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

leehuan · Mar 31, 2016

Can someone check my working? I lost motivation to do it (because I'm timing myself doing a past paper) and now I can't force the final answer out, but just incase anything happened I want to know if what I had done was correct.

$\\$Question: Let $ABCD $ be a quadrilateral in which $AD\parallel BC$. Let $M$ be the midpoint of $AB$ and $N$ be the midpoint of $CD$. Use vector methods to prove that $MN$ is parallel to $AD$ and that the length of $MN$ is the average of the lengths of $AD$ and $BC$

Need checked: Proof of parallel
Need help: Proving the length part.

$\\$Method: Construct the trapezium $ABCD$ where $AD\parallel BC$ as given, and mark $M$ and $N. \\ $Define the following vectors as such:$\\ \begin{align*} \overrightarrow{AD}&=\bf{a}\\\overrightarrow{DC}&=\bf{d}\\\overrightarrow{CB}&=\bf{c}\\\overrightarrow{BA}&=\bf{b}\end{align*}$

$\\$Since we know that $AD\parallel BC$, we also have $\textbf{c}=\alpha\textbf{a}, \, \alpha \in \mathbb R$

$\\$Then, $\overrightarrow{MA}=\frac{1}{2}\textbf{a}= \overrightarrow{BM} \\$ and, $\overrightarrow{DN}=\frac{1}{2}\textbf{d}= \overrightarrow{NC}$

$\\\overrightarrow{AN}= \overrightarrow{AD} + \overrightarrow{DN} \Rightarrow \overrightarrow{AN} = \textbf{a} + \frac{1}{2} \textbf{d}\\\therefore \overrightarrow{MN} = \overrightarrow{MA} + \overrightarrow{AN} \Rightarrow \overrightarrow{MN} = \frac{3}{2}\textbf{a}+\frac{1}{2}\textbf{d}\\$Similarly, by using $BN,\,BC,\,CN\\\overrightarrow{MN}=-\frac{1}{2}\textbf{a}-\textbf{c}-\frac{1}{2}\textbf{d}$

$$Equating:$\\ \frac{3}{2} \textbf{a}+\frac{1}{2}\textbf{d}=-\frac{1}{2}\textbf{a}-\textbf{c}-\frac{1}{2}\textbf{d}\\\Leftrightarrow \frac{1}{2}\textbf{d} = -\textbf{a} - \frac{1}{2}\textbf{c}$
$\therefore \overrightarrow{MN}=\frac{3}{2}\textbf{a}+\left(-\textbf{a} - \frac{1}{2}\textbf{c}\right)\\ \Leftrightarrow = \left(1-\alpha\right) \textbf{a}$

$\text{Hence as the vector MN is a scalar product of that for AD, they must be parallel}$

Wow that LaTeX took forever to type only to look so ugly

Drongoski · Mar 31, 2016

leehuan said:
Can someone check my working? I lost motivation to do it (because I'm timing myself doing a past paper) and now I can't force the final answer out, but just incase anything happened I want to know if what I had done was correct.

$\\$Question: Let $ABCD $ be a quadrilateral in which $AD\parallel BC$. Let $M$ be the midpoint of $AB$ and $N$ be the midpoint of $CD$. Use vector methods to prove that $MN$ is parallel to $AD$ and that the length of $MN$ is the average of the lengths of $AD$ and $BC$

Need checked: Proof of parallel
Need help: Proving the length part.

$\\$Method: Construct the trapezium $ABCD$ where $AD\parallel BC$ as given, and mark $M$ and $N. \\ $Define the following vectors as such:$\\ \begin{align*} \overrightarrow{AD}&=\bf{a}\\\overrightarrow{DC}&=\bf{d}\\\overrightarrow{CB}&=\bf{c}\\\overrightarrow{BA}&=\bf{b}\end{align*}$

$\\$Since we know that $AD\parallel BC$, we also have $\textbf{c}=\alpha\textbf{a}, \, \alpha \in \mathbb R$

$\\$Then, $\overrightarrow{MA}=\frac{1}{2}\textbf{a}= \overrightarrow{BM} \\$ and, $\overrightarrow{DN}=\frac{1}{2}\textbf{d}= \overrightarrow{NC}$

$\\\overrightarrow{AN}= \overrightarrow{AD} + \overrightarrow{DN} \Rightarrow \overrightarrow{AN} = \textbf{a} + \frac{1}{2} \textbf{d}\\\therefore \overrightarrow{MN} = \overrightarrow{MA} + \overrightarrow{AN} \Rightarrow \overrightarrow{MN} = \frac{3}{2}\textbf{a}+\frac{1}{2}\textbf{d}\\$Similarly, by using $BN,\,BC,\,CN\\\overrightarrow{MN}=-\frac{1}{2}\textbf{a}-\textbf{c}-\frac{1}{2}\textbf{d}$

$$Equating:$\\ \frac{3}{2} \textbf{a}+\frac{1}{2}\textbf{d}=-\frac{1}{2}\textbf{a}-\textbf{c}-\frac{1}{2}\textbf{d}\\\Leftrightarrow \frac{1}{2}\textbf{d} = -\textbf{a} - \frac{1}{2}\textbf{c}$
$\therefore \overrightarrow{MN}=\frac{3}{2}\textbf{a}+\left(-\textbf{a} - \frac{1}{2}\textbf{c}\right)\\ \Leftrightarrow = \left(1-\alpha\right) \textbf{a}$

$\text{Hence as the vector MN is a scalar product of that for AD, they must be parallel}$

Wow that LaTeX took forever to type only to look so ugly

It is not easy to follow your argument. Also the part vector MA = 0.5 a I just cannot see why. But let me show my attempt.

Drongoski · Mar 31, 2016

$\overrightarrow {AD} = k \overrightarrow {BC} $ for some real number k$\\ \\ \overrightarrow {OM} = 0.5 \overrightarrow {OA} + 0.5 \overrightarrow {OB}\\ \\ \overrightarrow {ON} = 0.5 \overrightarrow {OC} + 0.5\overrightarrow {OD}\\ \\ \therefore \overrightarrow {MN} = \overrightarrow {ON} - \overrightarrow {OM} = 0.5\overrightarrow {OA} + 0.5 \overrightarrow {OB} - 0.5 \overrightarrow {OC} - 0.5 \overrightarrow {OD}\\ \\ = 0.5(\overrightarrow {OD} - \overrightarrow {OA}) + 0.5(\overrightarrow {OC} - \overrightarrow {OB})\\ \\ = 0.5 \overrightarrow {AD} + 0.5 \overrightarrow {BC} $ . . . . . [*]$\\ \\ = (1 + k)\overrightarrow {BC} \\ \\ \therefore \overrightarrow {MN} // \overrightarrow {BC}\\ \\ $Hence $ MN // AD // BC $ \\ \\$

and by [*], length of MN = average of lengths of AD and BC.

InteGrand · Mar 31, 2016

Drongoski said:
It is not easy to follow your argument. Also the part vector MA = 0.5 a I just cannot see why. But let me show my attempt.

I think he typo'ed or got confused with his notation or something; the vector MA should be half the vector BA, which he labelled as b.

This problem can be done more easily if we define the origin to be at one of the vertices (say A), and then use the convention of calling the position vector for a point X the lower-case letter for that point (x) to avoid confusion.

leehuan · Mar 31, 2016

I realised the need to utilise the origin way too late; only after I was done.

But yes I felt like it was awkward to follow what I was doing. Can't blame anyone for that.

However the main thing was there was a mistake. Yep, thanks for pointing that out; that explains my struggle.

Next question: I muffled the epsilon-M definition. Need someone to check my working again

$It is given that $0<\epsilon <1$, and the solution to $\left|f(x)-1\right|<\epsilon$ is $\\ x \in \left(-\infty , - \sqrt{\frac{1}{\epsilon}-1} \right)\cup \left( \sqrt{\frac{1}{\epsilon}-1}, \infty \right) \\ $Does $\lim_{x \rightarrow \infty}{f(x)}$ exist? If so, what is it's value? Give reasons for your answer.$

$$Solution: From the given information we have$\\ \begin{align*}\left|x\right| & > \sqrt{\frac{1}{\epsilon}-1}\\ \Leftrightarrow {x}^{2}&>\frac{1}{\epsilon}-1 \\ \Leftrightarrow {x}^{2}-1 & > \frac{1}{\epsilon} \\ \Leftrightarrow \frac{1}{{x}^{2}-1} & < \epsilon \end{align*}$

Main thing I'm worried about is cause what I've done next is probably worded wrong and also out of order

$$Now, $\lim_{x\rightarrow\infty}{f(x)}$ exists and equals $L=1$ here, iff. $\\ \forall \, \epsilon>0, \exists x=M \Rightarrow \left|f(x)-L\right|<\epsilon$

$\therefore $ Pick $M=\sqrt{\frac{1}{\epsilon}-1}$

$$As the above calculations are reversible, for very large $x$ it would appear that the above criteria is indeed satisfied. Here $ \\ \left|f(x)-L\right| =\left|f(x)-1\right| = \frac{1}{{x}^{2}-1}$

$\therefore \lim_{x\rightarrow\infty}{f(x)}=1$

Edit - Justification to the reciprocating part - if x is really large than 1/(x^2-1) is positive

seanieg89 · Mar 31, 2016

It's kind of simpler than that.

All you need to show that the limit exists and is 1, is that for all e > 0 we can find M(e) > 0 such that |f(x)-1| < e for all x > M(e). Just take M(e)=sqrt(1/e-1).

InteGrand · Mar 31, 2016

leehuan said:
I realised the need to utilise the origin way too late; only after I was done.

But yes I felt like it was awkward to follow what I was doing. Can't blame anyone for that.

However the main thing was there was a mistake. Yep, thanks for pointing that out; that explains my struggle.

Next question: I muffled the epsilon-M definition. Need someone to check my working again

$It is given that $0<\epsilon <1$, and the solution to $\left|f(x)-1\right|<\epsilon$ is $\\ x \in \left(-\infty , - \sqrt{\frac{1}{\epsilon}-1} \right)\cup \left( \infty , \sqrt{\frac{1}{\epsilon}-1} \right) \\ $Does $\lim_{x \rightarrow \infty}{f(x)}$ exist? If so, what is it's value? Give reasons for your answer.$

$$Solution: From the given information we have$\\ \begin{align*}\left|x\right| & > \sqrt{\frac{1}{\epsilon}-1}\\ \Leftrightarrow {x}^{2}&>\frac{1}{\epsilon}-1 \\ \Leftrightarrow {x}^{2}-1 & > \frac{1}{\epsilon} \\ \Leftrightarrow \frac{1}{{x}^{2}-1} & < \epsilon \end{align*}$

Main thing I'm worried about is cause what I've done next is probably worded wrong and also out of order

$$Now, $\lim_{x\rightarrow\infty}{f(x)}$ exists and equals $L=1$ here, iff. $\\ \forall \, \epsilon>0, \exists x=M \Rightarrow \left|f(x)-L\right|<\epsilon$

$\therefore $ Pick $M=\sqrt{\frac{1}{\epsilon}-1}$

$$As the above calculations are reversible, for very large $x$ it would appear that the above criteria is indeed satisfied. Here $ \\ \left|f(x)-L\right| =\left|f(x)-1\right| = \frac{1}{{x}^{2}-1}$

$\therefore \lim_{x\rightarrow\infty}{f(x)}=1$

The limit exists because when they say "the solution" to the inequation, it means that |f(x) – 1| < ε iff x is in the union of those intervals. For the limit as x → +∞, the left interval is irrelevant (although that one actually shows that the limit as x → -∞ is also 1). The right one is relevant here. It tells us that for any ε in (0, 1), whenever x > sqrt(1/ε – 1), we have |f(x) – 1| < ε. Thus we can take M = sqrt(1/ε – 1), then whenever x > M, we have |f(x) – 1| < ε for any 0 < ε < 1. Hence the limit in question is 1.

(Although the definition of the limit says for all ε > 0, really it is equivalent to do it for any ε in (0, ε₀) for some fixed ε₀ > 0. This is because if we can show that there is an M (or ) that makes |f(x) – L| < ε for any ε in (0, ε₀) whenever x is in the correct interval for the limit (e.g. x > M, or 0 < |x - a| < ), we automatically can do it for any larger ε too.)

Edit: said above.

InteGrand · Mar 31, 2016

(Those question marks in my above post were meant to be a delta. Test: δ. When I try to edit that post, the stuff I wrote disappears.)

leehuan · Mar 31, 2016

InteGrand said:
(Those question marks in my above post were meant to be a delta. Test: δ. When I try to edit that post, the stuff I wrote disappears.)

Is it just convention or something how for limits to infinity we use M and for a specific point we use delta?

seanieg89 · Mar 31, 2016

leehuan said:
Is it just convention or something how for limits to infinity we use M and for a specific point we use delta?

M is often used to represent something large and delta is very often used to represent something small, so they are pretty natural to use. Wouldn't say the M one is really a convention though.

leehuan · Mar 31, 2016

seanieg89 said:
M is often used to represent something large and delta is very often used to represent something small, so they are pretty natural to use. Wouldn't say the M one is really a convention though.

Wait I'm having a mind blank. Delta being used for something small makes sense but when is M used for large things again?

InteGrand · Mar 31, 2016

Epsilon is also traditionally used to represent small things, which is basically why it's used in the limits, since we think of the "tolerance" as being made as small as we like. (There is this maths joke "Let epsilon be smaller than zero.")

InteGrand · Mar 31, 2016

leehuan said:
Wait I'm having a mind blank. Delta being used for something small makes sense but when is M used for large things again?

It was used in the 2011 HSC 4U paper question in the end about polynomials, referring the maximum of the moduli of the coefficients or something if I recall correctly. (And M is often used in other contexts to refer to something big or the max. of things etc.)

seanieg89 · Mar 31, 2016

leehuan said:
Wait I'm having a mind blank. Delta being used for something small makes sense but when is M used for large things again?

You probably wouldn't have encountered it as much but its common enough in first year calculus/analysis books. Like I said, its not realllly convention but its unsurprising that Integrand and I chose the same letter.

seanieg89 · Mar 31, 2016

Anyway, worry more about the ideas than the notation. As long as you can clearly express your arguments, that is what is more important. Have seen plenty of papers/proofs in which the roles of epsilon and delta are switched.

leehuan · Mar 31, 2016

InteGrand said:
Epsilon is also traditionally used to represent small things, which is basically why it's used in the limits, since we think of the "tolerance" as being made as small as we like. (These is this maths joke "Let epsilon be smaller than zero.")

Lol at the joke
_____________________________________________

Nice recalling. I remember being unable to get that question out when I studied for 4U.......... should really have a go at those questions I got stuck on

@sean Fair enough, I see I see

leehuan · Mar 31, 2016

seanieg89 said:
Anyway, worry more about the ideas than the notation. As long as you can clearly express your arguments, that is what is more important. Have seen plenty of papers/proofs in which the roles of epsilon and delta are switched.

Very well...

leehuan · Apr 1, 2016

Thoroughly confused over fractional powers of negative numbers now...

http://www.wolframalpha.com/input/?i=y=cbrt(x-8)
http://www.wolframalpha.com/input/?i=(x-8)^(1/3)

I personally thought that cbrt(x-8) and x^(1/3) meant the exact same thing. And today I learn that at least, when fed to Mathematica
- cbrt results in the real value root
- ^(1/3) results in the principal root

But I seem to have lead to the wronged belief that they would be the same. I always imagined that the principal root of x^3=-1 would be x=-1 but instead it's x=e^(i.π/3)

Which was why when someone told me the domain of f(x)=(x-8)^(1/3) was x≥8 I got completely stumped.

Did I miss a lesson on the principal root? Does it have to be the one with the lowest positive argument or something?

Paradoxica · Apr 1, 2016

leehuan said:
Thoroughly confused over fractional powers of negative numbers now...

http://www.wolframalpha.com/input/?i=y=cbrt(x-8)
http://www.wolframalpha.com/input/?i=(x-8)^(1/3)

I personally thought that cbrt(x-8) and x^(1/3) meant the exact same thing. And today I learn that at least, when fed to Mathematica
- cbrt results in the real value root
- ^(1/3) results in the principal root

But I seem to have lead to the wronged belief that they would be the same. I always imagined that the principal root of x^3=-1 would be x=-1 but instead it's x=e^(i.π/3)

Which was why when someone told me the domain of f(x)=(x-8)^(1/3) was x≥8 I got completely stumped.

Did I miss a lesson on the principal root? Does it have to be the one with the lowest positive argument or something?

The principle nth root of a number is always the smallest positive complex argument.

leehuan · Apr 1, 2016

Disappointing. What would be the inverse to:

f: R->R, f(x) = x^3
then, if g(x) = x^(1/3) is undefined for x<0?

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

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