Author Topic: 4U Maths Question Thread (Read 665232 times) Tweet Share

massive · « **Reply #600 on:** October 16, 2016, 02:27:19 pm »

hey guys how do you part b, im so close yet so far :/ i keep getting (-1)^1-k instead of (-1)^k+1

RuiAce · « **Reply #601 on:** October 16, 2016, 03:03:08 pm »

Quote from: massive on October 16, 2016, 02:27:19 pm

hey guys how do you part b, im so close yet so far :/ i keep getting (-1)^1-k instead of (-1)^k+1

$(-1)^{-1}=-1 \therefore (-1)^{1-k}=[(-1)^{-1}]^{k-1}=(-1)^{k-1}$

DaCoon · « **Reply #602 on:** October 16, 2016, 06:15:42 pm »

Just curious but is it possible to self learn 4U maths?

I've heard mixed rumors from people saying things such as "4U requires critical thinking", "you need a maths tutor for 4U or else you'll do bad", "yes 4U is self learnable" etc but the thing is I don't know what is right and what is wrong

RuiAce · « **Reply #603 on:** October 16, 2016, 06:30:14 pm »

Quote from: DaCoon on October 16, 2016, 06:15:42 pm

Just curious but is it possible to self learn 4U maths?

I've heard mixed rumors from people saying things such as "4U requires critical thinking", "you need a maths tutor for 4U or else you'll do bad", "yes 4U is self learnable" etc but the thing is I don't know what is right and what is wrong

Things I successfully self learnt:

Integration: 100% of it
Graphs: 100% of it
Complex numbers: All of it except for the complex locus section
Conics: Mostly all of it
Polynomials: 100% of it
Harder 3U: Induction

Things I failed to self learn:

Everything else

So that sounded pretty arrogant admittedly, just saying I learnt this and that. But here's the moral of the story.

It varies from person to person. Some people will have it in them to be able to teach themselves a certain amount of the 4U course. Some may even be able to self learn the entire thing. Whereas it's not possible for others.

What I will say, however, is that you do not NEED tutoring. Whilst having one is most certainly beneficial, this "need" word is. I got my mark in 4U without ANY tutoring whatsoever. (I just offer tutoring simply because I have somewhat of a passion for teaching.)

massive · « **Reply #604 on:** October 16, 2016, 07:53:05 pm »

Guys how do you prove the attached question by mathematical induction :S

RuiAce · « **Reply #605 on:** October 16, 2016, 08:18:26 pm »

Quote from: massive on October 16, 2016, 07:53:05 pm

Guys how do you prove the attached question by mathematical induction :S

$\text{The statement is clearly true when }n=1\text{ as the RHS equals to 8.}\\ \text{So assume it is true for some }n=k:\\ 8+88+\dots+\underbrace{88...88}_{k\text{ digits}}=\frac{8}{81}(10^{k+1}-9k-10)$
$\text{We need to use this assumption to show that}\\ 8+88+\dots+\underbrace{88...88}_{k\text{ digits}}+\underbrace{88...88}_{k+1\text{ digits}}=\frac{8}{81}(10^{k+2}-9k-19)$
$\text{From the assumption, we have}\\ \begin{align*}LHS&=\frac{8}{81}(10^{k+1}-9k-10)+\underbrace{88...88}_{k+1\text{ digits}}\\ &= \frac{8}{81}(81\times \underbrace{11...11}_{k+1\text{ digits}}+10^{k+1}-9k-10)\end{align*}$
$\text{Noting the very well hidden }\textbf{geometric progression:}\\ \begin{align*} \underbrace{11...11}_{k+1\text{ digits}}&=1+10+100+\dots+10^{k}\\ &= \frac{10^{k+1}-1}{10-1}\\ &= \frac{1}{9}(10^{k+1}-1)\end{align*}$
$\text{Now we can continue from earlier:}\\ \begin{align*}LHS&=\frac{8}{81}\left(81\times \frac{1}{9}(10^{k+1}-1)+10^{k+1}-9k-10\right)\\ &= \frac{8}{81}(10\times 10^{k+1}-9k-19)\\ &= \frac{8}{81}(10^{k+2}-9k-19)\end{align*}$

massive · « **Reply #606 on:** October 16, 2016, 09:45:50 pm »

Quote from: RuiAce on October 16, 2016, 08:18:26 pm

$\text{The statement is clearly true when }n=1\text{ as the RHS equals to 8.}\\ \text{So assume it is true for some }n=k:\\ 8+88+\dots+\underbrace{88...88}_{k\text{ digits}}=\frac{8}{81}(10^{k+1}-9k-10)$
$\text{We need to use this assumption to show that}\\ 8+88+\dots+\underbrace{88...88}_{k\text{ digits}}+\underbrace{88...88}_{k+1\text{ digits}}=\frac{8}{81}(10^{k+2}-9k-19)$
$\text{From the assumption, we have}\\ \begin{align*}LHS&=\frac{8}{81}(10^{k+1}-9k-10)+\underbrace{88...88}_{k+1\text{ digits}}\\ &= \frac{8}{81}(81\times \underbrace{11...11}_{k+1\text{ digits}}+10^{k+1}-9k-10)\end{align*}$
$\text{Noting the very well hidden }\textbf{geometric progression:}\\ \begin{align*} \underbrace{11...11}_{k+1\text{ digits}}&=1+10+100+\dots+10^{k}\\ &= \frac{10^{k+1}-1}{10-1}\\ &= \frac{1}{9}(10^{k+1}-1)\end{align*}$
$\text{Now we can continue from earlier:}\\ \begin{align*}LHS&=\frac{8}{81}\left(81\times \frac{1}{9}(10^{k+1}-1)+10^{k+1}-9k-10\right)\\ &= \frac{8}{81}(10\times 10^{k+1}-9k-19)\\ &= \frac{8}{81}(10^{k+2}-9k-19)\end{align*}$

thankyou <3

Blissfulmelodii · « **Reply #607 on:** October 17, 2016, 03:46:49 pm »

Just wondering if anyone has any tips on how to recognise which substitution to use in Integration for 4U.
In 3U integration is the easiest topic for me because majority of the time they give you the substitution to use and it just becomes a matter of algebra but with 4U clearly they don't and I always get stuck in recognising what substitution to use especially when trig is involved. Are there any little markers to recognise or patterns that i am just not seeing?

massive · « **Reply #608 on:** October 17, 2016, 04:44:35 pm »

Guys how do you do this?:

Prove by mathematical induction that the sum of the exterior angles of a n sided convex polygon is 360 degrees

Also in general is there any technique to do geometry induction proofs?

RuiAce · « **Reply #609 on:** October 17, 2016, 05:15:09 pm »

Quote from: Blissfulmelodii on October 17, 2016, 03:46:49 pm

Just wondering if anyone has any tips on how to recognise which substitution to use in Integration for 4U.
In 3U integration is the easiest topic for me because majority of the time they give you the substitution to use and it just becomes a matter of algebra but with 4U clearly they don't and I always get stuck in recognising what substitution to use especially when trig is involved. Are there any little markers to recognise or patterns that i am just not seeing?

$\text{Firstly, note that substitution is NOT required if you can reverse the chain rule.}\\ \text{E.g. For something such as }\int\tan^3x\sec^2x\\ \text{The substitution }u=\tan x\text{ busts it to }\int u^3 \,du$
$\text{But if you can just inspect that }\int \tan^3x\sec^2x\,dx = \frac{\tan^4x}{4}+C\\ \text{Then you probably should.}$
_________________
$\text{Now, in general, integration by substitution is used}\\ \text{when the integrand can be reduced to a more }\textbf{simplified form}.\\ \text{This means that often when it is used, it does not have to be. But it should.}$
$\text{The above example of reversing the chain rule is the first application of integration by sub.}$
$\text{Take }\int x\sqrt{x+1}\,dx\\ \text{There are many ways to approach it, but by using }u=x+1\\ \text{We have }\int(u-1)\sqrt{u}du\\ \text{Note that this is only favourable because we can }\textbf{expand}.\\ \text{That's one thing worth looking for.}$
$\text{Then consider an integral such as this one: }\int \frac{u}{(u+1)^3}\\ \text{We don't want to expand denominators, but we are NOT scared of expanding numerators}\\ \text{Hence by using }s=u+1\text{ we have }\int\frac{(s+1)^3}{s}ds\\ \text{Which is easier to manage.}$
$\text{Integration by substitution is also worth using when you are confused.}\\ \text{Take the 2014 HSC final question: }\int \frac{\ln x}{(1+\ln x)^2}dx\\ \text{When you don't know what you're doing, trying }x=e^u\text{ just makes sense.}$

RuiAce · « **Reply #610 on:** October 17, 2016, 05:17:41 pm »

RuiAce · « **Reply #611 on:** October 17, 2016, 05:42:04 pm »

Quote from: massive on October 17, 2016, 04:44:35 pm

Guys how do you do this?:

Prove by mathematical induction that the sum of the exterior angles of a n sided convex polygon is 360 degrees

Also in general is there any technique to do geometry induction proofs?

Let me say this now, there is a 0% chance of a geometric induction
$\text{Note that }n\ge 3\text{ or we have no polygon}\\ \text{When }n=3\text{, the sum of all interior angles is }180^\circ$ $\text{Recall the definition of an exterior angle.}\\ \text{For a convex polygon, the exterior angle is just made by extending a side, and drawing the adjacent angle on that new straight line.}$
See the mathsisfun link
$\text{A triangle has 3 vertices.}\\ \text{3 straight angles sum up to }540^\circ$
$540^\circ - 180^\circ = 360^\circ\\ \text{Hence the base case is true.}$
__________________________________
$\text{Suppose, for some integer }k\text{, the exterior angle sum}\\ \text{of the convex }k\text{-gon is }360^\circ\\ \text{We want to prove that the exterior angle sum}\\ \text{of the }(k+1)\text{-gon is }360^\circ$
$\text{Draw a diagram of the }(k+1)\text{-gon.}\\ \text{Join two sides on the }(k+1)\text{-gon, so that:}\\ \text{On one hand, you have a }k\text{-gon}\\ \text{On the other hand, you have a triangle.}$
$\text{The exterior angle sum of the }k\text{-gon is }360^\circ.\\ \text{But obviously we can't just add on the exterior angle sum of the triangle.}$
__________________________________
$\text{Name the sides of the }k\text{-gon:}\\ A_1A_2A_3\dots A_k\\ \text{and let your triangle be }A_1A_kA_{k+1}$
$\text{Produce the sides }A_{k-1}A_k\text{ and }A_2A_1\\ \text{You will realise that if you just did }360^\circ + 360^\circ\\ \text{You end up double counting some angles.}$
$\text{Post up a diagram, and we can continue from there.}$

Blissfulmelodii · « **Reply #612 on:** October 17, 2016, 07:38:48 pm »

Quote from: RuiAce on October 17, 2016, 05:17:41 pm

$\text{A trigonometric substitution is a special case of integration by substitution}\\ \text{so I have chosen to make a small split post for it.}$
$\text{Trigonometric substitutions should ONLY be considered when you have a quadratic}\\ \text{under a square root, and you can't bust the square root.}$
$\text{If you have something in the form }\sqrt{Ax^2+Bx+C}\\ \text{Almost always will you have to }\textbf{complete the square.}\\ \text{This brings us back to either one of:}\\ \sqrt{a^2-x^2}\\ \sqrt{x^2-a^2}\\ \sqrt{x^2+a^2}\\ \text{to which the substitutions }x=a\sin \theta, a\sec \theta, a\tan\theta\text{ you just must know.}$
$\text{Post some examples for walkthroughs if you like. Maybe 2 examples at a time helps here for contrast.}$

OHH that makes so much sense, I had not thought of it in that way. Thank you!!
I've given up on maths for the night but I definitely will post up some questions tomorrow (cause there were a few where the worked solutions only left me more confused)

RuiAce · « **Reply #613 on:** October 17, 2016, 07:59:55 pm »

Quote from: Blissfulmelodii on October 17, 2016, 07:38:48 pm

OHH that makes so much sense, I had not thought of it in that way. Thank you!!
I've given up on maths for the night but I definitely will post up some questions tomorrow (cause there were a few where the worked solutions only left me more confused)

Post anytime you want and I'll be there as soon as I'm free (or Jake will)

massive · « **Reply #614 on:** October 17, 2016, 10:10:20 pm »

how does this line: e^1/(n+1) < 1 + 1/n < e^1/n become this line: (1+1/n)ⁿ < e < (1+1/n)ⁿ⁺¹
Like in the solutions there's no other lines in between, so i was just wondering how they got that second line :S

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