Author Topic: 4U Maths Question Thread (Read 665491 times) Tweet Share

RuiAce · « **Reply #1635 on:** January 04, 2018, 06:09:29 pm »

Quote from: Jeeffffffffffffff on January 04, 2018, 05:46:20 pm

Use De Moivre’s theorem to express cos5theta and sin5theta in powers of costheta and sintheta. Hence, express tan5theta as a rational function of t where t=tantheta and deduce that tan(Pi/5)tan(2pi/5)tan(3pi/5)tan(4pi/5)=5.

I’ve gotten up to having
Tan5theta= (t⁵-10t³+5t)/(5t⁴-10t²+1)
But I don’t really know how I’m supposed to use that to deduce the last line in the question. Any help would be appreciated, thanks.

$\text{Consider the equation }x(x^4-10x^2+5)=0.$
The expression is basically your numerator from the previous part.
$\text{Dividing both sides by }5x^4-10x^2+1\text{ gives the equation}\\ \frac{x(x^4-10x^2+5)}{5x^4-10x^2+1}=0.$
Note that this is safe, because we haven't divided by 0 at the "important" points. This is as a consequence of the fact that the numerator and denominator have no common factors.
$\text{Let }x=\tan \theta.\text{ The equation then becomes}\\ \frac{\tan\theta(\tan^4\theta-10\tan^2\theta+5)}{5\tan^4\theta-10\tan^2\theta+1} = 0.\\ \text{Using what you've already proven, the equation now becomes}\\ \tan 5\theta = 0.$
$\text{Hence, since we had a degree 5 polynomial, we take the 5 consecutive solutions}\\ \begin{align*}5\theta &= 0, \pi, 2\pi, 3\pi, 4\pi\\ \theta &= 0, \frac\pi5, \frac{2\pi}5, \frac{3\pi}5, \frac{4\pi}5\end{align*}$
$\text{So our solutions to }x(x^4-10x^2+5) = 0\text{ are}\\ x=0, \tan \frac\pi5, \tan \frac{2\pi}5, \tan \frac{3\pi}5, \tan \frac{4\pi}5.$
$\text{If we divide }x\text{ out, we lose the solution }x=0,\\ \text{Thus, the solutions to }x^4-10x^2+5=0\text{ are just}\\ x=\tan \frac\pi5, \tan \frac{2\pi}5, \tan \frac{3\pi}5, \tan \frac{4\pi}5$
$\text{It now immediately follows from the product of roots that}\\ \tan\frac\pi5\tan \frac{2\pi}5\tan\frac{3\pi}5\tan\frac{4\pi}5=5.$

RustyWasTaken · « **Reply #1636 on:** January 05, 2018, 08:13:04 pm »

b) How do I identify which roots belong to z^6 + z^3 + 1 = 0
Cheers

RuiAce · « **Reply #1637 on:** January 05, 2018, 10:37:34 pm »

Quote from: RustyWasTaken on January 05, 2018, 08:13:04 pm

b) How do I identify which roots belong to z^6 + z^3 + 1 = 0
Cheers

$\text{Since }z^9-1 = (z^3-1)(z^6+z^3+1)\\ \text{and the equation }z^3-1=0\text{ has roots}\\ z = 1, \text{ cis }\frac{2\pi}3, \text{ cis }\frac{4\pi}{3},$
$\text{all of the }\textbf{other}\text{ roots of }z^9-1\\ \text{must be the roots of }z^6+z^3+1=0$
These should be $\text{ cis }\frac{2\pi}{9},\text{ cis }\frac{4\pi}{9},\text{ cis }\frac{8\pi}{9},\text{ cis }\frac{10\pi}{9},\text{ cis }\frac{14\pi}{9},\text{ cis }\frac{16\pi}{9} $

maria.micale · « **Reply #1638 on:** January 06, 2018, 11:38:56 am »

Quote from: RuiAce on January 04, 2018, 01:05:34 pm

So that gives you the weird expression involving $C-k$ and according to your answers we can stop there. Which is reasonable, because we now have something that links $m$ and $k$ together.

The second point is actually the fixed point. If you let the first, second and third points have $x$-coordinate $\alpha-d, \alpha, \alpha+d$, you'll see that $\alpha = -\frac{A}{3} $

I think I’m confused, could you please explain further because I don’t know how to get to the answer still. Thanks

RuiAce · « **Reply #1639 on:** January 06, 2018, 11:42:01 am »

maria.micale · « **Reply #1640 on:** January 06, 2018, 12:19:43 pm »

Ok thanks heaps Rui!!!

Lefkiiii6 · « **Reply #1641 on:** January 06, 2018, 12:23:53 pm »

Quote from: RuiAce on January 06, 2018, 11:42:01 am

$\text{As the points of intersection are in arithmetic progression,}\\ \text{let the points of intersection be }\alpha-d, \alpha, \alpha+d$
$\text{So using the sum of roots of }x^3+Ax^2+(B-m)x+(C-k)\\ (\alpha-d)+\alpha+(\alpha+d) = A \implies \alpha = \frac{A}{3}$
$\text{Hence, the second point of intersection must be at }x=\frac{A}{3}\\ \text{which is constant.}$

Sorry I'm looking at this question too and I don't understand how to find that c-k = A/27 (9(B-m) - 2A^2)

RuiAce · « **Reply #1642 on:** January 06, 2018, 12:41:48 pm »

Quote from: Lefkiiii6 on January 06, 2018, 12:23:53 pm

Sorry I'm looking at this question too and I don't understand how to find that c-k = A/27 (9(B-m) - 2A^2)

That part was addressed a few posts back when she first posted this question

Lefkiiii6 · « **Reply #1643 on:** January 06, 2018, 01:20:45 pm »

Quote from: RuiAce on January 06, 2018, 12:41:48 pm

That part was addressed a few posts back when she first posted this question

Yeah I read this but I am still not sure.

RuiAce · « **Reply #1644 on:** January 06, 2018, 02:17:50 pm »

Quote from: Lefkiiii6 on January 06, 2018, 01:20:45 pm

Yeah I read this but I am still not sure.

Please elaborate further on the confusion.

(Pretty much all I did was consider what Maria had proven in part a), and demonstrated the link between that and part b). At that point, all that was left was basically just substituting a = A, b = B-m, c = C-k)

itssona · « **Reply #1645 on:** January 07, 2018, 06:04:25 pm »

how would you go about graphing x^3 + y^3 =1
as in what process would you follow

RuiAce · « **Reply #1646 on:** January 07, 2018, 09:11:40 pm »

Quote from: sssona09 on January 07, 2018, 06:04:25 pm

how would you go about graphing x^3 + y^3 =1
as in what process would you follow

$\text{Starting with the usual,}\\ 3x^2+3y^2\frac{dy}{dx}=0\implies \frac{dy}{dx} = -\frac{x^2}{y^2}$
$\text{So we must have stationary points when }x=0\\ \implies y^3 = 1 \implies y=1\\ \text{and vertical tangents when }y=0\\ \implies x^3=1\implies x=1$
i.e. a S.P. at $ (0,1) $ and a point where the derivative is undefined at $ (1,0) $.
$\text{Observe, however, that the equation rearranges to give}\\ y=(1-x^3)^{1/3}\\ \text{which is perfectly well defined as }x\to \infty$
As opposed to the semicircle $ y=(1-x^2)^{1/2} $. Square rooting a negative number is problematic but cube rooting is not.
$\text{As }x\to \infty,\\ \text{notice how the behaviour of }x^3\text{ is going to}\\ \textbf{dominate}\text{ the behaviour of the constant, 1.}$
Think about it visually. If you graphed $y=1$ it's just gonna be a horizontal line, because a constant doesn't change. If you graphed $y=x^3$ however, that thing is gonna eventually blow up. Whatever blows up will have a much more significant impact on anything that does not blow up, as you let $x$ become really really large.
$\text{So effectively, as }x\text{ tends to }\infty\\ (1-x^3)^{1/3}\text{ is really going to tend towards }(-x^3)^{1/3}\\ \text{because the constant pretty much does nothing.}$
$\text{Of course, that's just equivalent to }-x\\ \text{so as }x\to \infty, y\to -x.\\ \text{Note that this is also true as }x\to -\infty.$
$\text{And thus, we have an oblique asymptote in }y=-x.\\ \text{Those are all the ingredients required.}$

RuiAce · « **Reply #1647 on:** January 11, 2018, 02:01:46 pm »

Hey Sona, I wanna come back to this one. I've found the formal way to prove where the asymptote is.

Quote from: sssona09 on January 07, 2018, 06:04:25 pm

how would you go about graphing x^3 + y^3 =1
as in what process would you follow

$\text{What we're interested in is what happens as }x\to \infty\\ \text{for }y=(1-x^3)^{1/3}.$
$\text{The above explanation demonstrates that ultimately,}\\ y=(1-x^3)^{1/3}\text{ must grow at the same rate as }y=-x.\\ \textbf{But it doesn't tell us anything about constants offsetting it.}$
$\text{Indeed, it is possible that instead, as }x\to \infty,\\ y \to -x-1\text{ or }-x+1\text{ for example.}\\ \text{The following method will guarantee that no constants float around here and there.}$
Of course, it's also a bit advanced. But at least it's a legit proof now.
____________________________________________
$\text{If as }x\to \infty\text{, we're anticipating that }(1-x^3)^{1/3}\to -x,\\ \text{we should be able to }\textbf{move that }-x\textbf{ across to the left,}\\\text{and presume that }(1-x^3)^{1/3} + x \to 0\text{ as }x\to \infty.$
$\text{We now have to compute the limit}\\ \lim_{x\to \infty} \left[(1-x^3)^{1/3} + x\right].$
____________________________________________
$\text{To compute this limit, we need to rationalise the numerator.}\\ \text{We do this, with an old friend; the sum of two cubes.}\\ \text{The formula is }(a+b)(a^2-ab+b^2) = a^3+b^3\\ \text{which rearranges to }a+b = \frac{a^3+b^3}{a^2-ab+b^2}$
$\text{Let }a = (1-x^3)^{1/3}\text{ and }b = x.\\ \text{Upon substitution, we arrive at}\\ \begin{align*}(1-x^3)^{1/3} + x &= \frac{(1-x^3)+x^3}{(1-x^3)^{2/3} - x (1-x^3)^{1/3} + x^2}\\ &= \frac1{(1-x^3)^{2/3} - x (1-x^3)^{1/3} + x^2}\end{align*}$
$\text{From here, we just compute an ordinary 3U limit.}\\ \text{Note that the highest power on the }x\text{'s is going to be 2}.\\ \begin{align*}\lim_{x\to \infty} \left[(1-x^3)^{1/3} + x\right]&= \lim_{x\to \infty}\frac1{(1-x^3)^{2/3} - x (1-x^3)^{1/3} + x^2}\\ &= \lim_{x\to \infty}\frac{\frac{1}{x^2}}{\frac{(1-x^2)^{2/3}}{x^2}- \frac{(1-x^3)^{1/3}}{x}+1} \\ &= \lim_{x\to \infty} \frac{\frac{1}{x^2}}{\frac{(1-x^2)^{2/3}}{(x^3)^{2/3}}- \frac{(1-x^2)^{1/3}}{(x^3)^{1/3}} + 1}\\ &= \lim_{x\to \infty} \frac{x^{-2}}{(x^{-3}-x^{-1})^{2/3} - (x^{-3}-x^{-1})^{1/3} + 1}\\ &= \frac{0}{0-0+1}\\ &=0\end{align*}$
Note: $ \sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}} $ and similarly $ \sqrt[3]{\frac{a}{b}} = \frac{\sqrt[3]{a}}{\sqrt[3]{b}}$, which has been used here. Some steps could've been skipped if you're really good at tackling algebra.
$\text{So since this gives you 0,}\\ \text{this must give you the asymptote of }y=-x.$

RustyWasTaken · « **Reply #1648 on:** January 16, 2018, 01:05:36 pm »

Cheers

RuiAce · « **Reply #1649 on:** January 16, 2018, 01:18:56 pm »

Quote from: RustyWasTaken on January 16, 2018, 01:05:36 pm

Cheers

$\text{The second region is known to just be}\\ \text{the exterior of the circle, radius 1 and centre }(0,1)$
$\text{For the first region, let }z=x+y.\\ \text{We see that }z-\overline{z} = 2iy.\text{ Hence,}\\ \begin{align*}|z-\overline{z}|&<2\\ |2iy|&<2\\ 2y&<2\\ y&<1\end{align*}$

$\text{The region required will be in purple}\\ \text{and is essentially the region below }y<1\\ \text{but without the circle.}$

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