HSC 2012 MX1 Marathon #2 (archive) (2 Viewers)

Sy123 · Oct 21, 2012

Re: HSC 2012 Marathon

Just going to check my step 1 here to see if I actually understand the question, Ill do the rest tomorrow too tired right now lol

$$Define the polynomial $ u=a_0 x^n+a_1 x^{n-1}+...+a_n$

$$Step 1 $: n=0 \\ \\ u=a_n \\ \therefore u'=0 \ \ $for all n in sequence$ \\ \\ \therefore u'=(0,0,0...)$

Correct or way off?

seanieg89 · Oct 21, 2012

Re: HSC 2012 Marathon

Sy123 said:
Just going to check my step 1 here to see if I actually understand the question, Ill do the rest tomorrow too tired right now lol

$$Define the polynomial $ u=a_0 x^n+a_1 x^{n-1}+...+a_n$

$$Step 1 $: n=0 \\ \\ u=a_n \\ \therefore u'=0 \ \ $for all n in sequence$ \\ \\ \therefore u'=(0,0,0...)$

Correct or way off?

Yep, notation is a little funky but the idea is write, a constant sequence will get killed straight away by the operator '. Compare this to the behaviour of a constant function being differentiated

.

seanieg89 · Oct 21, 2012

Re: HSC 2012 Marathon

Sy123 said:
Just going to check my step 1 here to see if I actually understand the question, Ill do the rest tomorrow too tired right now lol

$$Define the polynomial $ u=a_0 x^n+a_1 x^{n-1}+...+a_n$

$$Step 1 $: n=0 \\ \\ u=a_n \\ \therefore u'=0 \ \ $for all n in sequence$ \\ \\ \therefore u'=(0,0,0...)$

Correct or way off?

Yep, notation is a little funky but the idea is right, a constant sequence will get killed straight away by the operator '. Compare this to the behaviour of a constant function being differentiated

.

Sy123 · Oct 21, 2012

Re: HSC 2012 Marathon

seanieg89 said:
Yep, notation is a little funky but the idea is write, a constant sequence will get killed straight away by the operator '. Compare this to the behaviour of a constant function being differentiated .

Alright nice I will see if I can complete this tomorrow

(inb4 detroyed)

seanieg89 · Oct 21, 2012

Re: HSC 2012 Marathon

Sy123 said:
Alright nice I will see if I can complete this tomorrow

(inb4 detroyed)

Honestly not that difficult, maybe think about how you would try to prove the analogue of this theorem for functions: f is killed by taking finitely many derivatives if and only if f is a polynomial

. Because differentiation is more familiar this should be quite easy, then try to mimic this approach for the less familiar operation going on here.

Sy123 · Oct 22, 2012

Re: HSC 2012 Marathon

Progress

Thats just my progress on Step 3, but I keep running into that barrier, I want to be able to prove that
$v^{(k)}=0 \ \ \ OR \ \ \ C \ \ \ $(constant C)$$

But I cant because I just end up with:

$v^{(k)}=\sum_{i=0}^{k-1}u^{(k-1)}$

Am I missing something here?

seanieg89 · Oct 22, 2012

Re: HSC 2012 Marathon

Hint: polynomial degree. Gtg now but will look closer at your work later if you don't get it yet.

lgnorance · Oct 28, 2012

Re: HSC 2012 Marathon

eskimogenius said:
A point P is marked inside a regular hexagon ABCDEF so that angleBAP=angleDCP=50deg. Find angle ABP.

GL.

PS. Answer is 80deg.

Fwark, this is a good question.

Just did it using complex numbers.

Sy123 · Oct 28, 2012

Re: HSC 2012 Marathon

Sy123 said:
Another question I made up:

$$A projectile is fired from the origin with velocity V, under the conditions of gravity g and no air resistance. The projectile is fired such that the maximum height is h, and the horizontal distance from the origin to the maximum height is h.$ \\ \\ $Show that $ V=\sqrt{\frac{5gh}{2}}$

$$ A projectile is fired from the origin with velocity V, under the conditions of gravity g and no air resistance. The projectile is fired such that the maximum height is half the range of the projectile. Taking maximum height as h. A particle is starting Simple Harmonic Motion where it's centre of motion is where the projectile hits the ground. It is initially at the centre of motion, with period of $ 2\pi $ . The projectile hits the particle while it is at its centre of motion, right when the particle has completed a Period.$ \\ \\ $Show that $ h=\frac{g \pi^2}{2}$

Use the results from the previous question

HSC 2012 MX1 Marathon #2 (archive) (2 Viewers)

Sy123

This too shall pass

seanieg89

Well-Known Member

seanieg89

Well-Known Member

Sy123

This too shall pass

seanieg89

Well-Known Member

Sy123

This too shall pass

seanieg89

Well-Known Member

lgnorance

Member

Sy123

This too shall pass

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