HSC 2013 MX2 Marathon (archive) (1 Viewer)

GoldyOrNugget · Dec 27, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
$$Consider the Triangular Numbers, defined by the recurrence sequence $ \\ \\ T_n=n+T_{n-1} \ \ \ \ T_1=1$

$$i) By considering the identity $ \\ \\ j^2(j+1)^2-j^2(j-1)^2 = 4j^3 \\ \\ $Show that$ \ \ \ \ \sum_{k=1}^n k^3=\frac{n^2}{4}(n+1)^2$

$$ii) Hence show that $ \ \ \ \ T_n^2 = \sum_{k=1}^n k^3$

(Just thought it was a cool result)

$i) $\sum_{k=1}^{n} k^3 = \frac{1}{4} \sum_{k=1}^{n} 4k^3 = \frac{1}{4} \sum_{k=1}^{n} k^2(k+1)^2 - k^2(k-1)^2$. On expansion, all terms except the last will cancel, leaving the RHS.$

$ii) T_n = $\frac{n(n+1)}{2}$ (using arithmetic series). \therefore \quad $T_n^2$ = the RHS from part i).$

RealiseNothing · Dec 27, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
$$Consider the Triangular Numbers, defined by the recurrence sequence $ \\ \\ T_n=n+T_{n-1} \ \ \ \ T_1=1$

$$i) By considering the identity $ \\ \\ j^2(j+1)^2-j^2(j-1)^2 = 4j^3 \\ \\ $Show that$ \ \ \ \ \sum_{k=1}^n k^3=\frac{n^2}{4}(n+1)^2$

$$ii) Hence show that $ \ \ \ \ T_n^2 = \sum_{k=1}^n k^3$

$iii) Consider a word made up of 'n' distinct letters. If one particular letter must always precede another particular letter, show that the amount of possible words is$

$\frac{n!}{2}$

$iv) Using the same scenario, but by first arranging the two particular letters then arranging the rest, show that the amount of possible words is also$

$T_{n-1}(n-2)!$

$v)$Hence explain why$$

$n!=2T_{n-1}(n-2)!$

$vi) Deduce that$

$T_n = \frac{n}{2}(n+1)$

$vii) Hence show that$

$n^4 + 2n^3 + n^2 = 4\sum_{k=1}^{n}k^3$

$viii) Hence using previous results, deduce that$

$(2m)!=2^m \prod_{k=1}^{m}\sqrt{\sum_{j=1}^{2k-1}j^3}$

Really just some random results, might have done a few parts before.

RealiseNothing · Dec 27, 2012

Re: HSC 2013 4U Marathon

Consider the quadratic:

$x^2 + Ax + 1$

If one of the roots of the quadratic is $\alpha$ where $\alpha$ is not real.

Find the range of values for $A$

I got my answer which I found interesting but I'm not sure if it's 100% correct. I want to see if anyone gets the same as me.

RealiseNothing · Dec 27, 2012

Re: HSC 2013 4U Marathon

RealiseNothing said:
Consider the quadratic:

$x^2 + Ax + 1$

If one of the roots of the quadratic is $\alpha$ where $\alpha$ is not real.

Find the range of values for $A$

I got my answer which I found interesting but I'm not sure if it's 100% correct. I want to see if anyone gets the same as me.

Like I know what the answer should be, but I did it a different way and got something different. Trying to work out why that is.

seanieg89 · Dec 27, 2012

Re: HSC 2013 4U Marathon

RealiseNothing said:
Consider the quadratic:

$x^2 + Ax + 1$

If one of the roots of the quadratic is $\alpha$ where $\alpha$ is not real.

Find the range of values for $A$

I got my answer which I found interesting but I'm not sure if it's 100% correct. I want to see if anyone gets the same as me.

$|A|\geq 2$

Sy123 · Dec 27, 2012

Re: HSC 2013 4U Marathon

RealiseNothing said:
$iii) Consider a word made up of 'n' distinct letters. If one particular letter must always precede another particular letter, show that the amount of possible words is$

$\frac{n!}{2}$

$iv) Using the same scenario, but by first arranging the two particular letters then arranging the rest, show that the amount of possible words is also$

$T_{n-1}(n-2)!$

$v)$Hence explain why$$

$n!=2T_{n-1}(n-2)!$

$vi) Deduce that$

$T_n = \frac{n}{2}(n+1)$

$vii) Hence show that$

$n^4 + 2n^3 + n^2 = 4\sum_{k=1}^{n}k^3$

$viii) Hence using previous results, deduce that$

$(2m)!=2^m \prod_{k=1}^{m}\sqrt{\sum_{j=1}^{2k-1}j^3}$

Really just some random results, might have done a few parts before.

Great question there, here is my attempt:

iii)

Two letters k, and j, can only be arranged in a set of n letters, either that k is BEFORE j, or k is AFTER j.
The number of arrangements of k after j and k before j are equal.

Hence the number of arrangements are total/2
$\frac{n!}{2}$

iv)

We must arrange n letters into n slots.

If k and j have no spaces between them, there are n-1 ways they can be arranged ( i.e. _ _ _ k j _ _ .. _ _ )
If k and j hve 1 space between them, there are n-2 ways to arrange them (i.e. _ _ k _ j _ _ .. _ _ )

and so on.

So if we consider the number of ways we can arrange k and j, such that k is before j, when they are both together, 1 space apart, 2 spaces ... n-2 spaces apart the number of arrangements is 1 + 2 + 3 + ... + n-1 = T_(n-1). Arrange other letters in (n-2)! ways
Hence total arrangemetns is: T_(n-1) (n-2)!

v)

They both are the number of arrangements of same scenario hence they are equal:

$\therefore n!=2T_{n-1}(n-2)!$

vi)

$T_{n-1}= \frac{n!}{(n-2)! 2} = \frac{n(n-1)}{2} \\ \therefore T_n=n/2 (n+1)$

vii) Using the result from part ii

$\sum = T_n ^2 = \frac{n^4+2n^3+n^2}{4} \\ \therefore 4\sum= n^4+2n^3+n^2$

viii) From part v

$(2m)! = 2T_{2m-1}(2m-2)!$

From part ii

$T_{2m-1} = \sqrt{\sum_{k=1}^{2m-1} k^3}$

$(2m-2)!=2T_{2m-3}(2m-4)!$

Hence when we sub it in, we get recurrence.

$(2m)! = 2^m (T_{2m-1} T_{2m-3} ... T_{1}) = 2^m \prod_{k=1}^{m} T_{2k-1} = 2^m \prod_{k=1}^m \sqrt{\sum_{j=1}^{2k-1}k^3}$

seanieg89 · Dec 27, 2012

Re: HSC 2013 4U Marathon

What was your method that produced an incorrect answer?

RealiseNothing · Dec 27, 2012

Re: HSC 2013 4U Marathon

seanieg89 said:
$|A|\geq 2$

Wait, isn't it $|A| < 2$

seanieg89 · Dec 27, 2012

Re: HSC 2013 4U Marathon

Oh right you said not real, yeah, then just take the complement.

RealiseNothing · Dec 27, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
Great question there, here is my attempt:

iii)

Two letters k, and j, can only be arranged in a set of n letters, either that k is BEFORE j, or k is AFTER j.
The number of arrangements of k after j and k before j are equal.

Hence the number of arrangements are total/2
$\frac{n!}{2}$

iv)

We must arrange n letters into n slots.

If k and j have no spaces between them, there are n-1 ways they can be arranged ( i.e. _ _ _ k j _ _ .. _ _ )
If k and j hve 1 space between them, there are n-2 ways to arrange them (i.e. _ _ k _ j _ _ .. _ _ )

and so on.

So if we consider the number of ways we can arrange k and j, such that k is before j, when they are both together, 1 space apart, 2 spaces ... n-2 spaces apart the number of arrangements is 1 + 2 + 3 + ... + n-1 = T_(n-1). Arrange other letters in (n-2)! ways
Hence total arrangemetns is: T_(n-1) (n-2)!

v)

They both are the number of arrangements of same scenario hence they are equal:

$\therefore n!=2T_{n-1}(n-2)!$

vi)

$T_{n-1}= \frac{n!}{(n-2)! 2} = \frac{n(n-1)}{2} \\ \therefore T_n=n/2 (n+1)$

vii) Using the result from part ii

$\sum = T_n ^2 = \frac{n^4+2n^3+n^2}{4} \\ \therefore 4\sum= n^4+2n^3+n^2$

viii) From part v

$(2m)! = 2T_{2m-1}(2m-2)!$

From part ii

$T_{2m-1} = \sqrt{\sum_{k=1}^{2m-1} k^3}$

$(2m-2)!=2T_{2m-3}(2m-4)!$

Hence when we sub it in, we get recurrence.

$(2m)! = 2^m (T_{2m-1} T_{2m-3} ... T_{1}) = 2^m \prod_{k=1}^{m} T_{2k-1} = 2^m \prod_{k=1}^m \sqrt{\sum_{j=1}^{2k-1}k^3}$

That was quick lol, very nice.

GoldyOrNugget · Dec 27, 2012

Re: HSC 2013 4U Marathon

RealiseNothing said:
$iii) Consider a word made up of 'n' distinct letters. If one particular letter must always precede another particular letter, show that the amount of possible words is$

$\frac{n!}{2}$

$iv) Using the same scenario, but by first arranging the two particular letters then arranging the rest, show that the amount of possible words is also$

$T_{n-1}(n-2)!$

$v)$Hence explain why$$

$n!=2T_{n-1}(n-2)!$

$vi) Deduce that$

$T_n = \frac{n}{2}(n+1)$

$vii) Hence show that$

$n^4 + 2n^3 + n^2 = 4\sum_{k=1}^{n}k^3$

$viii) Hence using previous results, deduce that$

$(2m)!=2^m \prod_{k=1}^{m}\sqrt{\sum_{j=1}^{2k-1}j^3}$

Really just some random results, might have done a few parts before.

iii) Let the two letters be a and b. In exactly half the permutations, a will come before b. There are n! total permutations, therefore a can come before b in n!/2 ways. (this seems very obvious to me, so I'm not sure how much more mathematically it must be phrased. If you imagine that a comes before b in more than half the permutations, then since a and b can refer to any pair of letters, we can swap their assignments so that a now refers to the b letter and vice versa, but now you'd have a coming before b in less than half the permutations, which is a contradiction. Similarly if you start off by assuming that a comes before b in less than half the permutations.)

iv) Choose the 2 spots where we will place a and b in that order, then permute the others in the remaining spots: ⁿC₂ * (n-2)!. Now ⁿC₂ can be visualised by picking one item from n, then picking the second one from the remaning (n-1) items, then dividing by 2 to eliminate duplicates -- i.e. n(n-1)/2, which we know is T_n-1 (from arithmetic series, or i and ii), so an equivalent expression for iii) is T_n-1 * ⁿC₂ as required.

v) The two methods solve the same problem, so the expressions can be equated. Then multiply both sides by 2.

$vi) $(n+1)!= 2T_n(n-1)!$ \\ $n(n+1) = 2T_n$ \\ $T_n = \frac{n}{2}(n+1)$$

$vii) From i), $4\sum_{k=1}^{n}k^3$ = $n^2(n+1)^2$ = $n^4 + 2n^3 + n^2$ as req.$

$viii) From v), $n! = 2T_{n-1}(n-2)! = 2T_{n-1} \times 2T_{n-3} \times (n-4)! = 2T_{n-1} \times 2T_{n-3} \times ... \times 1 = 2^{\frac{n}{2}} T_{n-1}T_{n-3}T_{n-5}...$ Similarly, $(2m)! = 2^mT_{2m-1}T_{2m-3}T_{2m-5}...$ Now from ii), $T_n = \sqrt{\sum_{k=1}^{n}k^3$. Substituting this in, $(2m)!=2^m \prod_{k=1}^{m}\sqrt{\sum_{j=1}^{2k-1}j^3}$$

EDIT: aww Sy already answered it

it takes so long to type this shit up...

RealiseNothing · Dec 27, 2012

Re: HSC 2013 4U Marathon

seanieg89 said:
Oh right you said not real, yeah, then just take the complement.

Here's what I did which ended up wrong (I think?)

Since one root is $\alpha$ , the other must be $\frac{1}{\alpha}$

So we get:

$\alpha^2 + A\alpha + 1 = 0$

$\alpha^{-2} + A\alpha^{-1} + 1 = 0$

Using:

$z^n + z^-n = 2cos(n\theta)$ and adding the two equation together we obtain:

$2cos2\theta + 2Acos\theta + 1 = 0$

Using the identity:

$cos2\theta =2cos^2\theta -1$

We get a quadratic in $cos\theta$ which when solved using the quadratic formula gives:

$\theta = cos^{-1}(\frac{A}{2})$ or $\theta = cos^{-1}(\frac{-3A}{4})$

Now the first solution works for all $|A| < 2$ , but the second solution doesn't as $\frac{-3A}{4} > 1$ for certain values of A within the domain of $|A| < 2$

Hence solving for A:

$-1 \leq \frac{-3A}{4} \leq 1$

$\frac{4}{3} \geq A \geq \frac{-4}{3}$

Which is a different result to $|A| < 2$

(note I forgot to add the fact that $|\alpha| = 1$ )

What's wrong with this solution?

Sy123 · Dec 27, 2012

Re: HSC 2013 4U Marathon

RealiseNothing said:
Here's what I did which ended up wrong (I think?)

Since one root is $\alpha$ , the other must be $\frac{1}{\alpha}$

So we get:

$\alpha^2 + A\alpha + 1 = 0$

$\alpha^{-2} + A\alpha^{-1} + 1 = 0$

Using:

$z^n + z^-n = 2cos(n\theta)$ and adding the two equation together we obtain:

2cos2\theta + 2Acos\theta + 1 = 0

Using the identity:

$cos2\theta =2cos^2\theta -1$

We get a quadratic in $cos\theta$ which when solved using the quadratic formula gives:

$\theta = cos^{-1}(\frac{A}{2})$ or $\theta = cos^{-1}(\frac{-3A}{4})$

Now the first solution works for all $|A| < 2$ , but the second solution doesn't as $\frac{-3A}{4} > 1$ for certain values of A within the domain of $|A| < 2$

Hence solving for A:

$-1 \leq \frac{-3A}{4} \leq 1$

$\frac{4}{3} \geq A \geq \frac{-4}{3}$

Which is a different result to $|A| < 2$

(note I forgot to add the fact that $|\alpha| = 1$ )

What's wrong with this solution?

Bolded what was wrong, when you add the two equations, you get:

$\alpha^2+\alpha^{-2}+A(\alpha+1/\alpha)+2 = 0$

when you use that formula, you get:

$2\cos 2\theta+ A2\cos \theta +2 = 0 \\ \\ \cos 2 \theta + A \cos \theta + 1 =0$

Intuitive method though

seanieg89 · Dec 27, 2012

Re: HSC 2013 4U Marathon

$z^n+z^{-n}=2\cos(n\theta)$

? Assuming theta is meant to be the argument of z, this is not true.

RealiseNothing · Dec 27, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
Bolded what was wrong, when you add the two equations, you get:

$\alpha^2+\alpha^{-2}+A(\alpha+1/\alpha)+2 = 0$

when you use that formula, you get:

$2\cos 2\theta+ A2\cos \theta +2 = 0 \\ \\ \cos 2 \theta + A \cos \theta + 1 =0$

Intuitive method though

Woops that was a typo, hit 1 instead of 2. In my working I did have +2, and it worked out to be what I had.

seanieg89 · Dec 27, 2012

Re: HSC 2013 4U Marathon

Shortest way I can think of is:

1. Whatever we choose A to be the poly has two complex roots counting multiplicity.

2. One of them being non-real is the opposite of both of them being real.

3. Both of them are real <=> A=-(r+1/r) for some real r.

4. The range of -(r+1/r) is |A| >= 2, A real. (By AM-GM or calculus or w/e).

5. Take complements.

6. Profit.

RealiseNothing · Dec 27, 2012

Re: HSC 2013 4U Marathon

seanieg89 said:
$z^n+z^{-n}=2\cos(n\theta)$

? Assuming theta is meant to be the argument of z, this is not true.

Why isn't it true? Am I just being a massive retard or...

Sy123 · Dec 27, 2012

Re: HSC 2013 4U Marathon

RealiseNothing said:
Why isn't it true? Am I just being a massive retard or...

I'm not sure why its untrue either.

$|z|=1$

$z^n+z^{-n}=\cos n\theta + \sin n\theta + \cos (-n\theta) + \sin (-n\theta) = 2\cos n\theta$

seanieg89 · Dec 27, 2012

Re: HSC 2013 4U Marathon

How do we know the roots are of unit modulus?

RealiseNothing · Dec 27, 2012

Re: HSC 2013 4U Marathon

seanieg89 said:
How do we know the roots are of unit modulus?

I stated that they were lol (well it was a late edit lol).

HSC 2013 MX2 Marathon (archive) (1 Viewer)

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