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)
Last year's maths is easy, this year's maths is hard and next year's maths is impossible.
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)
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 of positive elements in the closed unit ball of
such that
for all
.
Let be a Hilbert space with infinite orthonormal basis
. The C*-algebra
is now non-unital. If
is the projection onto
, then the increasing sequence
is an approximate unit for
. It will suffice to show that
if
, since
is dense in
. Now if
, there exist
,
such that:
.
Hence,
.
Since for all
, therefore for each $k$:
.
Hence, .
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.
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.
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.
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.
In week 2 we began to look at homogenous second order linear differential equations (with constant coefficients). These are differential equations of the form
, (*)
where are real constants and the solution is
. We showed that to solve these equations we look at the auxiliary equation
.
In general, if the roots of of this equation are and
then the solution to (*) is given by
where and
are real constants.
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.
.
————————————————————————————————————————————————–
.
.
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.
Multiplying the first and last coefficients: . Now the factors of
are
. So I want to rewrite the middle term
in terms of these:
Taking out the common factor :
.
Multiplying the first and last coefficients: . The factors of
are
. I want to rewrite the middle term
in terms of one of these factors:
.
Let be a Banach algebra such that for all
the implication
or
holds. Let ,
be linear mappings from
to itself such that for all
,
,
, and
.
Show that and
are necessarily continuous.
Let be a unital C*-algebra.
If are positive elements of
, show that
.
For elements of a unital algebra
:
If then
so that
Now if , for any
,
. Hence
and the result follows (note that
need not be hermitian)
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.
We must choose a and
— where we choose
according to the LIATE rule. Here there are no logs, no inverses but there is an algebraic (sum of powers of
). Hence let
,
(find
by integrating:
):
; and
.
Hence by the integration by parts formula:
We check our solution by differentiating (using the product rule):
,
as required.
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.
We solved some more linear systems and began a study of matrices. Updated notes.
I will give the solution to the first part of a question and then the answers to remaining parts.
Suppose that we want to invest € over
years with an interest rate of
(if the interest rate is 4%, then
).
Initially we have € – called the principal. After 1 year however, we will have the principal plus the interest: which is
of
. Hence the value after one year will be:
.
So to get the value of the investment you multiply the previous years value by . So after three years, for example, the investment has value:
,
and we can generalise to any number of years.
The final value of an investment of a principal after
years at an interest rate of
,
, is given by:
.
p. 30 of the tables – Compound Interest
But what about inflation?
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