Author Topic: brightsky's Maths Thread (Read 57786 times) Tweet Share

brightsky · « **Reply #165 on:** October 02, 2013, 08:55:20 pm »

Apologies for the double post, but the questions in this one are of a different kind, so I thought I'd make a new post. To all UMEP/Melbourne Uni students on AN, does Melbourne Uni post study designs or course outlines for any of the subjects they offer? I'm always a bit confused as to what is and what isn't on the 'course' (this is with regard to Extension Maths at Melbourne U). If there is a course outline, is there any way I can access it?

Thanks heaps!!

Aurelian · « **Reply #166 on:** October 02, 2013, 09:20:37 pm »

Quote from: brightsky on October 02, 2013, 08:55:20 pm

Apologies for the double post, but the questions in this one are of a different kind, so I thought I'd make a new post. To all UMEP/Melbourne Uni students on AN, does Melbourne Uni post study designs or course outlines for any of the subjects they offer? I'm always a bit confused as to what is and what isn't on the 'course' (this is with regard to Extension Maths at Melbourne U). If there is a course outline, is there any way I can access it?

Thanks heaps!!

For regular students, the syllabus (or something resembling a syllabus) is typically posted on the LMS. If you guys have access to the LMS, that probably where you'll find it. If not, unfortunately you won't find it published anywhere for the general public; your best bet would probably be to email your instructor (or ask someone who's done previously UMEP Extension Maths specifically).

stolenclay · « **Reply #167 on:** October 03, 2013, 02:02:40 am »

Quote from: brightsky on October 02, 2013, 12:41:17 pm

1. Calculate the dimensions of the image and kernel of the linear transformation T:R^3 -> R^3 by reflection in the plane x+y+z=1. Is T even a linear transformation? If we reflect (0,0,0) in the plane x+y+z=1, it becomes something else...

Haha, I agree with you on that one... I don't think it's linear...

Quote from: brightsky on October 02, 2013, 12:41:17 pm

2. T(x,y) = (2x+y,x+y,x-y,x-2y). Explain why the linear transformation T^2 isn't defined. I just wrote: matrix multiplication doesn't work, but I don't feel as though that is adequate.

Yeah, that's probably it. Or probably something like the codomain of T's being disjoint with the domain of T.

Quote from: brightsky on October 02, 2013, 12:41:17 pm

3. What's the transformation matrix for a reflection in the line which makes an angle theta with the positive x-axis? I was thinking [cos2x, sin2x; sin2x, cos2x] but I don't think that's right...

Is it $\begin{bmatrix} \cos{2\theta} & \sin{2\theta} \\ \sin{2\theta} & -\cos{2\theta} \end{bmatrix}$ ? Because I get $(1,0)$ going to $(\cos{2\theta}, \sin{2\theta})$ and $(0,1)$ going to $(\sin{2\theta}, -\cos{2\theta})$ .

lzxnl · « **Reply #168 on:** October 03, 2013, 11:33:55 am »

Quote from: brightsky on October 02, 2013, 12:41:17 pm

3. What's the transformation matrix for a reflection in the line which makes an angle theta with the positive x-axis? I was thinking [cos2x, sin2x; sin2x, cos2x] but I don't think that's right...

Thanks!

I did that once using vectors and I think I got something on those lines.

brightsky · « **Reply #169 on:** October 03, 2013, 04:02:17 pm »

i see I see thanks guys!

also, is there are difference between the upper bound of S and the greatest element of S. both are technical terms, but there doesn't appear to be any difference between them...

and while we're at it, can someone explain what infimum and supremum are, in laymen's terms? I've yet to read a definition of supS and infS that isn't laden with jargon and confusing as heck...

thanks!

TrueTears · « **Reply #170 on:** October 03, 2013, 06:01:08 pm »

Probably best illustrated with an example.

Consider the set $S = \{x|x^2<11\}$ which implies $S = \{x|-\sqrt{11} < x < \sqrt{11}\}$ , let $b$ be a value such that for all $x \in S$ there exists a $b$ such that $x \le b$ , then the set $B$ is defined as $B = \{b | b \ge \sqrt{11}\}$ . We call the set $B$ , the upper bound.

The supremum of $S$ is simply the least upper bound, ie $\sup(S) = \sqrt{11}$ .

Note: in this case $\sup(S) \notin S$ . Infimum is defined similarly, I'm sure you can work it out.

brightsky · « **Reply #171 on:** October 03, 2013, 06:55:14 pm »

ahhhhhhhhh make sense.......thanks so much tt!!

Alwin · « **Reply #172 on:** October 04, 2013, 11:00:08 am »

Quote from: brightsky on October 02, 2013, 08:55:20 pm

Apologies for the double post, but the questions in this one are of a different kind, so I thought I'd make a new post. To all UMEP/Melbourne Uni students on AN, does Melbourne Uni post study designs or course outlines for any of the subjects they offer? I'm always a bit confused as to what is and what isn't on the 'course' (this is with regard to Extension Maths at Melbourne U). If there is a course outline, is there any way I can access it?

Thanks heaps!!

There is a course outline in the notes (if you purchased them that is). Like Aurelian said, it is on the LMS too, although I kinda forgotten my password which is pretty damn awkward =w=

Also, just one thing with the you previous question about reflection in a line angle theta from the x-axis, how did you tackle it? recognition where e₁ and e₂ would rotate to?
I've always approached such problems like this (using a set up move) and never with vectors @nliu or other methods...

brightsky · « **Reply #173 on:** October 04, 2013, 03:05:42 pm »

Quote from: Alwin on October 04, 2013, 11:00:08 am

There is a course outline in the notes (if you purchased them that is). Like Aurelian said, it is on the LMS too, although I kinda forgotten my password which is pretty damn awkward =w=

Also, just one thing with the you previous question about reflection in a line angle theta from the x-axis, how did you tackle it? recognition where e₁ and e₂ would rotate to?
I've always approached such problems like this (using a set up move) and never with vectors @nliu or other methods...

yeah but the course outline on lms is very, very, VERY brief. nothing like the sort of study designs which vcaa releases.

dunno what exactly nliu is talking about, but I presume we're all talking about the same technique. I still have a feeling however that the answer is wrong...

stolenclay · « **Reply #174 on:** October 04, 2013, 04:06:40 pm »

Well apart from finding the image of a basis, if you want to do it as a composition of transformations, then it could be

Rotation by $\theta$ clockwise
Reflection in X-axis
Rotation by $\theta$ anti-clockwise

And that gives the transformation matrix as
$\begin{aligned} & \begin{bmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} \cos(-\theta) & -\sin(-\theta) \\ \sin(-\theta) & \cos(-\theta) \end{bmatrix} \\ = & \begin{bmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix} \begin{bmatrix} \cos(-\theta) & -\sin(-\theta) \\ -\sin(-\theta) & -\cos(-\theta) \end{bmatrix} \\ = & \begin{bmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix} \begin{bmatrix} \cos(\theta) & \sin(\theta) \\ \sin(\theta) & -\cos(\theta) \end{bmatrix} \\ = & \begin{bmatrix} \cos ^2(\theta )-\sin ^2(\theta ) & 2 \cos (\theta ) \sin (\theta ) \\ 2 \cos (\theta ) \sin (\theta ) & \sin ^2(\theta )-\cos ^2(\theta ) \end{bmatrix} \\ = & \begin{bmatrix} \cos(2\theta ) & \sin(2\theta ) \\ \sin(2\theta ) & -\cos(2\theta ) \end{bmatrix} \end{aligned} $
Fixed@below; thanks b^3!

I also am not sure what nliu1995 is talking about, but I suppose this is a vector method.
$\begin{aligned} \mathbf{p}&=\cos(\theta)\mathbf{i}+\sin(\theta)\mathbf{j} \\ \mathbf{r}&=x\mathbf{i}+y\mathbf{j} \text{ (Position vector of a point we are trying to transform)} \end{aligned}$
The position vector of the image would be $\left( \mathbf{r} \cdot \mathbf{p} \right) \mathbf{p}- \left[ \mathbf{r} - \left( \mathbf{r} \cdot \mathbf{p} \right) \mathbf{p} \right]$ aka "vector resolute of $\mathbf{r}$ in the direction of $\mathbf{p}$ " - "vector resolute of $\mathbf{r}$ in the direction perpendicular to $\mathbf{p}$ ".
I believe the result is $\left( x\cos(2\theta) + y\sin(2\theta) \right) \mathbf{i}+\left( x\sin(2\theta) - y\cos(2\theta) \right) \mathbf{j}$ .

b^3 · « **Reply #175 on:** October 04, 2013, 04:12:20 pm »

Quote from: heaiyuo on October 04, 2013, 04:06:40 pm

Well apart from finding the image of a basis, if you want to do it as a composition of transformations, then it could be
Rotation by $\theta$ clockwise
Reflection in X-axis
Rotation by $\theta$ anti-clockwise
And that gives the transformation matrix as
$\begin{aligned} & \begin{bmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} \cos(-\theta) & -\sin(-\theta) \\ \sin(-\theta) & \cos(-\theta) \end{bmatrix} \\ = & \begin{bmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix} \begin{bmatrix} \cos(-\theta) & -\sin(-\theta) \\ -\sin(-\theta) & -\cos(-\theta) \end{bmatrix} \\ = & \begin{bmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix} \begin{bmatrix} \cos(\theta) & \sin(\theta) \\ \sin(\theta) & -\cos(\theta) \end{bmatrix} \\ = & \begin{bmatrix} \cos ^2(\theta )-\sin ^2(\theta ) & 2 \cos (\theta ) \sin (\theta ) \\ 2 \cos (\theta ) \sin (\theta ) & \sin ^2(\theta )-\cos ^2(\theta ) \end{bmatrix} \\ = & \begin{bmatrix} \cos(2\theta ) & \sin(2\theta ) \\ \sin(2\theta ) & -\cos(2\theta ) \end{bmatrix} \end{aligned} $
OK, spare me; I do not know how to fix that LaTeX.

I also am not sure what nliu1995 is talking about, but I suppose this is a vector method.
$\begin{aligned} \mathbf{p}&=\cos(\theta)\mathbf{i}+\sin(\theta)\mathbf{j} \\ \mathbf{r}&=x\mathbf{i}+y\mathbf{j} \text{ (Position vector of a point we are trying to transform)} \end{aligned}$
The position vector of the image would be $\left( \mathbf{r} \cdot \mathbf{p} \right) \mathbf{p}- \left[ \mathbf{r} - \left( \mathbf{r} \cdot \mathbf{p} \right) \mathbf{p} \right]$ aka "vector resolute of $\mathbf{r}$ in the direction of $\mathbf{p}$ " - "vector resolute of $\mathbf{r}$ in the direction perpendicular to $\mathbf{p}$ ".
I believe the result is $\left( x\cos(2\theta) + y\sin(2\theta) \right) \mathbf{i}+\left( x\sin(2\theta) - y\cos(2\theta) \right) \mathbf{j}$ .

To fix the , put all the "\\" on the next lines (it's a little weird and not like most latex renderers, but it works.
e.g.
$\begin{aligned} & \begin{bmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} \cos(-\theta) & -\sin(-\theta) \\ \sin(-\theta) & \cos(-\theta) \end{bmatrix} \\ = & \begin{bmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix} \begin{bmatrix} \cos(-\theta) & -\sin(-\theta) \\ -\sin(-\theta) & -\cos(-\theta) \end{bmatrix} \\ = & \begin{bmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix} \begin{bmatrix} \cos(\theta) & \sin(\theta) \\ \sin(\theta) & -\cos(\theta) \end{bmatrix} \\ = & \begin{bmatrix} \cos ^2(\theta )-\sin ^2(\theta ) & 2 \cos (\theta ) \sin (\theta ) \\ 2 \cos (\theta ) \sin (\theta ) & \sin ^2(\theta )-\cos ^2(\theta ) \end{bmatrix} \\ = & \begin{bmatrix} \cos(2\theta ) & \sin(2\theta ) \\ \sin(2\theta ) & -\cos(2\theta ) \end{bmatrix} \end{aligned} $

Code:

Code: [Select]

[tex]\begin{aligned}
& \begin{bmatrix}
\cos(\theta) & -\sin(\theta) 
\\ \sin(\theta) & \cos(\theta)
\end{bmatrix}
\begin{bmatrix}
1 & 0 
\\ 0 & -1
\end{bmatrix}
\begin{bmatrix}
\cos(-\theta) & -\sin(-\theta) 
\\ \sin(-\theta) & \cos(-\theta)
\end{bmatrix} 
\\ = & \begin{bmatrix}
\cos(\theta) & -\sin(\theta) 
\\ \sin(\theta) & \cos(\theta)
\end{bmatrix}
\begin{bmatrix}
\cos(-\theta) & -\sin(-\theta) 
\\ -\sin(-\theta) & -\cos(-\theta)
\end{bmatrix} 
\\ = & \begin{bmatrix}
\cos(\theta) & -\sin(\theta) 
\\ \sin(\theta) & \cos(\theta)
\end{bmatrix}
\begin{bmatrix}
\cos(\theta) & \sin(\theta) 
\\ \sin(\theta) & -\cos(\theta)
\end{bmatrix} 
\\ = & \begin{bmatrix}
\cos ^2(\theta )-\sin ^2(\theta ) & 2 \cos (\theta ) \sin (\theta ) 
\\ 2 \cos (\theta ) \sin (\theta ) & \sin ^2(\theta )-\cos ^2(\theta )
\end{bmatrix} 
\\ = & \begin{bmatrix}
\cos(2\theta ) & \sin(2\theta ) 
\\ \sin(2\theta ) & -\cos(2\theta )
\end{bmatrix}
\end{aligned}
[/tex]

Alwin · « **Reply #176 on:** October 04, 2013, 06:27:30 pm »

Quote from: heaiyuo on October 04, 2013, 04:06:40 pm

Rotation by $\theta$ clockwise
Reflection in X-axis
Rotation by $\theta$ anti-clockwise

Not particularly related to anything, but this process is called conjugation (a term sometimes applied to solving rubiks cubes too).
Usually I just call it a "setup" move of rotating clockwise, reflection (the desired transformation) then undoing the "setup" move, rotating anticlock. Formally,
$T(\mathbf{x}) = A B A^{-1}\mathbf{x}$

Quote from: heaiyuo on October 04, 2013, 04:06:40 pm

I also am not sure what nliu1995 is talking about, but I suppose this is a vector method.
...

Hmm, makes sense. nliu when you get on AN care to elaborate?

Quote

Quote from: heaiyuo on October 03, 2013, 02:02:40 am

Is it $\begin{bmatrix} \cos{2\theta} & \sin{2\theta} \\ \sin{2\theta} & -\cos{2\theta} \end{bmatrix}$ ? Because I get $(1,0)$ going to $(\cos{2\theta}, \sin{2\theta})$ and $(0,1)$ going to $(\sin{2\theta}, -\cos{2\theta})$ .

There actually is also a graphical approach that can be used, although I dunno how many marks this might gain in an exam. If only I'd been able to go to the UMEP sesh yesterday *shakes fist at sky*

(thanks for the pm brightsky, ill reply later... if I don't forget again sorry!)

The rotation of e₁ can be described in the following diagram:

The 'x-value' in red can be found by solving a simple trig equation, cos(2θ) = x / 1 --> x = cos(2θ)
The 'y-value' in green can be found by solving a simple trig equation, sin(2θ) = y / 1 --> y = sin(2θ)

$\implies \text{We get } \mathbf{e_1} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \mapsto \begin{bmatrix} \cos(2\theta) \\ \sin(2\theta) \end{bmatrix} \text{ ie } T(\mathbf{e_1})= \begin{bmatrix} \cos(2\theta) \\ \sin(2\theta) \end{bmatrix}$

We can create a similar diagram for the transformation of e₂ and use the trig identity
$\sin\left(\frac{\pi}{2} - \theta \right) = \cos(\theta)$ to get: $\mathbf{e_2} = \begin{bmatrix} 0 \\ 1 \end{bmatrix} \mapsto \begin{bmatrix} \sin(2\theta) \\ -\cos(2\theta) \end{bmatrix} \text{ ie } T(\mathbf{e_2})= \begin{bmatrix} \sin(2\theta) \\ -\cos(2\theta) \end{bmatrix}$

When we combine the two, we get:

$T(\mathbf{x}) = \begin{bmatrix} \cos{2\theta} & \sin{2\theta} \\ \sin{2\theta} & -\cos{2\theta} \end{bmatrix} \mathbf{x}$

Just a different method that relies less heavily on matrices and more on diagrams and (some) knowledge of bases
EDIT: I dunno if this is an acceptable method for UMEP, I think I'll double check it. However, it is a good check if you're unsure of your answer like you were brightsky.

lzxnl · « **Reply #177 on:** October 04, 2013, 07:26:22 pm »

OK...so according to my rather limited knowledge of reflections (lol I know nothing about matrices and transformations in that regard

)

Reflecting OX in the line OM produces the point OY such that the vector resolute of OX perpendicular to OM is the opposite of the analogous vector resolute for OY.

So...the vector resolute of OX perpendicular to OM is $OX-OM\frac{OX.OM}{OM.OM}$ sorry I don't know how to type in those vector arrows.
We need to add twice the negative of this to OX for OY. $2OM\frac{OX.OM}{OM.OM}-OX$ sorry I don't know how to type in those vector arrows.

Letting OM be the vector $\cos\theta i + \sin\theta j$
And letting OX be the vector $xi+yj$
We find that OY= $2OM(x\cos\theta + Y\sin\theta) - OX$
$= i(2x\cos^2\theta+2Y\sin\theta \cos\theta-x) + j(2x\sin\theta \cos\theta + 2y\sin^2\theta-y)$
Which simplifies to $i(x\cos{2\theta}+y*\sin{2\theta}) + j(x\sin{2\theta}-y\cos{2\theta})$

As you can, I suck at vectors. And latex. But that's my proof and pretty much coincides with what heaiyuo said.

brightsky · « **Reply #178 on:** November 16, 2013, 08:57:15 pm »

a few questions:

1. is there are proper definition for the term 'hyperplane'? it is bandied about in maths circles and often used to refer to different things so interested to know whether there is a precise definition.
2. does the phrase '3 linearly independent vectors' make sense? I've always been under the impression that linearly dependent/independent were adjectives that modified SETS rather than vectors. would it be more correct to say 'a linearly independent set of 3 vectors', as in 'a linearly independent set of 3 vectors is required to span the whole of R^3'?

thanks in advance!

rife168 · « **Reply #179 on:** November 17, 2013, 06:34:08 pm »

My understanding is that a hyperplane is an n-1 dimensional subspace of some n dimensional space.
For example in $\mathbb{R}^2$ any line through the origin could be considered a hyperplane since it is the span of 1 (=2-1) vector.
In $\mathbb{R}^3$ any plane through the origin could be considered a hyperplane since it is the span of 2 (=3-1) vectors.
The way I think about dimension with this sort of stuff is just the least number of vectors needed to span the object in question. i.e. a two dimensional subspace (e.g. plane in $\mathbb{R}^3$ ) requires at least 2 spanning vectors.
These may be enlightening:
http://en.wikipedia.org/wiki/Half-space_(geometry)
http://en.wikipedia.org/wiki/Ham_sandwich_theorem

'3 linearly independent vectors' makes sense and means exactly what you would expect.
You could take a set S of m vectors in a vector space V of dimension n (where m<n) which are linearly independent, and then find another vector v which is linearly independent to the set (perhaps it should be linearly independent 'of' the set, but 'to' is the standard terminology). In this case you have used linearly independent to describe both a set and a vector, and it is well defined in both scenarios.

More precisely the above would be interpreted as:
$\operatorname{span}\{S\}\subset V, \operatorname{dim} \operatorname{span}\{S\}=m$ and $v\in V, v\notin \operatorname{span}\{S\}$ .

(I feel that may have made my explanation even more unclear...

)

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