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HSC 2013 MX2 Marathon (archive) (3 Viewers)

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nrlwinner

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Re: HSC 2013 4U Marathon

Yeah I did it again and you are correct



Ah yes reading the solution again it is incorrect, what I had in mind was


... (1)

.... (2)

This is done by subbing in the AM-GM inequality, all substitutions are positive since x_k <= 1/2

Dividing (2) by (1), which we can do since all things are positive.

We get (P is what we want)

Then flip everything, when we do that the inequality sign switches, yeilding P > (that)

But I was skimming through her solution and wasn't really paying attention, which is why I couldn't see that.

=====


I'm making way too many mistakes :(

(there is no mistake in this one)






Here's my go at the question. I'll just do the left inequality and presume the right inequality is similar.





 

Sy123

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Re: HSC 2013 4U Marathon

Here's my go at the question. I'll just do the left inequality and presume the right inequality is similar.





Yep, the right inequality is kinda similar.

========







 

seanieg89

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Re: HSC 2013 4U Marathon



If one was very bored, they could expand this final expression out in terms of the trigonometrically defined f.

(Many such formulae for primes exist, but the elementary ones are all basically algorithms for checking primality in disguise. The algorithm in this case is the process of checking divisibility by all smaller natural numbers.)
 

RealiseNothing

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Re: HSC 2013 4U Marathon

If they have formulae for primes, then what is preventing them from finding very very very large prime numbers larger than the largest known prime? Is it how advanced technology and computational programs are or what?
 

Sy123

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Re: HSC 2013 4U Marathon

Did you have a different way of solving it?
I modified a hsc question, it first asked to prove |a| = 1 and Re(a) = (1-p)/2, So I decided to omit this and and ask for the whole number since it can be found with those properties



If one was very bored, they could expand this final expression out in terms of the trigonometrically defined f.

(Many such formulae for primes exist, but the elementary ones are all basically algorithms for checking primality in disguise. The algorithm in this case is the process of checking divisibility by all smaller natural numbers.)
I see, thanks for taking the time to write your solution out.

============



 

RealiseNothing

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Re: HSC 2013 4U Marathon

I modified a hsc question, it first asked to prove |a| = 1 and Re(a) = (1-p)/2, So I decided to omit this and and ask for the whole number since it can be found with those properties



I see, thanks for taking the time to write your solution out.

============



The second part is easily done without the first as and the summation in part (ii) contains all of the terms in this sum.
 

seanieg89

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Re: HSC 2013 4U Marathon

If they have formulae for primes, then what is preventing them from finding very very very large prime numbers larger than the largest known prime? Is it how advanced technology and computational programs are or what?
Computational complexity. This algorithm is horrendously inefficient.

More efficient algorithms are exactly how they found the current largest known primes.
 

Galactic Jam

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Re: HSC 2013 4U Marathon

I only just found this thread, these questions are so scary :S
Hoping I can at least figure some of these out.
 

Sy123

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Re: HSC 2013 4U Marathon

A good Sydney grammar question



 

seanieg89

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Re: HSC 2013 4U Marathon

A good Sydney grammar question



Was this the only part of the question? If so it will be interesting seeing how people justify that u_n tends to infinity. The easiest way I can see uses monotone convergence.
 

Sy123

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Re: HSC 2013 4U Marathon

Was this the only part of the question? If so it will be interesting seeing how people justify that u_n tends to infinity. The easiest way I can see uses monotone convergence.
Yeah this was it, I was able to show that and , is this enough to justify convergence given the of the final step in this problem? (which I don't want to type out here just yet)
I guess this is probably a form of monotone convegence
 

seanieg89

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Re: HSC 2013 4U Marathon

Yeah this was it, I was able to show that and , is this enough to justify convergence given the of the final step in this problem? (which I don't want to type out here just yet)
I guess this is probably a form of monotone convegence
Nope, sequences satisfying the properties you provided can either diverge or converge. An example of a convergent one being 1-1/n.
 

seanieg89

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Re: HSC 2013 4U Marathon

Note: I am sure it can be done without monotone convergence, I just think that it is a stiff ask for nearly any mx2 student.
 
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