HSC 2013-14 MX1 Marathon (archive) (4 Viewers)

asianese · Sep 23, 2013

Re: HSC 2013 3U Marathon Thread

Just because I am in a LaTexy mood:

$\frac{d}{dx}\sin x = \lim_{h\to 0} \dfrac{\sin(x+h)-\sin x}{h} \\ = \lim_{h\to 0} \dfrac{\sin x \cos h + \sin h \cos x -\sin x}{h} \\ =\lim_{h\to 0} \dfrac{\sin x(\cos h-1)+\sin h\cos x}{h}\\=\sin x \lim_{h\to 0}\frac{\cos h-1}{h} + \cos x \\ \sin x \lim_{h\to 0}\frac{1-2\sin^2(h/2)-1}{h}+\cos x \\ = -2\sin x\lim_{h\to 0} \frac{\sin^2 h}{h^2} h +\cos x \\ =0+\cos x \\ =\cos x \qed$

ehhh the last few lines cbb fixing up the constants

VBN2470 · Sep 26, 2013

Re: HSC 2013 3U Marathon Thread

Could someone please help me with the following induction question? Having some trouble here

seanieg89 · Sep 26, 2013

Re: HSC 2013 3U Marathon Thread

VBN2470 said:
Could someone please help me with the following induction question? Having some trouble here
View attachment 28706

How many terms are you adding to the LHS every time you increase n by one?

What are these terms all bigger than?

VBN2470 · Sep 26, 2013

Re: HSC 2013 3U Marathon Thread

Please explain further ^

obliviousninja · Sep 26, 2013

Re: HSC 2013 3U Marathon Thread

I think im doing it wrong lol

Do you prove,

1/ 2^(k+1)-1 > 1/2

Sy123 · Sep 26, 2013

Re: HSC 2013 3U Marathon Thread

VBN2470 said:
Please explain further ^

Say I have a sequence of terms

$s_1 = a_1 + a_2$

$s_2 = a_1 + a_2 + a_3 + a_4$

$s_3 = a_1 + a_2 + a_3 + a_4 + a_5 + a_6$

and so on, so that means:

$s_n = a_1 + a_2 + \dots + a_{2n-1} + a_{2n}$

So when we do $s_{n+1}$ then, how many extra terms are we adding on?

Remember how, normally when we do induction with series, we normally add 1 term to both sides of the step 2 expression to end up with the LHS of the Step 3 expression
Remember that this is the reason why we add to both sides of the Step 2, so we can get something similar to the Step 3 expression to end up with maybe a proof.

Similarly for the above example I made, when trying to manipulate the Step 2 expression, to make it look like the Step 3 LHS, how many terms would we have to add on?

Likewise, how many terms would we have to add on to your induction inequality's Step 2 expression, to make it look like the Step 3 expression?

VBN2470 · Sep 26, 2013

Re: HSC 2013 3U Marathon Thread

It makes more sense when you explain it that way because usually it is always adding one term, not multiple terms

VBN2470 · Sep 26, 2013

Re: HSC 2013 3U Marathon Thread

Would you have to add an extra 2^n terms to the LHS of the inequality?

Sy123 · Sep 26, 2013

Re: HSC 2013 3U Marathon Thread

VBN2470 said:
Would you have to add an extra 2^n terms to the LHS of the inequality?

Yep

VBN2470 · Sep 26, 2013

Re: HSC 2013 3U Marathon Thread

But do you still take the last term of the series as the nth term or is an extra independent term in which you add it on the series leading up to 1/n? This is what is confusing me, where is the general term of the series?

Sy123 · Sep 26, 2013

Re: HSC 2013 3U Marathon Thread

VBN2470 said:
But do you still take the last term of the series as the nth term or is an extra independent term in which you add it on the series leading up to 1/n? This is what is confusing me, where is the general term of the series?

If you write it in summation notation it becomes easier to understand, so this is the general sum:

$S_{n} = \sum_{m=1}^{2^{n} -1 } \frac{1}{m}$

We need to prove that

$S_n > \frac{n}{2}$

For n = k (just like normal)

$$assume true$ \ \ S_k = \sum_{m=1}^{2^{k} -1} \frac{1}{m}> \frac{k}{2}$

And we need to prove that for n=k+1

$S_{k+1} = \sum_{m=1}^{2^{k+1} - 1} \frac{1}{m} > \frac{k+1}{2}$

And we know that

$S_{k} + \frac{1}{2^k} + \frac{1}{2^k + 1} + \dots + \frac{1}{2^{k+1} - 2} + \frac{1}{2^{k+1} -1} = S_{k+1}$

VBN2470 · Sep 26, 2013

Re: HSC 2013 3U Marathon Thread

You made it so much clearer for me now, thanks!

Sy123 · Sep 26, 2013

Re: HSC 2013 3U Marathon Thread

$By taking the fact that$

$(1+x)^n = (1+x)(1+x)(1+x) \dots (1+x)$

$Prove the Binomial Theorem$

Sy123 · Oct 5, 2013

Re: HSC 2013 3U Marathon Thread

$i) Use the substitution$ \ \ u = \sin x \ \ $to evaluate$

$\int_0^{\pi /2} \sin^{2k} x \cos x \ dx = \frac{1}{2k+1}$

$Hence evaluate$

$\int_0^{\pi /2} \cos^{2n+1} x \ dx$

$Hence by substitution find$

$\int_0^{2} x^{2n} \sqrt{4-x^2} \ dx$

mahmoudali · Oct 7, 2013

Re: HSC 2013 3U Marathon Thread

i)

$\int_0^{\pi /2} \sin^{2k} x \cos x \ dx = \frac{1}{2k+1}$

$\int_0^{1} \ u^{2k} \ \ du = \left(\frac{x^{2k+1}}{2k+1}\right)_0^1 = \frac{1}{2k+1}$

ii)

$\int_0^{\pi /2} \cos^{2N+1} x\ dx$

$can go fuck itself with its binomial 7 marker bullshit$

Sy123 · Oct 7, 2013

Re: HSC 2013 3U Marathon Thread

mahmoudali said:
i)

$\int_0^{\pi /2} \sin^{2k} x \cos x \ dx = \frac{1}{2k+1}$

$\int_0^{1} \ u^{2k} \ \ du = \left(\frac{x^{2k+1}}{2k+1}\right)_0^1 = \frac{1}{2k+1}$

ii)

$\int_0^{\pi /2} \cos^{2N+1} x\ dx$

$can go fuck itself with its binomial 7 marker bullshit$

Says the guy who takes 10 minutes to write 2 lines of latex

Menomaths · Oct 7, 2013

Re: HSC 2013 3U Marathon Thread

Moey, is that you?

mahmoudali · Oct 8, 2013

Re: HSC 2013 3U Marathon Thread

$$$$$ Hurry the hell up sy make another question that we can all benefit from not only the 4Uers $$$$$

Sy123 · Oct 8, 2013

Re: HSC 2013 3U Marathon Thread

$How many ways are there to arrange 4 married couples around$

$a round table so that no person is seated next to their partner?$

Disclaimer: I can't do this

mahmoudali · Oct 8, 2013

Re: HSC 2013 3U Marathon Thread

Sy123 said:
$How many ways are there to arrange 4 married couples around$

$a round table so that no person is seated next to their partner?$

Disclaimer: I can't do this

my first method was totally wrong i realised after 2 hours

total number of ways they can be arranged is 7! =5040
number of ways all couples sit together = 3!2!2!2!2! =96
number of ways 2 couples sit together but other 2 random = 5!2!2! =480
number of ways only 1 couple sit together = 6!2!=1440
therefore number of ways none sit together = 5040-96-480-1440 =3024
once again i dont know for sure

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