HSC 2013 MX2 Marathon (archive) (1 Viewer)

nrlwinner · Aug 5, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Yeah I did it again and you are correct

Ah yes reading the solution again it is incorrect, what I had in mind was

$(n- \frac{1}{2} \right )^n \geq (1-x_1)(1-x_2) \dots (1-x_n)$ ... (1)

$(n+1/2)^n \geq (1+x_1)(1+x_2) \dots (1+x_n)$ .... (2)

This is done by subbing in the AM-GM inequality, all substitutions are positive since x_k <= 1/2

Dividing (2) by (1), which we can do since all things are positive.

We get $\frac{(n+1/2)^n}{(n-1/2)^n} \geq 1/P$ (P is what we want)

Then flip everything, when we do that the inequality sign switches, yeilding P > (that)

But I was skimming through her solution and wasn't really paying attention, which is why I couldn't see that.

=====

I'm making way too many mistakes

(there is no mistake in this one)

$$For some real numbers$ \ x >y > 0 \ $and integer$ \ n$

$Prove$

$(n+1)y^n < \frac{x^{n+1} - y^{n+1}}{x-y} < (n+1) x^n$

Here's my go at the question. I'll just do the left inequality and presume the right inequality is similar.

$\frac{x^{n+1}-y^{n+1}}{x-y}=x^n+x^{n+1}y+...+xy^{n-1}+y^n$

$=y^n(1+\frac{x}{y}+\frac{x^2}{y^2}+...+\frac{x^n}{y^n})>y^n(n+1)$

$As \left ( \frac{x}{y} \right )^k>1$

Sy123 · Aug 5, 2013

Re: HSC 2013 4U Marathon

nrlwinner said:
Here's my go at the question. I'll just do the left inequality and presume the right inequality is similar.

$\frac{x^{n+1}-y^{n+1}}{x-y}=x^n+x^{n+1}y+...+xy^{n-1}+y^n$

$=y^n(1+\frac{x}{y}+\frac{x^2}{y^2}+...+\frac{x^n}{y^n})>y^n(n+1)$

$As \left ( \frac{x}{y} \right )^k>1$

Yep, the right inequality is kinda similar.

========

$$Suppose the polynomial$ \ P(x) = x^3+px^2+px+ 1 \ $has root$ \ - 1$

$$And complex root$ \ \alpha$

$Show that$

$\alpha = \frac{1-p}{2} + \frac{i}{\sqrt{2}} \sqrt{1+2p-p^2}$

RealiseNothing · Aug 5, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Yep, the right inequality is kinda similar.

========

$$Suppose the polynomial$ \ P(x) = x^3+px^2+px+ 1 \ $has root$ \ - 1$

$$And complex root$ \ \alpha$

$Show that$

$\alpha = \frac{1-p}{2} + \frac{i}{\sqrt{2}} \sqrt{1+2p-p^2}$

$P(x) = (x+1)(x^2+(p-1)x+1)$

$\alpha$ is a root of the quadratic, so we just use the quadratic formula to arrive at the answer

Sy123 · Aug 5, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
$P(x) = (x+1)(x^2+(p-1)x+1)$

$\alpha$ is a root of the quadratic, so we just use the quadratic formula to arrive at the answer

I'm running out of good questions :/

RealiseNothing · Aug 5, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
I'm running out of good questions :/

Did you have a different way of solving it?

seanieg89 · Aug 5, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
$1. For positive integers $d$ and $m$, evaluate:\\ $f(m,d)=\frac{1}{d}\sum_{j=1}^{d} \cos\left(\frac{2\pi jm}{d}\right).$\\ \\2. Use the above to find a function $g(n)$ defined in terms of MX2 functions such that $g(n)=1$ if $n$ is prime and $g(n)=0$ if $n$ is composite.\\ \\ 3. Use the function $g$ to construct a function $p(n)$ whose value at $n$ is the $n$-th prime number.$

$As Sy123 proved, $f(m,d)$ is $1$ if $d$ divides $m$ and $0$ otherwise.\\ \\ By the definition of a prime as a positive integer greater than $1$ that is indivisible by all smaller positive integers that are greater that $1$, we get:\\ \\ $g(n)=\prod_{d=2}^{n-1}(1-f(n,d))$ \\ \\ for positive integers greater than $1$. Note that the empty product is $1$ by convention.\\ \\ Hence the number of primes smaller than or equal to $n$ is given by:\\ $\pi(n)=\sum_{n=2}^n g(n)$.\\ \\ A number $p$ is the $n$-th prime if and only if:\\ \\ $\pi(p)=n \textrm{ and }g(p)=1$. \\ \\Using that $h(m)=\int_{-\pi}^\pi \cos(mx)\, dx$ is $1$ if $m=0$ and $0$ otherwise, one MX2 formula for the n-th prime is hence given by:\\ \\ $\sum_{p=2}^\infty p\cdot h(\pi(p)-n)h(g(p)-1)$\\ \\(Note that the sum is finite by the infinitude of primes.)$

If one was very bored, they could expand this final expression out in terms of the trigonometrically defined f.

(Many such formulae for primes exist, but the elementary ones are all basically algorithms for checking primality in disguise. The algorithm in this case is the process of checking divisibility by all smaller natural numbers.)

RealiseNothing · Aug 5, 2013

Re: HSC 2013 4U Marathon

If they have formulae for primes, then what is preventing them from finding very very very large prime numbers larger than the largest known prime? Is it how advanced technology and computational programs are or what?

Sy123 · Aug 5, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
Did you have a different way of solving it?

I modified a hsc question, it first asked to prove |a| = 1 and Re(a) = (1-p)/2, So I decided to omit this and and ask for the whole number since it can be found with those properties

seanieg89 said:
$As Sy123 proved, $f(m,d)$ is $1$ if $d$ divides $m$ and $0$ otherwise.\\ \\ By the definition of a prime as a positive integer greater than $1$ that is indivisible by all smaller positive integers that are greater that $1$, we get:\\ \\ $g(n)=\prod_{d=2}^{n-1}(1-f(n,d))$ \\ \\ for positive integers greater than $1$. Note that the empty product is $1$ by convention.\\ \\ Hence the number of primes smaller than or equal to $n$ is given by:\\ $\pi(n)=\sum_{n=2}^n g(n)$.\\ \\ A number $p$ is the $n$-th prime if and only if:\\ \\ $\pi(p)=n \textrm{ and }g(p)=1$. \\ \\Using that $h(m)=\int_{-\pi}^\pi \cos(mx)\, dx$ is $1$ if $m=0$ and $0$ otherwise, one MX2 formula for the n-th prime is hence given by:\\ \\ $\sum_{p=2}^\infty p\cdot h(\pi(p)-n)h(g(p)-1)$\\ \\(Note that the sum is finite by the infinitude of primes.)$

If one was very bored, they could expand this final expression out in terms of the trigonometrically defined f.

(Many such formulae for primes exist, but the elementary ones are all basically algorithms for checking primality in disguise. The algorithm in this case is the process of checking divisibility by all smaller natural numbers.)

I see, thanks for taking the time to write your solution out.

============

$$i) Prove$ \ 2k+1 \geq 2\sqrt{k(k+1)}$

$ii) Hence prove that$ \ \ \sum_{k=1}^n \frac{1}{\sqrt{k}} \ $increases without bound if$ \ n \ $increases without bound$

RealiseNothing · Aug 5, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
I modified a hsc question, it first asked to prove |a| = 1 and Re(a) = (1-p)/2, So I decided to omit this and and ask for the whole number since it can be found with those properties

I see, thanks for taking the time to write your solution out.

============

$$i) Prove$ \ 2k+1 \geq 2\sqrt{k(k+1)}$

$ii) Hence prove that$ \ \ \sum_{k=1}^n \frac{1}{\sqrt{k}} \ $increases without bound if$ \ n \ $increases without bound$

The second part is easily done without the first as $\sum_{k=1}^{\infty} \frac{1}{k} \to \infty$ and the summation in part (ii) contains all of the terms in this sum.

seanieg89 · Aug 5, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
If they have formulae for primes, then what is preventing them from finding very very very large prime numbers larger than the largest known prime? Is it how advanced technology and computational programs are or what?

Computational complexity. This algorithm is horrendously inefficient.

More efficient algorithms are exactly how they found the current largest known primes.

seanieg89 · Aug 7, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Yep nice work

===========

$You may use the following identity$

$\frac{a_1 + a_2 + a_3 + \dots + a_n}{n} \geq \sqrt[n]{a_1 a_2 a_3 \dots a_n}$

$$The sum of the positive real numbers$ \ x_1, x_2, x_3, \dots, x_n \ $is$ \ 1/2$

$Prove that$

$\frac{1-x_1}{1+x_1} \cdot \frac{1-x_2}{1+x_2} \cdot \frac{1-x_3}{1+x_3} \dots \frac{1-x_n}{1+x_n} \geq \frac{1}{3}$

$We might as well assume instead that the $x_j$ are non-negative, as this will imply our desired result anyway. \\ For $n=1$, the assertion is trivial. We proceed by induction.\\ \\ Let $P_k(x_1,\ldots,x_k)=\prod_{j=1}^k \frac{1-x_j}{1+x_j}$ for $k$-tuples with each $x_j\in [0,1/2]$ with sum $1/2$.\\ \\ We have: \\ \\ $\frac{P_{k+1}(x_1,\ldots,x_{k+1})}{P_k(x_1,\ldots,x_k+x_{k+1})}=\left(\frac{(1-x_k)(1-x_{k+1})}{(1+x_k)(1+x_{k+1})}\right)\div \left(\frac{1-x_k-x_{k+1}}{1+x_k+x_{k+1}}\right)\geq 1 \quad (*).$\\ \\ Whence by induction we have $P_n(x_1,\ldots,x_n)\geq 1/3$ for all $n$ and all non-negative $x_j$ with sum $1/2$.\\ \\ $(*)$ is a consequence of the fact that if $0\leq a \leq b\neq 0$ and $t\geq 0$ then $\frac{a}{b}\leq\frac{a+t}{b+t}$ which is itself clear by simply expanding.$

Galactic Jam · Aug 7, 2013

Re: HSC 2013 4U Marathon

I only just found this thread, these questions are so scary :S
Hoping I can at least figure some of these out.

Sy123 · Aug 8, 2013

Re: HSC 2013 4U Marathon

$Using the result or otherwise$

$\frac{a_1 + a_2 + \dots + a_n}{n} \geq \sqrt[n]{a_1 a_2 a_3 \dots a_n} \ \ $for positive$ \ a_k \ k=1,2, \dots , n$

$$Prove that for all positive integers$ \ n$

$n^n \geq (n+1)^{n-1}$

$for what value of$ \ n \ $does equality hold?$

Sy123 · Aug 9, 2013

Re: HSC 2013 4U Marathon

A good Sydney grammar question

$u_{n+1} = u_n + u_n^2, \ u_1 = \frac{1}{3}$

$$Find$ \ \sum_{n=1}^{\infty} \frac{1}{1+u_n}$

seanieg89 · Aug 9, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
A good Sydney grammar question

$u_{n+1} = u_n + u_n^2, \ u_1 = \frac{1}{3}$

$$Find$ \ \sum_{n=1}^{\infty} \frac{1}{1+u_n}$

Was this the only part of the question? If so it will be interesting seeing how people justify that u_n tends to infinity. The easiest way I can see uses monotone convergence.

Sy123 · Aug 9, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
Was this the only part of the question? If so it will be interesting seeing how people justify that u_n tends to infinity. The easiest way I can see uses monotone convergence.

Yeah this was it, I was able to show that $u_{n+1} > u_n$ and $u_n > 0$ , is this enough to justify convergence given the of the final step in this problem? (which I don't want to type out here just yet)
I guess this is probably a form of monotone convegence

seanieg89 · Aug 9, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Yeah this was it, I was able to show that $u_{n+1} > u_n$ and $u_n > 0$ , is this enough to justify convergence given the of the final step in this problem? (which I don't want to type out here just yet)
I guess this is probably a form of monotone convegence

Nope, sequences satisfying the properties you provided can either diverge or converge. An example of a convergent one being 1-1/n.

seanieg89 · Aug 9, 2013

Re: HSC 2013 4U Marathon

Note: I am sure it can be done without monotone convergence, I just think that it is a stiff ask for nearly any mx2 student.

Sy123 · Aug 9, 2013

Re: HSC 2013 4U Marathon

To be fair, I'll put this first part:

$$A series$ \ a_1, a_2, a_3, \dots \ \ $diverges if$ \ \lim_{n \to \infty} \left | \frac{a_{n+1}}{a_n} \right | > 1$
$Hence or otherwise$

Sy123 said:
A good Sydney grammar question

$u_{n+1} = u_n + u_n^2, \ u_1 = \frac{1}{3}$

$$Find$ \ \sum_{n=1}^{\infty} \frac{1}{1+u_n}$

Sy123 · Aug 11, 2013

Re: HSC 2013 4U Marathon

$$Prove for some integer$ \ n$

$\binom{n}{1} - \frac{1}{2} \binom{n}{2} + \frac{1}{3} \binom{n}{3} - \dots + \frac{(-1)^{n-1}}{n} \binom{n}{n} = \sum_{k=1}^n \frac{1}{k}$

HSC 2013 MX2 Marathon (archive) (1 Viewer)

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