HSC 2013 MX2 Marathon (archive) (1 Viewer)

Sy123 · Jan 2, 2013

Re: HSC 2013 4U Marathon

Carrotsticks said:
0

You are thinking too hard.

EDIT: Just saw your x =/= x, but I think that was the intended result.

I have seen the joke before and that is my response most of the time with the reasoning $x \neq $x$$ , just to be a smartass

RealiseNothing · Jan 2, 2013

Re: HSC 2013 4U Marathon

Let:

$S(n) = \sum_{k=2}^{n}(\sqrt{k}+\frac{1}{k})$

Show that:

$\frac{n}{2}+1 > \frac{S(n)}{n-1} > 1$

$for~n\geq 2$

Sy123 · Jan 2, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
Let:

$S(n) = \sum_{k=2}^{n}(\sqrt{k}+\frac{1}{k})$

Show that:

$\frac{n}{2}+1 > \frac{S(n)}{n-1} > 1$

$for~n\geq 2$

My way is pretty weird but it works, probably not the intended way but here it goes:

The diagram is needed in order to understand solution (or not)

Ok, first sketch the graphs:

$f(x)=\sqrt{x}+\frac{1}{x}$

$g(x)=x$

$h(x)=1$

Via calculus, I can show that $x > \sqrt{x}+\frac{1}{x}$ for $x \geq 2$
I can also show via calculus that $\sqrt{x}+ \frac{1}{x} > 1$

Now, draw the lines x=2, x=3, x=4, x=5 ... , x= n-1, x= n, x= n+1

We will now use the rectangles formed by:

(2, f(2) , (3, f(3) , (4, f(4) ... (n, f(n))

same for h, and same for g.
These rectangles are shown in the diagram, and it is obvious that, the sum of the areas of the rectangles for g > f > h

Hence we can construct the inequality:

$2+3+4+...+n > S(n) > 1+1+1+...+1$

$\frac{n}{2}(n+1) -1 > S(n) > n-1$

$\frac{(n+2)(n-1)}{2} > S(n) > n-1$

$\frac{n}{2} + 1 > \frac{S(n)}{n-1} > 1$

$\square$

RealiseNothing · Jan 2, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
My way is pretty weird but it works, probably not the intended way but here it goes:

The diagram is needed in order to understand solution (or not)

Ok, first sketch the graphs:

$f(x)=\sqrt{x}+\frac{1}{x}$

$g(x)=x$

$h(x)=1$

Via calculus, I can show that $x > \sqrt{x}+\frac{1}{x}$ for $x \geq 2$
I can also show via calculus that $\sqrt{x}+ \frac{1}{x} > 1$

Now, draw the lines x=2, x=3, x=4, x=5 ... , x= n-1, x= n, x= n+1

We will now use the rectangles formed by:

(2, f(2) , (3, f(3) , (4, f(4) ... (n, f(n))

same for h, and same for g.
These rectangles are shown in the diagram, and it is obvious that, the sum of the areas of the rectangles for g > f > h

Hence we can construct the inequality:

$2+3+4+...+n > S(n) > 1+1+1+...+1$

$\frac{n}{2}(n+1) -1 > S(n) > n-1$

$\frac{(n+2)(n-1)}{2} > S(n) > n-1$

$\frac{n}{2} + 1 > \frac{S(n)}{n-1} > 1$

$\square$

What I did was prove that:

$x>\sqrt{x}+\frac{1}{x}>1~for~x\geq2$

Then:

$2>\sqrt{2}+\frac{1}{2}>1$

$3>\sqrt{3}+\frac{1}{3}>1$

...

$n>\sqrt{n}+\frac{1}{n}>1$

Add them all up, then simplify to get the required result.

RealiseNothing · Jan 2, 2013

Re: HSC 2013 4U Marathon

courtesy of this thread:

asianese · Jan 2, 2013

Re: HSC 2013 4U Marathon

You guys need to take a break! But damn, dat maffs.

Sy123 · Jan 2, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
courtesy of this thread:

Haha nice

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

$$Prove that the product of n consecutive integers is divisible by $ \ \ n!$

RealiseNothing · Jan 2, 2013

Re: HSC 2013 4U Marathon

asianese said:
You guys need to take a break! But damn, dat maffs.

dats only ~50% of it over the last 2-3 months.

RealiseNothing · Jan 2, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$$Prove that the product of n consecutive integers is divisible by $ \ \ n!$

lolwat

aren't they the same thing.

RealiseNothing · Jan 2, 2013

Re: HSC 2013 4U Marathon

Also I'm not sure if this question is easy/hard/has an answer, but it just popped into my mind so let's see if we can find an answer:

Find a closed expression for:

$\sum_{k=2}^{n} \frac{t_{k-1}}{t_k}$

$t_k$ is the k'th triangular number etc etc we all know what it denotes by now lol.

Sy123 · Jan 2, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
lolwat

aren't they the same thing.

Not necessarily starting from 1, 2, 3
Generally from r, r+1 etc. But its not worth answering lol its very simple, I just thought it was a cool result.

RealiseNothing said:
Also I'm not sure if this question is easy/hard/has an answer, but it just popped into my mind so let's see if we can find an answer:

Find a closed expression for:

$\sum_{k=2}^{n} \frac{t_{k-1}}{t_k}$

$t_k$ is the k'th triangular number etc etc we all know what it denotes by now lol.

I was able to arrive at:

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

And I'm not sure if there exists a closed form for the harmonic numbers.

RealiseNothing · Jan 2, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Not necessarily starting from 1, 2, 3
Generally from r, r+1 etc. But its not worth answering lol its very simple, I just thought it was a cool result.

Oh right I see what you mean now. Well it's pretty simple, so I might leave it for some one else to try since everyone seems scared off by some of these questions lol.

twinklegal19 · Jan 2, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
Oh right I see what you mean now. Well it's pretty simple, so I might leave it for some one else to try since everyone seems scared off by some of these questions lol.

you don't say

this thread is now called the "realisenothing and sy123 marathon thread" heh

seanieg89 · Jan 2, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Not necessarily starting from 1, 2, 3
Generally from r, r+1 etc. But its not worth answering lol its very simple, I just thought it was a cool result.

I was able to arrive at:

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

And I'm not sure if there exists a closed form for the harmonic numbers.

There is not one in terms of elementary functions.

seanieg89 · Jan 2, 2013

Re: HSC 2013 4U Marathon

Here is a polynomial question that is a past IMO question. Don't let that scare you though as it is from a while ago and is certainly do-able by a good mx2 student.

Prove that the solution to the inequality:

$\sum_{k=1}^{70}\frac{k}{x-k}\geq 5/4$

is the union of disjoint intervals, the sum of whose lengths is 1988.

My hint is to think of to first think of what the function on the left looks like intuitively before going about rigorously proving things.

tywebb · Jan 2, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
How would I prove that at m infinite the inequality becomes equality?

Well the 2002 HSC question 8a can be extended to solve the basel problem:

I also have a solution to seanieg89's imo question (1988 IMO Q4 actually):

tywebb · Jan 2, 2013

Re: HSC 2013 4U Marathon

Furthermore, in 2002 Bill Pender declared that the basel problem is beyond the 4 unit syllabus:

http://members.optusnet.com.au/limkw/SGS_harder3U_2002.zip (see annotation in the middle of page 8)

But the extension of the 2002 question shows that it isn't.

Also see what happened in the 2010 HSC exam? A full proof (without the need to extend the question!)

There is a danger when teachers say something is beyond the syllabus.

There was a time when teachers would say the proofs of irrationality of e and pi were beyond the syllabus. Then the proofs appeared in HSC exams!

Sy123 · Jan 2, 2013

Re: HSC 2013 4U Marathon

tywebb said:
Well the 2002 HSC question 8a can be extended to solve the basel problem:

I also have a solution to seanieg89's imo question (1988 IMO Q4 actually):

Yeah I know that about the Basel Problem, someone helped me prove it as an extension (but I didn't post about it)

- And I would appreciate it if you didn't post the solution to the question straight away, maybe after 4-5 hours or something.

Fus Ro Dah · Jan 2, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$$Prove that the product of n consecutive integers is divisible by $ \ \ n!$

$\\ k(k+1)(k+2)...(k+n) = \frac{(k+n)!}{(k-1)!}$ , where k>1 and we want to prove that $\frac{(k+n)!}{(k-1)!n!}$ is an integer. Consider $\binom{k+n}{k-1}$ , which we know is an integer K. $\binom{k+n}{k-1} = \frac{(k+n)!}{(k-1)!(n+1)!}=K$ so $\frac{(k+n)!}{(k-1)!n!}=K(n+1) \in \mathbb{N}\qquad \square$

Sy123 · Jan 2, 2013

Re: HSC 2013 4U Marathon

Fus Ro Dah said:
$\\ k(k+1)(k+2)...(k+n) = \frac{(k+n)!}{(k-1)!}$ , where k>1 and we want to prove that $\frac{(k+n)!}{(k-1)!n!}$ is an integer. Consider $\binom{k+n}{k-1}$ , which we know is an integer K. $\binom{k+n}{k-1} = \frac{(k+n)!}{(k-1)!(n+1)!}=K$ so $\frac{(k+n)!}{(k-1)!n!}=K(n+1) \in \mathbb{N}\qquad \square$

Your first product is the product of n+1 consecutive integers, then you go on and prove that the product of n+1 consecutive integers is divisible by n!.
Though the idea is correct

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