1st Year University Mathematics Thread (2 Viewers)

HeroicPandas · Nov 26, 2014

RealiseNothing said:
RHS is defined to be integer lol.

True that, LOL

RenegadeMx · Nov 26, 2014

RealiseNothing said:
RHS is defined to be integer lol.

ye but if and only if proof so need to be proven both ways

RenegadeMx · Nov 26, 2014

HeroicPandas said:
IF x is an integer, then it can be odd or even

Suppose x is odd, then there exists an integer M such that x = 2M + 1

$x=\left \lfloor \frac{2M+1}{2} \right \rfloor + \left \lceil \frac{2M+1}{2} \right \rceil$

$= \left \lfloor M+\frac{1}{2} \right \rfloor + \left \lceil M + \frac{1}{2}\right \rceil$

$= M + (M+1)$

$= 2M+1$

$= x$

Suppose x is even, then there exists an integer M such that x = 2M

So
$x=\left \lfloor \frac{2M}{2} \right \rfloor + \left \lceil \frac{2M}{2} \right \rceil = \left \lfloor M \right \rfloor + \left \lceil M\right \rceil = M + M= 2M= x$

ONLY IF, not sure atm, thinking about the contrapositive

I have this:

Let x be p/q such that p and q are integers and q is not 0 or 1

LHS = x = p/q

RHS = floor(x/2) + ceil(x/2)
= floor(p/2q) + ceil(p/2q)

yeah this was on of the questions in the discrete final, did exactly like you so far but cant get the opposite way

seanieg89 · Nov 26, 2014

RealiseNothing said:
RHS is defined to be integer lol.

So what? The claim is not that the RHS function outputs an integer iff x is an integer, the claim is that the RHS expression is EQUAL to x iff x is an integer.

We write x/2=m+d with m an integer and 0 =< d < 1.

If d=0, then RHS=m+m=2m=x=LHS.

If d is nonzero, then RHS-LHS=(2m+1)-(2m+2d).

Which is zero iff d=1/2.

This means that the identity in the question is true iff x/2 is a half-integer, that is iff x is an integer.

anomalousdecay · Nov 26, 2014

Carrotsticks said:
$\\ $Find two unit vectors in $ \mathbb{R}^3 $ that make an angle of $ \frac{\pi}{4} $ with $ 4\bold{i}+3\bold{j}$

$Using:$

$\frac{ {\begin{pmatrix} 4 \\ 3 \\ 0 \end{pmatrix}} \cdot { \begin{pmatrix} x \\ y \\ z \end{pmatrix}} }{ {\left | \begin{pmatrix} 4 \\ 3 \\ 0 \end{pmatrix} \right |} \cdot {\left | \begin{pmatrix} x \\ y \\ z \end{pmatrix} \right |} } \ = \ \cos{45 $^{\circ}$ \degree} \ = \ \frac{1}{\sqrt{2}} \ , \ \ $ Where $ x \in \mathbb{R}, \ \ y \in \mathbb{R} $ and $ z \in \mathbb{R}$

$$Letting z = 0 and expanding we get:$ \\$

$\frac{4x + 3y}{5 \sqrt{x^2 + y^2} } = \frac{1}{\sqrt{2}} \ \\ \\$

$$Solving:$ \\$

$x = - 7y , \ \ 7x = y \\ \\ $In both cases, $ z = 0 . $ If $ z \neq 0 , $ then we obtain solutions too difficult to solve for finding a unit vector. If I have time later I might try and find an expression for this where $z$ is some arbitrary constant, such that : $ z = \lambda, \ \lambda \in \mathbb{R}$

$$Now we need unit vectors, so we need:$ \\$

$\sqrt{x^2 + y^2 + z^2} = 1 \\ \\ \text{For the case where } x = - 7y, \\ \\ y = \frac{1}{\sqrt{50}} \ , \ \ x = \frac{ - 7}{\sqrt{50}} \ , \ \ z = 0 \\ \\ $Or, due to the negative case: $ y = \frac{ - 1}{\sqrt{50}} \ , \ \ x = \frac{ 7}{\sqrt{50}} \ , \ \ z = 0 \\ \\ \\ $For the other case where $ 7x = y, \\ \\ x = \frac{1}{\sqrt{50}} \ , \ \ y = \frac{7}{\sqrt{50}} \ , \ \ z = 0 \\ \\ $Or: $ x = \frac{ - 1}{\sqrt{50}} \ , \ \ y = \frac{ - 7}{\sqrt{50}} \ , \ \ z = 0 \\ \\$

$Noticing that we need to make sure that we obtained the acute 45 degree angle in both cases, obtain the dot product, and those two vectors which give us a positive dot product are the vectors we want.$

$\text{So two unit vectors subtending at an angle from } \begin{pmatrix} 4 \\ 3 \\ 0 \end{pmatrix} $ are: $ \\ \\ \\ \frac{1}{\sqrt{50}} \begin{pmatrix} 7 \\ $ $ - 1 $ $ \\ $ $ 0 $ $ \end{pmatrix} \\ \\ \\ $And$ \\ \\ \frac{1}{\sqrt{50}} \begin{pmatrix} \ 1 \ \\ \ 7 \ \\ \ 0 \ \end{pmatrix} \\$

Will try out the DE that FrankXie put up a while back later tonight when I get the time. Now its dinner time.

nightweaver066 · Nov 26, 2014

Carrotsticks said:
$\\ $Find two unit vectors in $ \mathbb{R}^3 $ that make an angle of $ \frac{\pi}{4} $ with $ 4\bold{i}+3\bold{j}$

Alternatively, a simpler way to do it would be to use complex numbers.

Consider the case where z = 0. Then, it boils down to the vectors formed by z = 1/5(4 + 3i)exp(i pi/4), w = 1/5(4 + 3i)exp(-i pi/4)

So we get our two solutions, $(\frac{4}{5\sqrt{2}} - \frac{3}{5\sqrt{2}}, \frac{4}{5\sqrt{2}} + \frac{3}{5\sqrt{2}}, 0) \; \; \& \; \; (\frac{4}{5\sqrt{2}} + \frac{3}{5\sqrt{2}}, \frac{3}{5\sqrt{2}} - \frac{4}{5\sqrt{2}}, 0)$

$(\frac{1}{5\sqrt{2}}, \frac{7}{5\sqrt{2}}, 0) \; \; \& \; \; (\frac{7}{5\sqrt{2}}, -\frac{1}{5\sqrt{2}}, 0)$

Alternatively alternatively, rotate the vector using a rotation matrix.

emilios · Nov 26, 2014

hey y'all. how would you say 1st year maths stacks up against HSC 4U? harder? about the same? more crammed syllabus?

anomalousdecay · Nov 27, 2014

FrankXie said:
solve or find the ananytic solution of $y^\prime+y^2=y\cos x$

Ok I gave this a crack but nah got nowhere. Non-linear, non-separable ODE's is something I've never encountered before. Tried substitutions, not exact, not separable, integrating factor does not work. Can anyone give me a tip on what can be done for this (do keep in mind I'm only in first year so I have never touched an ODE like this before)?

emilios said:
hey y'all. how would you say 1st year maths stacks up against HSC 4U? harder? about the same? more crammed syllabus?

Make a thread on this if you feel necessary.

The workload is a bit more in my opinion. However, the maths requires a lot more rigourous and well defined rules and definitions. There is quite a degree of abstraction to it as well.

You'll be fine though. I found first year maths more interesting than what we learnt in high school.

HeroicPandas · Nov 27, 2014

anomalousdecay said:
Ok I gave this a crack but nah got nowhere. Non-linear, non-separable ODE's is something I've never encountered before. Tried substitutions, not exact, not separable, integrating factor does not work. Can anyone give me a tip on what can be done for this (do keep in mind I'm only in first year so I have never touched an ODE like this before)?

Dividing both sides by 'y', we can use some kind of substitution (involving that fact that y'/y = d(lny)/dy) to reduce the LHS into something easier (I guess it might a substitution involving logs)

anomalousdecay · Nov 27, 2014

HeroicPandas said:
Dividing both sides by 'y', we can use some kind of substitution (involving that fact that y'/y = d(lny)/dy) to reduce the LHS into something easier (I guess it might a substitution involving logs)

I'll try it again now.

EDIT: Got this now:

$\frac{du}{dx} + e^u = \cos x$

Keep going around in circles getting more and more ODE's which I can't solve with these substitutions

Updating my substitutions so far lol:

u = ln y
u = 1/y
sqrt(u) = y
u = dy/dx
y = du/dx

Every single time I just get something that can't be solved using first year methods using these substitutions. I guess someone else can confirm if any of these substitutions work or not as I may have made a mistake somewhere.

HeroicPandas · Nov 27, 2014

Awwww, damn

nightweaver066 · Nov 27, 2014

With that DE, I don't think a substitution is necessary, and that the solution can be implicit.

A tip, assuming I did it correctly, is just aim to solve the DE, don't solely rely on methods you learnt in uni.

1st Year University Mathematics Thread (2 Viewers)

HeroicPandas

Heroic!

RenegadeMx

Kosovo is Serbian

RenegadeMx

Kosovo is Serbian

seanieg89

Well-Known Member

anomalousdecay

Premium Member

nightweaver066

Well-Known Member

emilios

Well-Known Member

anomalousdecay

Premium Member

HeroicPandas

Heroic!

anomalousdecay

Premium Member

HeroicPandas

Heroic!

nightweaver066

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 2)