HSC 2015 MX2 Marathon (archive) (1 Viewer)

Sy123 · Jul 4, 2015

Re: HSC 2015 4U Marathon

$\\ $Consider the equation$ \ z^3 + az^2 + bz +c = 0 \ (*) \\ \\ $i) Find an appropriate substitution of the form$ \ z = x + A \ $that transforms the equation $ \ (*) \ $into one of the form$ \ x^3 + px + q = 0 \ (**) \\ \\ $ii) Find an appropriate substitution of the form$ \ x = t + \frac{B}{t} \ $to transform the equation$ \ (**) \ $into one of the form$ \ t^6 + \alpha t^3 + \beta = 0 \\ \\ $iii) Hence, solve the cubic equation$ \ 12x^3 - 80x^2 + 163x - 105 = 0$

Sy123 · Jul 4, 2015

Re: HSC 2015 4U Marathon

$\\ $i) Show that if$ \ I_n = \int_0^{\pi /2} \sin^{2n}x \ dx \ $then$ \ \frac{I_{n}}{I_{n-1}} = \frac{2n-1}{2n}$

$\\ $ii) Hence show that$ \ I_n = \frac{\pi}{2} \frac{1}{4^n} \binom{2n}{n}$

$\\ $iii) Find$ \ \sum_{n=0}^{\infty} \frac{1}{16^n} \binom{2n}{n}$

Sy123 · Jul 6, 2015

Re: HSC 2015 4U Marathon

$\\ $Let$ \ F_n = 2^{2^n} + 1 \\ \\ $i) Show by mathematical induction that$ \ F_0F_1\dots F_{n-1} = F_n - 2 \ $for$ \ n \geq 1 \\ \\ $ii) Hence show that there is no$ \ F_i \ $and$ \ F_j \ $with$ \ i <j \ $such that there is an integer that divides into both$ \ F_i \ $and$ \ F_j$

porcupinetree · Jul 6, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
$\\ $i) Show that if$ \ I_n = \int_0^{\pi /2} \sin^{2n}x \ dx \ $then$ \ \frac{I_{n}}{I_{n-1}} = \frac{2n-1}{2n}$

$\\ $ii) Hence show that$ \ I_n = \frac{\pi}{2} \frac{1}{4^n} \binom{2n}{n}$

$\\ $iii) Find$ \ \sum_{n=0}^{\infty} \frac{1}{16^n} \binom{2n}{n}$

https://imgur.com/wTEEPbf

Don't know how to do part (iii)

Ekman · Jul 6, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
$\\ $i) Show that if$ \ I_n = \int_0^{\pi /2} \sin^{2n}x \ dx \ $then$ \ \frac{I_{n}}{I_{n-1}} = \frac{2n-1}{2n}$

$\\ $ii) Hence show that$ \ I_n = \frac{\pi}{2} \frac{1}{4^n} \binom{2n}{n}$

$\\ $iii) Find$ \ \sum_{n=0}^{\infty} \frac{1}{16^n} \binom{2n}{n}$

$$ I) $ I_{n} = \int_{0}^{\frac{\pi}{2}}\sin^{2n-2}x (1-\cos^{2}x) dx \Rightarrow I_{n-1} - \int_{0}^{\frac{\pi}{2}}\sin^{2n-2}x \cos^{2}x dx$

$\therefore $ Via IBP: $ I_{n} = I_{n-1} - \frac{I_n}{2n-1} \Rightarrow I_n=\frac{2n-1}{2n}I_{n-1} $ for $ n\geq1$

$$ II) $ I_n = \frac{2n-1}{2n}I_{n-1} ... I_{n-1} = \frac{2n-3}{2(n-1)}I_{n-2} ... I_{0} = \frac{\pi}{2}$

$\therefore I_n= \frac{\pi}{2} \frac{\frac{(2n)(2n-1)(2n-2)(2n-3)(2n-4)...4\times3\times2}{(2n)(2n-2)(2n-4)....2}}{2^n (n)(n-1)(n-2)(n-3)...1} = \frac{\pi}{2} \frac{(2n)!}{4^n (n!)^2} = \frac{\pi}{2} \frac{1}{4^n} \binom{2n}{n}$

$$ III) $ \sum_{n=0}^{\infty} \frac{1}{16^n} \binom{2n}{n} = \sum_{n=0}^{\infty} \frac{2}{4^n\pi}I_{n}$

$= \frac{2}{\pi} \sum_{n=0}^{\infty} \frac{I_{n}}{4^n} = \frac{2}{\pi} \int_{0}^{\frac{\pi}{2}} \frac{1}{1-\frac{\sin^2{x}}{4}}$

$= \frac{2}{\pi} \times \frac{\pi}{\sqrt{3}} = \frac{2}{\sqrt{3}}$

porcupinetree · Jul 6, 2015

Re: HSC 2015 4U Marathon

Ekman said:
$= \frac{2}{\pi} \sum_{n=0}^{\infty} \frac{I_{n}}{4^n} = \frac{2}{\pi} \int_{0}^{\frac{\pi}{2}} \frac{1}{1-\frac{\sin^2{x}}{4}}$

How did you do this?

Ekman · Jul 6, 2015

Re: HSC 2015 4U Marathon

porcupinetree said:
How did you do this?

I converted the sum into integral format:

$\frac{2}{\pi} \sum_{n=0}^{\infty} \frac{I_{n}}{4^n} = \frac{2}{\pi} \int_{0}^{\frac{\pi}{2}} \frac{sin^{0}x}{4^0} + \frac{\sin^{2}x}{4^1} + \frac{\sin^{4}x}{4^2} + .... =\frac{2}{\pi} \int_{0}^{\frac{\pi}{2}} \sum_{n=0}^{\infty} \frac{\sin^{2n}x}{4^n} dx$

Then I used limiting sum to get:

$\frac{2}{\pi} \int_{0}^{\frac{\pi}{2}} \frac{1}{1-\frac{\sin^{2}x}{4}} dx$

porcupinetree · Jul 6, 2015

Re: HSC 2015 4U Marathon

Ekman said:
I converted the sum into integral format:

$\frac{2}{\pi} \sum_{n=0}^{\infty} \frac{I_{n}}{4^n} = \frac{2}{\pi} \int_{0}^{\frac{\pi}{2}} \frac{sin^{0}x}{4^0} + \frac{\sin^{2}x}{4^1} + \frac{\sin^{4}x}{4^2} + .... =\frac{2}{\pi} \int_{0}^{\frac{\pi}{2}} \sum_{n=0}^{\infty} \frac{\sin^{2n}x}{4^n} dx$

Then I used limiting sum to get:

$\frac{2}{\pi} \int_{0}^{\frac{\pi}{2}} \frac{1}{1-\frac{\sin^{2}x}{4}} dx$

And then how do you evaluate that integral? t method seems a little tedious, is there a simpler way?

leehuan · Jul 6, 2015

Re: HSC 2015 4U Marathon

Multiply by 4/4 and also multiply by sec^2x/sec^2x
Bottom can all be converted into tangents. Then just reverse chain rule.

Sy123 · Jul 6, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
$\\ $Consider the equation$ \ z^3 + az^2 + bz +c = 0 \ (*) \\ \\ $i) Find an appropriate substitution of the form$ \ z = x + A \ $that transforms the equation $ \ (*) \ $into one of the form$ \ x^3 + px + q = 0 \ (**) \\ \\ $ii) Find an appropriate substitution of the form$ \ x = t + \frac{B}{t} \ $to transform the equation$ \ (**) \ $into one of the form$ \ t^6 + \alpha t^3 + \beta = 0 \\ \\ $iii) Hence, solve the cubic equation$ \ 12x^3 - 80x^2 + 163x - 105 = 0$

$\\ $i) Let$ \ z = x + A \ $for some constant$ \ A \ $so that$ \ (x+A)^3 + a(x+A)^2 + b(x+A) + c = 0 \\ $The coefficient of$ \ x^2 \ $is zero$$

$\\ \therefore 3x^2A + ax^2 = 0 \Rightarrow A = \frac{-a}{3}$

$\\ \therefore \ z = x - \frac{a}{3} \ $is a sufficient substitution to get equation$ \ (*) \ $into the form$ \ (**)$

$\\ $Moreover$ \ p = b - \frac{a^2}{3} \ $and$ \ q = \frac{2a^3}{27} - \frac{ab}{3} + c \ $(for future reference)$$

$$ii) Let$ \ x = t + \frac{B}{t} \ $so$ \ \left(t + \frac{B}{t} \right)^3 + p\left(t + \frac{B}{t} \right) + q = 0$

$(and multiplying both sides by$ \ t^3)$

$t^6 + qt^3 + B^3 + t^4(3B + p) + t^2B(3B +p) \equiv t^6 + \alpha t^3 + \beta$

$\\ \Rightarrow B = \frac{-p}{3}$

$\\ \therefore x = t - \frac{p}{3t} \ $is a suitable substitution$$

$\\ $Moreover$ \ \alpha = q \ $and$ \ \beta = \frac{-p^3}{27}$

$\\ $iii) You just apply these methods to the previous parts to find$ \ \alpha, \beta, p, q, a, b, c \ $to then solve a quadratic and then finally find$ \ z \ $after resubstituting. The numbers turn out much uglier than I expected, so there's no need to write it all down$$

Sy123 · Jul 6, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
$\\ $Let$ \ F_n = 2^{2^n} + 1 \\ \\ $i) Show by mathematical induction that$ \ F_0F_1\dots F_{n-1} = F_n - 2 \ $for$ \ n \geq 1 \\ \\ $ii) Hence show that there is no$ \ F_i \ $and$ \ F_j \ $with$ \ i <j \ $such that there is an integer that divides into both$ \ F_i \ $and$ \ F_j$

bump

A translation of part (ii) is simply "show that no two members of the sequence F_n share a common factor"

InteGrand · Jul 6, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
$\\ $i) Let$ \ z = x + A \ $for some constant$ \ A \ $so that$ \ (x+A)^3 + a(x+A)^2 + b(x+A) + c = 0 \\ $The coefficient of$ \ x^2 \ $is zero$$

$\\ \therefore 3x^2A + ax^2 = 0 \Rightarrow A = \frac{-a}{3}$

$\\ \therefore \ z = x - \frac{a}{3} \ $is a sufficient substitution to get equation$ \ (*) \ $into the form$ \ (**)$

$\\ $Moreover$ \ p = b - \frac{a^2}{3} \ $and$ \ q = \frac{2a^3}{27} - \frac{ab}{3} + c \ $(for future reference)$$

$$ii) Let$ \ x = t + \frac{B}{t} \ $so$ \ \left(t + \frac{B}{t} \right)^3 + p\left(t + \frac{B}{t} \right) + q = 0$

$(and multiplying both sides by$ \ t^3)$

$t^6 + qt^3 + B^3 + t^4(3B + p) + t^2B(3B +p) \equiv t^6 + \alpha t^3 + \beta$

$\\ \Rightarrow B = \frac{-p}{3}$

$\\ \therefore x = t - \frac{p}{3t} \ $is a suitable substitution$$

$\\ $Moreover$ \ \alpha = q \ $and$ \ \beta = \frac{-p^3}{27}$

$\\ $iii) You just apply these methods to the previous parts to find$ \ \alpha, \beta, p, q, a, b, c \ $to then solve a quadratic and then finally find$ \ z \ $after resubstituting. The numbers turn out much uglier than I expected, so there's no need to write it all down$$

Isn't this essentially a derivation of the cubic formula? That is famously messy lol.

leehuan · Jul 6, 2015

Re: HSC 2015 4U Marathon

(i)
$$When $n=1\\ LHS={ F }_{ 0 }={ 2 }^{ { 2 }^{ 0 } }+1=3\\ RHS={ F }_{ 1 }-2={ 2 }^{ { 2 }^{ 1 } }+1-2=3\\ LHS=RHS$ so the statement holds for $n=1$

$$Assume true for $n=k \\ { F }_{ 0 }{ F }_{ 1 }\dots { F }_{ k-1 }={ F }_{ k }-2\quad \quad k\in \mathbb Z ,\quad k\ge 1$

$$When $n=k+1 \\ LHS={ F }_{ 0 }{ F }_{ 1 }\dots { F }_{ k-1 }{ F }_{ k }\\ =\left( { F }_{ k }-2 \right) { F }_{ k }\quad by\quad assumption\\ =\left( { 2 }^{ { 2 }^{ k } }-1 \right) \left( { 2 }^{ { 2 }^{ k } }+1 \right) \\ ={ 2 }^{ 2\cdot { 2 }^{ k } }-1\\ =\left( { 2 }^{ { 2 }^{ k+1 } }+1 \right) -2\\ ={ F }_{ n }-2\\ =RHS\\$

$Hence the statement holds for $n=k+1$ if it holds for $n=k. \\ $So since it holds for $n=1$, by the principle of mathematical induction it also holds for $n=2, 3, \dots $ i.e. all $n\ge 1$

Drsoccerball · Jul 6, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
bump

A translation of part (ii) is simply "show that no two members of the sequence F_n share a common factor"

Theyre all prime numbers

InteGrand · Jul 6, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
Theyre all prime numbers

Not all of them are; the first several are, but the pattern breaks down after a point.

E.g. $2^{2^5} + 1 = 4294967297 = 641\times 6700417$ .

Ekman · Jul 6, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
$\\ $Let$ \ F_n = 2^{2^n} + 1 \\ \\ $i) Show by mathematical induction that$ \ F_0F_1\dots F_{n-1} = F_n - 2 \ $for$ \ n \geq 1 \\ \\ $ii) Hence show that there is no$ \ F_i \ $and$ \ F_j \ $with$ \ i <j \ $such that there is an integer that divides into both$ \ F_i \ $and$ \ F_j$

For part ii:

$F_n $ is an odd integer for $ n\geq0 $ $ \therefore \ F_0F_1\dots F_{n-1} $ and $ F_n - 2 $ are odd integers as well$$

$$ Since: $ \ F_0F_1\dots F_{I} \dots F_{j-1} = F_j - 2 \Rightarrow \frac{\ F_0F_1\dots F_{I} \dots F_{j-1}}{F_j} = 1-\frac{2}{F_j}$

$Since $ 1-\frac{2}{F_j} $ is not an integer (because even integers cannot be divided by odd integers) $ \therefore \frac{\ F_0F_1\dots F_{I} \dots F_{j-1}}{F_j} $ is not an integer$

Edit: Nevermind, I only proved that they are not divisible, I didn't prove that they didn't have a common factor.

RealiseNothing · Jul 6, 2015

Re: HSC 2015 4U Marathon

Ekman said:
For part ii:

$F_n $ is an odd integer for $ n\geq0 $ $ \therefore \ F_0F_1\dots F_{n-1} $ and $ F_n - 2 $ are odd integers as well$$

$$ Since: $ \ F_0F_1\dots F_{I} \dots F_{j-1} = F_j - 2 \Rightarrow \frac{\ F_0F_1\dots F_{I} \dots F_{j-1}}{F_j} = 1-\frac{2}{F_j}$

$Since $ 1-\frac{2}{F_j} $ is not an integer (because even integers cannot be divided by odd integers) $ \therefore \frac{\ F_0F_1\dots F_{I} \dots F_{j-1}}{F_j} $ is not an integer$

How does that show the result?

RealiseNothing · Jul 6, 2015

Re: HSC 2015 4U Marathon

Ekman said:
Edit: Nevermind, I only proved that they are not divisible, I didn't prove that they didn't have a common factor.

You're pretty close though.

Sy123 · Jul 6, 2015

Re: HSC 2015 4U Marathon

Ekman said:
For part ii:

$F_n $ is an odd integer for $ n\geq0 $ $ \therefore \ F_0F_1\dots F_{n-1} $ and $ F_n - 2 $ are odd integers as well$$

$$ Since: $ \ F_0F_1\dots F_{I} \dots F_{j-1} = F_j - 2 \Rightarrow \frac{\ F_0F_1\dots F_{I} \dots F_{j-1}}{F_j} = 1-\frac{2}{F_j}$

$Since $ 1-\frac{2}{F_j} $ is not an integer (because even integers cannot be divided by odd integers) $ \therefore \frac{\ F_0F_1\dots F_{I} \dots F_{j-1}}{F_j} $ is not an integer$

Edit: Nevermind, I only proved that they are not divisible, I didn't prove that they didn't have a common factor.

Try proving via contradiction

Ekman · Jul 6, 2015

Re: HSC 2015 4U Marathon

RealiseNothing said:
You're pretty close though.

Sy123 said:
Try proving via contradiction

I'm assuming you would use a similar method but instead of using the numbers F_n, you would use the factors of those integers, and prove that the factors cannot be integers, hence proving they do not share common factors.

HSC 2015 MX2 Marathon (archive) (1 Viewer)

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