1st Year University Mathematics Thread (1 Viewer)

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 (1 Viewer)

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