HSC 2013 MX2 Marathon (archive) (49 Viewers)

seanieg89 · Jul 15, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
First find the function f(t) such that the equality holds:

$f(t) = A + B\int_0^t f(s) \ ds$

$f'(t) = B(F'(t) - F'(0))$

For F the primitive of f:

$\frac{f'(t)}{f(t)} = B$

$\ln f(t) = Bt + c$

$f(0) = A + B \int_0^0 f(s) \ ds = A$

$c = \ln A$

$\therefore f(t) = e^c e^{Bt}$

$f(t) = Ae^{Bt}$

$$Now fix for some function$ \ \ g(t) \leq f(t)$

$\therefore \ $if$ \ \ g(t) < A + B \int_0^t g(s) \ ds$

$g(t) \leq Ae^{Bt}$

I have a feeling this argument is not valid but I cant see it right now, its a bit late

Your f is indeed the unique function which gives us equality for all t >= 0 (*), but I don't know how you are deducing that any g satisfying the integral inequality must be smaller than f everywhere (which is what we are aiming for).

(*) Your method of showing this is mostly okay, but the division by something that is potentially (a priori) negative or zero does raise some niggling potential problems...thankfully none of them crop up here so don't worry about this.

Sy123 · Jul 15, 2013

Re: HSC 2013 4U Marathon

$Prove$

$x - \frac{1}{3}x^3+ \frac{1}{5}x^5 - \dots - \frac{1}{4k-1}x^{4k-1} < \tan^{-1} x < x - \frac{1}{3}x^3 + \frac{1}{5}x^5 - \dots + \frac{1}{4k+1}x^{4k+1}$

$And hence prove that$

$\tan^{-1} x = \sum_{n=1}^{\infty} \frac{(-1)^{k-1}x^{2k-1}}{2k-1}$

RealiseNothing · Jul 15, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$Prove$

$x - \frac{1}{3}x^3+ \frac{1}{5}x^5 - \dots - \frac{1}{4k-1}x^{4k-1} < \tan^{-1} x < x - \frac{1}{3}x^3 + \frac{1}{5}x^5 - \dots + \frac{1}{4k+1}x^{4k+1}$

$And hence prove that$

$\tan^{-1} x = \sum_{n=1}^{\infty} \frac{(-1)^{k-1}x^{2k-1}}{2k-1}$

$\frac{1-x^{4k}}{1+x^2} < \frac{1}{1+x^2} < \frac{1+x^{4k+2}}{1+x^2}$

$1-x^2+x^4-...-x^{4k-2} < \frac{1}{1+x^2} < 1-x^2+x^4-...+x^{4k}$

Integrating everything gives:

$x-\frac{1}{3}x^3+\frac{1}{5}x^5-...-\frac{1}{4k-1}x^{4k-1} < tan^{-1}x < x-\frac{1}{3}x^3+\frac{1}{5}x^5-...+\frac{1}{4k+1}x^{4k+1}$

$0 < tan^{-1}x - \sum_{j=1}^{\infty} \frac{(-1)^{j-1}x^{2j-1}}{2j-1} < \frac{1}{4k+1}x^{4k+1}$

Take $\lim_{k \to \infty}$

$0 < tan^{-1}x - \sum_{k=1}^{\infty} \frac{(-1)^{k-1}x^{2k-1}}{2k-1} < 0$

$tan^{-1}x = \sum_{k=1}^{\infty} \frac{(-1)^{k-1}x^{2k-1}}{2k-1}$

Sy123 · Jul 15, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
$\frac{1-x^{4k}}{1+x^2} < \frac{1}{1+x^2} < \frac{1+x^{4k+2}}{1+x^2}$

$1-x^2+x^4-...-x^{4k-2} < \frac{1}{1+x^2} < 1-x^2+x^4-...+x^{4k}$

Integrating everything gives:

$x-\frac{1}{3}x^3+\frac{1}{5}x^5-...-\frac{1}{4k-1}x^{4k-1} < tan^{-1}x < x-\frac{1}{3}x^3+\frac{1}{5}x^5-...+\frac{1}{4k+1}x^{4k+1}$

$0 < tan^{-1}x - \sum_{j=1}^{\infty} \frac{(-1)^{j-1}x^{2j-1}}{2j-1} < \frac{1}{4k+1}x^{4k+1}$

Take $\lim_{k \to \infty}$

$0 < tan^{-1}x - \sum_{k=1}^{\infty} \frac{(-1)^{k-1}x^{2k-1}}{2k-1} < 0$

$tan^{-1}x = \sum_{k=1}^{\infty} \frac{(-1)^{k-1}x^{2k-1}}{2k-1}$

Yep this is pretty much it, although I would suggest that for your first step you use a dummy variable such as z, then integrating from 0 to x.

EDIT: I probably should of put the conditions on x

seanieg89 · Jul 16, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
$$g(t)$ is a continuous function and $A,B$ are positive constants such that:\\ \\ $g(t)\leq A+B\int_0^t g(s)\, ds $\\ \\ for all $t\geq 0.$\\ \\Prove that $g(t)\leq Ae^{Bt}$\\ \\for all $t\geq 0.$$

$\frac{d}{dt}\left(e^{-Bt}\int_0^t g(s)\, ds\right)=\left(g(t)-B\int_0^t g(s)\,ds\right)e^{-Bt}\leq Ae^{-Bt}$

by the product rule, fundamental theorem of calculus, and our assumed integral inequality.

Replacing t with r and integrating this expression from r=0 to r=t gives:

$e^{-Bt}\int_0^t g(s)\, ds \leq AB^{-1}(1-e^{-Bt}).$

Hence we can conclude that:

$g(t)\leq A+B\int_0^t g(s)\, ds \leq A+B(AB^{-1}(e^{Bt}-1))=Ae^{Bt}.$

Our working is valid for all non-negative t.

Sy123 · Jul 16, 2013

Re: HSC 2013 4U Marathon

$$Sum the series by expanding out the sum$ \ \ \sum_{k=1}^{n} (k - (k-1))$

$Do something similar to prove without induction$

$\sum_{j=1}^{n} \tan^{-1} \left(\frac{1}{2j^2}\right) = \tan^{-1} \left(\frac{n}{n+1} \right)$

$Hence find the limiting sum as$ \ n \ $increases without bound$

rural juror · Jul 16, 2013

Re: HSC 2013 4U Marathon

tan-1(n/n+1) - tan-1(n-1/n) = tan-1(1/2j^2)

Sy123 · Jul 16, 2013

Re: HSC 2013 4U Marathon

rural juror said:
tan-1(n/n+1) - tan-1(n-1/n) = tan-1(1/2j^2)

Yep,

$By considering the cases whereby the positive numbers$ \ \ a, b, c \ \ $are sides of a triangle and when they are not sides of a triangle, prove the inequality$

$(-a+b+c)(a-b+c)(a+b-c) \leq abc$

Sy123 · Jul 17, 2013

Re: HSC 2013 4U Marathon

$a, b, c \ $are positive integers such that$$

$a+b+c+ab+ac+bc+abc= 7\cdot 11 \cdot 13 - 1$

$$Find$ \ \ a,b,c$

AnimeX · Jul 17, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$a, b, c \ $are positive integers such that$$

$a+b+c+ab+ac+bc+abc= 7\cdot 11 \cdot 13 - 1$

$$Find$ \ \ a,b,c$

sorry, but are the dots like 7.11 meant to be multiplication signs?

RealiseNothing · Jul 17, 2013

Re: HSC 2013 4U Marathon

AnimeX said:
sorry, but are the dots like 7.11 meant to be multiplication signs?

Yep.

AnimeX · Jul 17, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$a, b, c \ $are positive integers such that$$

$a+b+c+ab+ac+bc+abc= 7\cdot 11 \cdot 13 - 1$

$$Find$ \ \ a,b,c$

Oh yay got it haha, never done a question like this before.

What topic is this?

AnimeX · Jul 17, 2013

Re: HSC 2013 4U Marathon

rolpsy · Jul 17, 2013

Re: HSC 2013 4U Marathon

$\\\text{Let } a, b, c \text{ be real numbers such that the roots of the cubic equation }\\[2pt]x^3 + ax^2 + bx + c = 0 \text{ are real}\\[5pt]\text{Prove these roots are bounded above by } \frac{2\sqrt{a^2 - 3b} - a}{3}$

RealiseNothing · Jul 17, 2013

Re: HSC 2013 4U Marathon

rolpsy said:
$\\\text{Let } a, b, c \text{ be real numbers such that the roots of the cubic equation }\\[2pt]x^3 + bx^2 + cx + d = 0 \text{ are real}\\[5pt]\text{Prove these roots are bounded above by } \frac{2\sqrt{a^2 - 3b} - a}{3}$

I'm guessing it's meant to be:

$x^3+ax^2+bx+c=0$

Not sure though, can you clarify before I attempt it?

rolpsy · Jul 17, 2013

Re: HSC 2013 4U Marathon

oops
haha yep

twinklegal19 · Jul 18, 2013

Re: HSC 2013 3U Marathon Thread

An interesting volumes question

It took me ages to visualise everything lol

albertcamus · Jul 18, 2013

Re: HSC 2013 3U Marathon Thread

twinklegal19 said:
An interesting volumes question

It took me ages to visualise everything lol

"Cambridge 4 unit Year 12" lawl

Carrotsticks · Jul 18, 2013

Re: HSC 2013 3U Marathon Thread

albertcamus said:
"Cambridge 4 unit Year 12" lawl

Yep, Bill Pender worked on an Extension 2 version of his well-known Extension 1 book.

albertcamus · Jul 18, 2013

Re: HSC 2013 3U Marathon Thread

Carrotsticks said:
Yep, Bill Pender worked on an Extension 2 version of his well-known Extension 1 book.

Nah as in isn't this the 3U thread?

nvm

HSC 2013 MX2 Marathon (archive) (49 Viewers)

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