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HSC 2012 MX1 Marathon #2 (archive) (2 Viewers)

deswa1

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Re: HSC 2012 Marathon :)

This is the problem with your argument Sy123, you cannot treat such a series "term by term". There isn't really a proof of this equality that a 3U student could be expected to produce...I will post a sandwich argument a bit later.
You can prove this identity with entirely 3U methods. I remember an extension question in Cambridge went through it (it used the upper and lower bound rectangles)- I'll try and find the question.
 

Carrotsticks

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Re: HSC 2012 Marathon :)

What part of his solution would a 3u student not be expected to produce? Just curious
Everything he did was within the scope of the Syllabus for Extension 1, so a good Extension 1 student would be able to 'understand' everything he did.

HOWEVER, he assumed that the behaviour of the binomial expansion is the same for infinitely large values of n, which CANNOT be assumed. It has to be justified, refer to seanieg89's post about treating an infinite number of terms.

Also deswa1, the proof for it has been in the HSC beforehand (or were you referring to the series expansion? If so, it's in the BOS MX2 Seminar booklet):

 

seanieg89

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Re: HSC 2012 Marathon :)

You can prove this identity with entirely 3U methods. I remember an extension question in Cambridge went through it (it used the upper and lower bound rectangles)- I'll try and find the question.
Yep, the methods are 3U but I more meant that we cannot expect a 3U student to answer this kind of question without being led through a specific argument...there are too many stumbling blocks.
 

Sy123

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Re: HSC 2012 Marathon :)

Time for another question I guess.



Also Im trusting the fact that the values for the summation notation, are beside the sigma because of how I put them in a fraction?


Ok its all fine, you can now attempt the question, the typo was corrected.

Also please post a question after your done...
 
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Sanjeet

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Re: HSC 2012 Marathon :)

Time for another question I guess.



Also Im trusting the fact that the values for the summation notation, are beside the sigma because of how I put them in a fraction?


Ok its all fine, you can now attempt the question, the typo was corrected.

Also please post a question after your done...

I'm guessing you wanted it like this? If so the tex code is \sum_{r=1}^{n}
 

Sy123

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Re: HSC 2012 Marathon :)

Thats pretty much what I did, except Im guessing that because I condensed it so much, the latex editor decided to put the values there.
 

OMGITzJustin

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Re: HSC 2012 Marathon :)

integral of (a+(1/a))^n da, a is a complex number (over C - unit circle)
 

seanieg89

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Re: HSC 2012 Marathon :)

integral of (a+(1/a))^n da, a is a complex number (over C - unit circle)
Haha how is this anywhere near HSC level?

Assuming the integration is counterclockwise, the answer is:

I=0 if n is an even non-negative integer
I=2*pi*i*nC_{(n+1)/2} if n is an odd non-negative integer.
 

Aesytic

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Re: HSC 2012 Marathon :)

Using cosine rule,
cosA = [b^2 + c^2 - a^2]/2bc
2cos^2(A/2) - 1 = [b^2 + c^2 - a^2]/2bc
2cos^2(A/2) = [b^2 + c^2 - a^2]/2bc + 1
= [b^2 + c^2 - a^2 + 2bc]/2bc
= [(b+c)^2 - a^2]/2bc
= [(b+c+a)(b+c-a)]/2bc
= [p((p-a)-a)]/2bc
= [p(p-2a)]/2bc
cos^2(A/2) = [p(p-2a)]/4bc
cos(A/2) = +- sqrt{[p(p-2a)]/4bc}
= +- 1/2*sqrt{[p(p-2a)]/bc}
but 0 < A < 180
0 < A/2 < 90
.'. cos(A/2) > 0
.'. cos(A/2) = 1/2*sqrt{[p(p-2a)]/bc}
 

barbernator

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Re: HSC 2012 Marathon :)

<a href="http://www.codecogs.com/eqnedit.php?latex=cos(A)=\frac{b^{2}@plus;c^{2}-a^{2}}{2bc}\\ 2cos^{2}(\frac{A}{2})-1=\frac{b^{2}@plus;c^{2}-a^{2}}{2bc}\\ cos^{2}(\frac{A}{2})=\frac{b^{2}@plus;c^{2}-a^{2}@plus;2bc}{4bc}\\ cos(\frac{A}{2})=\sqrt{\frac{b^{2}@plus;c^{2}-a^{2}@plus;2bc}{4bc}}~(taking~@plus;ve)\\ ~~~~~~~~~~~=\sqrt{\frac{(b@plus;c)^{2}-a^{2}}{4bc}}\\ ~~~~~~~~~~~=\sqrt{\frac{(b@plus;c-a)(b@plus;c@plus;a)}{4bc}}\\ ~~~~~~~~~~~=\frac{1}{2}\sqrt{\frac{(p)(p-2a)}{bc}}~as~required\\" target="_blank"><img src="http://latex.codecogs.com/gif.latex?cos(A)=\frac{b^{2}+c^{2}-a^{2}}{2bc}\\ 2cos^{2}(\frac{A}{2})-1=\frac{b^{2}+c^{2}-a^{2}}{2bc}\\ cos^{2}(\frac{A}{2})=\frac{b^{2}+c^{2}-a^{2}+2bc}{4bc}\\ cos(\frac{A}{2})=\sqrt{\frac{b^{2}+c^{2}-a^{2}+2bc}{4bc}}~(taking~+ve)\\ ~~~~~~~~~~~=\sqrt{\frac{(b+c)^{2}-a^{2}}{4bc}}\\ ~~~~~~~~~~~=\sqrt{\frac{(b+c-a)(b+c+a)}{4bc}}\\ ~~~~~~~~~~~=\frac{1}{2}\sqrt{\frac{(p)(p-2a)}{bc}}~as~required\\" title="cos(A)=\frac{b^{2}+c^{2}-a^{2}}{2bc}\\ 2cos^{2}(\frac{A}{2})-1=\frac{b^{2}+c^{2}-a^{2}}{2bc}\\ cos^{2}(\frac{A}{2})=\frac{b^{2}+c^{2}-a^{2}+2bc}{4bc}\\ cos(\frac{A}{2})=\sqrt{\frac{b^{2}+c^{2}-a^{2}+2bc}{4bc}}~(taking~+ve)\\ ~~~~~~~~~~~=\sqrt{\frac{(b+c)^{2}-a^{2}}{4bc}}\\ ~~~~~~~~~~~=\sqrt{\frac{(b+c-a)(b+c+a)}{4bc}}\\ ~~~~~~~~~~~=\frac{1}{2}\sqrt{\frac{(p)(p-2a)}{bc}}~as~required\\" /></a>
 

OMGITzJustin

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Re: HSC 2012 Marathon :)

oh sorry boys, wrong thread hahahaha (i usually type in "marathon" in my search bar) saw seanieg and typed away :p

lets take a break from the harder ones, find the integral of sin^2 (1/2x)
 

barbernator

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Re: HSC 2012 Marathon :)

oh sorry boys, wrong thread hahahaha (i usually type in "marathon" in my search bar) saw seanieg and typed away :p

lets take a break from the harder ones, find the integral of sin^2 (1/2x)
again, assuming u didn't type out the question wrongly, this is 3 unit. did you mean the integral of sin^2(x/2)?
 

Timske

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Re: HSC 2012 Marathon :)

lets do some 3d pm
 

Sy123

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Re: HSC 2012 Marathon :)



Enjoy.

Typo has been fixed.
 
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