HSC 2013 MX2 Marathon (archive) (3 Viewers)

HeroicPandas · Aug 29, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
Since it is for $a>0$ , then by making that transformation, when $a>1$ we get $\frac{1}{a} - 1 < 0$

thanks! i now understand

Sy123 · Aug 29, 2013

Re: HSC 2013 4U Marathon

$$Carefully prove for positive real$ \ \ x \ \ $and integer$ \ n$

$\lim_{n \to \infty} \frac{x^n}{n!} = 0$

AnimeX · Aug 29, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$$i) Prove$ \ \ \frac{\sin x}{x} = \cos \left( \frac{x}{2} \right ) \cdot \cos \left( \frac{x}{4} \right) \cdot \cos \left( \frac{x}{8} \right ) \cdot \cos \left( \frac{x}{16} \right ) \cdot \dots$

$ii) Hence prove that$

$\pi = 2\cdot \frac{2}{\sqrt{2}} \cdot \frac{2}{\sqrt{2+\sqrt{2}}} \cdot \frac{2}{\sqrt{2 + \sqrt{2+\sqrt{2}}}} \cdot \frac{2}{\sqrt{2+ \sqrt{2 + \sqrt{2+ \sqrt{2}}}}} \cdot \dots$

haha any hints for this one? something to do with small angles?

Sy123 · Aug 29, 2013

Re: HSC 2013 4U Marathon

AnimeX said:
haha any hints for this one? something to do with small angles?

If by small angles you mean the limit

$\lim_{a\to 0} \frac{\sin (a)}{a} = 1$ then yes you need to use this limit

asianese · Aug 30, 2013

Re: HSC 2013 4U Marathon

Think of why the square root appears.

seanieg89 · Aug 30, 2013

Re: HSC 2013 4U Marathon

And the fact that you are taking the sine/cosine of a geometric sequence of angles.

Sy123 · Aug 31, 2013

Re: HSC 2013 4U Marathon

$$Prove for real$ \ x,y$

$x^4+y^4+8 \geq 8xy$

alexandred · Sep 1, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$$Prove for real$ \ x,y$

$x^4+y^4+8 \geq 8xy$

(1) (x^2 - y^2)^2 >= 0
(2) 2(xy - 2)^2 >= 0
Add (1) and (2)

Sy123 · Sep 1, 2013

Re: HSC 2013 4U Marathon

$$A polynomial$ \ \ P(x) \ \ $has roots$ \ \ a_1, a_2, a_3, \dots , a_n$

$$Find the polynomial with roots$ \ \ \pm a_1, \pm a_2, \pm a_3 , \dots , \pm a_n$

HeroicPandas · Sep 1, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$$A polynomial$ \ \ P(x) \ \ $has roots$ \ \ a_1, a_2, a_3, \dots , a_n$

$$Find the polynomial with roots$ \ \ \pm a_1, \pm a_2, \pm a_3 , \dots , \pm a_n$

Can i let polynomial P(x) be $P(x) = \sum_{k=0}^{n}a_kx^k$ with roots a_1, ...,a_n?

Sy123 · Sep 1, 2013

Re: HSC 2013 4U Marathon

HeroicPandas said:
Can i let polynomial P(x) be $P(x) = \sum_{k=0}^{n}a_kx^k$ with roots a_1, ...,a_n?

You can, but it isn't necessary, however do whatever method you can find.

asianese · Sep 1, 2013

Re: HSC 2013 4U Marathon

It's actually simpler than that. A hint would be to use the root form of the polynomial. (Express it as a product of factors)

asianese · Sep 1, 2013

Re: HSC 2013 4U Marathon

Um...

$P(x) = (x-a_1)(x-a_2)\cdots(x-a_n)$

Trebla · Sep 1, 2013

Re: HSC 2013 4U Marathon

asianese said:
Um...

$P(x) = k(x-a_1)(x-a_2)\cdots(x-a_n)$

ftfy

Sy123 · Sep 1, 2013

Re: HSC 2013 4U Marathon

$$Prove carefully$ \ \lim_{x\to \infty} \frac{\ln x}{x} = 0$

rural juror · Sep 1, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$$A polynomial$ \ \ P(x) \ \ $has roots$ \ \ a_1, a_2, a_3, \dots , a_n$

$$Find the polynomial with roots$ \ \ \pm a_1, \pm a_2, \pm a_3 , \dots , \pm a_n$

just tell me its not P(|x|), before i go any further with this lol

rural juror · Sep 1, 2013

Re: HSC 2013 4U Marathon

oh actually that wouldn't work would it, ok then

Sy123 · Sep 1, 2013

Re: HSC 2013 4U Marathon

rural juror said:
just tell me its not P(|x|), before i go any further with this lol

That is indeed a function with those roots, its not a polynomial however

seanieg89 · Sep 1, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
That is indeed a function with those roots, its not a polynomial however

It doesn't even have to have those roots. Eg P(x)=x+1.

Sy123 · Sep 1, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
It doesn't even have to have those roots. Eg P(x)=x+1.

Ah ok my bad

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