HSC 2013 MX2 Marathon (archive) (6 Viewers)

Sy123 · Aug 11, 2013

Re: HSC 2013 4U Marathon

$$Prove for some integers$ \ m, n \ m>n$

$\binom{m}{0} - \binom{m}{1} + \binom{m}{2} - \dots + (-1)^n \binom{m}{n} = (-1)^n \binom{m-1}{n}$

(above results and the one in the 3U thread are verified since they come from the book, Summation of Series - Jolley)

Sy123 · Aug 11, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$Using the result or otherwise$

$\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 \ $does equality hold?$

First transforming the wanted inequality so its easier to solve:

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

$\left ( \frac{n}{n+1} \right )^n \geq \frac{1}{n+1} \ \ \ (*)$

This is what we need to prove, transforming the above AM-GM inequality:

$\left( \frac{a_1 + a_2 + \dots + a_n}{n} \right )^n \geq a_1 a_2 \dots a_n$

So we simply need to select numbers a_1, a_2, ... , a_n so that we get (*)

For simplicity sake, let $a_2 = a_3 = a_3 = .... = a_n = 1$

Now we find a_1 so that we yield (*)

$\frac{n}{n+1} = a_1 + (n-1)$

$a_1 = \frac{1}{n+1}$

Yielding $(n/(n+1))^n > 1/(n+1)$ yielding the wanted inequality.

Trebla · Aug 11, 2013

Re: HSC 2013 4U Marathon

I like how you answer your own questions

Sy123 · Aug 11, 2013

Re: HSC 2013 4U Marathon

Trebla said:
I like how you answer your own questions

Because some people want to know solutions of these questions?

In the past people have asked for solutions to a lot of these questions, and I thought this thread would be of little benefit to people if they don't know the solutions to these questions.

asianese · Aug 11, 2013

Re: HSC 2013 4U Marathon

Trebla said:
I like how you answer your own questions

^.

Sy123 · Aug 11, 2013

Re: HSC 2013 4U Marathon

Sure I'll just leave all questions here that are old to be unanswered rather than post the solutions for other people to benefit
No need to be sarcastic about it =)

Makematics · Aug 11, 2013

Re: HSC 2013 4U Marathon

I appreciate da solutions!!

Fred2013 · Aug 13, 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}$

Post solutions that does not involve divergence/convergence? Thx.

Sy123 · Aug 13, 2013

Re: HSC 2013 4U Marathon

Fred2013 said:
Post solutions that does not involve divergence/convergence? Thx.

I'm afraid you would have to prove the sum converges in order to find it, however I doubt that the Sydney grammar teachers who made this question expected students to do so.

$u_{n+1} = u_n + u_n^2 \ \Rightarrow \ \frac{u_{n}}{u_{n+1}} = \frac{1}{1+u_n}$

$\frac{u_n}{u_{n+1}} = \frac{u_n^2}{u_{n+1}u_n} = \frac{u_{n+1} - u_{n}}{u_{n+1} u_n} = \frac{1}{u_n} - \frac{1}{u_{n+1}}$

$\therefore \ \frac{1}{1+u_n} = \frac{1}{u_n} - \frac{1}{u_{n+1}}$

$\sum_{n=1}^N \frac{1}{1+u_n} = \sum_{n=1}^N \left( \frac{1}{u_n} - \frac{1}{u_{n+1}} \right ) = \frac{1}{u_1} - \frac{1}{u_{N+1}}$

-> A Sydney Grammar teacher may be satisfied with simply the statement that u_n is increasing therefore 1/u_n converges to zero (I could be wrong, I don't know the marking criteria, however that is my guess), however a rigorous proof of the limit is given below:

$$Now, we must find$ \ \lim_{N \to \infty} \frac{1}{u_{N+1}}$

$Using the 'hint' I gave in the next post$

$\lim_{n \to \infty} \left|\frac{u_{n+1}}{u_n} \right | = 1 + \lim_{n \to \infty} u_n$

$$Since$ \ u_n^2 > 0 \ \therefore u_{n+1} > u_n$

$\therefore \lim_{n \to \infty} \frac{u_{n+1}}{u_n} = 1 + \lim_{n \to \infty} u_n > 1 + u_1 > 1$

$\therefore \ u_n \ $diverges$ \ \ \ $hence$ \ \frac{1}{u_n} \ \ $converges to zero$$

$\therefore \ \lim_{N \to \infty} \frac{1}{u_{N+1}} = 0$

$\therefore \sum_{n=1}^{\infty} \frac{1}{1+u_n} = \frac{1}{u_1} = 3$

seanieg89 · Aug 13, 2013

Re: HSC 2013 4U Marathon

A point P is chosen inside the unit square at random. What is the probability that P is closer to the centre of the square than any side thereof?

Bonus: Solve this for a cube, replacing the word "side" with "face".

Fred2013 · Aug 13, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
I'm afraid you would have to prove the sum converges in order to find it, however I doubt that the Sydney grammar teachers who made this question expected students to do so.

$u_{n+1} = u_n + u_n^2 \ \Rightarrow \ \frac{u_{n}}{u_{n+1}} = \frac{1}{1+u_n}$

$\frac{u_n}{u_{n+1}} = \frac{u_n^2}{u_{n+1}u_n} = \frac{u_{n+1} - u_{n}}{u_{n+1} u_n} = \frac{1}{u_n} - \frac{1}{u_{n+1}}$

$\therefore \ \frac{1}{1+u_n} = \frac{1}{u_n} - \frac{1}{u_{n+1}}$

$\sum_{n=1}^N \frac{1}{1+u_n} = \sum_{n=1}^N \left( \frac{1}{u_n} - \frac{1}{u_{n+1}} \right ) = \frac{1}{u_1} - \frac{1}{u_{N+1}}$

-> A Sydney Grammar teacher may be satisfied with simply the statement that u_n is increasing therefore 1/u_n converges to zero (I could be wrong, I don't know the marking criteria, however that is my guess), however a rigorous proof of the limit is given below:

$$Now, we must find$ \ \lim_{N \to \infty} \frac{1}{u_{N+1}}$

$Using the 'hint' I gave in the next post$

$\lim_{n \to \infty} \left|\frac{u_{n+1}}{u_n} \right | = 1 + \lim_{n \to \infty} u_n$

$$Since$ \ u_n^2 > 0 \ \therefore u_{n+1} > u_n$

$\therefore \lim_{n \to \infty} \frac{u_{n+1}}{u_n} = 1 + \lim_{n \to \infty} u_n > 1 + u_1 > 1$

$\therefore \ u_n \ $diverges$ \ \ \ $hence$ \ \frac{1}{u_n} \ \ $converges to zero$$

$\therefore \ \lim_{N \to \infty} \frac{1}{u_{N+1}} = 0$

$\therefore \sum_{n=1}^{\infty} \frac{1}{1+u_n} = \frac{1}{u_1} = 3$

Thx Sy123.

RealiseNothing · Aug 13, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
A point P is chosen inside the unit square at random. What is the probability that P is closer to the centre of the square than any side thereof?

Bonus: Solve this for a cube, replacing the word "side" with "face".

On first thoughts, you would just construct a circle within the square which has a radius a quarter that of the side length of the square. Then do the area of the circle divided by the area of the square. Similarly for the bonus, you construct a sphere and do the volume of the sphere divided by the volume of the cube.

Sy123 · Aug 13, 2013

Re: HSC 2013 4U Marathon

I got, $\frac{13\sqrt{2}}{6} - \frac{7}{3} = 0.73079....$

I want to figure out the bonus before I post my solution for this one.

seanieg89 · Aug 13, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
On first thoughts, you would just construct a circle within the square which has a radius a quarter that of the side length of the square. Then do the area of the circle divided by the area of the square. Similarly for the bonus, you construct a sphere and do the volume of the sphere divided by the volume of the cube.

Nah, the regions won't be circles and spheres.

seanieg89 · Aug 14, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
I got, $\frac{13\sqrt{2}}{6} - \frac{7}{3} = 0.73079....$

I want to figure out the bonus before I post my solution for this one.

Will check now...I didn't actually do the calculation.

Sy123 · Aug 14, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
Will check now...I didn't actually do the calculation.

Its just 4 times the area bounded by the region (in the Cartesian plane where the square's co-ordinates are (0,0) (1,0) (1,1) and (0,1) ) , $y >x, \ \ y<(1-x) , (y-1/2)^2 > x-\frac{1}{4}$ ??

seanieg89 · Aug 14, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Its just 4 times the area bounded by the region (in the Cartesian plane where the square's co-ordinates are (0,0) (1,0) (1,1) and (0,1) ) , $y >x, \ \ y<(1-x) , (y-1/2)^2 > x-\frac{1}{4}$ ??

yeah that sounds about right

seanieg89 · Aug 14, 2013

Re: HSC 2013 4U Marathon

I got:

(4*sqrt(2)-5)/3 and am pretty confident about it...

Sy123 · Aug 14, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
I got:

(4*sqrt(2)-5)/3 and am pretty confident about it...

Well firstly I found the complementary probability, the probability that its closer to the face instead :/
And even then I still get the wrong answer because I don't know to find the area of a triangle properly and used the wrong base length.

I think it should be, and I'm not bothered to do the calculation again

$1 - 4 \left(\left(1-\frac{1}{\sqrt{2}} \right )^2 + \int_{1-1/\sqrt{2}}^{1/\sqrt{2}} \frac{1}{4} + (y- \frac{1}{2} )^2 \ dy \right)$

The above is supposed to be 4 times the area of the region bounded by the cartesian equations:

$y>x$
$y<1-x$

$(y-1/2)^2 < x-1/4$

dem mistakes

seanieg89 · Aug 14, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Well firstly I found the complementary probability, the probability that its closer to the face instead :/
And even then I still get the wrong answer because I don't know to find the area of a triangle properly and used the wrong base length.

I think it should be, and I'm not bothered to do the calculation again

$1 - 4 \left(\left(1-\frac{1}{\sqrt{2}} \right )^2 + \int_{1-1/\sqrt{2}}^{1/\sqrt{2}} \frac{1}{4} + (y- \frac{1}{2} )^2 \ dy \right)$

The above is supposed to be 4 times the area of the region bounded by the cartesian equations:

$y>x$
$y<1-x$

$(y-1/2)^2 < x-1/4$

dem mistakes

Yeah your equations are pretty similar to mine. Did mine quite carefully so am pretty sure its right. Will try the 3d one in a sec.

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