HSC 2013 MX2 Marathon (archive) (1 Viewer)

Sy123 · May 10, 2013

Re: HSC 2013 4U Marathon

$Prove that$

$$i)$ \ \ \binom{2n}{n} = \binom{n}{0}^2 + \binom{n}{1}^2 + \binom{n}{2}^2 + \dots + \binom{n}{n}^2$

$$ii) Find$ \ \ \sum_{k=0}^{n} (-1)^k k \binom{n}{k}^2$

seanieg89 · May 10, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Nice work, it can also be done by considering the geometric series:

$z+z^2+ \dots +z^n = \frac{z(z^n-1)}{z-1}$

Substituting cis theta, equating real parts for a series in cosine. Then integrating with bounds alpha and pi yields the LHS and the RHS plus a limit to infinity of an integral which one must prove converges to zero.

Ah alright, is that not a definition of an algebraic number? To be a root of a rational polynomial?

Also your method is very clever.

Yes but the existence and uniqueness of a MINIMAL polynomial is not immediately clear and must be proven. And thanks

.

Registered User · May 10, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Using the result from the previous question:

$\sum_{k=1}^{n} \frac{\sin(k\theta)}{k} = 0.5(\pi -\theta) + \int_{\pi}^{\theta} \frac{\cos(n\alpha) + \cos(n+1)\alpha}{4\sin^2 \alpha/2} \ d\alpha$

Integrating both sides from pi to 2, and using the double angle rule, the result from the Basel problem and so on, we arrive at:

$2\sum_{k=1}^{\infty} \frac{\sin^2 k}{k^2} = \pi - 1+ \lim_{n \to \infty} \varepsilon(\theta, \alpha)$

Then noticing that $\frac{\sin^2 (-k)}{(-k)^2} = \frac{\sin^2 k}{k^2}$

We can notice that the LHS of what I have right now, is the LHS of the question - 1 (since at k=0 the function approaches 1). This then cancels out the term on the RHS, therefore leaving it with:

$\sum_{k = -\infty}^{\infty} \frac{\sin^2 k}{k^2} = \pi + \lim_{n \to \infty}\varepsilon(\theta, \alpha)$

$\varepsilon(\theta, \alpha) = \int_{\pi}^2 \int_{\pi}^{\theta} \frac{\cos(n\alpha) + \cos(n+1)\alpha}{4\sin^2 \alpha /2} \ d\alpha \ d\theta$

I know that it should evaluate to zero, however the same problem from when I encountered trying the Basel problem comes up where I cannot justify interchanging the theta integral and the limit to infinity as the alpha integral converges to zero due to by parts. However I cannot justify putting it in there unfortunately :/

Nice work but I don't know how we can prove that the limit goes to 0. There is a more rigorous method to prove this, will post it sometime next week.
I found this question in a book called $\pi$ $Unleashed$ .

RealiseNothing · May 10, 2013

Re: HSC 2013 4U Marathon

$\int ln(Ax)~ln(Bx)~dx$

Registered User · May 10, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$Prove that$

$$i)$ \ \ \binom{2n}{n} = \binom{n}{0}^2 + \binom{n}{1}^2 + \binom{n}{2}^2 + \dots + \binom{n}{n}^2$

Sy123 said:
$$ii) Find$ \ \ \sum_{k=0}^{n} (-1)^k k \binom{n}{k}^2$

Working on this.

Registered User · May 10, 2013

Re: HSC 2013 4U Marathon

$Show that$

$\sum_{k=0}^{m} \frac{m!(n-k)!}{n!(m-k)!} = \frac{n+1}{n-m+1}$

Edit: $For$ $n\geq m$

Registered User · May 10, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
$\int ln(Ax)~ln(Bx)~dx$

$=ln(A)ln(B)\int dx+ ln(A)\int ln(x)dx+ln(B)\int ln(x) dx+\int (ln(x))^{2}dx$

$=ln(A)ln(B)x+ln(A)x(ln(x)-1)$

$+ln(B)x(ln(x)-1)+(x(lnx)^{2}-2x(lnx-1))$

$=ln(A)ln(B)x+ln(A)x(ln(x)-1)$

$ln(B)x(ln(x)-1)+x((lnx)^{2}-2lnx+2)$

Sy123 · May 10, 2013

Re: HSC 2013 4U Marathon

Registered User said:
$Show that$

$\sum_{k=0}^{m} \frac{m!(n-k)!}{n!(m-k)!} = \frac{n+1}{n-m+1}$

It becomes:

$\sum_{k=0}^{m } \frac{\binom{m}{k}}{\binom{n}{k}}$

And I don't know how to do that :/
Telescoping, pairing, binomial theorem don't seem to work

Possibly a combinatoric proof?
I suck at those but yeah

EDIT:

Considering the cards of numbers 1, 2, 3, 4, ... , m, .... , n on the table.
We must pick k cards from the n on the table.

The probability that we pick these k cards from the first m is

$P(k) = \frac{\binom{m}{k}}{\binom{n}{k}}$

So summing these probabilities will give us the LHS.

Now is there a nice way of doing this probability a different way to yield the RHS?

RealiseNothing · May 10, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
It becomes:

$\sum_{k=0}^{m } \frac{\binom{m}{k}}{\binom{n}{k}}$

And I don't know how to do that :/
Telescoping, pairing, binomial theorem don't seem to work

Possibly a combinatoric proof?
I suck at those but yeah

EDIT:

Considering the cards of numbers 1, 2, 3, 4, ... , m, .... , n on the table.
We must pick k cards from the n on the table.

The probability that we pick these k cards from the first m is

$P(k) = \frac{\binom{m}{k}}{\binom{n}{k}}$

So summing these probabilities will give us the LHS.

Now is there a nice way of doing this probability a different way to yield the RHS?

What if $m>n$

Although I think the question should state that $n>m$ otherwise it won't work.

Sy123 · May 10, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
What if $m>n$

Although I think the question should state that $n>m$ otherwise it won't work.

Well yeah if m > n then we arrive at negative factorials so it should be discounted.

RealiseNothing · May 10, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Well yeah if m > n then we arrive at negative factorials so it should be discounted.

The RHS is equal to (n+1)P(m) / (n)P(m) (idk how to do permutations in latex)

maybe you can choose the cards in this way some how.

Which is equal to:

$\frac{\binom{n+1}{m}}{\binom{n}{m}}$

Sy123 · May 10, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
The RHS is equal to (n+1)P(m) / (n)P(m) (idk how to do permutations in latex)

maybe you can choose the cards in this way some how.

Which is equal to:

$\frac{\binom{n+1}{m}}{\binom{n}{m}}$

Nice observation.

So that can mean that when counting the probability the different all the cards needs to be on the table at once. And we need to find the probability of selecting the cards in that fashion among the m sets of (1,2,...,n) cards in one go (because there is only one fraction).

And for permutation, I guess you can do:

$^{n} P _k$

Registered User · May 11, 2013

Re: HSC 2013 4U Marathon

Registered User said:
$Show that$

$\sum_{k=0}^{m} \frac{m!(n-k)!}{n!(m-k)!} = \frac{n+1}{n-m+1}$

Edit: $For$ $n\geq m$

$For$ $n\geq m$ $\sum_{k=0}^{m}\frac{m!(n-k)!}{n!(m-k)!}=\frac{m!(n-m)!}{n!}\sum_{k=0}^{m}\frac{(n-k!)}{(n-m)!(m-k)!}$

$=\frac{1}{\binom{n}{m}}\sum_{k=0}^{m}\binom{n-k}{n-m}$

$\sum_{k=0}^m\frac{\binom {{1+}m}{k}}{\binom {{1+}n}{k}}$

$&=\frac1{\binom {{1+}n}{m}}\sum_{k=0}^m\binom {n-k}{n-m}\binom {k}{0}$

$&=\frac{\binom {n+1}{n-m+1}}{\binom {{1+}n}{m}}$

$&=\frac{\frac{n+1}{n-m+1}\binom{\vphantom{1+}n}{n-m}}{\binom{\vphantom{1+}n}{m}}$

$&=\frac{n+1}{n-m+1}$

Registered User · May 11, 2013

Re: HSC 2013 4U Marathon

$Prove that$

$1) \frac{bc}{a}+\frac{ac}{b}+\frac{ab}{c}\geqslant (a+b+c)$

$2) \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geqslant \frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}$ $+\frac{1}{\sqrt{ab}}$

bleakarcher · May 11, 2013

Re: HSC 2013 4U Marathon

Registered User said:
$Prove that$

$1) \frac{bc}{a}+\frac{ac}{b}+\frac{ab}{c}\geqslant (a+b+c)$

$2) \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geqslant \frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}$ $+\frac{1}{\sqrt{ab}}$

1) [(1/a)-(1/b)]^2 + [(1/a)-(1/c)]^2 + [(1/b)-(1/c)]^2 >=0 (equality occurs iff a=b=c)
2/a^2 + 2/b^2 + 2/c^2 -2[(1/ab)+(1/bc)+(1/ac)] >=0
=> (1/a)^2 + (1/b)^2 + (1/c)^2 >= 1/bc + 1/ac + 1/ab for all real a,b,c=/=0 (*)
Hence, bc/a + ac/b + ab/c = abc[(1/a^2)+(1/b^2) +(1/c^2)]>=abc[(1/ab)+(1/bc)+(1/ac)]=a+b+c as req.
2) Refer back to (*), replace a with sqrt(a), b with sqrt(b) and c with sqrt(c) and the second inequality immediately falls out for a,b,c>0

Sorry if it's hard to read, still haven't learnt latex.

Sy123 · May 11, 2013

Re: HSC 2013 4U Marathon

$I_n = \int_0^{\frac{\pi}{2}} \cos^{2n} x \ dx$

$Prove that$

$\sum_{k=1}^{n} I_k = (2n+1)I_n - I_0$

Its integration but its more about series than anything.

shongaponga · May 11, 2013

Re: HSC 2013 4U Marathon

bleakarcher said:
1) [(1/a)-(1/b)]^2 + [(1/a)-(1/c)]^2 + [(1/b)-(1/c)]^2 >=0 (equality occurs iff a=b=c)
2/a^2 + 2/b^2 + 2/c^2 -2[(1/ab)+(1/bc)+(1/ac)] >=0
=> (1/a)^2 + (1/b)^2 + (1/c)^2 >= 1/bc + 1/ac + 1/ab for all real a,b,c=/=0 (*)
Hence, bc/a + ac/b + ab/c = abc[(1/a^2)+(1/b^2) +(1/c^2)]>=abc[(1/ab)+(1/bc)+(1/ac)]=a+b+c as req.
2) Refer back to (*), replace a with sqrt(a), b with sqrt(b) and c with sqrt(c) and the second inequality immediately falls out for a,b,c>0

Sorry if it's hard to read, still haven't learnt latex.

http://www.codecogs.com/latex/eqneditor.php Use this website, then simply copy and paste the code generated and wrap it around [ tex ] "CODE" [ / tex ] (without the spaces).

bleakarcher · May 11, 2013

Re: HSC 2013 4U Marathon

$\int x^3$

bleakarcher · May 11, 2013

Re: HSC 2013 4U Marathon

shongaponga said:
http://www.codecogs.com/latex/eqneditor.php Use this website, then simply copy and paste the code generated and wrap it around [ tex ] "CODE" [ / tex ] (without the spaces).

Did it. ^ That happened.

Makematics · May 11, 2013

Re: HSC 2013 4U Marathon

oooo i also wanna learn how to use LaTex

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