MATH6014: Test 2 Results

May 4, 2011 in MATH6014 | Leave a comment

Updated on May 24th 2011.

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 2 (out of 15).

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

St No	One	Two	Three	Four	Five	Six	/40	%	CA/15
390	10	5	10	3	5	2	35	88	13.2
000	10	5	10	3	3	4	35	88	13.2
928	7	4	9	5	5	5	35	88	13.2
136	10	5	10	2	0	4	31	78	11.7
669	10	5	8	3	1	0	27	68	10.2
784	9	5	10	0	2	0	26	65	9.75
051	10	5	5	3	1	2	26	65	9.75
817	9	5	5	3	0	2	24	60	9
417	9	5	4	3	2	1	24	60	9
933	9	2	10	0	0	0	21	53	7.95
175	5	5	10	0	0	0	20	50	7.5
465	6	2	9	0	0	0	17	43	6.45
000	9	2	2	0	0	0	13	33	4.95
917	0	2	4	0	0	0	6	15	2.25
812	3	0	3	0	0	0	6	15	2.25
828	2	1	5	0	0	0	8	20	3
163	5	4	5	3	0	4	21	53	7.95
953	3	2	0	0	0	0	5	13	1.95
513	2	0	2	0	0	0	4	10	1.5
000	2	0	0	0	0	0	2	5	0.75
Ave	6.5	3.2	6.05	1.4	0.95	1	19	49	7.275
Ave %	65	64	60.5	28	19	24

MATH6037: Final Exam Sample

April 15, 2011 in MATH6037 | Leave a comment

Sample

MATH6014: Test 2 Sample

April 15, 2011 in MATH6014 | Leave a comment

The MATH6014 Test 2 will be held at 5 p.m. Tuesday 03/05/11. The test is worth 15% of your final mark. The test will be 50 minutes long and you must answer all questions. The marks each question carries will be stated on the test and I will attach a set of tables.

Sample to be found here.

MATH7019: Test 2 Sample

April 11, 2011 in MATH7019 | 1 comment

Test 2 on Thursday 19 April at 8.25 p.m

Please find a sample test here.

Note that the format will be the same as this.

Write down the bending moment for three loadings.
Exam Standard Beam I
Exam Standard Beam II
Binomial Distribution
Poisson Distribution

Ideals and Positive Functionals: Exercises

April 7, 2011 in C*-Algebras & Operator Theory, Research | Leave a comment

Question 1

Let $a,b$ be normal elements of a C*-algebra $A$ , and $c$ an element of $A$ such that $ac=cb$ . Show that $a^*c=cb^*$ , using Fuglede’s theorem and the fact that the element

$d=\left(\begin{array}{cc}a &0\\ 0&b\end{array}\right)$

is normal in $M_2(A)$ and commutes with

$d'=\left(\begin{array}{cc} 0&c\\ 0&0\end{array}\right)$ .

This more general result is called the Putnam-Fuglede theorem.

Solution

Fuglede’s theorem states that if $a$ is a normal element commuting with some $b\in A$ , then $b^*$ also commutes with $a$ . Now we can show that $d^*d=d^*d$ using the normality of $a$ and $b$ . We can also show that $d$ and $d'$ commute. Hence by the theorem $d$ and $d^*$ commute. This yields:

$bc^*=c^*a$ .

Taking conjugates:

$cb^*=a^*c$ ,

as required $\bullet$

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MATH6037: Inverse Laplace Transforms — Selected Solutions

March 31, 2011 in MATH6037 | Leave a comment

Find the inverse Laplace transforms of the following functions:

Note that as soon as we step inside the exam hall we write down the following rule that is on the tables but not nearly in as nice a form:

$\mathcal{L}^{-1}\left\{\frac{1}{s^n}\right\}=\frac{t^{n-1}}{(n-1)!}$ .

(i)

$\frac{(s+1)^3}{s^4}$ .

Solution

This looks quite tricky but all it needs is a little trick. Multiply out $(s+1)^3$ by hand (or by memory or by the Binomial Theorem (in tables)) to get:

$\frac{s^3+3s^2+3s+1}{s^4}=\frac{s^3}{s^4}+3\frac{s^2}{s^4}+3\frac{s}{s^4}+\frac{1}{s^4}=\frac{1}{s}+3\frac{1}{s^2}+3\frac{1}{s^3}+\frac{1}{s^4}$ .

Now taking advantage of the linearity of the inverse Laplace transform (I’m just going to write $K$ instead of $\mathcal{L}^{-1}$ . To get a nice inverse Laplace transform symbol I have to type \mathcal{L}^{-1} — along with some dollar signs and I’m sick of it! — no keyboard shortcuts on this page — the notes have shortcuts.)

$K\left\{\frac{1}{s}\right\}+3 K\left\{\frac{1}{s^2}\right\}+3K\left\{\frac{1}{s^3}\right\}+K\left\{\frac{1}{s^4}\right\}$ .

Now using the tables;

$=1+3t+3\frac{t^2}{2}+\frac{t^3}{6}$ .

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MATH6037: Properties of the Laplace Transform — Selected Exercise Solutions

March 31, 2011 in MATH6037 | 2 comments

Question 1

Find the Laplace transform of the following functions.

(i)

$6\cos 4t+t^3$

Solution

Using linearity:

$\mathcal{L}\{6\cos 4t+t^3\}=6\mathcal{L}\{\cos 4t\}+\mathcal{L}\{t^3\}$ .

Now using the tables:

$=6\frac{s}{s^2+4^2}+\frac{3!}{s^4}=\frac{6s}{s^2+16}+\frac{6}{s^4}$ .

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MATH6037: The Laplace Transform of Basic Functions — Selected Exercise Solutions

March 31, 2011 in MATH6037 | Leave a comment

Question 3

Find, from first principles, $\mathcal{L}\{t^2\}$ .

Solution

Using the definition:

$\mathcal{L}\{t^2\}=\int_0^\infty t^2e^{-st}\,dt$

To ease the steps, first we will evaluate the indefinite integral:

$I= \int t^2e^{-st}\,dt$ .

This integral needs to be integrated by parts (this hint would be proffered in an exam situation). Using the LIATE rule, choose $u=t^2$ and $dv=e^{-st}\,dt$ . Now

$\frac{du}{dt}=2t\Rightarrow du=2t\,dt$ , and

$v=\int dv=\int e^{-st}\,dt=\frac{e^{-st}}{-s}=-\frac{1}{s}e^{-st}$ .

In this context $-1/s$ is a constant and it will be handy to have it out the front so it can be easily taken out of an integral ( $\int kf(x)=k\int f(x)$ .) Hence using the integration by parts formula:

$I=t^2\left(-\frac{1}{s}e^{-st}\right)-\int \left(-\frac{1}{s}e^{-st}2t\,dt\right)=-\frac{1}{s}t^2e^{-st}+\frac{2}{s}\underbrace{\int te^{-st}}_{=:J}$ .

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MATH6037: Definition of the Laplace Transform Selected Exercise Solutions

March 31, 2011 in MATH6037 | Leave a comment

Question 1

Determine whether the integrals are convergent or divergent. Evaluate those that are convergent.

This exercise is more to get us used to evaluating infinite integrals — none of these bar (iii) are necessarily related to Laplace transforms.

(i)

$\int_1^\infty \frac{1}{(3x+1)^2}\,dx$

Solution

We first try and evaluate

$\int \frac{1}{(3x+1)^2}\,dx$ .

Try the substitution $u=3x+1$ :

$\frac{du}{dx}=3\Rightarrow dx =\frac{du}{3}$ .

$I=\int\frac{1}{u^2}\cdot\frac{du}{3}=\frac{1}{3}\int u^{-2}\,du$

$=\frac{1}{3}\frac{u^{-1}}{-1}=-\frac{1}{3u}=-\frac{1}{3(3x+1)}$ .

Now evaluating $I$ between the limits $x=1$ and $x=R$ :

$\left(-\frac{1}{3(3R+1)}\right)-\left(-\frac{1}{3(3+1)}\right)=\frac{1}{12}-\frac{1}{9R+3}$ .

Now taking the limit as $R\rightarrow\infty$ :

$I=\lim_{R\rightarrow\infty}\left(\frac{1}{12}-\underbrace{\frac{1}{9R+3}}_{\rightarrow 0}\right)=\frac{1}{12}$ .

The integral is convergent with value $1/12$ .

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MA 1008: Summer 2010, Q. 9 (b)

March 30, 2011 in MA 1008 | Leave a comment

Made a slip somewhere in the tutorial and rather looking for it I said I’d put it up here — the point is that the $u(x)$ term disappears so we have a differential equation in $u'(x)$ and $u''(x)$ — in other words a function and it’s derivative. This can then be integrated in the usual way.

Verify that $y_1(x)=x-1$ is a solution to the second order ODE

$(x-1)^2\frac{d^2y}{dx^2}-(x^2-1)\frac{dy}{dx}+(x+1)y=0.$ (*)

Solution

We have that $y'_1(x)=1$ and $y''_1(x)=0$ . Hence

$(x-1)^2y''_1(x)-(x^2-1)y'_1(x)+(x+1)y_1(x)$

$=-(x^2-1)+(x+1)(x-1)=0$

as required. That is $y_1(x)$ is a solution.

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MATH6014: Test 2 Results

MATH6037: Final Exam Sample

MATH6014: Test 2 Sample

MATH7019: Test 2 Sample

Ideals and Positive Functionals: Exercises

Question 1

Solution

Read the rest of this entry »

MATH6037: Inverse Laplace Transforms — Selected Solutions

Find the inverse Laplace transforms of the following functions:

(i)

Solution

MATH6037: Properties of the Laplace Transform — Selected Exercise Solutions

Question 1

(i)

Solution

Read the rest of this entry »

MATH6037: The Laplace Transform of Basic Functions — Selected Exercise Solutions

Question 3

Solution

MATH6037: Definition of the Laplace Transform Selected Exercise Solutions

Question 1

(i)

Solution

MA 1008: Summer 2010, Q. 9 (b)

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