HSC 2012 MX2 Marathon (archive) (2 Viewers)

bleakarcher

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Re: 2012 HSC MX2 Marathon

Exactly lol, just be aware that you can integrate inequalities (provided you use definite integrals) without running into too many complications but that's not the case when you differentiate them.
Only if the inequality holds in the domain of integration of course.
 

Trebla

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Re: 2012 HSC MX2 Marathon

I was using the "racetrack principle".

Since in each step I thought my way was right? Unless that's only for integrating, in which case it would stilll be right wouldn't it since you can just go backwards through my proof by starting at for x>0 and integrating.
It's also not correct in the sense that you are assuming the inequality you are trying to prove already holds then using it arrive at some statement.
e.g. If you want to prove x2 + y2 > 2xy then you should start with (x - y)2 > 0 to arrive at x2 + y2 > 2xy NOT start with x2 + y2 > 2xy and arrive at (x - y)2 > 0 since the latter assumes the very result you are trying to prove. The latter approach is something you would use as a technique to work backwards to try to find your starting point.
 

RealiseNothing

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Re: 2012 HSC MX2 Marathon

It's also not correct in the sense that you are assuming the inequality you are trying to prove already holds then using it arrive at some statement.
e.g. If you want to prove x2 + y2 > 2xy then you should start with (x - y)2 > 0 to arrive at x2 + y2 > 2xy NOT start with x2 + y2 > 2xy and arrive at (x - y)2 > 0 since the latter assumes the very result you are trying to prove. The latter approach is something you would use as a technique to work backwards to try to find your starting point.
That's what I'm wondering, I assumed the first inequality then came to my starting point which was a correct inequality itself. Wouldn't I then just have to reverse this process to complete the proof?
 

cutemouse

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Re: 2012 HSC MX2 Marathon

It's also not correct in the sense that you are assuming the inequality you are trying to prove already holds then using it arrive at some statement.
NOT start with x2 + y2 > 2xy and arrive at (x - y)2 > 0 since the latter assumes the very result you are trying to prove.

x2 + y2 > 2xy

iff x2 - 2xy + y2 > 0

iff (x-y)2 > 0, which is true for all real x,y ...

Nothing wrong with that I don't think...
 

Carrotsticks

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Re: 2012 HSC MX2 Marathon

x2 + y2 > 2xy

iff x2 - 2xy + y2 > 0

iff (x-y)2 > 0, which is true for all real x,y ...

Nothing wrong with that I don't think...
I wouldn't say it's 'wrong' per se, but rather it's just not as nice =)
 

Carrotsticks

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Re: 2012 HSC MX2 Marathon

Just a matter of personal preference, but I would rather take the harder (sometimes longer) way to prove something from scratch, rather than start with the known identity and work backwards.

(also deja vu)
 

math man

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Re: 2012 HSC MX2 Marathon

x2 + y2 > 2xy

iff x2 - 2xy + y2 > 0

iff (x-y)2 > 0, which is true for all real x,y ...

Nothing wrong with that I don't think...
This is how you properly execute the greek method:
greek method.png

How the greek method (assuming answer and going in reverse order) works is
you assume the inequality you want to prove then manipulate it till you get something
you defs knows is true, but this is NOT a proof. All the greek method does is allows
you a cheap way of getting inequality questions if your stuck, desperate and running
out of time. To execute it properly, once you reduce down to somethign you know is true
you must rewrite it in reverse order and its best to rub out previous part just so teacher
doesnt know you dont really know how to do question.

Note: i do not recommend this method, it does not highlight your skill as a mathematician,
but it can grab you a precious mark or two.
 

Carrotsticks

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Re: 2012 HSC MX2 Marathon

I presume you asked that question because you expected me to start with the usual prove a_k - a_{k-1} >= 0 for all k>1, to which you would then argue that I started with the required result.

However. the example you have here is very different to the above example you gave above with the iff arguments going backwards.

To me, the equivalent of starting with the given identity then playing around with it until you reach a trivial result, would be something like starting your proof with:



And then working your way down.

I would rather see something like:

"Consider a_k - a_{k-1}" etc etc, THEN working with it. I know it seems very similar to the above, but the example you provided doesn't really leave much room to differentiate between the two methods.

You can use the given identity to have an idea of where to START the proof, but not to always actually use the given identity, then work your way down to a trivial result.

So suppose wanted to prove x^2 + y^2 >= 2xy.

Sure, you could do what you had there with the iff going backwards etc, but I wouldn't think it's as nice as say starting from (x-y)^2 >= 0, then acquiring it because (I know this is a basic example but ofc there are harder ones) thinking of starting with (x-y)^2 >= 0 requires a certain level of ingenuity, whereas anybody can just re-arrange the given expression and play around with it until a trivial result is held, then claim end of proof.

Both are correct of course, and indeed for many inequality problems it's easier to start with the given result, then work down to trivial result (usually something like (x-y)^2 >= 0), but I don't think using that method really develops that 'sixth sense' in Maths that I think is so important.

When I say '6th sense', I mean that feeling when you just 'know' what to do but can't really explain it too easily, I am sure you know what I mean.

I see the learning experience beyond just 'doing the inequality questions to get marks' because that kinda defeats the purpose of the education of Mathematics. I see it more as 'developing your intuition', which I think would be MUCH better done by starting with a known result (or using the required identity to guess where to start), then deducing the answer, as opposed to starting with the answer then deducing something obvious.
 

seanieg89

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Re: 2012 HSC MX2 Marathon

I presume you asked that question because you expected me to start with the usual prove a_k - a_{k-1} >= 0 for all k>1, to which you would then argue that I started with the required result.

However. the example you have here is very different to the above example you gave above with the iff arguments going backwards.

To me, the equivalent of starting with the given identity then playing around with it until you reach a trivial result, would be something like starting your proof with:



And then working your way down.

I would rather see something like:

"Consider a_k - a_{k-1}" etc etc, THEN working with it. I know it seems very similar to the above, but the example you provided doesn't really leave much room to differentiate between the two methods.

You can use the given identity to have an idea of where to START the proof, but not to always actually use the given identity, then work your way down to a trivial result.

So suppose wanted to prove x^2 + y^2 >= 2xy.

Sure, you could do what you had there with the iff going backwards etc, but I wouldn't think it's as nice as say starting from (x-y)^2 >= 0, then acquiring it because (I know this is a basic example but ofc there are harder ones) thinking of starting with (x-y)^2 >= 0 requires a certain level of ingenuity, whereas anybody can just re-arrange the given expression and play around with it until a trivial result is held, then claim end of proof.

Both are correct of course, and indeed for many inequality problems it's easier to start with the given result, then work down to trivial result (usually something like (x-y)^2 >= 0), but I don't think using that method really develops that 'sixth sense' in Maths that I think is so important.

When I say '6th sense', I mean that feeling when you just 'know' what to do but can't really explain it too easily, I am sure you know what I mean.

I see the learning experience beyond just 'doing the inequality questions to get marks' because that kinda defeats the purpose of the education of Mathematics. I see it more as 'developing your intuition', which I think would be MUCH better done by starting with a known result (or using the required identity to guess where to start), then deducing the answer, as opposed to starting with the answer then deducing something obvious.

Seconded. Not only does the whole "working from the answer" approach potentially run into logical problems when we try to reverse the argument,

(eg.
-1=1
(-1)^2=1^2
the final line is true so the first line must be true....WRONG.)

this approach moreover kills creativity in mathematics...it hinders the development of the skill of seeing what you can deduce about objects given the properties you already know they possess etc. This is a more important logical skill to pick up in high school mathematics than remembering how to solve random equations, it is universally useful.
 
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Re: 2012 HSC MX2 Marathon

Wait...so the 'better' approach would be for eg.



??
 

Godmode

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Re: 2012 HSC MX2 Marathon

Can graphs be used in inequality type questions to show how 1 curve may be larger than the other or are always greater than a specific value?

For example, in cutemouse's question, Just by manipulating the values of and and then using the fact that (for k>= 1) I have come up with this expression:



In order to show I need to prove

I can simply graph the RHS of my result for a_k/a_(k+1) and this will show that it is always greater than 1 for k>= 1, but am I allowed to? If this question (or something similar) popped up in the HSC I don't think I would bother with the graphs due to time constraints. I could of course use limits, but is there anything wrong with a graphical solution?


And here is another question:

Find the derivative of
 
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Godmode

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Re: 2012 HSC MX2 Marathon

Ok so you know how multiplication can be considered repeated addition of a number (eg 5x = x + x + x + x + x), and exponentiation can be considered repeated multiplication(eg x^5 = x * x * x * x * x)? Well repeated exponentiation (eg x^x^x^x^x^....^x is called tetration, and is denoted by

Using the fact that the derivative of is which is equal to ,
Find the derivative of:

i.
ii.
iii. I am trying to find some sort of general pattern for the derivative of . If you keep deriving higher and higher tetrations of x, you may see a pattern emerge. Try and help me piece this pattern together in order to find the derivative of .

btw, I use ^{n}\textrm{x} for tetration notation, just doing ^{n}x doesn't seem to work for me.
 
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Re: 2012 HSC MX2 Marathon

Oh ic, even easier lols...
 

cutemouse

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Re: 2012 HSC MX2 Marathon

but this is NOT a proof.
Uhh, it is.

I get what people are saying about creativity etc. But my point is that sometimes the easiest way is to work backwards using iff arguments. Also if you square both sides of an inequality (or equation) then iff doesn't hold.
 

Trebla

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Re: 2012 HSC MX2 Marathon

x2 + y2 > 2xy

iff x2 - 2xy + y2 > 0

iff (x-y)2 > 0, which is true for all real x,y ...

Nothing wrong with that I don't think...
Whilst there is nothing wrong with the iff arguments in a logical sense, I don't think they constitute a well presented 'proof' of the statement (though if there are pure mathematicians amongst us specially those who've studied logic and foundations, please offer your insights). A proof is a way to demonstrate the truth of statement using a set of axioms.

So starting from the 'axiom' (x - y)2 > 0 you make the logical deduction to arrive at the result. If we start with the result, you're effectively taking the x2 + y2 > 2xy as your 'axiom' to show that
(x - y)2 > 0.

Looking at it more carefully:

x2 + y2 > 2xy

iff x2 - 2xy + y2 > 0

iff (x-y)2 > 0, which is true for all real x,y ...

You then conclude that x2 + y2 > 2xy. Why are you able to conclude that? (the flow of the main argument is fine but its the point that you make the conclusion that I'm a bit concerned about)

This is because (x - y)2 > 0 is a true statement (an axiom) which then leads us to the result we want by a set of deductions (notice that the direction of the logical argument is different to the order you've presented it in). So what you've presented here in relation to the logical argument isn't in the 'logical order'.

In other words, a mathematical proof is where we use axiom A to demonstrate the truth of statement S (i.e. A --> S) so when you present them, naturally you should use the axiom to arrive at the statement by the conclusion.

Anyway, back to my original point, the working backwards approach isn't a valid proof in that question because for some domain D, if



then



but the converse isn't necessarily true
 
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seanieg89

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Re: 2012 HSC MX2 Marathon

You are right of course cutemouse, I just like to stress the dangers of working backwards as many high school students don't tend to think too much about which directions arguments work. The squaring example is a rather trivial one, but can be well disguised sometimes, it is bizarre how often I see students make mistakes when solving inequalities involving absolute value signs for instance.

Once one is comfortable with what he is doing and has a sense for when an argument is "valid" I have no problem with them using any heuristics they like to arrive at a proof.
 

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