HSC 2012 MX2 Marathon (archive) (6 Viewers)
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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.
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
Find the derivative of:
i.
ii.
iii. I am trying to find some sort of general pattern for 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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seanieg89
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Re: 2012 HSC MX2 Marathon
Don't need the 'put'. Expand (x^{1/2}-y^{1/2})^2>=0.Wait...so the 'better' approach would be for eg.
??
Re: 2012 HSC MX2 Marathon
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.
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.
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Re: 2012 HSC MX2 Marathon
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
 > g(x))
then
\,dx > \displaystyle\int_D g(x)\,dx)
but the converse isn't necessarily true
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.
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.
seanieg89
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Re: 2012 HSC MX2 Marathon
Let f(x)=x*arcsin(x). As the product of two odd functions f is even, so it suffices to prove the assertion for 0<=x<1.
f is the product of two positive and increasing functions on the interval 0<=x<=1. Hence 0=f(0)<=f(x) < f(1)=pi/2 for all -1 < x < 1.
Let f(x)=x*arcsin(x). As the product of two odd functions f is even, so it suffices to prove the assertion for 0<=x<1.
f is the product of two positive and increasing functions on the interval 0<=x<=1. Hence 0=f(0)<=f(x) < f(1)=pi/2 for all -1 < x < 1.
seanieg89
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Re: 2012 HSC MX2 Marathon
The obvious method that comes to mind is looking at the sector contour with one side being part of the positive real axis, and an angle of pi/4 radians. The integral along the circular arc tends to zero which implies that half of the Gaussian integral is equal to the negative integral of e^(-z^2) over the diagonal ray. Which after applying Euler's identity gives us what we want.
Clearly this is beyond the scope of mx2 so perhaps you had another solution in mind.
The obvious method that comes to mind is looking at the sector contour with one side being part of the positive real axis, and an angle of pi/4 radians. The integral along the circular arc tends to zero which implies that half of the Gaussian integral is equal to the negative integral of e^(-z^2) over the diagonal ray. Which after applying Euler's identity gives us what we want.
Clearly this is beyond the scope of mx2 so perhaps you had another solution in mind.
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Johnstan
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seanieg89
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Re: 2012 HSC MX2 Marathon
He explained his method to me through PM and I don't think it is valid, but we will see if someone can post a valid solution using MX2 only techniques (I don't think this is possible).
Carrotsticks
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Re: 2012 HSC MX2 Marathon
Would you mind going over the second example? Didn't really understand the method.Could have generalised the problem by using the equation xy = k^2 instead =p
Oh and there is actually a nice and cheap way of calculating (or checking) the Implicit Derivative.
So for the equation xy= k^2, we have:
I know it doesn't seem worth it for this particular example, but it really helps when it comes to implicitly differentiating things like:
Carrotsticks
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Re: 2012 HSC MX2 Marathon
Similarly for Partial f / Partial Y.
Partial f / Partial X means you differentiate f(x,y) purely with respect to x, and you treat Y as a constant.
Similarly for Partial f / Partial Y.