HSC 2014 MX2 Marathon ADVANCED (archive) (1 Viewer)

seanieg89 · Mar 23, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Carrotsticks said:
Yep, I figured that Sy was going for this approach of reduction formula --> factorials --> substitute in value.

Now that we've 'guessed' what the value of (1/2) factorial is, I think it'd be good to have a follow up question actually proving that (1/2)! = root(pi)/2 using the Gamma function. Might be a bit difficult (or even impossible) within MX2 constraints though since the reduction formulae used for the integral x^{n-1}*e^{-x} from 0 to 1 is not considered to be continuous.

Well, evaluating Gamma(3/2) is the same integral as the Gaussian. Just from 0 to infinity, and with a possible constant factor floating around somewhere. This can be done (not rigorously of course, but as rigorous as the mx2 course gets.)

I don't think there is any mx2-able method to prove the gamma-beta relation which would be an alternate method of proof.

Sy123 · Mar 23, 2014

Re: HSC 2014 4U Marathon - Advanced Level

seanieg89 said:
$For $m,n\geq 1$, let:\\ \\ $B(m,n)=\int_0^1 x^m(1-x)^n \, dx=\int_0^1 \frac{d}{dx}\left(\frac{x^{m+1}}{m+1}\right)(1-x)^n\, dx\\ \\ = \frac{n}{m+1}\int_0^1 x^{m+1}(1-x)^{n-1}\, dx = \frac{m}{n+1}B(m+1,n-1).\\ \\ $So, for positive integers m,n: $\\ \\ B(m,n)=\frac{m!n!}{(m+n)!}I(m+n,0)=\frac{m!n!}{(m+n+1)!}.\\ \\ $So using this expression formally to chuck in $m=n=1/2$, we get: $\\ \\ (1/2)!^2=2I(1/2,1/2)=\pi/4\\ \\ \Rightarrow \left(\frac{1}{2}\right)!=\frac{\sqrt{\pi}}{2}\\ \\ $ by a routine integration.$$

Yep this is pretty much what I was aiming for

Sy123 · Mar 25, 2014

Re: HSC 2014 4U Marathon - Advanced Level

$\\ $Denote that$ \ \ \frac{1}{N} \ \ $be called a unit fraction for natural number$ \ N \\ \\ $i) Show that there exists distinct$ \ a,b, \ $natural numbers such that$ \\ \frac{1}{N} = \frac{1}{a} + \frac{1}{b} \\ \\ $ii) Show that if$ \ N \ $is prime, then$ \ \frac{1}{N} \ $can be expressed by the sum of 2 distinct unit fractions in only ONE way$$

seanieg89 · Mar 25, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
$\\ $Denote that$ \ \ \frac{1}{N} \ \ $be called a unit fraction for natural number$ \ N \\ \\ $i) Show that there exists$ \ a,b, \ $natural numbers such that$ \\ \frac{1}{N} = \frac{1}{a} + \frac{1}{b} \\ \\ $ii) Show that if$ \ N \ $is prime, then$ \ \frac{1}{N} \ $can be expressed by the sum of 2 unit fractions in only ONE way$$

i) a = b = 2N is an example, for each N.

ii) Are there meant to be additional assumptions or something?

1/3 = 1/6 + 1/6 = 1/4 + 1/12 is a counter-example to the claim.

RealiseNothing · Mar 25, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Without using the AM-GM for 3 real numbers, show that:

$\frac{a}{b} + \frac{b}{c} + \frac{c}{a} \geq 3$

seanieg89 · Mar 25, 2014

Re: HSC 2014 4U Marathon - Advanced Level

RealiseNothing said:
Without using the AM-GM for 3 real numbers, show that:

$\frac{a}{b} + \frac{b}{c} + \frac{c}{a} \geq 3$

(You should state that a,b,c >0.)

RealiseNothing · Mar 25, 2014

Re: HSC 2014 4U Marathon - Advanced Level

seanieg89 said:
(You should state that a,b,c >0.)

seanieg89 · Mar 25, 2014

Re: HSC 2014 4U Marathon - Advanced Level

RealiseNothing said:

The inequality is not necessarily true if we don't also assume something like a,b,c >0. Eg a=-1, b=c=1.

RealiseNothing · Mar 25, 2014

Re: HSC 2014 4U Marathon - Advanced Level

seanieg89 said:
The inequality is not necessarily true if we don't also assume something like a,b,c >0. Eg a=-1, b=c=1.

Note to self, don't stay logged in around iBibah

Sy123 · Mar 25, 2014

Re: HSC 2014 4U Marathon - Advanced Level

seanieg89 said:
i) a = b = 2N is an example, for each N.

ii) Are there meant to be additional assumptions or something?

1/3 = 1/6 + 1/6 = 1/4 + 1/12 is a counter-example to the claim.

I apologize throughout the entire question a, b are distinct.

seanieg89 · Mar 26, 2014

Re: HSC 2014 4U Marathon

seanieg89 said:
$Prove that not all roots of the polynomial:\\ \\ $P(x):=x^n-5x^{n-1}+12x^{n-2}-15x^{n-3}+a_{n-4}x^{n-4}+\ldots+a_0\\ \\$are positive real numbers.$

bump.

Sy123 · Mar 26, 2014

Re: HSC 2014 4U Marathon

seanieg89 said:
bump.

Hopefully my proof is correct.

$\\ $Letting the roots of$ \ P(x) \ $be$ \ \alpha_k \ , \ k=1,2,\dots,n \\ $It is clear that$ \\ \sum \alpha_k = 5 \ , \sum \alpha_i \alpha_j = 12 \Rightarrow \ \sum a_k^2 = 1 \\ \\ $Now assume that$ \ P(x) \ $has all real positive roots$ \ \ \ (*) \\ \therefore \ 0 \leq \alpha_k \leq 1 \\ \\ $Now, consider the polynomial$ \ \ S(x) \ \ $with roots$ \ \frac{1}{\alpha_k} \ $for$ \ k=1,2,\dots,n \ \ $it follows$ \\ S(x) = a_0x^n + \dots -5x + 1 \ \ \ (a_0\neq 0) \\ \\ $Since the roots of$ \ P(x) \ $is bounded by 0 and 1, then, all the roots of$ \ S(x) \ $must be greater than or equal to 1, due to the assumption$ \ \ (*)$

$\\ $Letting the roots of$ \ S(x) \ $be$ \ \beta_k \ \ $due to product of roots it follows that$ \\ \left|\prod \beta_k \right| = 1 \\ \\ $However all$ \ \beta_k \geq 1 \ \ $this is a clear contradiction, unless$ \ \beta_k = 1 \ $for all$ \ k \\ \\ $If this is the case then$ \ P(x) = (x-1)^n \ \ $however the contradiction arises since the co-efficients of$ \ x^{n-1} \ $are not the same$ \\ \\ $On the case$ \ a_0 = 0 \ $then simply apply the same method to the polynomial$ \ \ Q(x) = \frac{P(x)}{x^m} \ \ $for repeated$ \ m \ $zeroes of zero$ \\ \\ $Therefore the contradiction is clear in all cases$ \\ $Therefore it cannot be true that all roots are positive real numbers$ \\ \square$

Ikki · Mar 26, 2014

Re: HSC 2014 4U Marathon - Advanced Level

What does that pi looking thing mean? (Noob alert hahaha)

Sy123 · Mar 26, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Ikki said:
What does that pi looking thing mean? (Noob alert hahaha)

It is the same thing as $\sum$ but for products =)

i.e. $\\ \sum_{k=1}^n a_k = a_1 + a_2 + \dots + a_n \\ \\ \prod_{k=1}^n a_k = a_1 \cdot a_2 \cdot \dots \cdot a_n$

seanieg89 · Mar 26, 2014

Re: HSC 2014 4U Marathon

Sy123 said:
Hopefully my proof is correct.

$\\ $Letting the roots of$ \ P(x) \ $be$ \ \alpha_k \ , \ k=1,2,\dots,n \\ $It is clear that$ \\ \sum \alpha_k = 5 \ , \sum \alpha_i \alpha_j = 12 \Rightarrow \ \sum a_k^2 = 1 \\ \\ $Now assume that$ \ P(x) \ $has all real positive roots$ \ \ \ (*) \\ \therefore \ 0 \leq \alpha_k \leq 1 \\ \\ $Now, consider the polynomial$ \ \ S(x) \ \ $with roots$ \ \frac{1}{\alpha_k} \ $for$ \ k=1,2,\dots,n \ \ $it follows$ \\ S(x) = a_0x^n + \dots -5x + 1 \ \ \ (a_0\neq 0) \\ \\ $Since the roots of$ \ P(x) \ $is bounded by 0 and 1, then, all the roots of$ \ S(x) \ $must be greater than or equal to 1, due to the assumption$ \ \ (*)$

$\\ $Letting the roots of$ \ S(x) \ $be$ \ \beta_k \ \ $due to product of roots it follows that$ \\ \left|\prod \beta_k \right| = 1 \\ \\ $However all$ \ \beta_k \geq 1 \ \ $this is a clear contradiction, unless$ \ \beta_k = 1 \ $for all$ \ k \\ \\ $If this is the case then$ \ P(x) = (x-1)^n \ \ $however the contradiction arises since the co-efficients of$ \ x^{n-1} \ $are not the same$ \\ \\ $On the case$ \ a_0 = 0 \ $then simply apply the same method to the polynomial$ \ \ Q(x) = \frac{P(x)}{x^m} \ \ $for repeated$ \ m \ $zeroes of zero$ \\ \\ $Therefore the contradiction is clear in all cases$ \\ $Therefore it cannot be true that all roots are positive real numbers$ \\ \square$

I don't see how you get the product of beta_k as 1, since we don't know that |a_0|=1?

Sy123 · Mar 26, 2014

Re: HSC 2014 4U Marathon - Advanced Level

oops

Sy123 · Mar 27, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Attempt #2

$\\ $Letting the roots of$ \ P \ $be$ \ \alpha_k \ (k=1,2,\dots ,n) \\ $It is clear that$ \ \ \sum \alpha_k = 5 \ , \ \sum \alpha_i \alpha_j = 12 \\ \Rightarrow \ \sum \alpha_k^2 = 1 \\ $Assume that$ \ \alpha_k > 0 \ \ $then it is clear that$ \ \ \ 0 < \alpha_k \leq 1 \\ $If 1 root$ \ \alpha_m = 1 \ $then$ \ P(x) = x^{n-1} (x-1) \ \ $which is not the same as the original polynomial therefore$ \ 1 \ $cannot be a root$$

$\\ $Therefore the roots of the polynomial$ \ P \ $are strictly bounded by$ \ 0 \ $and$ \ 1 \ $for ANY co-efficient of the sequence$ \ \{a_k \} \\$Therefore if we find a set of$ \ a_k \ $such that the polynomial has a root outside of the above bounds, then we arrive at a contradiction caused by the assumption that$ \ \alpha_k \ $is positive and real$ \\ \\ $Setting$ \ a_0 = 0 \ $it is clear that for all other values of$ \ a_k \ $then$ \ P(0) = 0 \ $and$ \ \alpha_k \neq 0 \\ \\ $Contradiction$ \\ $Therefore not all roots are real and positive$$

seanieg89 · Mar 27, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
Attempt #2

$\\ $Letting the roots of$ \ P \ $be$ \ \alpha_k \ (k=1,2,\dots ,n) \\ $It is clear that$ \ \ \sum \alpha_k = 5 \ , \ \sum \alpha_i \alpha_j = 12 \\ \Rightarrow \ \sum \alpha_k^2 = 1 \\ $Assume that$ \ \alpha_k > 0 \ \ $then it is clear that$ \ \ \ 0 < \alpha_k \leq 1 \\ $If 1 root$ \ \alpha_m = 1 \ $then$ \ P(x) = x^{n-1} (x-1) \ \ $which is not the same as the original polynomial therefore$ \ 1 \ $cannot be a root$$

$\\ $Therefore the roots of the polynomial$ \ P \ $are strictly bounded by$ \ 0 \ $and$ \ 1 \ $for ANY co-efficient of the sequence$ \ \{a_k \} \\$Therefore if we find a set of$ \ a_k \ $such that the polynomial has a root outside of the above bounds, then we arrive at a contradiction caused by the assumption that$ \ \alpha_k \ $is positive and real$ \\ \\ $Setting$ \ a_0 = 0 \ $it is clear that for all other values of$ \ a_k \ $then$ \ P(0) = 0 \ $and$ \ \alpha_k \neq 0 \\ \\ $Contradiction$ \\ $Therefore not all roots are real and positive$$

There is a bit of a logical flaw here, remember that we are trying to show that NO polynomial of the prescribed form has all roots positive and real.

You have just provided one example of a polynomial of the prescribed form that does not have all roots positive (any poly that is of the prescribed form and additionally has a_0=0), this is not what the question is asking for.

You are on the right track dealing with inequalities and symmetric sums, seeking a contradiction.

RealiseNothing · Mar 31, 2014

Re: HSC 2014 4U Marathon - Advanced Level

$\lim_{x \to \infty} \sqrt{x+m} - \sqrt{x+n}$

braintic · Mar 31, 2014

Re: HSC 2014 4U Marathon - Advanced Level

RealiseNothing said:
$\lim_{x \to \infty} \sqrt{x+m} - \sqrt{x+n}$

Surely its zero? Or better: 0+ if m>n, 0- if m<n

Not sure how to show working though (at least not without making the same sort of assumptions I have made in my head already).

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