HSC 2012 MX2 Marathon (archive) (2 Viewers)
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RealiseNothing
what is that?It is Cowpea
Re: 2012 HSC MX2 Marathon
lol so I pretty much did the reverse of the proof.
lol so I pretty much did the reverse of the proof.
bleakarcher
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Re: 2012 HSC MX2 Marathon
I know fully comprehend Trebla's comment about you going in the wrong direction.
RealiseNothing
what is that?It is Cowpea
Re: 2012 HSC MX2 Marathon
Since in each step = g(0))
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.
I was using the "racetrack principle".You cannot differentiate both sides of an inequality and expect know for sure what the inequality sign is
e.g. x2 + 1 > x2
If you differentiate both sides you get: 2x > 2x which is clearly false
Though you are on the right track...just not in the right direction
Since in each step
RealiseNothing
what is that?It is Cowpea
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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.
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Carrotsticks
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bleakarcher
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Re: 2012 HSC MX2 Marathon
Only if the inequality holds in the domain of integration of course.
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Re: 2012 HSC MX2 Marathon
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.
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.I was using the "racetrack principle".
Since in each step
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
what is that?It is Cowpea
Re: 2012 HSC MX2 Marathon
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?
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...
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
I wouldn't say it's 'wrong' per se, but rather it's just not as nice =)
Re: 2012 HSC MX2 Marathon
Well, that's the easiest way to prove some inequalities.
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)
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)
Re: 2012 HSC MX2 Marathon
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.
This is how you properly execute the greek method:
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:
k} \geq \frac{2^{k+1}}{(3^{k+1}+1)(k+1)})
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.
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
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.
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
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:
}{2(3^k + 1)} + \frac{(3^{k+1} + 1)}{2(3^k + 1)k} )
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
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
In order to show
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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