first deck used on the table is .
Now, from , generate a quasi-random sequence:
,
using the recurrence:
.
is the remainder when is divided by . For example,
, , , , etc.
Here , , and have to be chosen carefully. For example,
,
using seed gives a sequence of 64 quasi-random (random ‘looking’) numbers:
So an online poker site can start with a random deck, and using a quasi-random number generator recurrence, generate a list of random-looking decks.
However, a serious error was made by the those in charge of one particular site. The way they stored the deck orders was not enough to encode all possible orders and in fact (using 32 bits rather than the 200+ necessary) only a tiny, tiny proportion of all possible deck orders were stored.
This resulted in carnage: carnage that you can read more about here.
How much would you pay to spin this wheel (the numerical values are prizes):
Such games should be analysed using expected average. Imagine playing it 80 times. You expect to hit each of the eight ‘prizes’ ten times:
winnings from 80 plays = 10*20+10*5+10*10+10*5=400,
and so dividing this by 80 we get an expected average of 5. This means, on average, when you play you win 5. Therefore if you intend on playing regularly, if you pay less than €5 you will win on average.
If we divide the above sum by 80 we see:
expected winnings = .
This sum here is ‘add up prize by probability of prize’.
When the spinning machine has a ‘spin again’ it is also possible to analyse it. Consider the following:
Using the expected winnings equal to , we have
expected winnings = .
But the spin again prize is the same as the expected winnings so we have, if we denoted by the expected winnings:
.
Value the following games:
You are invited to give your feedback on this module here.
I hope to have these corrected at some stage next week. I will also send out the solutions at some point.
We finished our study of Graph Theory by looking at Eulerian graphs, Fleury’s Algorithm, Hamiltonian graphs, and Dirac’s Theorem. We then began the last chapter on recursion. We had our test on Friday
In Week 12 we will finish our study of recursion and perhaps do a little revision.
In Week 13, perhaps we will have five tutorials (normal rooms and times) of which you are invited to up to four (your own tutorial slot plus the up to three of the lecture slots).
Some students need to do extra work outside tutorials. Please feel free to ask me questions about the exercises via email or even better on this webpage.
Anyone who is missing notes is to email me.
Please see the Student Resources tab on the top of this page for information on the Academic Learning Centre, etc. There are some excellent notes on Blackboard for MATH6055.
You are invited to give your feedback on this module here.
I hope to have these corrected at some stage next week. I will also send out the solutions at some point.
We looked at completing the square and work.
We will look at centroids and centres of gravity.
There is an exam paper at the back of your notes — I will go through this on the board in the lecture times (in the usual venues):
We will also have tutorial time in the tutorial slots. You can come to as many tutorials as you like.
Please feel free to ask me questions about the exercises via email or even better on this webpage.
Please see the Student Resources tab on the top of this page for information on the Academic Learning Centre, etc.
You are invited to give your feedback on this module here.
Results have been emailed to you. You have a chance to see your work this Friday in tutorial. Some comments here.
We looked at more general Taylor Series: not just near and also for functions of several variables.
We will finish off Chapter 4 by looking at Error Analysis.
We will go through last year’s exam on the board and then I will answer your questions if there are any. If there are none I will help one-to-one. Usual class times and locations.
We will also have tutorials on Friday 8 December in the usual times and venues.
Please feel free to ask me questions about the exercises via email or even better on this webpage.
Please see the Student Resources tab on the top of this page for information on the Academic Learning Centre, etc.
,
that .
,
where and , prove that for large :
Recall the Doubling Mapping given by:
At the end of the last post we showed that this dynamical system displays sensitivity to initial conditions. Now we show that it displays topological mixing (a chaotic orbit) and density of periodic points.
First we must talk about periodic points.
Consider, for example, the initial state . The orbit of is given by:
Here we see repeats itself and so gets ‘stuck’ in a repeating pattern:
The orbit of .
The orbit of any fraction, e.g. , must be periodic, because is either equal to of and so the orbit consists only of states of the form:
,
and there are only 243 of these and so after 244 iterations, some state must be repeated and so we get locked into a periodic cycle.
If we accept the following:
A fraction has a recurring binary expansion:
,
then this is another way to see that fractions are (eventually) periodic. Take for example,
.
Then, recalling chops off the first binary digit:
,
,
,
we see that if is recurring (after a point), then it is (eventually) periodic.
The (eventually) periodic points of are those with a recurring binary expansion (aka the fractions).
Show that
.
From the above, and the previous post, we know to find an initial state that has a chaotic orbit (never repeating) it cannot be recurring. We also want it to get close to every possible point. Consider the following:
Make a state that agrees to these to one, two, three binary places (recall that numbers are close when they agree to a number of binary/decimal places)
.
Now this state has a terminating binary expansion and so is a fraction. However note it’s orbit gets close to everything:
This orbit gets close to everything: draw a horizontal line and some iterate is close to it… but is eventually fixed at .
If instead we keep the pattern going, with all the four digit binaries, the five digit binaries, etc., etc., we get an initial state that is not periodic and gets close to every possible state in , because it eventually agrees to every state to an arbitrary number of binary digits.
Such an initial state has a chaotic orbit.
Have we got close to every state a periodic state. Yes! Take any , for example
.
Note that is not a fraction. Then the initial state is periodic, and close to (because they agree to five binary places).
This is the density of periodic points and the mega-sensitivity to initial conditions mentioned in the previous post. I believe that has a chaotic orbit, yet it is very close to the periodic point which has very different behaviour:
, but their orbits have very different behaviour: chaotic (I think) in red vs periodic in green.
Therefore is a chaotic mapping.
In the real world, measurement come with an error. Suppose, for example, that we measure many aspects of the weather and feed them into our computer models that predict temperature.
Now the problem is, if we think the temperature today is 15.5, and measurement error of is present, and , and have very different orbits, what temperature should be put into our computer model? The answer is we should put all of them in! We should put in an ensemble of initial states. So rather than we should look at perhaps 100 starting values:
.
Here we see the orbits of ten initial states close to . Note that after three days there is already quite a large variance between the predicted temperatures. After five days there isn’t really any pattern at all. The best we can do is predict, perhaps , and . After this it is a lottery.
Has now been submitted. I am going to do my utmost to get these corrected ASAP.
We looked at Hypothesis Testing and began Chapter 4 with a Revision of Differentiation. We looked then at Maclaurin Series.
We will look at more general Taylor Series: not just near and also for functions of several variables.
We will finish off Chapter 4 by looking at Error Analysis.
We will go through last year’s exam on the board and then I will answer your questions if there are any. If there are none I will help one-to-one. Usual class times and locations.
We will also have tutorials on Friday 8 December in the usual times and venues.
Please feel free to ask me questions about the exercises via email or even better on this webpage.
Please see the Student Resources tab on the top of this page for information on the Academic Learning Centre, etc.