HSC 2016 MX2 Marathon ADVANCED (archive) (1 Viewer)

dan964 · Feb 24, 2016

Re: HSC 2016 4U Marathon - Advanced Level

Subtle, possibly very easy, not entirely sure. Without graphing the thing:

Prove thoroughly that
lim (1/n^R) as n --> infinity, and where R is a member of the reals, e.g. pi or e, or 3, etc. (and yes R>0)
is 0.

and R>0

InteGrand · Feb 24, 2016

Re: HSC 2016 4U Marathon - Advanced Level

dan964 said:
Subtle, possibly very easy, not entirely sure. Without graphing the thing:

Prove thoroughly that
lim (1/n^R) as n --> infinity, and where R is a member of the reals, e.g. pi or e, or 3, etc.
is 0.

$$\noindent (Note that this is the case if and only if the real number $R$ is a \emph{positive} real.)$

KingOfActing · Feb 24, 2016

Re: HSC 2016 4U Marathon - Advanced Level

$\text{Looking at the definition of a limit, } \lim_{n\to\infty}\frac{1}{n^R} := \forall\varepsilon>0\,\exists M\forall n > M : |\frac{1}{n^R} - 0| < \varepsilon\\\text{Note that if we let } M = \frac{1}{\sqrt[R]\varepsilon}\\n > \frac{1}{\sqrt[R]{\varepsilon}} \Longrightarrow \left| \frac{1}{n^R}-0\right | < \varepsilon$

However, if we're allowed to use L'Hospitals law:

$\lim_{n\to\infty}\frac{1}{n^R} = \lim_{n\to\infty}\frac{0}{Rn^{R-1}} = 0$

Finally, another way which I just don't feel like typing up would be to prove that the function is monotone decreasing on (1,infinity) and note that it is bounded below by 0 (which is trivial, just note that the division of two positive numbers is positive).

EDIT: Note that R > 0

Paradoxica · Feb 24, 2016

Re: HSC 2016 4U Marathon - Advanced Level

KingOfActing said:
$\text{Looking at the definition of a limit, } \lim_{n\to\infty}\frac{1}{n^R} := \forall\varepsilon>0\,\exists M\forall n > M : |\frac{1}{n^R} - 0| < \varepsilon\\\text{Note that if we let } M = \frac{1}{\sqrt[R]\varepsilon}\\n > \frac{1}{\sqrt[R]{\varepsilon}} \Longrightarrow \left| \frac{1}{n^R}-0\right | < \varepsilon$

However, if we're allowed to use L'Hospitals law:

$\lim_{n\to\infty}\frac{1}{n^R} = \lim_{n\to\infty}\frac{0}{Rn^{R-1}} = 0$

Finally, another way which I just don't feel like typing up would be to prove that the function is monotone decreasing on (1,infinity) and note that it is bounded below by 0 (which is trivial, just note that the division of two positive numbers is positive).

EDIT: Note that R > 0

L'Hôpital's rule doesn't work here, because the expression does not reach indeterminate form.

KingOfActing · Feb 25, 2016

Re: HSC 2016 4U Marathon - Advanced Level

Paradoxica said:
L'Hôpital's rule doesn't work here, because the expression does not reach indeterminate form.

Actually, L'Hospital's rule is applicable for f(x)/g(x) when f and g tend to 0, or |g(x)| tends to infinity.

dan964 · Feb 25, 2016

Re: HSC 2016 4U Marathon - Advanced Level

InteGrand said:
$$\noindent (Note that this is the case if and only if the real number $R$ is a \emph{positive} real.)$

yes. sorry forgot to add that.

Paradoxica · Feb 26, 2016

Re: HSC 2016 4U Marathon - Advanced Level

$$\noindent The monic cubic polynomial $x^3 + ax^2 + bx +c$ is one to one over $\mathbb{R}$ (\textit{injective}). Determine all restrictions and relations on the co-efficients.$

KingOfActing · Feb 27, 2016

Re: HSC 2016 4U Marathon - Advanced Level

Paradoxica said:
$$\noindent The monic cubic polynomial $x^3 + ax^2 + bx +c$ is one to one over $\mathbb{R}$ (\textit{injective}). Determine all restrictions and relations on the co-efficients.$

$$Considering the graph of a cubic function, it is clear the function will be injective if it has no turning point, i.e. the function is monotone, meaning the derivative doesn't change sign.$\\f'(x) = 3x^2 + 2ax + b\\$For a quadratic to be positive-semidefinite or negative-semidefinite, $ \Delta \leq 0 \\4a^2-12b \leq 0\\a^2 \leq 3b$

Paradoxica · Feb 28, 2016

Re: HSC 2016 4U Marathon - Advanced Level

$$\noindent By considering the expansions of $\cos{k\theta}$ for $k \in \{2,3,4,5,6,7,8\} \\ $Solve for zeroes of the polynomial equation $\\32x^8 - 56x^6 + 30x^4 - 5x^2 = 16x^7 - 24x^5 + 10x^3 - x$

agha · Mar 10, 2016

Re: HSC 2016 4U Marathon - Advanced Level

$\\ x \ $is a real number randomly chosen from the interval$ \ 0 < x < 1$

$\\ $Let$ \ f(x) = \left\{\frac{1}{x} \right\} \ $where$ \ \{A\} \ $is the fractional part of$ \ A. \\ $Examples of this function include,$ \ \{\pi \} = \pi - 3 \ $or$ \ \{1.5 \} = 0.5$

$\\ $Find an expression for the probability that$ \ x > f(x)$

agha · Mar 10, 2016

Re: HSC 2016 4U Marathon - Advanced Level

$\\ $Prove that there are no positive integers$ \ a,b \ $satisfying$ \ \frac{a}{b} + \frac{b}{a} = 4$

Paradoxica · Mar 10, 2016

Re: HSC 2016 4U Marathon - Advanced Level

agha said:
$\\ $Prove that there are no positive integers$ \ a,b \ $satisfying$ \ \frac{a}{b} + \frac{b}{a} = 4$

$\noindent This is equivalent to the claim that $x+\frac{1}{x} = 4$ has rational solutions in $x$. But the only two real solutions are irrational. Thus, there is no such $x$, and hence, no integer pairs $(a,b)$ which satisfy the claim.$

InteGrand · Mar 10, 2016

Re: HSC 2016 4U Marathon - Advanced Level

agha said:
$\\ x \ $is a real number randomly chosen from the interval$ \ 0 < x < 1$

$\\ $Let$ \ f(x) = \left\{\frac{1}{x} \right\} \ $where$ \ \{A\} \ $is the fractional part of$ \ A. \\ $Examples of this function include,$ \ \{\pi \} = \pi - 3 \ $or$ \ \{1.5 \} = 0.5$

$\\ $Find an expression for the probability that$ \ x > f(x)$

Do you know whether there is a closed-form answer, or can we just give our answer as an infinite series?

Paradoxica · Mar 10, 2016

Re: HSC 2016 4U Marathon - Advanced Level

InteGrand said:
Do you know whether there is a closed-form answer, or can we just give our answer as an infinite series?

Show me your infinite series. I'm looking through my materials for a possible closed form.

InteGrand · Mar 10, 2016

Re: HSC 2016 4U Marathon - Advanced Level

Paradoxica said:
Show me your infinite series. I'm looking through my materials for a possible closed form.

$\sum _{n=1}^{\infty} \left[ 1-\frac{n}{2} \left(\sqrt{n^2 + 4}-n \right) \right] \cdot \frac{1}{n} =0.5355 \ldots$

Paradoxica · Mar 10, 2016

Re: HSC 2016 4U Marathon - Advanced Level

InteGrand said:
$\sum _{n=1}^{\infty} \left[ 1-\frac{n}{2} \left(\sqrt{n^2 + 4}-n \right) \right] \cdot \frac{1}{n} =0.5355 \ldots$

May I ask how you derived this? My line of thinking did not lead to surds...

seanieg89 · Mar 11, 2016

Re: HSC 2016 4U Marathon - Advanced Level

I got an equivalent series and the same numerics as Integrand. Will spend a little time trying to evaluate it before posting.

InteGrand · Mar 11, 2016

Re: HSC 2016 4U Marathon - Advanced Level

Paradoxica said:
May I ask how you derived this? My line of thinking did not lead to surds...

Here was my method (for obtaining the series).

$\noindent Let $X$ be uniformly distributed on $[0,1]$, i.e. $X\sim \mathcal{U}(0,1)$. Let $A_n$ be the event that $\frac{1}{n+1}<X<\frac{1}{n}$ for $n=1,2,3,\ldots$. It is easy to see that the $A_n$'s are mutually exclusive and exhaustive. Now, suppose $\frac{1}{n+1}<x<\frac{1}{n}$, then we have $x=\frac{1}{n+1}+\varepsilon$, for some $\varepsilon \equiv \varepsilon _n \in \left(0, \underbrace{\frac{1}{n}-\frac{1}{n+1}}_{\frac{1}{n(n+1)}}\right)$ We ask: for which $\varepsilon$ is $x>\left \{ \frac{1}{x} \right \}?$$

$\noindent We note that $x=\frac{1}{n+1}+\varepsilon = \frac{1+\varepsilon (n+1)}{n+1}$, so $\frac{1}{x} = \frac{n+1}{1+\varepsilon (n+1)}$. Now, as $\frac{1}{n+1}<x<\frac{1}{n}$, we have $n<\frac{1}{x}<n+1$, so$

$\begin{align*} \left \{ \frac{1}{x} \right \} &= \frac{1}{x}-n \\ &= \frac{n+1}{1+\varepsilon (n+1)}-n.\end{align*}$

$\noindent So $x>\left \{ \frac{1}{x} \right \}$ iff$

$\begin{align*}\frac{1+\varepsilon (n+1)}{n+1} &> \frac{n+1}{1+\varepsilon (n+1)}-n \\ \Longleftrightarrow \frac{1+\varepsilon (n+1)+n(n+1)}{n+1}&> \frac{n+1}{1+\varepsilon (n+1)} \\ \Longleftrightarrow \left(1+\varepsilon (n+1)\right) \left(1+\varepsilon (n+1) + n(n+1)\right) &> (n+1)^2 \\ \Longleftrightarrow \varrho\left(\varrho + n(n+1)\right) &> (n+1)^2 \quad (\text{writing }\varrho \equiv 1 + \varepsilon (n+1)) \\ \Longleftrightarrow \varrho^2 + n(n+1)\varrho -(n+1)^2 &> 0. \end{align*}$

$\noindent Note that this quadratic has discriminant $\Delta = n^2 (n+1)^2 +4(n+1)^2 = (n+1)^2 \left(n^2 + 4\right)$, so $\sqrt{\Delta} = (n+1)\sqrt{n^2 + 4}$. So the roots of the quadratic are $\varrho _{\pm} = \frac{-n(n+1) \pm (n+1)\sqrt{n^2 + 4}}{2}$ (clearly $\varrho_{-} < \varrho_{+}$. In fact, $\varrho_{+} >0$ and $\varrho_{-} <0$, since $\sqrt{n^2 + 4}>n$). So the inequation above will hold iff $\varrho < \varrho_{-}$ or $\varrho_{-} > \varrho_{+}$. But $\varrho = 1 + \varepsilon (n+1) > 0$, so we ignore the negative root $\varrho_{-}$ and only have $\varrho > \varrho_{+}$, i.e.$

$\begin{align*}1+\varepsilon (n+1) &>\frac{-n(n+1) +(n+1)\sqrt{n^2 + 4}}{2} \\ \Longleftrightarrow \varepsilon > \frac{\sqrt{n^2 + 4}-n}{2}-\frac{1}{n+1}. \end{align*}$

$\noindent Now recall that $\varepsilon < \frac{1}{n}-\frac{1}{n+1}$. So the answer to our question ``for which $\varepsilon$ is $x>\left \{ \frac{1}{x} \right \}$?'' is $\frac{ \sqrt{n^2 + 4}-n}{2}-\frac{1}{n+1}< \varepsilon _n < \frac{1}{n}-\frac{1}{n+1}$.$

$\noindent Now, since we are dealing with uniform distributions, our probabilities are all just the lengths of relevant intervals divided by the length of the sample space for that random variable. For each $n$, $\varepsilon_n$ is uniformly distributed on $\left[\frac{1}{n+1},\frac{1}{n}\right]$ $\Big{(}$this is because the conditional distribution of $X$ given that it is between $\frac{1}{n+1}$ and $\frac{1}{n}$ is a new uniform distribution on this interval$\Big{)}$. Hence$

$\begin{align*} \mathbb{P} \left( X > \left \{ \frac{1}{X} \right \} \Big{|} A_n \right)&= \mathbb{P}\left(\frac{ \sqrt{n^2 + 4}-n}{2}-\frac{1}{n+1} < \varepsilon _n < \frac{1}{n}-\frac{1}{n+1} \right) \\ &= \frac{\frac{1}{n}-\frac{1}{n+1}-\left(\frac{ \sqrt{n^2 + 4}-n}{2}-\frac{1}{n+1}\right)}{\frac{1}{n}-\frac{1}{n+1}}\\ &= \frac{\frac{1}{n} - \frac{1}{2} \left(\sqrt{n^2 +4}-n\right)}{\mathbb{P}\left(A_n \right)} \\ \Rightarrow \mathbb{P} \left( X > \left \{ \frac{1}{X} \right \} \Big{|} A_n \right)\mathbb{P}\left(A_n \right) &= \frac{1}{n} - \frac{1}{2} \left(\sqrt{n^2 +4}-n\right) \end{align*}$

$\noindent So by the law of total probability, we have$

$\begin{align*}\mathbb{P}\left( X > \left \{ \frac{1}{X} \right \} \right) &= \sum_{n=1}^{\infty } \mathbb{P} \left( X > \left \{ \frac{1}{X} \right \} \Big{|} A_n \right) \mathbb{P} \left(A_n\right) \\ &= \sum_{n=1}^{\infty }\left(\frac{1}{n} - \frac{1}{2} \left(\sqrt{n^2 +4}-n\right)\right). \end{align*}$

agha · Mar 11, 2016

Re: HSC 2016 4U Marathon - Advanced Level

InteGrand said:
$\sum _{n=1}^{\infty} \left[ 1-\frac{n}{2} \left(\sqrt{n^2 + 4}-n \right) \right] \cdot \frac{1}{n} =0.5355 \ldots$

Yes this is what I got as well, I wasn't able to find a closed form either which is why I simply asked for an expression

agha · Mar 11, 2016

Re: HSC 2016 4U Marathon - Advanced Level

$\\ $Is$ \ 2^n + 3^n $, for positive integers$ \ n $, ever a square number?$$

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