leehuan's All-Levels-Of-Maths SOS thread (1 Viewer)

Drsoccerball · Mar 18, 2016

InteGrand said:
$\noindent That would be a plane through the origin (assuming the two vectors are non-parallel and non-zero of course).$

hmmm How can you spot the difference sorry can you write up something plz

InteGrand · Mar 18, 2016

Drsoccerball said:
hmmm How can you spot the difference sorry can you write up something plz

$\noindent If it was something like this: $\vec{x} = t_1 (1,0,0) + t_2 (2,0,0)$, then it's just a \emph{line} through the origin $\big{(}$in $\mathbb{R}^3$ of course$\big{)}$, because the two vectors are just scalar multiples of each other. Something like $\vec{x} = t_1 (1,0,0) + t_2 (2,1,0)$ however would be a plane, since the two vectors are non-parallel.$

Drsoccerball · Mar 18, 2016

InteGrand said:
$\noindent If it was something like this: $\vec{x} = t_1 (1,0,0) + t_2 (2,0,0)$, then it's just a \emph{line} through the origin $\big{(}$in $\mathbb{R}^3$ of course$\big{)}$, because the two vectors are just scalar multiples of each other. Something like $\vec{x} = t_1 (1,0,0) + t_2 (2,1,0)$ however would be a plane, since the two vectors are non-parallel.$

So assume all vector equations are a plane unless the vectors are scalar multiples of each other.

InteGrand · Mar 18, 2016

Drsoccerball said:
So assume all vector equations are a plane unless the vectors are scalar multiples of each other.

Well no. For example, if there were three vectors there and they were what is called linearly independent (which is equivalent to (although not defined as) saying that none of them is in the span of the other two ), then they would end up spanning the whole of ℝ³, rather than just a plane.

InteGrand · Mar 18, 2016

Drsoccerball said:
So assume all vector equations are a plane unless the vectors are scalar multiples of each other.

$\noindent But yes, if there are just \emph{two} direction vectors in the parametric equation and they are non-zero and non-parallel, it will be a plane (and conversely, any plane can be written in such a parametric form).$

Drsoccerball · Mar 18, 2016

hmmm :/ If I pick a different starting point for a vector I get a different answer is it wrong?

InteGrand · Mar 18, 2016

Drsoccerball said:
hmmm :/ If I pick a different starting point for a vector I get a different answer is it wrong?

No, it's not wrong; you can use any point that lies on the plane, so there are (infinitely) many choices for it. The parametric equation is not unique.

leehuan · Mar 19, 2016

I'm new to picking functions.

$\\$Use the squeeze theorem to find $\lim_{x\rightarrow 0}x\sin{\frac{1}{x}}$

Drsoccerball · Mar 19, 2016

leehuan said:
I'm new to picking functions.

$\\$Use the squeeze theorem to find $\lim_{x\rightarrow 0}x\sin{\frac{1}{x}}$

$-1 \leq \sin (\frac{1}{x}) \leq 1$

$-x \leq x\sin (\frac{1}{x}) \leq x$

$x \rightarrow 0 \ 0 \leq $ Desired Limit \leq 0$

$\therefore $ Limit $ =0$

leehuan · Mar 19, 2016

Drsoccerball said:
$-1 \leq \sin (\frac{1}{x}) \leq 1$

$-x \leq x\sin (\frac{1}{x}) \leq x$

$x \rightarrow 0 \ 0 \leq $ Desired Limit \leq 0$

$\therefore $ Limit $ =0$

But when multiplying by x how do we know x is positive?

Or does it like not matter cause we have a three sided inequality

InteGrand · Mar 19, 2016

leehuan said:
But when multiplying by x how do we know x is positive?

Or does it like not matter cause we have a three sided inequality

$\noindent We can do it like this to make sure the inequalities are correct: $-|x|\leq x\sin \frac{1}{x}\leq |x|$.$

Drsoccerball · Mar 19, 2016

leehuan said:
But when multiplying by x how do we know x is positive?

Or does it like not matter cause we have a three sided inequality

It really doesn't matter if x is positive it doesnt change anything as you said if its negative the inequalities switch and you are still left with :

$0 \leq Limit \leq 0$

leehuan · Mar 19, 2016

InteGrand said:
$\noindent We can do it like this to make sure the inequalities are correct: $-|x|\leq x\sin \frac{1}{x}\leq |x|$.$

$$That makes sense, but why is the MHS allowed to not be $\left|x\right|\sin{\frac{1}{x}}?$

It only makes sense logically - I just want a way to justify it.

Paradoxica · Mar 19, 2016

leehuan said:
But when multiplying by x how do we know x is positive?

Or does it like not matter cause we have a three sided inequality

To brush up on that technical matter, you have to multiply everything by x^2 to prevent change in sign.

Drsoccerball · Mar 19, 2016

Paradoxica said:
To brush up on that technical matter, you have to multiply everything by x^2 to prevent change in sign.

Then how are you going to obtain the required limit?

InteGrand · Mar 19, 2016

leehuan said:
$$That makes sense, but why is the MHS allowed to not be $\left|x\right|\sin{\frac{1}{x}}?$

It only makes sense logically - I just want a way to justify it.

$\noindent We shall prove the following general result, which is what justifies it (it's not hard to see is true or prove). Let $a> 0$ and $u,v \in \mathbb{R}$. Suppose $-a \leq u \leq a$ and $|u| = |v|$. Then $-a\leq v\leq a$.$

$\noindent \emph{Proof.} Note $-a \leq u \leq a$ is equivalent to $|u|\leq a$. But we are given $|v|=|u|$, so we have $|v| \leq a$. But this is equivalent to $-a\leq v\leq a$. Q.E.D.$

InteGrand · Mar 19, 2016

leehuan · Mar 19, 2016

InteGrand · Mar 19, 2016

InteGrand · Mar 19, 2016

leehuan said:
$$What elementary trick did I forget (just a hint this time for now please) to be unable to do $\\ \lim_{x\rightarrow64}{\frac{{x}^{1/3}-4}{{x}^{1/2}-8}}$

Try using L'Hôpital's rule or a substitution.

leehuan's All-Levels-Of-Maths SOS thread (1 Viewer)

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