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HSC 2015 MX2 Marathon ADVANCED (archive) (2 Viewers)

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Drsoccerball

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Re: HSC 2015 4U Marathon - Advanced Level

just clarify: if the last term is +c, what is c? and what is \theta? do u mean finding n given that equation is true for all \theta? or do u mean finding both n and \theta ?
Theta is just a random value and c is a random constant just find n
 

seanieg89

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Re: HSC 2015 4U Marathon - Advanced Level

Self made question... goodluck;):


Can you also elaborate on the RHS using some kind of summation notation? It is not clear what fills the "..." given that there are cosines at the start but not the end.
 

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Re: HSC 2015 4U Marathon - Advanced Level

Can you also elaborate on the RHS using some kind of summation notation? It is not clear what fills the "..." given that there are cosines at the start but not the end.
i can answer your question
but im extremely confused about how theta and c are random
 

Drsoccerball

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Re: HSC 2015 4U Marathon - Advanced Level

Can you also elaborate on the RHS using some kind of summation notation? It is not clear what fills the "..." given that there are cosines at the start but not the end.
Oh shit it should have cos(n-n)theta at the end my bad
 

Drsoccerball

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Re: HSC 2015 4U Marathon - Advanced Level

should i give a hint?
 

seanieg89

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Re: HSC 2015 4U Marathon - Advanced Level

fixed version
Using standard complex numbers tricks (raise a complex number on the unit circle plus it's reciprocal to the power 2n in two different ways) and exploiting that cos is even we can write the RHS as:

.

Then .

Then for any n such that the left hand side is positive and does not exceed 1, we take the 2 (or 1) real (2n)-th roots, and apply the general formula to solve cos(x)=a.

This is the complete set of solutions for the problem as posed.
 

seanieg89

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Re: HSC 2015 4U Marathon - Advanced Level

You can find n without using in terms of c and theta
No, you certainly can't. Otherwise suppose the equation is true for a given n and for any c and any theta.

Then by setting (theta,c)=(0,1) and then (theta,c)=(0,2), we are able to use these two equations to deduce that 1=2.

As far as I know, 1 does not equal 2.
 

Drsoccerball

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Re: HSC 2015 4U Marathon - Advanced Level

Using standard complex numbers tricks (raise a complex number on the unit circle plus it's reciprocal to the power 2n in two different ways) and exploiting that cos is even we can write the RHS as:

.

Then .

Then for any n such that the left hand side is positive and does not exceed 1, we take the 2 (or 1) real (2n)-th roots, and apply the general formula to solve cos(x)=a.

This is the complete set of solutions for the problem as posed.
you do use:




 

seanieg89

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Re: HSC 2015 4U Marathon - Advanced Level

you do use:




Note that there are an ODD number of terms in (z+1/z)^(2n).

That is why the 2nCn is there in my work and there isn't just a division by 2.

Also note that my 2n is your n. I wrote it that way to incorporate the even condition immediately.
 

Drsoccerball

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Re: HSC 2015 4U Marathon - Advanced Level

Note that there are an ODD number of terms in (z+1/z)^(2n).

That is why the 2nCn is there in my work and there isn't just a division by 2.

Also note that my 2n is your n. I wrote it that way to incorporate the even condition immediately.
Squared both sides?
 

seanieg89

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Re: HSC 2015 4U Marathon - Advanced Level

Squared both sides?
No, it is just a relabelling so we don't have to always pay attention to n's parity. This is different to squaring both sides of your equation. (Which is generally invalid as a solution method.)
 

Sy123

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Re: HSC 2015 4U Marathon - Advanced Level

 
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