HSC 2012 MX2 Marathon (archive) (1 Viewer)

Carrotsticks · Jul 18, 2012

Re: 2012 HSC MX2 Marathon

This question is possibly more complex than you thought. My reasons:

1. In a game of blackjack, you also have to count the fact that cards are given to another player at the same time (so there are fewer cards in the deck).

2. Remember that you can 'hit' in Blackjack, and we also have to determine the probability of one of them 'busting out', thus guaranteeing a win for the other player assuming they haven't already Blackjacked.

Godmode · Jul 18, 2012

Re: 2012 HSC MX2 Marathon

Just to make it slightly easier, reduce the deck number to 1, and we are only considering it to be a win if both players are below 22. (im not sure if this makes it easier or harder)

EDIT: And yes, this is becoming worse and worse as I think about it.

What's the worst we can get in MX2 probability?

lolcakes52 · Jul 21, 2012

Re: 2012 HSC MX2 Marathon

$$find $ \int \frac{\sin{x}-\cos{x}}{2+\sin{2x}} dx$
I made this one up myself.... I hope its alright....

Aesytic · Jul 21, 2012

Re: 2012 HSC MX2 Marathon

integral[(sinx-cosx)/(2+2sin2x)] = integral[(sinx-cosx)/(1+2sinxcosx+sin^2x+cos^2x)]
= integral[(sinx-cosx)/(1+(sinx+cosx)^2)]
let u=sinx + cosx
du = cosx - sinx dx
.'. integral[(sinx-cosx)/(1+(sinx+cosx)^2)] = -integral[1/(1+u^2)]
= -arctan(u)
= -arctan(sinx + cosx) + C

RealiseNothing · Jul 21, 2012

Re: 2012 HSC MX2 Marathon

lolcakes52 said:
$$find $ \int \frac{\sin{x}-\cos{x}}{2+\sin{2x}} dx$
I made this one up myself.... I hope its alright....

$\int \frac{sinx - cosx}{2 + 2sinxcosx}dx$
$\frac{1}{2}\int \frac{sinx - cosx}{1 + sinxcosx}dx$

Notice that this is the expansion of $tan(\alpha - \beta)$

$\int tan[tan^{-1}(sinx) - tan^{-1}(cosx)]dx$

Curious whether this could take you somewhere?

Carrotsticks · Jul 21, 2012

Re: 2012 HSC MX2 Marathon

RealiseNothing said:
$\int \frac{sinx - cosx}{2 + 2sinxcosx}dx$
$\frac{1}{2}\int \frac{sinx - cosx}{1 + sinxcosx}dx$

Notice that this is the expansion of $tan(\alpha - \beta)$

$\int tan[tan^{-1}(sinx) - tan^{-1}(cosx)]dx$

Curious whether this could take you somewhere?

I guess you could try using triangles to simplify the expression.

Trebla · Jul 21, 2012

Re: 2012 HSC MX2 Marathon

Show that for x > 0:

$\cos x < 1 - \dfrac{x^2}{2} + \dfrac{x^4}{24}$

Carrotsticks · Jul 21, 2012

Re: 2012 HSC MX2 Marathon

Delicious Taylor Polynomials.

bleakarcher · Jul 21, 2012

Re: 2012 HSC MX2 Marathon

Trebla said:
Show that for x > 0:

$\cos x < 1 - \dfrac{x^2}{2} + \dfrac{x^4}{24}$

Let f(x)=cos(x)-1+(x^2/2)-(x^4/24)
When x=0, f(x)=f(0)=0
f '(x)=-sin(x)+x-(x^3/6)
f ''(x)=-cos(x)+1-(x^2/2)
f '''(x)=sin(x)-x
Note that y=x is the tangent to y=sin(x) at x=0 meaning that x>sin(x) <=> sin(x)-x<0 for all x>0.
In order to prove the statement for x>0, which is equivalent to the statement "Prove f(x)<0 for x>0", we must consider the fact that if f(x)<0 for x>0 then f(0)<=0 and f '(x)<0 for all x>0. Since f ''(0)=0 and f '''(x)<0 for all x>0 it follows by the theorem stated above that f ''(x)<0 for all x>0. Since f '(0)=0 and f ''(x)<0 for all x>0 it follows again by the theorem that f '(x)<0 for all x>0. Applying the theorem once more we deduce that f(x)<0 for all x>0 i.e. cos(x)<1-(x^2/2)+(x^4/24) for x>0

RealiseNothing · Jul 21, 2012

Re: 2012 HSC MX2 Marathon

Can you do it this way?

$cosx < 1 - \frac{x^2}{2} + \frac{x^4}{24} = 1 - \frac{x^2}{2} + \frac{x^4}{4!}$

Since substituting in x=0 leaves both sides equal, we can say that $f'(x) < g'(x)$ for x>0.

So we end up with by differentiating:

$-sinx < -x + \frac{x^3}{3!}$

Therefore by multiplying both sides by -1:

$sinx > x - \frac{x^3}{3!}$

Sub in x=0 and both sides are equal, so we can do the same again:

$cosx > 1 - \frac{x^2}{2!}$

Sub in x=0 and both sides are equal so we differentiate again:

$-sinx > -x$

Therefore by once again multiplying both sides by -1:

$sinx < x$

Sub x=0 and both sides are equal, so differentiating again gives:

$cosx < 1$

Which is true for all x>0, and hence we conclude that:

$cosx < 1 - \frac{x^2}{2} + \frac{x^4}{24}$

Alternatively by using taylor series though, you would only have to prove that:

$-\frac{x^6}{6!} + \frac{x^8}{8!} - \frac{x^{10}}{10!} + \frac{x^{12}}{12!} - ..... < 0$

Trebla · Jul 21, 2012

Re: 2012 HSC MX2 Marathon

You cannot differentiate both sides of an inequality and expect know for sure what the inequality sign is
e.g. x² + 1 > x²
If you differentiate both sides you get: 2x > 2x which is clearly false

Though you are on the right track...just not in the right direction

bleakarcher · Jul 21, 2012

Re: 2012 HSC MX2 Marathon

Do you have another good question Trebla?

Trebla · Jul 21, 2012

Re: 2012 HSC MX2 Marathon

bleakarcher's solution is good but wasn't what I had in mind which is probably a simpler approach conceptually but requires a little messier working.

We know that for x > 0 and for some a > 0

$\cos x \leq 1\\\\\Rightarrow \displaystyle\int_0^a \cos x\,dx \leq \displaystyle\int_0^a dx\\\\\Rightarrow \sin a \leq a\\\\\Rightarrow \sin x < x\quad ($since $a$ can be considered a dummy variable and equality doesn't hold for $x > 0$)\\\\\Rightarrow \displaystyle\int_0^a \sin x\,dx < \displaystyle\int_0^a x\,dx\\\\\Rightarrow -\cos a + 1 < \dfrac{a^2}{2}\\\\\Rightarrow -\cos x + 1 < \dfrac{x^2}{2}\\\\\Rightarrow \displaystyle\int_0^a (-\cos x + 1)\,dx < \displaystyle\int_0^a \dfrac{x^2}{2}\,dx\\\\\Rightarrow -\sin a + a < \dfrac{a^3}{6}\\\\\Rightarrow -\sin x + x < \dfrac{x^3}{6}\\\\\Rightarrow \displaystyle\int_0^a (-\sin x + x)\,dx < \displaystyle\int_0^a \dfrac{x^3}{6}\,dx\\\\\Rightarrow \cos a - 1 + \dfrac{a^2}{2} < \dfrac{a^4}{24}\\\\\Rightarrow \cos x - 1 + \dfrac{x^2}{2} < \dfrac{x^4}{24}\\\\\therefore \cos x < 1 - \dfrac{x^2}{2} + \dfrac{x^4}{24}$

asianese · Jul 21, 2012

Re: 2012 HSC MX2 Marathon

Oh wow - insightful!

RealiseNothing · Jul 21, 2012

Re: 2012 HSC MX2 Marathon

lol so I pretty much did the reverse of the proof.

bleakarcher · Jul 21, 2012

Re: 2012 HSC MX2 Marathon

RealiseNothing said:
lol so I pretty much did the reverse of the proof.

I know fully comprehend Trebla's comment about you going in the wrong direction.

RealiseNothing · Jul 21, 2012

Re: 2012 HSC MX2 Marathon

Trebla said:
You cannot differentiate both sides of an inequality and expect know for sure what the inequality sign is
e.g. x² + 1 > x²
If you differentiate both sides you get: 2x > 2x which is clearly false

Though you are on the right track...just not in the right direction

I was using the "racetrack principle".

Since in each step $f(0) = g(0)$ I thought my way was right? Unless that's only for integrating, in which case it would stilll be right wouldn't it since you can just go backwards through my proof by starting at $cosx < 1$ for x>0 and integrating.

RealiseNothing · Jul 21, 2012

Re: 2012 HSC MX2 Marathon

bleakarcher said:
I know fully comprehend Trebla's comment about you going in the wrong direction.

Yer I lol'd when I read that.

Trebla · Jul 21, 2012

Re: 2012 HSC MX2 Marathon

RealiseNothing said:
lol so I pretty much did the reverse of the proof.

Exactly lol, just be aware that you can integrate inequalities (provided you use definite integrals) without running into too many complications but that's not the case when you differentiate them.

Carrotsticks · Jul 21, 2012

Re: 2012 HSC MX2 Marathon

bleakarcher said:
I know fully comprehend Trebla's comment about you going in the wrong direction.

I don't fully comprehend your comment here....

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