HSC 2012 MX1 Marathon #2 (archive) (3 Viewers)

Sy123 · Jul 14, 2012

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.

asianese · Jul 14, 2012

Re: HSC 2012 Marathon

Not sure if this question works but oh well

$Find the general solution to $ \cos(3\theta +2)=\sin54 \cos26-\sin26 \sin36. $ Give the answer in degrees.$

Timske · Jul 14, 2012

Re: HSC 2012 Marathon

asianese said:
Not sure if this question works but oh well

$Find the general solution to $ \cos(3\theta +2)=\sin54 \cos26-\sin26 \sin36. $ Give the answer in degrees.$

is it meant to be sin54cos26?

asianese · Jul 14, 2012

Re: HSC 2012 Marathon

Yes.

Timske · Jul 14, 2012

Re: HSC 2012 Marathon

wow never seen that b4

Sy123 · Jul 14, 2012

Re: HSC 2012 Marathon

asianese said:
Not sure if this question works but oh well

$Find the general solution to $ \cos(3\theta +2)=\sin54 \cos26-\sin26 \sin36. $ Give the answer in degrees.$

$\cos({3\theta +2})=\sin54 \cos26 -sin26 \cos54 \\ \\ \cos(3\theta +2)=\sin(28) \\ \\ 3\theta +2=360k \pm cos^{-1}(sin28) \\ \\ \therefore \theta=\frac{1}{3} (360k \pm cos^{-1}(sin28)-2)$

Ok time for my question:

$$Using Binomial Expansion and given that$ \ \ \ e=\lim_{n \to \infty}\left(1+\frac{1}{n}\right)^n \\ \\ $Prove that$ \\ \\ e=1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...$

Have fun!

asianese · Jul 14, 2012

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.

asianese · Jul 15, 2012

Re: HSC 2012 Marathon

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

$$Consider $ \left(1+\frac{1}{n}\right)^n\\ \begin{align*} \left( 1+\frac{1}{n} \right)^n &= \binom{n}{0}+\binom{n}{1} \left(\frac{1}{n}\right)+\binom{n}{2} \left(\frac{1}{n}\right)^2+\binom{n}{3} \left(\frac{1}{n}\right)^3+\ldots + \binom{n}{n} \left(\frac{1}{n}\right)^n\\ &1+\frac{n!}{1!(n-1)!n}+\frac{n!}{2!(n-2)!n^2}+\frac{n!}{3!(n-3)!n^3}+\ldots \ldots \\ &=1+\frac{1}{1!}+\frac{n-1}{2!n}+\frac{(n-1)(n-2)}{3!n^2}+\ldots \ldots \\ &=1+\frac{1}{1!}+\frac{1}{2!}\left(1+\frac{1}{n}\right)+\frac{1}{3!}\left(\frac{n^2-3n+2}{n^2}\right)+\ldots \ldots \\ &=1+\frac{1}{1!} + \frac{1}{2!}\left(1+\frac{1}{n}\right)+\frac{1}{3!}\left(1-\frac{3}{n}+\frac{2}{n^2}\right)+\ldots \ldots \\ \end{align*} \\ $Consider $ \lim_{n \to \infty}\left(1+\frac{1}{n}\right)^n= 1+\frac{1}{1!}+\frac{1}{2!}(1+0)+\frac{1}{3!}(1-0-0)+\ldots \ldots \\ = 1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\ldots \ldots$

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

Sanjeet · Jul 15, 2012

Re: HSC 2012 Marathon

^Invalid equation

seanieg89 · Jul 15, 2012

Re: HSC 2012 Marathon

Equation too long haha.

Carrotsticks · Jul 15, 2012

Re: HSC 2012 Marathon

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

$$Consider $ \left(1+\frac{1}{n}\right)^n\\ \begin{align*} \left( 1+\frac{1}{n} \right)^n &= \binom{n}{0}+\binom{n}{1} \left(\frac{1}{n}\right)+\binom{n}{2} \left(\frac{1}{n}\right)^2+\binom{n}{3} \left(\frac{1}{n}\right)^3+\ldots + \binom{n}{n} \left(\frac{1}{n}\right)^n\\ &= 1+\frac{n!}{1!(n-1)!n}+\frac{n!}{2!(n-2)!n^2}+\frac{n!}{3!(n-3)!n^3}+\ldots \ldots \\ &=1+\frac{1}{1!}+\frac{n-1}{2!n}+\frac{(n-1)(n-2)}{3!n^2}+\ldots \ldots \\ &=1+\frac{1}{1!}+\frac{1}{2!}\left(1+\frac{1}{n}\right)+\frac{1}{3!}\left(\frac{n^2-3n+2}{n^2}\right)+\ldots \ldots \\ &= 1+\frac{1}{1!} + \frac{1}{2!}\left(1+\frac{1}{n}\right)+\frac{1}{3!}\left(1-\frac{3}{n}+\frac{2}{n^2}\right)+\ldots \ldots \end{align*}$

$\\ $Consider $ \lim_{n \to \infty}\left(1+\frac{1}{n}\right)^n= 1+\frac{1}{1!}+\frac{1}{2!}(1+0)+\frac{1}{3!}(1-0-0)+\ldots \ldots \\ = 1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\ldots \ldots\\ $Hence $\begin{align*} e&=\lim_{n\to \infty}\left(1+\frac{1}{n} \right)^n &= 1+\frac{1}{1!}+\frac{1}{2!}+\ldots \ldots\end{align*}$

Tried to fix for you.

Too hard checking the first chunk of TeX =(

asianese · Jul 15, 2012

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.

asianese · Jul 15, 2012

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 · Jul 15, 2012

Re: HSC 2012 Marathon

asianese said:
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.

Carrotsticks · Jul 15, 2012

Re: HSC 2012 Marathon

Sy123 said:
$$Using Binomial Expansion and given that$ \ \ \ e=\lim_{n \to \infty}\left(1+\frac{1}{n}\right)^n \\ \\ $Prove that$ \\ \\ e=1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...$

Have fun!

A little iffy about that question, because the NSW Syllabus only ever defines integral values of n (for the Binomial Expansion)...

seanieg89 · Jul 15, 2012

Re: HSC 2012 Marathon

Carrotsticks said:
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 · Jul 15, 2012

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'

$\binom{n}{r}\frac{1}{n^r} \\ \\ \frac{n!}{(n-r)!r!} \times \frac{1}{n^r} \\ \\ \frac{n(n-1)(n-2)...(n-r+1)}{r! n^r} \\ \\ \frac{(n-1)(n-2)...(n-r+1)}{r!n^{r-1}} \\ \\ \frac{n^{r-1}}{r! n^{r-1}}+\frac{k n^{r-2}}{r! n^{r-1}}+\frac{k_1 n^{r-3}}{r! n^{r-1}}+...+\frac{k_2}{r! n^{r-1}} \\ \\ as \ \ n \to \infty \\ \\ \frac{1}{r!}+0+0+... \\ \\ \therefore $The binomial expansion of the limit, will work out for the rth term of any 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!

asianese · Jul 15, 2012

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 · Jul 15, 2012

Re: HSC 2012 Marathon

seanieg89 said:
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 · Jul 15, 2012

Re: HSC 2012 Marathon

asianese's solution made complete sense to me.

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