HSC 2012 MX1 Marathon #2 (archive) (1 Viewer)

Sy123

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Re: HSC 2012 Marathon :)

Solution

I figured it would just take way too long to do on latex, so I just did it on paper

My Question will be posted if I find out that it is possible within 3unit boundaries.
 
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Re: HSC 2012 Marathon :)

Not sure if this question works but oh well

 

Timske

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Re: HSC 2012 Marathon :)

wow never seen that b4
 
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Re: HSC 2012 Marathon :)

Sy123, good spotting! I'd actually intended the sin28 to turn back into a cos62 which would give a slightly nicer final answer, but same same it's fine.
 
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Re: HSC 2012 Marathon :)

Having a stab at that one...It's probably fudged...but just trying.





Ok the equation works on codecogs perfectly but doesn't here. WTF
 

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Re: HSC 2012 Marathon :)

Don't worry carrot lol. For some reason lately my upload speed has been terrible so I can't even upload a simple gif i downloaded off codecogs.

ANYWAY

What I did was Consider (1+1/n)^n and expand with binomial theorem: Cn0 +Cn1(1/n)+Cn2(1/n^2)=...

Then expand out into factorial notation

= 1+ n!/1!(n-1)! * 1/n ...
=1+ 1/1! + (n-1)/2!*n +...
=1+1/1! + (1/2!)* (1+1/n) and the rest of the brackets end up with 1 +/- x/n, n^2 so they all disappear once we take the limit as n=> infinity since 1/n goes off to 0. So you're left with 1+1/1!+1/2!+1/3!... which is the required result.
 
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Re: HSC 2012 Marathon :)

The image showed up, have a look lol. I will lmao if it's wrong haha. spent a good 30 minutes.
 

seanieg89

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Re: HSC 2012 Marathon :)

Don't worry carrot lol. For some reason lately my upload speed has been terrible so I can't even upload a simple gif i downloaded off codecogs.

ANYWAY

What I did was Consider (1+1/n)^n and expand with binomial theorem: Cn0 +Cn1(1/n)+Cn2(1/n^2)=...

Then expand out into factorial notation

= 1+ n!/1!(n-1)! * 1/n ...
=1+ 1/1! + (n-1)/2!*n +...
=1+1/1! + (1/2!)* (1+1/n) and the rest of the brackets end up with 1 +/- x/n, n^2 so they all disappear once we take the limit as n=> infinity since 1/n goes off to 0. So you're left with 1+1/1!+1/2!+1/3!... which is the required result.
A classic fallacious argument in analysis. The number of terms in this series is not fixed, so we cannot simply take the limits of each term individually and add them. If things like this were legit then we would have for example

lim (1+1/n)^n = lim 1^n = 1, as each term in the n-fold product (1+1/n)(1+1/n)...(1+1/n) tends to 1.
 

seanieg89

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Re: HSC 2012 Marathon :)

A little iffy about that question, because the NSW Syllabus only ever defines integral values of n (for the Binomial Expansion)...
That is all this question needs :), but it is sort of uni analysis-y.
 

Sy123

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Re: HSC 2012 Marathon :)

Im sorry if its not that good of a question, I just thought I would be a little creative rather than grabbing some question from a trial or textbook or something. :/

But either way, great work asianese, that is, if that proof is valid, I did something slightly differently.
I proved the limit of the binomial expansion's first few terms is 1+1/1!+1/2!

Then I proved it for integer 'r'



So I proved it for any integer r as well, however if any of the math pros could tell us the problem with our solution, and if so, is there a 3u solution possible for this proof, because this is the proof I had in mind when I asked it.

Thanks!
 
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Re: HSC 2012 Marathon :)

Yeah I also had a little suspicion because I didn't know what term to do it up to. If I did it to the nth term that would mean that the nth term disappears and there isn't any n! left off... Oh well, learn something everyday.
 

seanieg89

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Re: HSC 2012 Marathon :)

A classic fallacious argument in analysis. The number of terms in this series is not fixed, so we cannot simply take the limits of each term individually and add them. If things like this were legit then we would have for example

lim (1+1/n)^n = lim 1^n = 1, as each term in the n-fold product (1+1/n)(1+1/n)...(1+1/n) tends to 1.
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.
 

Sanjeet

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Re: HSC 2012 Marathon :)

asianese's solution made complete sense to me.
 

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