Author Topic: 4U Maths Question Thread (Read 665221 times) Tweet Share

RuiAce · « **Reply #1350 on:** July 20, 2017, 09:10:28 pm »

Quote from: Natmo243 on July 20, 2017, 07:53:04 pm

Could somebody please help me with this one? I'm trying to get an answer independent of a (their answer is sqrt(3)+pi/3). I've tried the substitution x=asinӨ and integration by parts. Any help would be much appreciated

Just looking at this I can tell that the answer will not be independent of a, due to the upper bound being unpleasant. Presumably the upper bound on the integral is incorrect and should be $a$.

Edit: According to Wolfram, if the upper bound is $a$ then the answer is what's given, except with a minus.

Natmo243 · « **Reply #1351 on:** July 21, 2017, 06:58:33 am »

Quote from: RuiAce on July 20, 2017, 09:10:28 pm

Just looking at this I can tell that the answer will not be independent of a, due to the upper bound being unpleasant. Presumably the upper bound on the integral is incorrect and should be $a$.

Edit: According to Wolfram, if the upper bound is $a$ then the answer is what's given, except with a minus.

My teacher and I could only come to that conclusion after changing the limits. It's not often that a mistake exists both in the question and the answer. Thankyou very much!

limtou · « **Reply #1352 on:** July 21, 2017, 09:16:24 pm »

Hey Rui,

Could you explain why y = log_e (e^x-2) has the asymptote y=x?

RuiAce · « **Reply #1353 on:** July 21, 2017, 09:21:36 pm »

Quote from: limtou on July 21, 2017, 09:16:24 pm

Hey Rui,

Could you explain why
$y=\log _e\left(e^x-2\right)$
has the asymptote y=x?

That one isn't easy to explain mathematically. Consider a more intuitive approach
$\text{As }x\text{ becomes really large, i.e. }x\to \infty\\ \text{The }-2\text{ bit stays as is; it is a constant}\\ \text{but }e^x\text{ is growing extremely rapidly}$
$\text{Effectively speaking, the behaviour of the exponential}\\ \text{'dominates' over the constant}$
$\text{Hence, as }x\to \infty\\ \ln \left(e^x - 2\right)\approx \ln \left(e^x\right)$
$\text{Of course, }\ln (e^x) = x\\ \text{so this is the equation of the asymptote}$
Also, LaTeX doesn't handle it well when you use double spaces.

limtou · « **Reply #1354 on:** July 21, 2017, 09:27:32 pm »

Quote from: RuiAce on July 21, 2017, 09:21:36 pm

That one isn't easy to explain mathematically. Consider a more intuitive approach
$\text{As }x\text{ becomes really large, i.e. }x\to \infty\\ \text{The }-2\text{ bit stays as is; it is a constant}\\ \text{but }e^x\text{ is growing extremely rapidly}$
$\text{Effectively speaking, the behaviour of the exponential}\\ \text{'dominates' over the constant}$
$\text{Hence, as }x\to \infty\\ \ln \left(e^x - 2\right)\approx \ln \left(e^x\right)$
$\text{Of course, }\ln (e^x) = x\\ \text{so this is the equation of the asymptote}$
Also, LaTeX doesn't handle it well when you use double spaces.

Thanks!
The latex killed me

Took me longer trying to get the latex than your explanation (and didn't even get the latex in the end).

RuiAce · « **Reply #1355 on:** July 21, 2017, 10:55:49 pm »

Quote from: limtou on July 21, 2017, 09:27:32 pm

Thanks!
The latex killed me Took me longer trying to get the latex than your explanation (and didn't even get the latex in the end).

Tbh, I think if you dropped all the spaces in your LaTeX (except for when splitting some functions up) it would've rendered properly.
$y=\log_e(e^x-2)$

beau77bro · « **Reply #1356 on:** July 23, 2017, 09:53:02 am »

ok so im getting really close to the answer, using u=x-a substitution, but i can't understand how if f(x) is odd that we treat f(u) as odd) because aren't we technically shifting the function?

p.s. i never really understood dumby variables - rushed lesson so an explanation of anything that answers this would be really appreciated

thanks

RuiAce · « **Reply #1357 on:** July 23, 2017, 10:08:28 am »

$\text{All the dummy variables in integration thing is saying is}\\ \int_a^b f(x)dx= \int_a^b f(u)du\\ \text{This makes perfect sense, as when performing the definite integration}\\ \text{in both cases, the }x\text{ and }u\text{ vanish.}$
$\text{To convince yourself it works, you should evaluate all three of these integrals}\\ \int_0^1 x^2\,dx, \quad \int_0^1 u^2\, du\quad \int_0^1 \theta^2\, d\theta\\ \text{and check they all equal each other.}$
____________________________
$\text{Let }u=x-a\implies du=dx\\ \text{Our integral then becomes }\int_{-a}^a f(u)\, du$
$\text{We were told that }f(x)\text{ is an odd function for }-a\le x \le a\\ \text{So }\textbf{over that domain}\text{, we know that }f(-x)=-f(x)$
$\text{Of course, this also means that }f(u)\text{ is an odd function for }-a\le u \le a\\ \text{So for that domain, we also know }f(-u)=-f(u)\\ \text{How does the shifting affect this?}$
$\text{So }\int_{-a}^a f(u)\,du = 0$
$\text{In fact, }f(x-a)\text{ is NOT an odd function}\\ f(x-a)\textbf{ does NOT}\text{ have rotational symmetry about the point }(0,0)\\ \text{rather, it has rotational symmetry about }(a,0)\text{ instead}\\ \text{(A part of why this happens is indeed because the domain got altered)}$

beau77bro · « **Reply #1358 on:** July 29, 2017, 09:33:29 am »

why dont we set one person in this and have to reduce the n part by another 1?

usually we say for n around a table it's (n-1)! because we set one person so they aren't the same. why don't we set one here? the answer is (n-4)!4! but what if n=4 then it's 4! and that's not right. is it?

RuiAce · « **Reply #1359 on:** July 29, 2017, 10:03:55 am »

Quote from: beau77bro on July 29, 2017, 09:33:29 am

why dont we set one person in this and have to reduce the n part by another 1?

usually we say for n around a table it's (n-1)! because we set one person so they aren't the same. why don't we set one here? the answer is (n-4)!4! but what if n=4 then it's 4! and that's not right. is it?

$\text{Technically we already }\textit{have}\text{ done the }(n-1)!\text{ thing.}$
$\text{Let's change the analogy to arrangement in a line to compare}\\ \text{If 4 people had to be together in a line, we would have }4!\text{ again.}\\ \text{But then we don't have }(n-4)!\text{, rather we have }(n-3)!$
$\text{This is because once we identify the 4 that must be grouped together}\\ \text{we have to arrange that group, }\textbf{with }n-4\text{ other people}\\ \text{i.e. we actually have }\boxed{n-3\text{ quantities}}\text{ to be arranged.}$
$\text{Whereas in the circular arrangement, we fix that group}\\ \text{and still have }n-4\text{ to be arranged.}\\ \text{The comparison of }4!(n-3)!\text{ and }4!(n-4)!\text{ shows us that}\\ \text{we did indeed do the }(n-1)!\text{ trick for circles}$
___________________________________________________________________
$\text{The apparent contradiction when }n=4\text{ is because that result}\\ \text{only works under the assumption that }\boxed{n > N}\\ \text{where }N\text{ is the actual amount of people }\textbf{BEING}\text{ arranged.}$
$\text{Here, }N=4\\ \text{So the result doesn't work when }n=4\\ \text{But it }\textbf{does}\text{ work when }n=5,6,\dots$
Compare the two cases. When n=4, what actually happens is that we've chosen everyone. This leaves nobody else left to use as, pretty much, a "basis" for the circular arrangement to handle the (n-1)! issue. So we actually have to cater for it manually.

Provided n>4, we have a way around it because there will always be one leftover guy. So as a result of the proof of the (n-1)! result, we address it in pretty much the way(s) explained above.

As far as 4U goes, it would probably have made sense to assume n>4 here. But pretty good point though.

beau77bro · « **Reply #1360 on:** July 29, 2017, 10:24:34 am »

Quote from: RuiAce on July 29, 2017, 10:03:55 am

$\text{Technically we already }\textit{have}\text{ done the }(n-1)!\text{ thing.}$
$\text{Let's change the analogy to arrangement in a line to compare}\\ \text{If 4 people had to be together in a line, we would have }4!\text{ again.}\\ \text{But then we don't have }(n-4)!\text{, rather we have }(n-3)!$
$\text{This is because once we identify the 4 that must be grouped together}\\ \text{we have to arrange that group, }\textbf{with }n-4\text{ other people}\\ \text{i.e. we actually have }\boxed{n-3\text{ quantities}}\text{ to be arranged.}$
$\text{Whereas in the circular arrangement, we fix that group}\\ \text{and still have }n-4\text{ to be arranged.}\\ \text{The comparison of }4!(n-3)!\text{ and }4!(n-4)!\text{ shows us that}\\ \text{we did indeed do the }(n-1)!\text{ trick for circles}$
___________________________________________________________________
$\text{The apparent contradiction when }n=4\text{ is because that result}\\ \text{only works under the assumption that }\boxed{n > N}\\ \text{where }N\text{ is the actual amount of people }\textbf{BEING}\text{ arranged.}$
$\text{Here, }N=4\\ \text{So the result doesn't work when }n=4\\ \text{But it }\textbf{does}\text{ work when }n=5,6,\dots$
Compare the two cases. When n=4, what actually happens is that we've chosen everyone. This leaves nobody else left to use as, pretty much, a "basis" for the circular arrangement to handle the (n-1)! issue. So we actually have to cater for it manually.

Provided n>4, we have a way around it because there will always be one leftover guy. So as a result of the proof of the (n-1)! result, we address it in pretty much the way(s) explained above.

As far as 4U goes, it would probably have made sense to assume n>4 here. But pretty good point though.

OHHHH YEA I FORGOT WHEN YOU GROUP THEM AS 4, they are still treated a just 1 person, so n-3, THEN U SET ONE. ok i see my bad. thanks rui.

Kle123 · « **Reply #1361 on:** July 29, 2017, 09:09:47 pm »

HI
It is true that (cis(2π/3))³=1 but why isn't it true when both sides are cube rooted? (cis2π/3≠1).
Thankss

RuiAce · « **Reply #1362 on:** July 29, 2017, 09:18:16 pm »

Quote from: Kle123 on July 29, 2017, 09:09:47 pm

HI
It is true that (cis(2π/3))³=1 but why isn't it true when both sides are cube rooted? (cis2π/3≠1).
Thankss

Simple reason: Cube rooting (or even square rooting) isn't that simple when you're dealing with the complex numbers, as opposed to the real numbers.

Recall when you computed elementary square, cube, fourth, fifth or whatever roots, you couldn't just pick one value of $ \theta$. e.g. To compute the fifth roots of $ i$, you couldn't just get away with $5\theta = \frac{\pi}2\implies \theta = \frac{\pi}{10}$. You had to do the whole $5\theta = \frac{\pi}2+2k\pi $ thing.

kiwiberry · « **Reply #1363 on:** July 31, 2017, 08:05:00 pm »

Heya, could someone please explain how they found the number of terms? I understand the rest of the question

Opengangs · « **Reply #1364 on:** July 31, 2017, 09:18:36 pm »

Quote from: kiwiberry on July 31, 2017, 08:05:00 pm

Heya, could someone please explain how they found the number of terms? I understand the rest of the question

Since we want to find when Alan eventually picks a black jellybean, we need to consider every case.

So, if Alan picks a black jellybean on his first go, it's simply 1/n
3rd go: (n - 1)/n * (n - 2)/(n - 1) * 1/(n - 2), since the first and second times, we don't want the black jellybean.
5th go: (n - 1)/n * (n - 2)/(n - 1) * (n - 3)/(n - 2) * (n - 4)/(n - 3) * 1/(n - 4), same reasoning.

So, we sum this up to get when Alan eventually picks a black jellybean.
P(Alan picks a black jellybean) = 1/n + (n - 1)/n * (n - 2)/(n - 1) * 1/(n - 2) + (n - 1)/n * (n - 2)/(n - 1) * (n - 3)/(n - 2) * (n - 4)/(n - 3) * 1/(n - 4) + ... + (n - 1)/n * (n - 2)/(n - 1) ... * 2/3 * 1/2 * 1

We can actually rewrite this as:
P(Alan picks a black jellybean) = 1/n + 1/n * (n - 1)/(n - 1) * (n - 2)/(n - 2) + 1/n * (n - 2)/(n - 1) * (n - 2)/(n - 2) * (n - 3)/(n - 3) * (n - 4)/(n - 4) + ... + 1/n * (n - 1)/(n - 1) ... * 3/3 * 2/2 * 1
= 1/n + 1/n + 1/n + ... 1/n

How do we know it's (n + 1)/2 terms?
Let's consider each term independently.

Case 1: 1st go -- 1/n
Case 2: 3rd go -- 1/n
Case 3: 5th go -- 1/n
Case 4: 7th go -- 1/n

We can develop a pattern here between the case number and the term we're interested in.
Case n: (2n - 1)th go -- 1/n - This is the general pattern.

Thus, the nth go is:
Case [(n + 1)/2]: nth go -- 1/n

This means that there are (n + 1)/2 terms.

Which means that the probability is (1/n) * (n + 1)/2 = (n + 1)/2n as required.

Search the forums now!

We have moved!

Author Topic: 4U Maths Question Thread (Read 665221 times) Tweet Share

RuiAce

Re: 4U Maths Question Thread

Natmo243

Re: 4U Maths Question Thread

limtou

Re: 4U Maths Question Thread

RuiAce

Re: 4U Maths Question Thread

limtou

Re: 4U Maths Question Thread

RuiAce

Re: 4U Maths Question Thread

beau77bro

Re: 4U Maths Question Thread

RuiAce

Re: 4U Maths Question Thread

beau77bro

Re: 4U Maths Question Thread

RuiAce

Re: 4U Maths Question Thread

beau77bro

Re: 4U Maths Question Thread

Kle123

Re: 4U Maths Question Thread

RuiAce

Re: 4U Maths Question Thread

kiwiberry

Re: 4U Maths Question Thread

Opengangs

Re: 4U Maths Question Thread

Recent Posts

User:
Password: