HSC 2014 MX2 Marathon ADVANCED (archive) (1 Viewer)

[jyu]

Re: HSC 2014 4U Marathon - Advanced Level

Because of the constants.
Change x+1 to X and y-1 to Y,

I have made a mistake in my last post.
Change x+1 to X and y+1 to Y,
Symmetry is preserved.
Your counterexample is valid.
Why does your "rule" apply for the question Sy asked, but not for the expression in this post? Have to think about this question. Because the symmetry idea can only apply to cyclic expressions?

 

[seanieg89]

Re: HSC 2014 4U Marathon - Advanced Level

So what is the point of the substitution then? You have just changed it from one symmetric counterexample to another.

There's no easy answer to that question. The "rule" you stated in your initial response to Sy's question is not valid unless we assume a lot more than just symmetry (something like convexity would probably suffice). We certainly can't blindly apply it to try to prove inequalities.

(And my counterexample is cyclic as well as symmetric. Cyclic and symmetric mean the same thing if n=2).

 

[jyu]

Re: HSC 2014 4U Marathon - Advanced Level

Great question!

Minor corrections:

Well done.
My solution:
a+b+c=abc when the triangle is equilateral, a=b=c=sqrt3
a+b+c is the perimeter of the triangle and abc is proportional to its area.
Symmetry gives minimum perimeter and maximum area.
When the triangle deviates from equilateral, perimeter increases and area decreases.
.: a+b+c>=abc

 

[jyu]

Re: HSC 2014 4U Marathon - Advanced Level

So what is the point of the substitution then? You have just changed it from one symmetric counterexample to another.

There's no easy answer to that question. The "rule" you stated in your initial response to Sy's question is not valid unless we assume a lot more than just symmetry (something like convexity would probably suffice). We certainly can't blindly apply it to try to prove inequalities.

(And my counterexample is cyclic as well as symmetric. Cyclic and symmetric mean the same thing if n=2).

Can you think of a counterexample for n=3?

 

[seanieg89]

Re: HSC 2014 4U Marathon - Advanced Level

Can you think of a counterexample for n=3?

The same idea obviously extends to any finite number of variables.

 

[jyu]

Re: HSC 2014 4U Marathon - Advanced Level

The same idea obviously extends to any finite number of variables.

Thanks to all your comments.
I can see the difference between your examples and Sy's.
Sy's is homogeneous but yours are not.
If your last example is changed to homogeneous by changing (x-1) to X and (x+1) to P etc
i.e. (X^2+Y^2+R^2)(X^2+Q^2+Z^2)(P^2+Y^2+Z^2), it is no longer symmetric/cyclic.
My reason is applicable to homogeneous cyclic polynomials.
When I said symmetry in a, b and c in the expression I should have said the expression is cyclic.
It does not matter when the polynomial is second degree.

 

[seanieg89]

Re: HSC 2014 4U Marathon - Advanced Level

I assume you mean homogeneous rational function rather than homogenous polynomial. (As that is what Sy's expression is, so we cannot consider x=y=z=0 for example.)

So the claim is: "If a homogeneous rational function with cyclic symmetry has a minima, it must occur where all variables are equal."

This is also not true.

Eg xy/(x^2+y^2) is minimised on the line y=-x, excluding the domain hole (0,0).

You should stop guessing and try to prove things if you think they are true. Such statements are worthless without proof.

 

[jyu]

Re: HSC 2014 4U Marathon - Advanced Level

I assume you mean homogeneous rational function rather than homogenous polynomial. (As that is what Sy's expression is, so we cannot consider x=y=z=0 for example.)

So the claim is: "If a homogeneous rational function with cyclic symmetry has a minima, it must occur where all variables are equal."

This is also not true.

Eg xy/(x^2+y^2) is minimised on the line y=-x, excluding the domain hole (0,0).

You should stop guessing and try to prove things if you think they are true. Such statements are worthless without proof.

I meant homogenous cyclic polynomials.
Sy's example is just that if multiplied by (a+b)(b+c)(c+a)
Forgotten to mention positive a, b and c, as stated in Sy's
Can apply to both max and min.
xy/(x^2+y^2) has a max of 1/2 for positive x, y

 

[seanieg89]

Re: HSC 2014 4U Marathon - Advanced Level

I meant homogenous cyclic polynomials.
Sy's example is just that if multiplied by (a+b)(b+c)(c+a)
Forgotten to mention positive a, b and c, as stated in Sy's
Can apply to both max and min.
xy/(x^2+y^2) has a max of 1/2 for positive x, y

Well most homogeneous polynomials won't even have global extrema, but even if we just look at those that do, your claim is still incorrect.

(x-2y)^2(2x-y)^2 is a homogeneous cyclic polynomial. It doesn't have a maximum, but it is minimised (and is zero) whenever we either have x=2y or y=2x. So in the quadrant with x and y positive, none of the extrema lie on the line y=x.

 

[jyu]

Re: HSC 2014 4U Marathon - Advanced Level

Thanks for the example.
I am beginning to see your point.
Will the reasoning apply to cyclic rational functions of degree 0 with positive variables?

 

[seanieg89]

Re: HSC 2014 4U Marathon - Advanced Level

I really don't think so. You should try to prove it if you think it's true though, thats the only way to actually know.

Will look for a counterexample when at home.

Sent from my GT-I9000 using Tapatalk 2

 

[jyu]

Re: HSC 2014 4U Marathon - Advanced Level

Thanks. Please look for one defined when a=b=c=..

 

[RealiseNothing]

Re: HSC 2014 4U Marathon - Advanced Level

Thanks. Please look for one defined when a=b=c=..

You should be trying to prove it rather than assuming something then searching for a counter-example.

 

[Carrotsticks]

Re: HSC 2014 4U Marathon - Advanced Level

You should be trying to prove it rather than assuming something then searching for a counter-example.

Finding a counter-example is a much faster (and often considerably) easier way of disproving something, than trying to prove that it doesn't work.

 

[RealiseNothing]

Re: HSC 2014 4U Marathon - Advanced Level

Finding a counter-example is a much faster (and often considerably) easier way of disproving something, than trying to prove that it doesn't work.

But what if you never find a counter-example? Doesn't necessarily disprove it.

 

[Carrotsticks]

Re: HSC 2014 4U Marathon - Advanced Level

But what if you never find a counter-example? Doesn't necessarily disprove it.

Oh most certainly. But if you have a 'hunch', the first plan of action would generally be to try to find counter-examples. Once you can't find any and have good reason to think that your claim is true, then you would try to prove it.

 

[seanieg89]

Re: HSC 2014 4U Marathon - Advanced Level

A hunch is nothing without evidence. In trying to prove something you really get to understand a statement. (And counterexamples usually stem from thinking about the step where a logical proof attempt fails).

I will provide a counterexample if I can when I get home tonight jyu, but it will be the last one unless you give me solid evidence for your next assertion. I am not going to keep spending my time and effort trying to explain something if there is no sign that you have put the same thought and effort in.

Sent from my GT-I9000 using Tapatalk 2

 

[seanieg89]

Re: HSC 2014 4U Marathon - Advanced Level

Oh most certainly. But if you have a 'hunch', the first plan of action would generally be to try to find counter-examples. Once you can't find any and have good reason to think that your claim is true, then you would try to prove it.

Emphasis on the "good reason" though. Simply tweaking assumptions and hypotheses by guessing is very counterproductive. If you read the last few posts, you will see this is what has been happening.

Also it is much more valuable to first at least try to sketch a proof than try to seek out pathological counterexamples imo.

Sent from my GT-I9000 using Tapatalk 2

 

[Carrotsticks]

Re: HSC 2014 4U Marathon - Advanced Level

Emphasis on the "good reason" though. Simply tweaking assumptions and hypotheses by guessing is very counterproductive. If you read the last few posts, you will see this is what has been happening.

Also it is much more valuable to first at least try to sketch a proof than try to seek out pathological counterexamples imo.

Sent from my GT-I9000 using Tapatalk 2

Generally, I wouldn't tweak any assumptions or hypotheses under risk of being overly specific to say a class of functions and then mistaking the property to being applicable over a broader spectrum (which looks like what happened above)

Perhaps a difference in style, I wouldn't dedicate too much effort in an attempt to seek out a disgustingly pathological counter-example (if by pathological you mean something that really tests the boundaries of the current assumptions a la Riemann's Pathological function versus the former definition of Riemann integrability). I personally think that is a waste of time because they can be very difficult to think of sometimes (the 'disgustingly pathological' counterexamples)

However, I do see some merit in spending a couple of minutes thinking about a quick counter-example before jumping into proofs.

 

[jyu]

Re: HSC 2014 4U Marathon - Advanced Level

Thank you all for the thoughts.
I certainly have put time in analysing the idea.
The idea is not baseless. Well known geometric facts support that, such as the question I posted a+b+c>=abc in a unit circle; area of a rectangle; volume of a cuboid.
What I was not able to do was using the right name for the type of expressions generated. This allowed seanieg89 to construct the counterexamples.
'Well most homogeneous cyclic polynomials won't even have global extrema'- This is true unless there is a constraint placed on a homogeneous
cyclic function of the variables in the original polynomial, which I also forgot to mention previously.
For example, (x-2y)^2(2x-y)^2 is a homogeneous cyclic polynomial, now plus the constraint xy=4, the minimum is 16 when x=y >0.