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

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Sy123

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Re: HSC 2013 3U Marathon Thread











 

RealiseNothing

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Re: HSC 2013 3U Marathon Thread





EDIT: Have I already posted this question? I have a feeling I already have.
by sum of a GP.



Equating the co-efficients of :



Note that in the co-efficient of is since you have to divide by which means you must take the term of the numerator.

Also in there is no co-efficient of as the highest power of the numerator is but you divide by , making the highest power of the whole expression only .
 

RealiseNothing

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Re: HSC 2013 3U Marathon Thread

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

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



Which can be expressed as such:



Re-arranging:













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

seanieg89

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Re: HSC 2013 3U Marathon Thread

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

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



Which can be expressed as such:



Re-arranging:













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

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Re: HSC 2013 3U Marathon Thread





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.
 

Sy123

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Re: HSC 2013 3U Marathon Thread

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

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



Which can be expressed as such:



Re-arranging:













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.
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

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funnytomato

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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
 
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seanieg89

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Re: HSC 2013 3U Marathon Thread

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

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Re: HSC 2013 3U Marathon Thread

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
 
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Sy123

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Re: HSC 2013 3U Marathon Thread

 

Capt Rifle

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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

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