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HSC 2013 MX2 Marathon (archive) (1 Viewer)

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RealiseNothing

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Re: HSC 2013 4U Marathon

I made this question just now:





Noting that:



By expansion:



Also using the first identity:



Collecting similar powers but opposite in sign:



Using the first identity again and applying the symmetry property of combinatorics:



Now we integrate both sides:





Now when we substitute in the limits, we end up with many:



Where is an integer. They are all equal to 0, so it all cancels out besides the last term, which leaves us with:



Now changing the combinatorics into factorial notion, we obtain:



Taking the power of 2 to the other side gives us the required result:

 
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RealiseNothing

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Re: HSC 2013 4U Marathon

Find and justify the value of the following in terms of :



When and is an integer.

Hence find:



When

Note this is more of a research kinda thing rather than an actual problem. I found an answer to this but it makes no sense at all besides all my working seeming legit. Curious to see what people get.
 
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RealiseNothing

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Re: HSC 2013 4U Marathon

^^^ this question is trippin' me out. I was playing around with something and found the above result. But the result doesnt make that much sense.
 

seanieg89

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Re: HSC 2013 4U Marathon

Find and justify the value of the following in terms of :



When and is an integer.

Hence find:



When

Note this is more of a research kinda thing rather than an actual problem. I found an answer to this but it makes no sense at all besides all my working seeming legit. Curious to see what people get.
Isn't the first thing undefined at theta=0?
 

RealiseNothing

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Re: HSC 2013 4U Marathon

Isn't the first thing undefined at theta=0?
You end up with

So you have to manipulate it so it doesn't end up like this (well that's what I'm intending). You could always L'Hopital's it but that's no fun.

Unless you can't do this? (because upon manipulation you get a value). I'm going to take a guess and say you can't.
 

seanieg89

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Re: HSC 2013 4U Marathon

You end up with

So you have to manipulate it so it doesn't end up like this (well that's what I'm intending). You could always L'Hopital's it but that's no fun.

Unless you can't do this? (because upon manipulation you get a value). I'm going to take a guess and say you can't.
Well firstly you asked for its value at theta=0 rather than its limit as theta->0. But assuming you meant the latter then your second question might not work out the way you want, as the two limiting processes (theta->0,n->inf) don't have to commute. I will think about it a little more properly now.
 

RealiseNothing

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Re: HSC 2013 4U Marathon

Well firstly you asked for its value at theta=0 rather than its limit as theta->0. But assuming you meant the latter then your second question might not work out the way you want, as the two limiting processes (theta->0,n->inf) don't have to commute. I will think about it a little more properly now.
Alright thanks. I was thinking more along the lines of but didn't really put that to paper.
 

seanieg89

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Re: HSC 2013 4U Marathon

Yep, so then the first question is fine and pretty straightforward using mx2 methods: 1/n.

The latter question turns out to be true taking limits in either order, ie:

 

RealiseNothing

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Re: HSC 2013 4U Marathon

You can use a really simple proof by contradiction to get the same result:

Assume that there is a finite amount of primes, and let there be amount, denoted by for the k'th prime.

Then we do:



This makes a prime which has not be accounted for, and hence there is more than amount of primes. Thus it is a contradiction, and the result follows.

I'll attempt your way though.
 

Sy123

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Re: HSC 2013 4U Marathon

You can use a really simple proof by contradiction to get the same result:

Assume that there is a finite amount of primes, and let there be amount, denoted by for the k'th prime.

Then we do:



This makes a prime which has not be accounted for, and hence there is more than amount of primes. Thus it is a contradiction, and the result follows.

I'll attempt your way though.
Wow, very good. This is just a 'filler' question while I find/make something else to ask, but I want to stick with Complex Numbers/Polynomials so others can contribute.

Also, this question is from another STEP paper but I forget which one.
 
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seanieg89

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Re: HSC 2013 4U Marathon

You can use a really simple proof by contradiction to get the same result:

Assume that there is a finite amount of primes, and let there be amount, denoted by for the k'th prime.

Then we do:



This makes a prime which has not be accounted for, and hence there is more than amount of primes. Thus it is a contradiction, and the result follows.

I'll attempt your way though.
Lies! p1p2...pn doesn't have to be prime!

But it does have to contain a prime not previously listed as a factor.
 

Sy123

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Re: HSC 2013 4U Marathon























 
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RivalryofTroll

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Re: HSC 2013 4U Marathon

Just double checking my answers for a question:

If (1+i) is a root of the equation:

x^3 - ax^2 + bx - 4 = 0

Find values a & b. Hence, solve the equation.

I got a = 4 and b = 6.

Then roots are (1+i), (1-i), 2?
 

Sy123

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Re: HSC 2013 4U Marathon

Just double checking my answers for a question:

If (1+i) is a root of the equation:

x^3 - ax^2 + bx - 4 = 0

Find values a & b. Hence, solve the equation.

I got a = 4 and b = 6.

Then roots are (1+i), (1-i), 2?
You can find all roots without finding a and b by considering conjugate root theorem and product of roots:



And then considering sum of roots 2 at a time and sum of roots we can then find a and b. You are correct.
 
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