This note is primarily about LC Project Maths but also contains details on the Old Syllabus

The Leaving Cert mocks are coming up very fast but don’t fret. There is a lot of time in between February and June to learn maths – but there will be no more chances to test your skills as an examinee. This short note will focus on five areas which you must address to get the maximum benefit from your mocks.

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I have continued to work through Murphy http://books.google.com/books?id=emNvQgAACAAJ&dq=gerald+murphy+c*+algebras+and+operator+theory&h

Before the Christmas break I finished off the chapter 1 exercises.

Chapter 2: C*-Algebras and Hilbert Space Operators.

2.1 C*-Algebras

Initially we defined a C*-algebra, A, as a complete normed algebra, together with a conjugate-linear involution * that satisfies the C*-equation:

\|a^*a\|=\|a\|^2, \forall\,a\in A

Self-adjoint or Hermitian elements are defined to have the property a^*=a. As a consequence of this, and the C*-equation, the spectral radius of a self-adjoint element, \nu(a), is equal to its norm, \|a\|. As a corollary of this, of all the norms that can be put on the *-algebra, only one makes it into a C*-algebra – i.e. satisfying the C*-equation.

In the previous chapter we have seen that an algebra, A, can be unitised to form a new algebra, \tilde{A}, which contains A as a subspace. In general, the norm got by extending the norm on A to a norm on \tilde{A} does not make \tilde{A} into a C*-algebra. However Theorem 2.1.6 shows that there does exist a (unique) norm on \tilde{A} making it a C*-algebra. In many examples we may now assume that a general C*-algebra is unital – replacing it with the unique unitisation, \tilde{A}, if necessary.

One such result which depends on this fact is that the the spectrum of a self-adjoint element is real.

A central result in this chapter is that all abelian C*-algebras are C_0(X), for some locally compact Hausdorff space, X. In fact X is the character space \Phi(A) (as with Belton, this is via the Gelfand transformation). This identification allows the development of the powerful functional calculus. Briefly, if a is a normal element of a C*-algebra A, (a^*a=aa^*), and z is the inclusion map from \sigma(a)\rightarrow \mathbb{C}, then there exists a unique *-homomorphism \varphi:C(\sigma(a))\rightarrow A such that \varphi(z)=a. This unique *-homorphism is called the functional calculus at a. This particular section ended with the Belton result that if X is a compact Hausdorff space, \Phi(C(X))\cong X (via x\mapsto \delta_x).

2.2 Positive Elements of C*-Algebras

This section introduces a partial order on A_{\text{SA}} (the set of self-adjoint elements of A). Namely, an element a\in A_{\text{SA}} is positive if \sigma(a)\subset \mathbb{R}^+. The partial order is defined in the obvious way.

As a consequence of the Gelfand transformation and the functional calculus, we can show that positive elements of a C*-algebra possess unique positive square roots.  Another prominent result is that for an arbitrary element a\in Aa^*a is positive.

2.3 Operators and Sesquilinear Forms

As a first move, we prove that bounded operators on Hilbert spaces have adjoints. Next projections are examined and partial isometries are examined. This leads onto the polar decomposition theorem. Namely, if T is a continuous linear operator on a Hilbert space H, there exists a unique partial isometry S such that T=S|T|; where |T|=(T^*T)^{1/2}. The rest of the section focusses on the connection between operators and sesquilinear forms.

2.4 Compact Hilbert Space Operators

At first this chapter looks at some of the basic properties of these objects – e.g. if T is compact so are |T| and T^*. Thus K(H) is self-adjoint and thus a C*-algebra (it is a closed ideal in B(H)). We see that normal compact operators are diagonalisable.

We look at the finite rank operators, F(H) and see that they are dense in K(H). Next the operator x\otimes y is examined:

(x\otimes y)(z)=\langle z,y\rangle x

These are rank-one, and the x\otimes x are rank-one projections if x is a unit vector. This leads on to the fact that F(H) is linearly spanned by these rank-one projections.

This is a synopsis of what I covered up until recently (up to p.56). As an experiment I am attempting to do my study of Murphy by way of fully presenting the details on this webpage. I am unsure of whether or not this is too time consuming. Presently I am on page 63 and I will have to cover the rest of the chapter material (10 pages) in one day or similar if I am going to consider this tactic feasible.

This question was asked at Monday’s tutorial (10/01/11) but the fire alarm went off mid-solution

Section 6.4, Q. 5

Evaluate the following integral:

\int\frac{(x-1)\,dx}{x^3-x^2-2x}

Solution

(Remarks in italics are by me and would not be required in an exam situation)

Simplify the integrand to get it into a usable form:

I=\int\frac{(x-1)\,dx}{x(x^2-x-2)}=\int\frac{(x-1)\,dx}{x(x-2)(x+1)}

Rule 1 (Section 6.4)

Given a rational function P(x)/Q(x) with \text{deg}(P)<\text{deg}(Q), such that Q(x) factors into non-repeated linear terms:

Q(x)=(a_1x+b_1)(a_2x+b_2)\cdots(a_nx+b_n)

(non-repeated means that no linear term  is equal to a constant multiple of another; e.g. (a_ix+b_i)=k(a_jx+b_j) for i\neq j, k\in \mathbb{C})

Then

\frac{P(x)}{Q(x)}=\frac{A_1}{a_1x+b_1}+\frac{A_2}{a_2x+b_2}+\cdots+\frac{A_n}{a_nx+b_n}

for some constants A_1,A_2,\dots,A_n\in\mathbb{C}.

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Section 8.8, Q. 4

Find the Taylor series expansion of the function \ln (1+x) about the point x=1.

(This question was asked at Friday’ tutorial but, with one eye on the answer given, I was unable to do it. Having looked at the problem again I’m sure that the question should have been:)

Find the Taylor series expansion of the function \ln x about the point x=1.

(I have indicated this issue to Prof. Stynes)

Solution

The Taylor series of any infinitely differentiable function about a point x=a is given by the power series:

f(x)\approx \sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!}(x-a)^n

Computing the first few derivatives of f(x)=\ln x:

\left.f^{(0)}(x)=\ln x\right|_{x=1}=0

\left.f'(x)=x^{-1}\right|_{x=1}=1

\left.f''(x)=(-1)x^{-2}\right|_{x=1}=-1

\left.f'''(x)=(-1)(-2)x^{-3}\right|_{x=1}=2

\left.f^{(iv)}(x)=(-1)(-2)(-3)x^{-4}\right|_{x=1}=-6

\vdots

\left.f^{(n)}(x)=(-1)^{n+1}(n-1)!x^{-n}\right|_{x=1}=(-1)^{n+1}(n-1)!

This is valid for n\geq 1. At n=0, f^{(0)}(1)=f(1)=0. Hence we have;

f(x)\approx \sum_{n=1}^\infty \frac{(-1)^{n+1}(n-1)!}{n!}(x-1)^n

f(x)\approx \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}(x-1)^n \Box

This is a note written to address an issue we had in Tueday’s tutorial. Specifically the method I have shown you for doing one of the types of differentiability questions is somewhat flawed. There are a number of further assumptions that we need to make in order to make the analysis correct. I apologise for this oversight. However all is not lost – we can still ‘fix’ our (easier) method by taking these additional assumptions into account.

This question will now not be examinable in your summer exam.

The Question

Let f:\mathbb{R}\rightarrow \mathbb{R} be defined by:

f(x):=\left\{\begin{array}{cc}x^2&\text{ if }x\geq0\\ 0&\text{ if }x<0\end{array}\right.

Show that f is differentiable but not twice differentiable.

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Firstly there are two tutorials next week – Monday and Tuesday at the same time and place as the the usual lecture/ tutorial.

Test Results

First of all results are down the bottom. You are identified by the last four digits of your student number. The scores are itemized as you can see. At the bottom there are some average scores. Finally the last column displays your total Continuous Assessment mark (out of 25).

If you would like to see your paper or have it discussed please email me.

Solutions

Test A and Test B.

Stud. Id Q 1(a)/3 Q 1(b)/2 Q 2/4 Q 3/3.5 Test 2 Percent CA Result
9705 2.5 2 4 1 9.5 76 18
1351 0.5 2 2 1 5.5 44 15
9822 3 2 4 1 10 80 19
2081 0.5 0.5 4 2 7 56 9
6454 1.5 2 3 2 8.5 68 13.5
7784 2.5 0.5 3.5 1 7.5 60 9.5
7238 1.5 2 4 1 8.5 68 12
8225 2.5 2 4 2 10.5 84 19.5
5757 2.5 2 4 3.5 12 96 24.5
2471 2 1.5 3 1 7.5 60 13
0869 1.5 2 0 1 4.5 36 10.5
1341 3 2 0 2 7 56 13
9056 2 0.5 4 1 7.5 60 14
7327 0 0 0 0 0 0 4
6188 2 2 4 2 10 80 17
7303 2.5 0.5 4 1 8 64 13.5
3831 2 2 2 1 7 56 16
3024 1 2 3 1 7 56 8
1947 0 0.5 0 0 0.5 4 2.5
2332 0 0.5 0 2 2.5 20 7.5
9423 2.5 2 0 1 5.5 44 10.5
5026 0 0.5 0 0 0.5 4 2.5
2366 2 2 4 0 8 64 18
2185 3 2 4 1 10 80 21
9014 1.5 2 4 1 8.5 68 14.5
3921 0 0 0 1 1 8 4
0166 0 0 0 0 0 0 7.5
8705 2.5 2 4 1 9.5 76 15.5
5321 2.5 0.5 4 2 9 72 10
1701 0 1 4 2 7 56 14
6218 0 2 0 1 3 24 11
4967 1 2 2.5 0 5.5 44 8.5
4761 1 2 3 0 6 48 12
5243 0*
1863 2.5 2 4 3.5 12 96 22
3995 0 1.5 4 1 6.5 52 8.5
5154 0.5 1.5 0 0 2 16 3
0385 2 2 2 2 8 64 15
9687 2.5 2 4 2 10.5 84 17.5
5642 0 2 2 2 6 48 17.5
7478 2 2 4 2 10 80 21
7029 3*
8026 0 0.5 0 1 1.5 12 4.5
4575 1 1.5 4 1 7.5 60 18
3845 1.5 2 2 2 7.5 60 14.5
0672 2.5 2 4 1 9.5 76 20.5
8793 2.5 2 4 1 9.5 76 15.5
7144 2.5 1.5 4 2 10 80 17.5
8108 0 0 0 0 0 0 5
3631 3 1 4 0 8 64 18
6302 1.5 1 4 0 6.5 52 12.5
1043 3 2 4 1 10 80 17.5
5904 2.5 1.5 0 2 6 48 15
4257 2 2 4 2 10 80 22.5
9063 2 2 4 1 9 72 21.5
3673 0.5 2 4 1 7.5 60 14
4482 0.5 2 0 1 3.5 28 9.5
4645 0.5 1.5 0 1 3 24 12.5
5527 0 0 0 0 0 0 1
8172 2 2 1 1 6 48 13
6838 0 0 0 0 0 0 6
1817 1.5 2 4 2 9.5 76 21.5
9738 3*
0511 1.5 2 4 1 8.5 68 18.5
7324 2 2 3 1 8 64 19
6511 0 0 0 0 0 0 3
0492 0 0.5 0 0 0.5 4 2.5
9501 2 0.5 0 1 3.5 28 4.5
0684 0 0 0 0 0 0 0
Ave 1.41 1.41 2.36 1.09 6.27 50.18 12.25
Ave % 46.97 70.45 59.09 31.17

On Monday we completed our introduction to horizontal asymptotes and vertical asymptotes.

On Wednesday we did some examples of curve sketching and applied max/ min problems.

Problems

You need to do exercises – all of the following you should be able to attempt. Do as many as you can/ want in the following order of most beneficial:

Wills’ Exercise Sheets

Stationary Points are points a\in\mathbb{R} where the derivative of a differentiable function f:\mathbb{R}\rightarrow\mathbb{R}, f'(a)=0.

When asked to find the critical points of a function defined on the entire real line (rather than just on a closed interval [a,b]), the ‘endpoints’, \pm\infty are not considered critical points.

Convex is concave up and concave is concave down.

Q. 11-17  from http://euclid.ucc.ie/pages/staff/wills/teaching/ms2001/exercise4.pdf

Other Exercise Sheets – Questions on Asymptotes

From section 4 Q. 5 from Problems, find the vertical asymptotes of the the functions 5(b) [(7-10), (13), (15-16), (23)] and Q. 5(c) [except (30-31)]

Past Exam Papers

Q. 1 (d) from http://booleweb.ucc.ie/ExamPapers/exams2008/Maths_Stds/MS2001Sum08.pdf

Q. 1(d) from http://booleweb.ucc.ie/ExamPapers/Exams2008/MathsStds/MS2001a08.pdf

Q. 2(a),  from http://booleweb.ucc.ie/ExamPapers/exams2006/Maths_Stds/MS2001Sum06.pdf

Q. 2(a), 6(b) from http://booleweb.ucc.ie/ExamPapers/exams2006/Maths_Stds/Autumn/ms2001Aut.pdf

Q. 2(a)  from http://booleweb.ucc.ie/ExamPapers/Exams2005/Maths_Stds/MS2001.pdf

Q. 2(a), 6(b) from http://booleweb.ucc.ie/ExamPapers/Exams2005/Maths_Stds/MS2001Aut05.pdf

Q. 6(b) from http://booleweb.ucc.ie/ExamPapers/exams2004/Maths_Stds/ms2001s2004.pdf

Q. 6(b) from http://booleweb.ucc.ie/ExamPapers/exams2004/Maths_Stds/MS2001aut.pdf

Q. 6(b) from http://booleweb.ucc.ie/ExamPapers/exams2003/Maths_Studies/MS2001.pdf

Q. 6(b) from http://booleweb.ucc.ie/ExamPapers/exams2003/Maths_Studies/ms2001aut.pdf

Q. 6(b) from http://booleweb.ucc.ie/ExamPapers/exams2002/Maths_Stds/ms2001.pdf

Q. 6(b) from http://booleweb.ucc.ie/ExamPapers/exams/Mathematical_Studies/MS2001.pdf

All of this except Q. 1(d) [this is the Autumn 2010 paper which wasn’t on the library website earlier in the year]

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

Example 1 (Asymptotes)

Let

f(x)=\frac{3x^2-2x+2}{x^2-x-2}

What is the domain of f? What are the roots and y-intercepts of f? Find the horizontal and vertical asymptotes of f.

Solution The domain of f is the set on which f is defined. As a quotient of continuous functions, f is defined when the denominator is not zero:

x^2-x-2\neq0,

\Rightarrow (x-2)(x+1)\neq 0.

That is the domain of f is \mathbb{R}\backslash\{-1,2\} (all the real numbers except -1 and 2. Note now that

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

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Just a chance for me to give some feedback on some of your most common… feedback. Of course your opinions are correct – they are your opinions. Here are my opinions on some of your opinions. The worst thing (and the failure of teaching evaluation), is that ye will never see the fruits of your criticisms 😦  – i.e. when I have made changes and improvements ye will no longer be my students.

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Firstly; there will be no MS 2001 lecture on Monday 6 December at 3 p.m. Instead you will have an MS 2003 lecture at this time in WG G 08. The 12 p.m. MS 2003 lecture on Wednesday December 1 in WG G 08 will now be an MS 2001 lecture.   Indeed it will be the final MS 2001 lecture as Wednesday 8 December is a test day and the week after is review week.The morning lecture at 9 a.m. on Wednesday 1 December will still go ahead.

The Tuesday 7 December tutorial goes ahead as normal.

In terms of revision week, there will be tutorials held on Monday 13 and Tuesday 14 December at the usual times in the usual places of the Monday 3 p.m lecture and Tuesday 1 p.m. tutorial respectively.

For those students who have not been able to attend Tuesday’s tutorial please see your email.