Leaving Cert Project Maths: The Average Vs The Expected Value

March 29, 2011 in General, Grinds, Leaving Cert | Leave a comment

The Average

The average or the mean of a finite set of numbers is, well, the average. For example, the average of the (multiset of) numbers \{2,3,4,4,5,7,11,12\} is given by:

\text{average}=\frac{2+3+4+4+5+7+11+12}{8}=\frac{48}{8}=6.

When we have some real-valued variable (a variable with real number values), for example the heights of the students in a class, that we know all about — i.e. we have the data or statistics of the variable — we can define it’s average or mean.

Definition

Let x be a real-valued variable with data \{x_1,x_2,\dots,x_n\}. The average or mean of x, denoted by \bar{x} is defined by:

\bar{x}=\frac{x_1+x_2+\cdots+x_n}{n}=\frac{\sum_{i=1}^nx_i}{n}.

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Leaving Cert Maths: Differentiation from First Principles

March 29, 2011 in General, Grinds, Leaving Cert | 1 comment

In Leaving Cert Maths we are often asked to differentiate from first principles. This means that we must use the definition of the derivative — which was defined by Newton/ Leibniz — the principles underpinning this definition are these first principles. You can follow the argument at the start of Chapter 8 of these notes:

https://jpmccarthymaths.com/wp-content/uploads/2010/07/lecture-notes.pdf,

to see where this definition comes from, namely:

f'(x)\equiv \frac{dy}{dx}=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}. (*)

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MATH6037: Maple Test

March 25, 2011 in MATH6037 | Leave a comment

Due to Miguel being away next Wednesday I have decided to postpone the Maple test until the week after.

MATH6037 Test Results

March 24, 2011 in MATH6037 | Leave a comment

The results are down the bottom. You are identified by the last three digits of your student number. If there is a blank it means that you did not write down a student number. If you want email me on jippo@campus.ie and I can confirm your score — otherwise you’ll be waiting until Wednesday.

The scores are itemized as you can see. At the bottom there are some average scores. Finally the last column displays your Continuous Assessment mark for the Test (out of 15).

If you would like to see your paper or have it discussed please email me at jippo@campus.ie

 

St No Q1 Q2 Q3 Q4 Q5 Q6 Q7 Total % CA
219 10 10 10 10 10 10 10 70 100 15
4 7 10 10 10 9 8 58 83 13
993 8 9 10 6 8 9 1 51 73 11
102 9 4 8 10 6 10 3 50 71 11
960 6 7 10 10 2 10 4 49 70 11
787 4 4 10 5 1 7 10 41 59 9
4 6 10 10 6 2 1 39 56 9
675 10 4 4 8 3 2 3 34 49 8
6 0 9 4 2 2 4 27 39 6
5 0 1 7 5 1 7 26 37 6
271 3 0 4 8 4 2 2 23 33 5
4 0 2 5 4 2 2 19 27 5
Ave 6.08 4.25 7.33 7.75 5.08 5.50 4.58 40.58 57.98 9.08

 

MATH6014 Test 1

March 16, 2011 in MATH6014 | Leave a comment

The results are down the bottom. You are identified by the last three digits of your student number. If there is a zero is means that you did not write down your student number. The results are in alphabetical order but if you are unsure email me and I will tell you what you got.

The scores are itemized as you can see. At the bottom there are some average scores. Finally the last column displays your Continuous Assessment mark for Test 1 (out of 15).

If you would like to see your paper or have it discussed please email me at jippo@campus.ie

St No 1a 1b 2a 2b 3a 3b % CA
784 5 1 15 3 20 20 64 10
136 15 4 15 15 20 18 87 14
933 10 0 1 15 20 17 63 10
0 0 0 0 0 20 0 20 3
390 13 1 3 8 20 20 65 10
0 12 4 7 6 20 20 69 11
828 0 0 0 0 20 0 20 3
917 0 0 0 0 20 4 24 4
051 8 4 7 3 20 17 59 9
918 0 0 0 1 20 14 35 6
465 2 0 2 3 20 13 40 6
817 8 0 3 2 10 12 35 6
162 0 0 3 0 20 19 42 7
417 15 0 7 12 10 15 59 9
0 0 0 2 0 20 12 34 6
953 2 0 1 0 5 3 11 2
175 0 0 0 0 20 4 24 4
498 0 0 1 0 10 6 17 3
812 6 1 0 0 20 14 41 7
513 5 0 3 0 20 11 39 6
669 4 10 10 15 20 20 79 12
928 9 5 6 0 20 18 58 9
440 0 0 1 0 20 14 35 6
092 3 0 1 0 5 5 14 3
163 4 0 0 8 20 12 44 7
Ave 4.84 1.20 3.52 3.64 17.60 12.32 43.12 6.92
Ave % 32.27 8.00 23.47 24.27 88.00 61.60

MATH6037: Approximate Integration Selected Exercise Solutions

March 14, 2011 in MATH6037 | Leave a comment

Remarks in italics are by me for extra explanation. These comments would not be necessary for full marks in an exam situation. Exercises taken from p.87 of these notes.

Question 1

Estimate \int_0^1\cos(x^2)\,dx using  (a) the Trapezoidal Rule and (b) the Midpoint Rule, each with n=4.

Solution

Part (a)

For the Trapezoidal Rule, we have \Delta x=0.25 and we have the points x_0=0, x_1=0.25, x_2=0.5, x_3=0.75, x_4=1. Hence using the formula:

\int_0^1\cos(x^2)\,dx\approx\frac{0.25}{2}[\cos(0)+2[\cos(0.25^2)+\cos(0.5^2)+\cos(0.75^2)]

+\cos(1)]\approx 0.895795.

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MATH6037: Root Approximation Exercise Selected Solutions

March 14, 2011 in MATH6037 | Leave a comment

Remarks in italics are by me for extra explanation. These comments would not be necessary for full marks in an exam situation. Exercises taken from p.75 of these notes.

Question 1

If f(x)=x^3-x^2+x, show that there is a number c such that f(c)=10.

Solution

Solving the equation f(x)=10 is equivalent to finding a root of the continuous function g(x)=f(x)-10=x^3-x^2+x-10. Note that this function is continuous hence we can use the Intermediate Value Theorem to find an interval with a root.

g(0)=-10<0,

g(1)=(1)^3-(1)^2+(1)-10=-9<0,

g(2)=(2)^3-(2)^2+(2)-10=-4<0,

g(3)=(3)^3-(3)^2+(3)-10=11>0.

Hence g changes sign between 2 and 3. Hence g has a root in (2,3), say at c. Hence g(c)=0\Leftrightarrow f(c)=10 \bullet

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MATH6037: Applications to Error Analysis Exercise Solutions

March 14, 2011 in MATH6037 | 2 comments

Remarks in italics are by me for extra explanation. These comments would not be necessary for full marks in an exam situation. Exercises taken from p.63 of these notes.

Question 1

Use differentials to estimate the amount of tin in a closed tin closed tin can with diameter 8 cm and height 12 cm if the can is 0.04 cm thick.

Solution

Assuming that the measurements of 8 cm and 12 cm are taken from the outside of the can, then we could estimate the change in volume of a cylinder if the radius were increased by 0.04 cm to 4 cm and the height increased by 0.08 cm to 12 cm (convince yourself of this with a picture.). Now the tin in the can comprises the difference between a (r,h)=(3.96,11.92) cylinder and a (r,h)=(3,12) cylinder.

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My Understanding of Non-Commutative Geometry

March 14, 2011 in General | 1 comment

This is intended to be the subject of a short postgraduate talk in UCC. At times there will be little attempt at rigour — mostly I am just concerned with ideas, motivation and giving a flavour of the philosophy. Also it is fully possible that I have got it completely wrong in my interpretation!

Introduction

It is a theme in mathematics that geometry and algebra are dual:

\text{Geometry }\leftrightarrow \text{ Algebra}

Arguably this theme began when Descartes began to answer questions about synthetic geometry using the (largely) algebraic methods of coordinate geometry. Since then this duality has been extended and refined to consider:

\text{Spaces }\leftrightarrow \text{ Algebra of Functions on the Space}

Here a space is a set of points with some additional structure, and the idea is that for a given space, there will be a canonical algebra of functions on the space. For example, given a compact, Hausdorff topological space X, the canonical algebra of functions is C(X) — the continuous functions on X.

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MATH6037: Sample Test

March 8, 2011 in MATH6037 | Leave a comment

The MATH6037 test will be held at 7 p.m. Wednesday 16/03/11. The test is worth 15% of your final mark. The test will be 60 minutes long and you must answer all questions. All questions carry equal marks and I will attach a set of tables. You will get a copy of these tables tomorrow night.

Please find a sample here.