EDIT: The following tables will be attached to your exam paper:

I had considered drafting a sample paper for ye however on mature reflection I’ve realised that this is unnecessary. The main reason for giving a sample is to show the layout of the paper so there are no nasty surprises and because the exam layout will be the same as in years 2007 — 2010, there is no need for a sample paper.

Past Papers

http://booleweb.ucc.ie/ExamPapers/exams2010/MathsStds/Autumn/MS2001Aut2010.pdf

— except Q. 1(d)

http://booleweb.ucc.ie/ExamPapers/exams2010/MathsStds/MS2001Sum2010.pdf

— except Q. 1(d)

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I’m getting the impression that the bra-ket notation is more useful for linear ON THE LEFT!

An approximate unit for a C*-algebra is an increasing net \{u_\lambda\}_{\lambda\in\Lambda} of positive elements in the closed unit ball of A such that a= \lim_{\lambda }au_\lambda=\lim_\lambda u_\lambda a for all a\in A.

Example

Let H be a Hilbert space with infinite orthonormal basis \{e_n\}. The C*-algebra K(H) is now non-unital. If P_n is the projection onto \langle e_1,\dots,e_n\rangle, then the increasing sequence \{P_n\}\subset K(H) is an approximate unit for K(H). It will suffice to show that T=\lim_np_nT if T\in F(H), since F(H) is dense in K(H). Now if  T\in F(H), there exist x_1,\dots,x_my_1,\dots,y_m\in H such that:

T=\sum_{k=1}^m|x_k\rangle\langle y_k|.

Hence,

P_nT=\sum_{k=1}^m|P_nx_k\rangle\langle y_k|.

Since \lim_n P_nx=x for all x\in H, therefore for each $k$:

\lim_{n\rightarrow \infty}\||P_nx_k\rangle\langle y_k-|x_k\rangle\langle y_k|\|=\lim_{n\rightarrow \infty} \|P_nx_k-x_k\|\|y_k\|=0.

Hence, \lim_{n}P_nT=T.

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I am emailing a link of this to everyone on the class list every week. If you are not receiving these emails or want to have them sent to another email address feel free to email me at jippo@campus.ie and I will add you to the mailing list.

This Week

We asked the question “when is a matrix invertible” and we answered the question by saying that a matrix is invertible when its determinant is non-zero. We didn’t give a definition of  a determinant but did see how to calculate it. Then we introduced Cramer’s Rule as yet another way of solving linear systems. We have a handout with some exercises on determinants and the chain rule.

We also did quite a lot of work on Maple in terms of the basics.

Exercise Solutions

I will give the solution to the first part of a question and then the answers to remaining parts. Some of code to write the matrices is very buggy so I’ve skipped some of them when they don’t compile straight away. For example, the answer below should have an extra line.

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I am emailing a link of this to everyone on the class list every week. If you are not receiving these emails or want to have them sent to another email address feel free to email me at jippo@campus.ie and I will add you to the mailing list.

Week 2

In week 2 we began to look at homogenous second order linear differential equations (with constant coefficients). These are differential equations of the form

a\frac{d^2y}{dx^2}+b\frac{dy}{dx}+cy=0,               (*)

where a,\,b,\,c are real constants and the solution is y=f(x). We showed that to solve these equations we look at the auxiliary equation

ar^2+br+c=0.

Fact 1

In general, if the roots of of this equation are \alpha and \beta then the solution to  (*) is given by

y(x)=Ae^{\alpha x}+Be^{\beta x}

where A and B are real constants.

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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.58 of these notes.

Question 1

(i)

\frac{\partial f}{\partial x}=3x^2-4y^2(1)+0.

————————————————————————————————————————————————–

\frac{\partial f}{\partial y}=-4x(2y)+4y^3=4y^4-8xy.

(ii)

\frac{\partial f}{\partial x}=e^y(2x)+0=2xe^y.

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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.48 of these notes.

Question 1

(i)

Multiplying the first and last coefficients: 1\times 5=5. Now the factors of 5 are 1\times 5. So I want to rewrite the middle term -4x in terms of these:

x^2-4x-5=x^2+x-5x-5=x(x+1)-5(x+1)=(x+1)(x-5).

(ii)

Taking out the common factor x: x(x-2).

(iii)

Multiplying the first and last coefficients: 15\times6=90. The factors of 90 are \{90,1\},\{45,2\},\{30,3\},\{18,5\},\{15,6\},\{10,9\}. I want to rewrite the middle term x in terms of one of these factors:

15x^2+10x-9x-6=5x(3x+2)-3(3x+2)=(3x+2)(5x-3).

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Question 1

Let A be a Banach algebra such that for all a\in A the implication

Aa=0 or aA=0 \Rightarrow a=0

holds. Let LR be linear mappings from A to itself such that for all a,b\in A,

L(ab)=L(a)bR(ab)=aR(b), and R(a)b=aL(b).

Show that L and R are necessarily continuous.

Question 2

Let A be a unital C*-algebra.

(a)

If a,b are positive elements of A, show that \sigma(ab)\subset \mathbb{R}^+.

Solution (Wills)

For elements a,b of a unital algebra A:

\sigma(ab)\cup\{0\}=\sigma(ba)\cup\{0\}

If a\in A^+ then a^{1/2}\in A^+ so that

\sigma(ab)\cup\{0\}=\sigma(a^{1/2}(a^{1/2}b))\cup\{0\}=\sigma(a^{1/2}ba^{1/2})\cup\{0\}

Now if b\in A^+, for any c\in Ac^*bc\in A^+. Hence \sigma(a^{1/2}ba^{1/2})\subset \mathbb{R^+} and the result follows (note that ab need not be hermitian) \bullet

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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.35 of https://jpmccarthymaths.wordpress.com/2011/02/01/math6037-general-information/

These questions all require integration by parts although they don’t explicitly say so.

Question 1

We must choose a u and dv — where we choose u according to the LIATE rule. Here there are no logs, no inverses but there is an algebraic (sum of powers of x). Hence let u=x, dv=\cos x\,dx (find v by integrating: \int \,dv=v):

\frac{du}{dx}=1\Rightarrow du=dx; and v=\sin x.

Hence by the integration by parts formula:

I=x\sin x-\int\sin x\,dx=x\sin x-(-\cos x)+c

=x\sin x+\cos x+c

We check our solution by differentiating I (using the product rule):

\frac{d}{dx}I=x\cos x+\sin x-\sin x=x\cos x,

as required.

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I am emailing a link of this to everyone on the class list every week. If you are not receiving these emails or want to have them sent to another email address feel free to email me at jippo@campus.ie and I will add you to the mailing list.

This Week

We solved some more linear systems and began a study of matrices. Updated notes.

Exercise Solutions

I will give the solution to the first part of a question and then the answers to remaining parts.

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Inflation Adjustment

Suppose that we want to invest €P over t years with an interest rate of i (if the interest rate is 4%, then i=0.04).

Initially we have €P – called the principal. After 1 year however, we will have the principal plus the interest: which is i of P. Hence the value after one year will be:

P+Pi=P(1+i).

So to get the value of the investment you multiply the previous years value by 1+i. So after three years, for example, the investment has value:

P(1+i)(1+i)(1+i)=P(1+i)^3,

and we can generalise to any number of years.

The final value of an investment of a principal P after t years at an interest rate of i, F, is given by:

F=P(1+i)^t.

p. 30 of the tables – Compound Interest

But what about inflation?

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