HSC 2013 MX2 Marathon (archive) (2 Viewers)

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dunjaaa

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Re: HSC 2014 4U Marathon

Oh... I common denominated. Shouldn't the inequality read < since A,B,C,D >0
 

Sy123

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Re: HSC 2014 4U Marathon

Oh... I common denominated. Shouldn't the inequality read < since A,B,C,D >0
Well we get





In order for the RHS to be positive, the numerator and denominator either both need to be positive ort both be negative, since hte numerator is always positive. Then the denominator must also always be positive.

 

seanieg89

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Re: HSC 2014 4U Marathon

This can fall under graphs

If you enforce A,B,C,D > 0, then this statement is only (technically) true because the implicitly defined curve is NEVER an ellipse. Your inequality in fact just gives a necessary (but not sufficient) condition for at least one point in the plane to satisfy your equation.

If you let A,B,C,D be real numbers (not necessarily positive) such that the curve implicitly defined by your equation is non-degenerate, then the OPPOSITE inequality to the one you posted is equivalent to being an ellipse. That is: the implicit equation Ax^2 + By^2 + Cxy + D = 0 is an ellipse if and only if the curve defined is non-degenerate and C^2-4AB < 0.

Degeneracy vs non-degeneracy is slightly more subtle and accounts for equations like x^2-y^2=0 whose graphs are in some sense exceptions to how second order polynomial curves usually look.
 
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bottleofyarn

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Re: HSC 2014 4U Marathon

Here's that inequality:


And one for 2014'ers (I think I've given enough to make it reasonable).

 

dunjaaa

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Re: HSC 2014 4U Marathon

Screen shot 2014-01-24 at 7.12.16 PM.png I think my U(k) expression is wrong, but I couldn't get the RHS result :( The expression I obtained looked like taylor's series expansion of e
 
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seanieg89

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Re: HSC 2014 4U Marathon

View attachment 29844 I think my U(k) expression is wrong, but I couldn't get the RHS result :( The expression I obtained looked like taylor's series expansion of e
Your expression for u_k is incorrect. Some of the (n-k)! permutations of the "non-fixed" points will fix points, so that method of counting doesn't work.
 

dunjaaa

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Re: HSC 2014 4U Marathon

sigh........, hopefully i got the u_k expression correct. Any tips on the next part? Screen shot 2014-01-25 at 4.34.20 PM.png
 

seanieg89

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Re: HSC 2014 4U Marathon

sigh........, hopefully i got the u_k expression correct. Any tips on the next part? View attachment 29849
This does seem roughly right, you definitely have the right idea for counting permutations that don't fix things (these are called derangements). Something seems to have gone wrong at some point because you have n-k as an upper limit in a sum indexed by k. This is probably just a slip in your notation earlier which would be easy to fix up.

My way was different (in fact my proof had very few equations), but your idea might still work if you can think of any clever ways to simplify the end summation. I will try to do it your way a little later to see if it works / how difficult it is.
 

dunjaaa

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Re: HSC 2014 4U Marathon

Screen shot 2014-01-29 at 10.48.34 PM.png Forgot about this question (forgot an 'up' after concave). I didn't include the other scenario when f(b)≤f(a) but either way it should still yield the same result
 
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seanieg89

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Re: HSC 2014 4U Marathon

General pattern recognition rather than a specific 4U topic. Difficulty 2/5.

Sum sec(k)sec(k+1) from k=0 to 88.

(Where sec is defined in degrees.)
 

seanieg89

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Re: HSC 2014 4U Marathon

Imagine a rectangular grid with n! rows (one for each permutation of the students), and n columns (one for each student.)

Put a counter on each square in the grid such that the corresponding student is fixed by the corresponding permutation.

Then the LHS is clearly equal to the number of counters on our grid, since each row corresponding to a permutation with k fixed points contains k counters.

On the other hand, each of the n columns contains (n-1)! counters (one for each permutation of the other (n-1)! students.

Hence LHS=n.(n-1)!=n!.



(My actual proof was essentially the above idea written up formally using sums and not needing the counters analogy, but the idea of counting the same thing in two different ways to prove identities is probably more valuable to mx2 students.)
 
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seanieg89

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Re: HSC 2014 4U Marathon

Let f be a real-valued function defined on R. For which values of the real constant p does the inequality below imply that f is twice differentiable?

|f(x)-f(y)| =< C|x-y|^p for all x,y. C a constant.

Justify your answer with proof.
 
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dunjaaa

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Re: HSC 2014 4U Marathon

Screen shot 2014-02-06 at 2.23.22 PM.png Felt so dumb after I realised how to do this :(
 
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