HSC 2016 MX2 Integration Marathon (archive) (6 Viewers)

InteGrand · Feb 4, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
See, I already knew this conceptually in my head, but no labels are ever placed on my ideas unless they are already labelled in the external world.

Of course, you being you refused to explicitly state anything, so even if I already had that knowledge in mind it was completely useless to me without any references.

New day, new pointer created.

This is why seanieg89 said he was sure you'd seen them before.

Paradoxica · Feb 4, 2016

Re: MX2 2016 Integration Marathon

InteGrand said:
This is why seanieg89 said he was sure you'd seen them before.

Yes, but I don't know that. My memory is mostly an inaccessible fuzz rendered blurry by my psychological defense mechanism.

seanieg89 · Feb 4, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
Of course, you being you refused to explicitly state anything, so even if I already had that knowledge in mind it was completely useless to me without any references.

Of course. I try to emphasise creativity with syllabus knowledge over recollection of previously known facts from outside syllabus when I write problems. I think this is a far more important thing for a high schooler to develop (it also makes questions fairer. the same principle is generally used in writing olympiad problems, where problems are generally designed so that niche knowledge does not provide an significant advantage such as a major shortcut for a student over an equally skilled student without this niche knowledge). And of course if I am too explicit that makes things googleable / stack exchangeable.

The relevance of these Taylor estimates should not be surprising, the quantity we are looking at is (x-sin(x))/(x*sin(x)). In order to analyse what this function and its derivative do at the origin (a potential place of bad behaviour), we need some way of comparing trigonometric functions with polynomial ones.

RealiseNothing · Feb 4, 2016

Re: MX2 2016 Integration Marathon

leehuan said:
Too bad that's outside MX2 scope

It was in a past HSC paper back in like 2010 lmao.

Drsoccerball · Feb 4, 2016

Re: MX2 2016 Integration Marathon

RealiseNothing said:
It was in a past HSC paper back in like 2010 lmao.

ah yes 2010 question 8 was a 15 mark integration question?

Also attempt anyone except paradoxica his insight is to good for such a crap question:

$\int \frac{(x+1)^4(x-1)^2\sqrt{x^4+x^2+1}}{(x^2+1)^2(x^6-1)} dx$

InteGrand · Feb 4, 2016

Re: MX2 2016 Integration Marathon

Drsoccerball said:
ah yes 2010 question 8 was a 15 mark integration question?

Also attempt anyone except paradoxica his insight is to good for such a crap question:

$\int \frac{(x+1)^4(x-1)^2\sqrt{x^4+x^2+1}}{(x^2+1)^2(x^6-1)} dx$

RealiseNothing was referring to what leehuan was saying was beyond MX2 scope, namely the Taylor polynomial bounds for sin / cos etc. So he was saying that this was in a past HSC (getting you to prove those inequalities). The 2010 HSC Integration Question 8 was unrelated to this if I recall correctly.

leehuan · Feb 4, 2016

Re: MX2 2016 Integration Marathon

RealiseNothing said:
It was in a past HSC paper back in like 2010 lmao.

Obviously in an HSC paper they gave a heck load of guidance. And I don't think it was specifically a Taylor series. http://boardofstudies.nsw.edu.au/hs...doc/2010-hsc-exam-mathematics-extension-2.pdf

Paradoxica · Feb 4, 2016

Re: MX2 2016 Integration Marathon

InteGrand said:
This is why seanieg89 said he was sure you'd seen them before.

furthermore, the only inequality I could conjure up was the inequality of sinx and cosx in their definite integrals.

$\int_0^x \cos{t} $d$t > \int_0^x \sin{t} $d$t\,\,\,\forall\,\,\, 0<x<\frac{\pi}{2}$

Paradoxica · Feb 5, 2016

Re: MX2 2016 Integration Marathon

$$\noindent By substituting for $(2n+1)x$, we obtain the original integral with the limits of integration now directly proportional to $n$. Taking the limit to infinity yields the integral we seek.$ \\ $Before proceeding, we exploit the properties of even definite integrals in order to avoid dealing with negative values of $x.\\ x-\frac{x^3}{6} < \sin{x} <x< x-\frac{x^3}{6}+\frac{x^5}{120} $ and $\\ 1- \frac{x^2}{2} + \frac{x^4}{24}-\frac{x^6}{720}<1-\frac{x^2}{2} < \cos{x} < 1- \frac{x^2}{2} + \frac{x^4}{24} \\ $The above inequalities hold true for $0<x<\frac{\pi}{2}$, and are provable through differentiation, or by iteratively integrating the principle inequalities $ \cos{x} \leq 1$ and $\sin{x}<x$. Irregardless (:$\mathcal{P}$), we will be using these to restrict $g(x)$. Separately shuffling around the inequalities for $\sin{x}$ yields $\\ \frac{x^3}{6}-\frac{x^5}{120}<x-\sin{x}<\frac{x^3}{6}.$

$$\noindent Different manipulations on the inequalities yields$\\ x^2\left(1-\frac{x^2}{6}\right)<x\sin{x}<x^2 \Leftrightarrow \frac{1}{x^2} < \frac{1}{x\sin{x}} < \frac{6}{x^2(6-x^2)} \\ $which is valid since $0<x<\frac{\pi}{2} < \sqrt{6} \\ $The order of restricted chained inequalities is preserved only if we multiply them in the same order/divide them in the opposite order. So we have $ \\ \frac{\frac{x^3}{6}-\frac{x^5}{120}}{x^2}<\frac{x-\sin{x}}{x\sin{x}}<\frac{x^3}{x^2(6-x^2)} \Leftrightarrow \frac{x}{6}-\frac{x^3}{120} <g(x)< \frac{x}{(6-x^2)} \\ $It is now clear by virtue of the squeeze theorem or the sandwich theorem or the sandwich rule or the squeeze lemma or whatever you choose to call it, we define $g(0)=0.$

$$\noindent By the double angle identities, $g'(x) = \frac{1-\cos{2x}-2x^2\cos{x}}{x^2(1-\cos{2x})} \\$By setting $x \mapsto 2x $ we obtain $ 1-2x^2+\frac{2}{3}x^4-\frac{4}{45}x^6 < \cos{2x} < 1-2x^2+\frac{2}{3}x^4 \\ \Leftrightarrow 2x^2 - \frac{2}{3}x^4 < 1-\cos{2x} < 2x^2 - \frac{2}{3}x^4 + \frac{4}{45}x^6\\ \Leftrightarrow 2x^4 - \frac{2}{3}x^6 < x^2(1-\cos{2x}) < 2x^4 - \frac{2}{3}x^6 + \frac{4}{45}x^8 \\ $We can also obtain $ 2x-x^4 < 2x^2\cos{x}<2x^2-x^4+\frac{x^6}{12} $ easily.$\\ $Combining these together in a manner identical to the previous part yields $\\ \frac{x^4 (\frac{1}{3} - \frac{x^2}{12})}{x^4 (2-\frac{2}{3}x^2 +\frac{4}{45}x^4)}<g'(x)<\frac{x^4 (\frac{1}{3}+\frac{4}{45}x^4)}{x^4 (2-\frac{2}{3}x^2)} \\ \Rightarrow \frac{\frac{1}{3} - \frac{x^2}{12}}{2-\frac{2}{3}x^2 +\frac{4}{45}x^4}<g'(x)<\frac{\frac{1}{3}+\frac{4}{45}x^4}{2-\frac{2}{3}x^2}$

$$\noindent By the whatever you choose to call it theorem/lemma, $g'(0)=\frac{1}{6}. \\$ We have now established that the function is well behaved near the origin, and now proceed to investigate the convergence of the integral.$\\\frac{J'_n - I'_n}{2} = \int_0^{\frac{\pi}{2}}g(x)\sin{(2n+1)x}\,\,$d$x\\$Integrating by parts, $\frac{J'_n-I'_n}{2}=-\left[\frac{g(x)}{2n+1}\cos{(2n+1)x}\right]_0^{\frac{\pi}{2}}+\cdots \\ \frac{J'_n-I'_n}{2} = \frac{1}{2n+1}\int_0^{\frac{\pi}{2}}g'(x)\cos{(2n+1)x}\,$d$x \\ \because -1 \leq \cos{\theta} \leq 1 \Rightarrow \, -\frac{1}{2n+1}\int_0^\frac{\pi}{2}g'(x)$d$x<\frac{J'_n-I'_n}{2}<\frac{1}{2n+1}\int_0^\frac{\pi}{2}g'(x)$d$x\\ -\frac{1-\frac{2}{\pi}}{2n+1}<\frac{J'_n - I'_n}{2}<-\frac{1-\frac{2}{\pi}}{2n+1}$

$\noindent By the whatever you choose to call it theorem/lemma, $J'_\infty = I'_\infty \\ $It is easy to show that $ I'_n = I'_{n-1}$, and so we conclude $I'_\infty = I'_1 = \frac{\pi}{2} \\\therefore \int_{-\infty}^\infty \frac{\sin{x}}{x} = \pi \\ $Note that $I'$ and $J'$ have been retroactively defined to be the identically valued integrals that avoid the negative domain of integration.$

seanieg89 · Feb 5, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$$\noindent By substituting for $(2n+1)x$, we obtain the original integral with the limits of integration now directly proportional to $n$. Taking the limit to infinity yields the integral we seek.$ \\ $Before proceeding, we exploit the properties of even definite integrals in order to avoid dealing with negative values of $x.\\ x-\frac{x^3}{6} < \sin{x} <x< x-\frac{x^3}{6}+\frac{x^5}{120} $ and $\\ 1- \frac{x^2}{2} + \frac{x^4}{24}-\frac{x^6}{720}<1-\frac{x^2}{2} < \cos{x} < 1- \frac{x^2}{2} + \frac{x^4}{24} \\ $The above inequalities hold true for $0<x<\frac{\pi}{2}$, and are provable through differentiation, or by iteratively integrating the principle inequalities $ \cos{x} \leq 1$ and $\sin{x}<x$. Irregardless (:$\mathcal{P}$), we will be using these to restrict $g(x)$. Separately shuffling around the inequalities for $\sin{x}$ yields $\\ \frac{x^3}{6}-\frac{x^5}{120}<x-\sin{x}<\frac{x^3}{6}.$

$$\noindent Different manipulations on the inequalities yields$\\ x^2\left(1-\frac{x^2}{6}\right)<x\sin{x}<x^2 \Leftrightarrow \frac{1}{x^2} < \frac{1}{x\sin{x}} < \frac{6}{x^2(6-x^2)} \\ $which is valid since $0<x<\frac{\pi}{2} < \sqrt{6} \\ $The order of restricted chained inequalities is preserved only if we multiply them in the same order/divide them in the opposite order. So we have $ \\ \frac{\frac{x^3}{6}-\frac{x^5}{120}}{x^2}<\frac{x-\sin{x}}{x\sin{x}}<\frac{x^3}{x^2(6-x^2)} \Leftrightarrow \frac{x}{6}-\frac{x^3}{120} <g(x)< \frac{x}{(6-x^2)} \\ $It is now clear by virtue of the squeeze theorem or the sandwich theorem or the sandwich rule or the squeeze lemma or whatever you choose to call it, we define $g(0)=0.$

$$\noindent By the double angle identities, $g'(x) = \frac{1-\cos{2x}-2x^2\cos{x}}{x^2(1-\cos{2x})} \\$By setting $x \mapsto 2x $ we obtain $ 1-2x^2+\frac{2}{3}x^4-\frac{4}{45}x^6 < \cos{2x} < 1-2x^2+\frac{2}{3}x^4 \\ \Leftrightarrow 2x^2 - \frac{2}{3}x^4 < 1-\cos{2x} < 2x^2 - \frac{2}{3}x^4 + \frac{4}{45}x^6\\ \Leftrightarrow 2x^4 - \frac{2}{3}x^6 < x^2(1-\cos{2x}) < 2x^4 - \frac{2}{3}x^6 + \frac{4}{45}x^8 \\ $We can also obtain $ 2x-x^4 < 2x^2\cos{x}<2x^2-x^4+\frac{x^6}{12} $ easily.$\\ $Combining these together in a manner identical to the previous part yields $\\ \frac{x^4 (\frac{1}{3} - \frac{x^2}{12})}{x^4 (2-\frac{2}{3}x^2 +\frac{4}{45}x^4)}<g'(x)<\frac{x^4 (\frac{1}{3}+\frac{4}{45}x^4)}{x^4 (2-\frac{2}{3}x^2)} \\ \Rightarrow \frac{\frac{1}{3} - \frac{x^2}{12}}{2-\frac{2}{3}x^2 +\frac{4}{45}x^4}<g'(x)<\frac{\frac{1}{3}+\frac{4}{45}x^4}{2-\frac{2}{3}x^2}$

$$\noindent By the whatever you choose to call it theorem/lemma, $g'(0)=\frac{1}{6}. \\$ We have now established that the function is well behaved near the origin, and now proceed to investigate the convergence of the integral.$\\\frac{J'_n - I'_n}{2} = \int_0^{\frac{\pi}{2}}g(x)\sin{(2n+1)x}\,\,$d$x\\$Integrating by parts, $\frac{J'_n-I'_n}{2}=-\left[\frac{g(x)}{2n+1}\cos{(2n+1)x}\right]_0^{\frac{\pi}{2}}+\cdots \\ \frac{J'_n-I'_n}{2} = \frac{1}{2n+1}\int_0^{\frac{\pi}{2}}g'(x)\cos{(2n+1)x}\,$d$x \\ \because -1 \leq \cos{\theta} \leq 1 \Rightarrow \, -\frac{1}{2n+1}\int_0^\frac{\pi}{2}g'(x)$d$x<\frac{J'_n-I'_n}{2}<\frac{1}{2n+1}\int_0^\frac{\pi}{2}g'(x)$d$x\\ -\frac{1-\frac{2}{\pi}}{2n+1}<\frac{J'_n - I'_n}{2}<-\frac{1-\frac{2}{\pi}}{2n+1}$

$\noindent By the whatever you choose to call it theorem/lemma, $J'_\infty = I'_\infty \\ $It is easy to show that $ I'_n = I'_{n-1}$, and so we conclude $I'_\infty = I'_1 = \frac{\pi}{2} \\\therefore \int_{-\infty}^\infty \frac{\sin{x}}{x} = \pi \\ $Note that $I'$ and $J'$ have been retroactively defined to be the identically valued integrals that avoid the negative domain of integration.$

Bingo

.

A classic example of how good IBP is for bounding things that oscillate fast.

P.s. with your discussion of g'(0), you have actually only found the limit of g'(x) as x->0, you have not established differentiability of g at the origin, nor have you shown that g' is continuous there.

This is not a major omission though, the same Taylor inequalities let us analyse (g(x)-g(0))/x=g(x)/x as x->0 which will be 1/6 as we need for continuity.

Paradoxica · Feb 5, 2016

Re: MX2 2016 Integration Marathon

Drsoccerball said:
ah yes 2010 question 8 was a 15 mark integration question?

Also attempt anyone except paradoxica his insight is to good for such a crap question:

$\int \frac{(x+1)^4(x-1)^2\sqrt{x^4+x^2+1}}{(x^2+1)^2(x^6-1)} dx$

$$By inspection, the integral is $\sec^{-1}\left(x+\frac{1}{x} \right)+2\sqrt{\frac{x^4+x^2+1}{(x^2+1)^2}}+C$

leehuan · Feb 5, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$$By inspection$

What

Paradoxica · Feb 5, 2016

Re: MX2 2016 Integration Marathon

leehuan said:
What

sarcastic reply to his claim of my insight.

porcupinetree · Feb 5, 2016

Re: MX2 2016 Integration Marathon

leehuan said:
What

Paradoxica said:
sarcastic reply to his claim of my insight.

Tbh, wouldn't have been surprised if you were being serious

InteGrand · Feb 5, 2016

Re: MX2 2016 Integration Marathon

porcupinetree said:
Tbh, wouldn't have been surprised if you were being serious

I thought he was being serious.

Paradoxica · Feb 5, 2016

Re: MX2 2016 Integration Marathon

InteGrand said:
I thought he was being serious.

I'm not THAT good.

InteGrand · Feb 5, 2016

Re: MX2 2016 Integration Marathon

$\nondent \textbf{A NEW QUESTION}$

$\noindent Let $x\geq 0$. Let $n$ be a positive integer.$

$\noindent (i) Show that$

$$-x^n \leq \frac{1}{1+x} -\left(1 - x + x^2 -x^3 +\cdots +(-1)^{n-1} x^{n-1} \right) \leq x^n$.$

$\noindent (ii) Hence, use integration to show that$

$$ -\frac{1}{n+1} \leq \ln 2 -\left(1-\frac{1}{2} +\frac{1}{3} -\frac{1}{4} +\cdots + (-1)^{n-1} \cdot \frac{1}{n} \right) \leq \frac{1}{n+1}$.$

$\noindent (iii) Justifying your answer, deduce the value of the infinite sum $1-\frac{1}{2} + \frac{1}{3} -\frac{1}{4} +\cdots$.$

RealiseNothing · Feb 5, 2016

Re: MX2 2016 Integration Marathon

leehuan said:
Obviously in an HSC paper they gave a heck load of guidance. And I don't think it was specifically a Taylor series. http://boardofstudies.nsw.edu.au/hs...doc/2010-hsc-exam-mathematics-extension-2.pdf

Just found it, question 7)a) from 2007.

http://www.boardofstudies.nsw.edu.au/hsc_exams/exam-papers-2007/index5.html

They give barely any guidance and it's EXACTLY what seanie posted.

InteGrand · Feb 5, 2016

Re: MX2 2016 Integration Marathon

RealiseNothing said:
Just found it, question 7)a) from 2007.

http://www.boardofstudies.nsw.edu.au/hsc_exams/exam-papers-2007/index5.html

They give barely any guidance and it's EXACTLY what seanie posted.

I suppose they give us "guidance" in the sense that they tell us what the inequality is we're supposed to prove, whereas seanieg89 wanted people tackling his question to come up with relevant inequalities etc. themselves.

seanieg89 · Feb 5, 2016

Re: MX2 2016 Integration Marathon

Oh lol that was actually my HSC year.

I remember seeing it earlier than that even, I swear something like it is in the Cambridge book but I don't have that on me and it's not really important.

HSC 2016 MX2 Integration Marathon (archive) (6 Viewers)

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what is that?It is Cowpea

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not actually a porcupine

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what is that?It is Cowpea

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