HSC 2012 MX2 Marathon (archive) (1 Viewer)

Godmode · Jul 13, 2012

Re: 2012 HSC MX2 Marathon

Well, at least I know it can be solved now.

Is it worth looking into the conjugates of each complex root?

bleakarcher · Jul 13, 2012

Re: 2012 HSC MX2 Marathon

seanieg89 said:
Might as well bring this back to life.

Prove that the polynomial

$p(x)=\sum_{k=0}^{2n} \frac{x^k}{k!}$

has no real roots. (Where n is a non-negative integer.)

If n is a non-negative integer, then n>=0. But say we find P(x) for n=0, we find that P(x)=0. This has infinite roots. Maybe you mean for all positive integers n, or is there something wrong with my logic?

Edit: P(x)=1 for n=0. Sorry, disregard.

seanieg89 · Jul 13, 2012

Re: 2012 HSC MX2 Marathon

Godmode said:
Well, at least I know it can be solved now.

Is it worth looking into the conjugates of each complex root?

It's not what I did but it might lead somewhere

.

RealiseNothing · Jul 13, 2012

Re: 2012 HSC MX2 Marathon

seanieg89 said:
Might as well bring this back to life.

Prove that the polynomial

$p(x)=\sum_{k=0}^{2n} \frac{x^k}{k!}$

has no real roots. (Where n is a non-negative integer.)

Trivially, for $x \geq 0$ it has no real roots due to the '1' being out the front of the expression, meaning it can never equal 0. So we only have to show that for $x < 0$ the statement is true.

Essentially what we want to prove is that $\frac{x^{2n}}{(2n)!} > |\frac{x^{2n-1}}{(2n-1)!}|$ for all $x < 0$

Edit: fixing something up.

seanieg89 · Jul 13, 2012

Re: 2012 HSC MX2 Marathon

Not quite, you are adding the two terms x^{2n}/(2n)! and x^{2n-1}/(2n-1)!. You would need the SUM of these to be positive for negative x to make your argument work. (Or equivalently you could show that their difference is positive for positive x).

And in fact this is not the case. It is easy to show that p'(0)=1, so the poly definitely gets a bit smaller as you move to the left of the origin. If what you are claiming is true, then the poly would have a minimum value of 1 at the origin.

RealiseNothing · Jul 13, 2012

Re: 2012 HSC MX2 Marathon

seanieg89 said:
Not quite, you are adding the two terms x^{2n}/(2n)! and x^{2n-1}/(2n-1)!. You would need the SUM of these to be positive for negative x to make your argument work. (Or equivalently you could show that their difference is positive for positive x).

And in fact this is not the case. It is easy to show that p'(0)=1, so the poly definitely gets a bit smaller as you move to the left of the origin. If what you are claiming is true, then the poly would have a minimum value of 1 at the origin.

Yep I just realised that.

seanieg89 · Jul 15, 2012

Re: 2012 HSC MX2 Marathon

seanieg89 said:
Might as well bring this back to life.

Prove that the polynomial

$p(x)=\sum_{k=0}^{2n} \frac{x^k}{k!}$

has no real roots. (Where n is a non-negative integer.)

Perhaps time for a hint...the derivatives of this polynomial look quite similar to it! A sign that some sort of induction will work.

asianese · Jul 15, 2012

Re: 2012 HSC MX2 Marathon

I'm not gonna latex something out and spend ages (like last night==) on it but I may have something if it is induction.

I wrote p(x) out as a sum to x^2k/(2k)!

Try the case for n=0, p(x)=1 which has no real roots

Assume that p(x) has no real roots for n=k, such that

p(x) = x^0/0! + x^1/1!+...+x^k/k! +... + x^2k/(2k)! has no real roots

Try the case for n=k+1

RTP: P(x) (new polynomial with extra term) = x^0/0! + x^1/1!+...+x^k/k! +... + x^2k/(2k)! + x^2k+1/(2k+1)! has no real roots.

Notice that P'(x) = 1+x/1!+...+x^k-1/(k-1)!+...+x^2k-1/(2k-1)!+x^2k/(2k)!.

Hence P(x)=P'(x)+x^2k/(2k)!

Now we know that P'(x)=p(x) and hence from our assumption, there are no real roots to P'(x).

The term x^2k/(2k)! is an even polynomial (and is completely positive so there is no worry about flipping into the x axis), and so that if this is added to a polynomial with no real roots, P'(x)=p(x), then the new polynomial (P(x)) will still have no real roots.

Hence p(x) has no real roots for n, non negative integer by mathematical induction

I thought about things like y=x^2+1 and if you add a even power you'll still get something with no real roots. lol probably soooo off track

seanieg89 · Jul 15, 2012

Re: 2012 HSC MX2 Marathon

Careful, you are not adding just one term to the poly every time n goes up by one.

asianese · Jul 15, 2012

Re: 2012 HSC MX2 Marathon

Hmm why is that?

seanieg89 · Jul 15, 2012

Re: 2012 HSC MX2 Marathon

Because 2(n+1)=2n+2

.

asianese · Jul 15, 2012

Re: 2012 HSC MX2 Marathon

HAHA wow. wowza. lolol sorry hahaa

barbernator · Jul 15, 2012

Re: 2012 HSC MX2 Marathon

<a href="http://www.codecogs.com/eqnedit.php?latex=P(x)=\sum_{r=1}^{2n}\frac{x^{r}}{r!}\\ P'(x)=\sum_{r=1}^{2n-1}\frac{x^{r}}{r!}\\ P''(x)=\sum_{r=1}^{2n-2}\frac{x^{r}}{r!}\\ We~will~now~show~by~induction~that~P(x)~has~no~real~roots.\\ Let~n=1~to~show,\\ P(x)=\frac{x^2}{2}@plus;x@plus;1\\ P'(x)=x@plus;1=0,~x=-1\\ P(-1)=\frac{1}{2}.~\therefore ~for~n=1,~P(x)~has~no~real~roots\\ P''(x)=1.~\therefore ~the~function~is~concave~up.\\ Assume~that~P(x)~has~no~real~roots~for~n=k.\\ i.e.~\sum_{r=1}^{2k}\frac{x^{r}}{r!}> 0\\ To~prove~true~for~n=k@plus;1\\ TBP:~P(x)=\sum_{r=1}^{2k@plus;2}\frac{x^{r}}{(r)!}> 0\\ LHS=\sum_{r=1}^{2k@plus;2}\frac{x^{r}}{(r)!}\\ =\sum_{r=1}^{2k}\frac{x^{r}}{r!}@plus;\frac{x^{2k@plus;1}}{(2k@plus;1)!}@plus;\frac{x^{2k@plus;2}}{(2k@plus;2)!}\\ taking~the~second~derivative\\ P''(x)=\sum_{r=1}^{2k}\frac{x^{r}}{r!}>0.\\ this~isnt~correct~cos~it~isnt~showing~that~the~min~stat~point~is~>~0~but~i~dunno~where~to~go." target="_blank"><img src="http://latex.codecogs.com/gif.latex?P(x)=\sum_{r=1}^{2n}\frac{x^{r}}{r!}\\ P'(x)=\sum_{r=1}^{2n-1}\frac{x^{r}}{r!}\\ P''(x)=\sum_{r=1}^{2n-2}\frac{x^{r}}{r!}\\ We~will~now~show~by~induction~that~P(x)~has~no~real~roots.\\ Let~n=1~to~show,\\ P(x)=\frac{x^2}{2}+x+1\\ P'(x)=x+1=0,~x=-1\\ P(-1)=\frac{1}{2}.~\therefore ~for~n=1,~P(x)~has~no~real~roots\\ P''(x)=1.~\therefore ~the~function~is~concave~up.\\ Assume~that~P(x)~has~no~real~roots~for~n=k.\\ i.e.~\sum_{r=1}^{2k}\frac{x^{r}}{r!}> 0\\ To~prove~true~for~n=k+1\\ TBP:~P(x)=\sum_{r=1}^{2k+2}\frac{x^{r}}{(r)!}> 0\\ LHS=\sum_{r=1}^{2k+2}\frac{x^{r}}{(r)!}\\ =\sum_{r=1}^{2k}\frac{x^{r}}{r!}+\frac{x^{2k+1}}{(2k+1)!}+\frac{x^{2k+2}}{(2k+2)!}\\ taking~the~second~derivative\\ P''(x)=\sum_{r=1}^{2k}\frac{x^{r}}{r!}>0.\\ this~isnt~correct~cos~it~isnt~showing~that~the~min~stat~point~is~>~0~but~i~dunno~where~to~go." title="P(x)=\sum_{r=1}^{2n}\frac{x^{r}}{r!}\\ P'(x)=\sum_{r=1}^{2n-1}\frac{x^{r}}{r!}\\ P''(x)=\sum_{r=1}^{2n-2}\frac{x^{r}}{r!}\\ We~will~now~show~by~induction~that~P(x)~has~no~real~roots.\\ Let~n=1~to~show,\\ P(x)=\frac{x^2}{2}+x+1\\ P'(x)=x+1=0,~x=-1\\ P(-1)=\frac{1}{2}.~\therefore ~for~n=1,~P(x)~has~no~real~roots\\ P''(x)=1.~\therefore ~the~function~is~concave~up.\\ Assume~that~P(x)~has~no~real~roots~for~n=k.\\ i.e.~\sum_{r=1}^{2k}\frac{x^{r}}{r!}> 0\\ To~prove~true~for~n=k+1\\ TBP:~P(x)=\sum_{r=1}^{2k+2}\frac{x^{r}}{(r)!}> 0\\ LHS=\sum_{r=1}^{2k+2}\frac{x^{r}}{(r)!}\\ =\sum_{r=1}^{2k}\frac{x^{r}}{r!}+\frac{x^{2k+1}}{(2k+1)!}+\frac{x^{2k+2}}{(2k+2)!}\\ taking~the~second~derivative\\ P''(x)=\sum_{r=1}^{2k}\frac{x^{r}}{r!}>0.\\ this~isnt~correct~cos~it~isnt~showing~that~the~min~stat~point~is~>~0~but~i~dunno~where~to~go." /></a>

EDIT: the r=1 are supposed to be r=0.

seanieg89 · Jul 15, 2012

Re: 2012 HSC MX2 Marathon

barbernator said:
$P(x)=\sum_{r=1}^{2n}\frac{x^{r}}{r!}\\ P'(x)=\sum_{r=1}^{2n-1}\frac{x^{r}}{r!}\\ P''(x)=\sum_{r=1}^{2n-2}\frac{x^{r}}{r!}\\ We~will~now~show~by~induction~that~P(x)~has~no~real~roots.\\ Let~n=1~to~show,\\ P(x)=\frac{x^2}{2}+x+1\\ P'(x)=x+1=0,~x=-1\\ P(-1)=\frac{1}{2}.~\therefore ~for~n=1,~P(x)~has~no~real~roots\\ P''(x)=1.~\therefore ~the~function~is~concave~up.\\ Assume~that~P(x)~has~no~real~roots~for~n=k.\\ i.e.~\sum_{r=1}^{2k}\frac{x^{r}}{r!}> 0\\ To~prove~true~for~n=k+1\\ TBP:~P(x)=\sum_{r=1}^{2k+2}\frac{x^{r}}{(r)!}> 0\\ LHS=\sum_{r=1}^{2k+2}\frac{x^{r}}{(r)!}\\ =\sum_{r=1}^{2k}\frac{x^{r}}{r!}+\frac{x^{2k+1}}{(2k+1)!}+\frac{x^{2k+2}}{(2k+2)!}\\ taking~the~second~derivative\\ P''(x)=\sum_{r=1}^{2k}\frac{x^{r}}{r!}>0.\\ this~isnt~correct~cos~it~isnt~showing~that~the~min~stat~point~is~>~0~but~i~dunno~where~to~go.$

Very good! Definitely on the right track using concavity. Keep working using this approach, you are actually very close to done.

sophisms · Jul 15, 2012

Re: 2012 HSC MX2 Marathon

i am so going to fail maths ext 2 after seeing this thread. I don't even get what's going on :/

asianese · Jul 15, 2012

Re: 2012 HSC MX2 Marathon

Dw sophisms I thought the exact same thing...but you'll get used to it! Promise!

barbernator · Jul 16, 2012

Re: 2012 HSC MX2 Marathon

still got no idea on how to finish it off lol

seanieg89 · Jul 16, 2012

Re: 2012 HSC MX2 Marathon

$We aim to prove that $p_n(x)$ is positive everywhere for each non-negative integer $n$. (The subscript simply tells us what we are taking n to be).\\So the argument can be structured as follows:\\i) $p_0(x)=1>0$ everywhere.\\ii) Observe that any real poly of odd degree with positive leading coefficient tends to $\pm\infty$ as $x \rightarrow \pm \infty$.\\iii) Assume $p_k>0$ everywhere for some non-negative integer $k$. From ii) and the fact that $p_{k+1}''=p_k>0$, we can show that $p_{k+1}$ has a unique stationary point and that this stationary point must be the minimum of p on the entire real line. (It is a little bit of work to prove this, but the fact is relatively obvious from graphing such a function).\\iv) Now if $\alpha$ is this stationary point, then $p_{k+1}'(\alpha)=0$. So $p_{k+1}(\alpha)=p_{k+1}(\alpha)-p_{k+1}'(\alpha)=\alpha^{2k}/(2k)!>0$. This completes the induction.$

sophisms · Jul 16, 2012

Re: 2012 HSC MX2 Marathon

asianese said:
Dw sophisms I thought the exact same thing...but you'll get used to it! Promise!

don't think i'll ever get use to it :L My maths must be really bad :'(

Godmode · Jul 18, 2012

Re: 2012 HSC MX2 Marathon

Alright! Time for a (possibly ridiculous) Probability question.

Jack and Michael are playing a game of blackjack, using 8 standard decks.

i. How many two card selections are there:
a) in total?
b) that will have blackjack?
ii. What is the probability of:
a) only one player being dealt blackjack?
b) both players being dealt blackjack?
iii. What is the probability of being dealt blackjack, where one card is a King?
iv. If Both Jack and Michael are dealt blackjack, what is the probability (out of all possible 2 card selections for both players, [blackjack or not] dealt to both players) that Jack will beat Michael (where 10 < Jack < Queen < King and let's say Spades = 4 points > Clubs = 3 > Diamonds = 2 > Hearts = 1, and the suit of both cards count, and most points wins)

For those who have never played Blackjack, (My teacher said he once had a Yr 12 class where barely anyone had even seen a deck of cards lol) Having Blackjack is when the sum of both card's numbers is 21 [Ace = 1 or 11, 10 = Jack = Queen = King]

Note: I have no idea how to do this question, I just made it up on the spot. Additionally, probability is not my forte, so don't rely on me to come up with the answers any time soon lol.

For part i. , I just mean a single hand, but for part iv. , I mean for both hands at the same time ( it is a bit confusing lol).

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