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

Sy123 · Mar 26, 2013

Re: HSC 2013 3U Marathon Thread

$Take a number line interval from 0 to 1$

$Remove the middle third of this interval, i.e. remove the interval from 1/3 to 2/3$

$We are left with 2 intervals. Remove the middle third of each of these intervals$

$We create 4 more intervals, remove the middle third of each one of these too$

$Repeat infinitely many times$

$Find the length of the interval removed$

asianese · Mar 26, 2013

Re: HSC 2013 3U Marathon Thread

Sy123 said:
$Take a number line interval from 0 to 1$

$Remove the middle third of this interval, i.e. remove the interval from 1/3 to 2/3$

$We are left with 2 intervals. Remove the middle third of each of these intervals$

$We create 4 more intervals, remove the middle third of each one of these too$

$Repeat infinitely many times$

$Find the length of the interval removed$

3h3h3h

RealiseNothing · Mar 26, 2013

Re: HSC 2013 3U Marathon Thread

Sy123 said:
$Prove using Binomial Theorem or Combinatorics or otherwise$

$\binom{k}{k} + \binom{k+1}{k} + \binom{k+2}{k} + \dots + \binom{n}{k} = \binom{n+1}{k+1}$

EDIT: Have I already posted this question? I have a feeling I already have.

$(1+x)^k + (1+x)^{k+1} + (1+x)^{k+2} + ... + (1+x)^n = \frac{(1+x)^k[(1+x)^{n-k+1}-1]}{x}$ by sum of a GP.

$(1+x)^k + (1+x)^{k+1} + (1+x)^{k+2} + ... + (1+x)^n = \frac{(1+x)^{n+1}}{x} - \frac{(1+x)^k}{x}$

Equating the co-efficients of $x^k$ :

$\binom{k}{k} + \binom{k+1}{k} + \binom{k+2}{k} + ... + \binom{n}{k} = \binom{n+1}{k+1}$

Note that in $\frac{(1+x)^{n+1}}{x}$ the co-efficient of $x^k$ is $\binom{n+1}{k+1}$ since you have to divide by $x$ which means you must take the $k+1~th~$ term of the numerator.

Also in $\frac{(1+x)^k}{x}$ there is no co-efficient of $x^k$ as the highest power of the numerator is $k$ but you divide by $x$ , making the highest power of the whole expression only $k-1$ .

RealiseNothing · Mar 26, 2013

Re: HSC 2013 3U Marathon Thread

Sy123 said:
$Take a number line interval from 0 to 1$

$Remove the middle third of this interval, i.e. remove the interval from 1/3 to 2/3$

$We are left with 2 intervals. Remove the middle third of each of these intervals$

$We create 4 more intervals, remove the middle third of each one of these too$

$Repeat infinitely many times$

$Find the length of the interval removed$

I kinda rushed this, so hope I haven't made a mistake:

Essentially if you continue, you end up removing this amount:

$\frac{1}{3} + 2(\frac{1}{3})^2 + 2^2(\frac{1}{3})^3 + 2^3(\frac{1}{3})^4 +...$

Which can be expressed as such:

$\sum_{k=0}^{\infty} 2^k(\frac{1}{3})^{k+1}$

Re-arranging:

$\sum_{k=0}^{\infty} \frac{2^k}{3^{k+1}}$

$\frac{1}{3}\sum_{k=0}^{\infty} \frac{2^k}{3^k}$

$\frac{1}{3}\sum_{k=0}^{\infty} (\frac{2}{3})^k$

$\frac{1}{3} \times \frac{1}{1-\frac{2}{3}}$

$\frac{1}{3} \times \frac{1}{\frac{1}{3}}$

$=1$

Although you could arrive at this intuitively, a proper mathematical proof is probably better.

seanieg89 · Mar 26, 2013

Re: HSC 2013 3U Marathon Thread

RealiseNothing said:
I kinda rushed this, so hope I haven't made a mistake:

Essentially if you continue, you end up removing this amount:

$\frac{1}{3} + 2(\frac{1}{3})^2 + 2^2(\frac{1}{3})^3 + 2^3(\frac{1}{3})^4 +...$

Which can be expressed as such:

$\sum_{k=0}^{\infty} 2^k(\frac{1}{3})^{k+1}$

Re-arranging:

$\sum_{k=0}^{\infty} \frac{2^k}{3^{k+1}}$

$\frac{1}{3}\sum_{k=0}^{\infty} \frac{2^k}{3^k}$

$\frac{1}{3}\sum_{k=0}^{\infty} (\frac{2}{3})^k$

$\frac{1}{3} \times \frac{1}{1-\frac{2}{3}}$

$\frac{1}{3} \times \frac{1}{\frac{1}{3}}$

$=1$

Although you could arrive at this intuitively, a proper mathematical proof is probably better.

Alternatively:

At each step, your figure is a finite union of intervals. By removing the middle third of each we are multiplying the sum of the lengths of the remaining intervals by 2/3. Hence after n iterations, the total "length" we have removed from our figure is 1-(2/3)^n. This trivially tends to 1 as n->inf.

seanieg89 · Mar 27, 2013

Re: HSC 2013 3U Marathon Thread

Sy123 said:
$Prove using Binomial Theorem or Combinatorics or otherwise$

$\binom{k}{k} + \binom{k+1}{k} + \binom{k+2}{k} + \dots + \binom{n}{k} = \binom{n+1}{k+1}$

EDIT: Have I already posted this question? I have a feeling I already have.

Here's a combinatorial proof for variety.

Consider picking k+1 numbers out of the numbers {1,2,...,n+1}.

The right hand side of your equation is clearly equal to the number of ways of doing this.

Now for any given choice of k+1 numbers, the highest number chosen must be some j with k+1 =< j =< n+1. In each of these cases, we must select the remaining k numbers to be chosen from the j-1 numbers smaller than j.

Summing up the number of ways of doing this for j=k+1,...,n+1 yields the LHS of your equation.

Capt Rifle · Mar 27, 2013

Re: HSC 2013 3U Marathon Thread

seanieg89 said:
$More generally, if $f(x)$ is an increasing function on $\mathbb{R}$ with an anti-derivative $F(x)$, explain why $f$ has an inverse and find a closed-form expression for: \\ \\ $\int_a^b f^{-1}(t)\,dt$\\ \\where $a$ and $b$ are arbitrary real numbers.\\ \\ Hence find anti-derivatives of the logarithm and arcsine functions.$

that is a good question...not sure where to start lol

Sy123 · Mar 27, 2013

Re: HSC 2013 3U Marathon Thread

RealiseNothing said:
I kinda rushed this, so hope I haven't made a mistake:

Essentially if you continue, you end up removing this amount:

$\frac{1}{3} + 2(\frac{1}{3})^2 + 2^2(\frac{1}{3})^3 + 2^3(\frac{1}{3})^4 +...$

Which can be expressed as such:

$\sum_{k=0}^{\infty} 2^k(\frac{1}{3})^{k+1}$

Re-arranging:

$\sum_{k=0}^{\infty} \frac{2^k}{3^{k+1}}$

$\frac{1}{3}\sum_{k=0}^{\infty} \frac{2^k}{3^k}$

$\frac{1}{3}\sum_{k=0}^{\infty} (\frac{2}{3})^k$

$\frac{1}{3} \times \frac{1}{1-\frac{2}{3}}$

$\frac{1}{3} \times \frac{1}{\frac{1}{3}}$

$=1$

Although you could arrive at this intuitively, a proper mathematical proof is probably better.

Yeah as you can see, it isn't necessarily hard, just a nice little result.

seanieg89 said:
Here's a combinatorial proof for variety.

Consider picking k+1 numbers out of the numbers {1,2,...,n+1}.

The right hand side of your equation is clearly equal to the number of ways of doing this.

Now for any given choice of k+1 numbers, the highest number chosen must be some j with k+1 =< j =< n+1. In each of these cases, we must select the remaining k numbers to be chosen from the j-1 numbers smaller than j.

Summing up the number of ways of doing this for j=k+1,...,n+1 yields the LHS of your equation.

Very nice. (the only reason why I ask for a possible combinatorial proof is because I want someone to show me one)

Sy123 · Mar 28, 2013

Re: HSC 2013 3U Marathon Thread

$Find$

$1+ \frac{4}{5} + \frac{7}{5^2} + \frac{10}{5^3} + \dots + \frac{3n-2}{5^{n-1}}$

funnytomato · Mar 28, 2013

Re: HSC 2013 3U Marathon Thread

Sy123 said:
$Find$

$1+ \frac{4}{5} + \frac{7}{5^2} + \frac{10}{5^3} + \dots + \frac{3n-2}{5^{n-1}}$

$S_n=1+\frac{4}{5}+\frac{7}{5^2}+\frac{10}{5^3}+ \dots +\frac{3n-5}{5^{n-2}}+\frac{3n-2}{5^{n-1}}$

$\frac{S_n}{5}=\frac{1}{5}+\frac{4}{5^2}+\frac{7}{5^3}+\frac{10}{5^4}+\dots+\frac{3n-5}{5^{n-1}}+\frac{3n-2}{5^n}$

$\frac{4S_n}{5}=1+\frac{3}{5}+\frac{3}{5^2}+\frac{3}{5^3}+\dots+\frac{3}{5^{n-1}}-\frac{3n-2}{5^n}$

$\frac{4S_n}{5}=1-\frac{3n-2}{5^n}+\frac{3}{4}(1-\frac{1}{5^{n-1}})$

$\vdots$

$S_n=\frac{7\cdot5^n-12n-7}{16\cdot5^{n-1}}$

funnytomato · Mar 28, 2013

Re: HSC 2013 3U Marathon Thread

Let M bet the midpoint of a chord PQ in a circle
Two chords AD and BC are drawn so that they intersect at M
AB and CD intersect the chord PQ at X and Y respectively
Show that M is the midpoint of XY

seanieg89 · Mar 28, 2013

Re: HSC 2013 3U Marathon Thread

funnytomato said:
Let M be the midpoint of chord PQ in a circle
AB and CD are two other chords which pass through M
AB and CD intersect the chord PQ at points X and Y respectively
Show that M is the midpoint of XY

As you have currently phrased the question, X and Y are both the point M itself...

funnytomato · Mar 28, 2013

Re: HSC 2013 3U Marathon Thread

seanieg89 said:
As you have currently phrased the question, X and Y are both the point M itself...

thanks for that , don't know what I was thinking

hopefully it makes some sense now

seanieg89 · Mar 28, 2013

Re: HSC 2013 3U Marathon Thread

funnytomato said:
thanks for that , don't know what I was thinking

hopefully it makes some sense now

It doesn't, as AB and CD are arbitrary the claim is false.

funnytomato · Mar 28, 2013

Re: HSC 2013 3U Marathon Thread

seanieg89 said:
It doesn't, as AB and CD are arbitrary the claim is false.

Could you check this?

Let M bet the midpoint of a chord PQ in a circle
Two chords AD and BC are drawn so that they intersect at M
AB and CD intersect the chord PQ at X and Y respectively
Show that M is the midpoint of XY

funnytomato · Mar 28, 2013

Re: HSC 2013 3U Marathon Thread

Sy123 · Mar 30, 2013

Re: HSC 2013 3U Marathon Thread

$\sum_{n=1}^{\infty} \frac{n-1}{n!}$

asianese · Mar 30, 2013

Re: HSC 2013 3U Marathon Thread

1.

Capt Rifle · Mar 30, 2013

Re: HSC 2013 3U Marathon Thread

Use mathematical induction and the product rule for differentiation to prove that
<a href="http://www.codecogs.com/eqnedit.php?latex=\frac{d}{dx}(x^n)=nx^{n-1}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\frac{d}{dx}(x^n)=nx^{n-1}" title="\frac{d}{dx}(x^n)=nx^{n-1}" /></a>

For all positive integers n.

I believe this is more 3u than 4u, anyone else agree?

Nws m8 · Mar 30, 2013

Re: HSC 2013 3U Marathon Thread

Capt Rifle said:
Use mathematical induction and the product rule for differentiation to prove that
<a href="http://www.codecogs.com/eqnedit.php?latex=\frac{d}{dx}(x^n)=nx^{n-1}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\frac{d}{dx}(x^n)=nx^{n-1}" title="\frac{d}{dx}(x^n)=nx^{n-1}" /></a>

For all positive integers n.

I believe this is more 3u than 4u, anyone else agree?

this IS 3u, I remember doing this in 3u, too easy for 4u m8

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

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