Math Is Fun Forum

It appears that you have done the bit to get f(0)=0 which is good, but the second bit where
you need to do a substitution like y = -x you appear not to have done correctly.

I will give you a few more lines of it to give you the idea of what I meant:

Let us assume that the result f(0)=0 is proven.
Now let y = -x

Using the original equation  f(x+y) = f(x) + f(y) + 2xy

We get:  f(x-x) = f(x) + f(-x) + 2(x)(-x)

Since x-x = 0 the bit in the left hand side can be simplified to f(0)
The bit where we have 2(x)(-x) can be written as -2x^2
So f(0) = f(x) + f(-x) - 2x^2
Using the result from earlier that f(0) = 0 we get:
0 = f(x) + f(-x) - 2x^2
This can be rearranged to give us:
f(x) = 2x^2 - f(-x)
[This was an interesting result, but we cannot eliminate f(-x).]

I am not sure at this point whether it is good style to change the letter that
is used like u = -x (as a substitution) or -u = x
[I now realise that this does not work. Sorry about that bit.]

Sorry about this. I have spotted a flaw in the final part of my proof.
This substitution does not appear to work, although it is confusing and
it is easy to accidentally flip the sign when writing it and think that it
proves that f(x) = f(-x) which would solve the problem.

Edit: Thanks Avon, also thanks to {7/3} for an interesting problem.
I have now read Avon's post, checked it to see that I agree, and yes it does work.
My result f(x) = 2x^2 - f(-x) is not needed so that was a blind alley.
However it is needed to do my y=-x substitution with the function g(x + y) to make the method that follows work.

I agree with bobbym's answer exactly.

I have decided to hide my comment about a possible method below.
It would be interesting to know how Agnishom got an answer that is very nearly
correct but slightly out. (Assuming of course that bobbym's was right - I think it is.)

Okay. I will think up some ideas and send you an email.

I have looked at the Wikipedia definition of a "hyperbolic paraboloid" and it gives an equation of:

My immediate intuitive thought is that the expression you gave cannot be rearranged into this.
However what about a rotated version of this ?

The equation that Wikipedia gives for this is:

If a = b then it is not exactly the same as your expression, but bares some resemblance with an extra x^2 term.

If a is not equal to b in magnitude then a y^2 term seems to be needed to make it a proper fit to the equation.

At the time of writing I am not sure whether the expression you gave is another rotation or not possible with any rotation.
My guess at the moment is that it is not possible with any rotation. I do not know how to rotate the equation by an angle
in the +z direction in general, but I would have thought the y^2 term would be non zero.
This is just a guess though so I wonder what other members think about this.

Yes that was the one I was talking about. Is it better to post to that one or continue the discussion here?
Also is anyone curious like me about the answer to that one?

There is a mathematical topic called "Linear Statistical Modelling" where I have an old text book from about 1999 approx.
I did do some studies to do with this some time ago, however my knowledge is a bit rusty and the software that goes with
the course I can no longer use because it was for my old computer and I only had one years use of the software allowed by
the license agreement anyway. There is a way that an approximate model is used in the textbook, but as I say it was a long
time ago that I studied this. Also if it were an academic question they would give out some test data with questions to guide
the student through it. (Obviously it would not be like that in real life.)

EDIT/FOOTNOTE:

I have looked at the book ("Statistical Modelling Using Genstat") and there is a piece on converting the multinomial
distribution into a Poisson generalized linear model. It is in chapter 12 called "Loglinear models for contingency tables".
A theorem is used where it is basically trying to say that Poisson models for categorical data behave like multinomial
models. At the time of writing I do not fully understand the proof of the theorem so I am not going to paraphrase the proof.
The hypothesis test would use this model with something like what the software calls:
"Stats|Regression Analysis|Generalized Linear| Log-linear" (using the menu system of Genstat)
Of course the software may have changed by now and other software may now be considered better now for some reason.
The software may have been chosen for educational reasons. However it is advisable to use software for this because otherwise
the calculations by hand would be too complicated. It strikes me that a control group that just has the before and after test
but with no induced (bad) taste stage would be needed to avoid confounding the experiment with the effect of the before
part of the test. However the hypothesis test I think would be the above test.

I realise (and did realise) that the victorious player is not necessarily superior.
In fact in my question what I meant was what is the probablility that this "common sense view" is wrong ?

The form of my question was confusing because I used the term "common sense superior" to mean "victorious".

In the long term the more games are played (assuming that the probability of each winning is constant) the better estimate
we get of one player being better than the other modelled by a probability P of one being victorious in a single game.

There may be some use in temporarily assigning the value P = 0.5 for some reason, but the Wikipedia entry I found confusing,
but it does sound quite sophisticated, and it is very unlikely that P is exactly 0.5, I agree with anonimnystefy that if we take
this literally then as the degree of accuracy increases the probability of this converges to zero as the number of decimal places
of the accuracy tends to infinity. In practice we could never have such perfect accuracy, but in that case an interval ought to be
given like: 0.45 to 0.55 or 0.495 to 0.505
However no such range was given. Do we really mean this ? [The range chosen would be an arbitary choice]
In other words would we be likely to want to ask the question in this way ?
Or is it better to ask the question: What is the probability that the non victorious player is better in terms of underlying ability ?

According to the method in the Wikipedia link by anonimnystefy what would the answer be ?

To be quite frank I did not follow the method given by the Wikipedia entry. I could plug in some values and see if I can get
an answer. This did refer to testing a coin for bias, but it might well be applicable to this exercise/puzzle.
Even a perfectly "fair" coin has some degree of bias. I think I remember reading once about an experiment that was done
which established that there was a small bias in a regular coin, however I cannot remember the details and frankly it can
only have been slightly weighted one way or the other.

Thanks Nehushtan, that was a much clearer explanation than mine. Basically it amounted to the same thing that the
case where k=1 does not allow the inductive step to work, because an intermediate horse is needed otherwise we
end up with two equations without a horse color on the other side of the equation, because if the set only has two
horses in it then removing one leaves just one, and of course the equation needs something on both sides, and
the horses from 2 to k are non-existent.

About the Bob Bundy point: anonimnystefy has answered this already, in that Bob has (perhaps like me in my first
post) not read the theorem very well and has interpreted it in a way where the set under consideration is just a
particular set (which could be thought of as being added to by one element each time one is added to n) rather than
ANY set of that size picked from the set of all horses. I realised after studying the question more closely and thinking
about it for longer, that it could not possibly be this problem that was being intended by the author of the book.
If it were that problem then it would be the inductive step that fails and in every case it would fail because if we are
adding a new horse from an outside source from which the theorem is not anything to do with. However this is as
I mentioned earlier not a correct interpretation of the problem, and as far as I know, does not lead to anything particularly
remarkable. It does however show that human error can occur both in producing a new mathematical idea, as well
as spotting the correct flaw (as I managed to get wrong in my post #2 when I jumped to the wrong conclusion regarding
the flaw). Also communicating the idea well can cause problems as I managed to show in my next post on when I had
realised what the flaw was in my head, but did not explain it very well.

To go back to the original problem though: The inductive step does work for k = 2 and all k > 2 as well.
However if we cannot get from case n=1 to case n=2 using the inductive step (which we can't) then this does not make any
difference because the case where n=3 can only be proven by induction if the n=2 case is proven.
The case n=2 really should be proven as a separate case (another basis case), but common sense tells us
that we cannot literally go through an infinite number of horses to show this for every horse set of size 2.
In mathematics a property of the mathematics might give us some way of proving something for an infinite set,
but obviously equal horse color is not likely to have such a property. If it were a real maths problem that we were trying
to solve and the inductive step was not valid for k=1 (but was for k>1) then we would try to solve this case as a special
case unless of course there was some other way round the problem.

I have thought of another idea for what is wrong:

Are we saying that the hypothesis is true for all sets of horses of that size of of just one particular horse set size?

There is a big difference between "I have one set of one horse which is the same color as itself"
to "All pairwise comparisons of pairs of horses have the same color".

Take the basis case n=1. If we take the first statement to be the interpretation then it is true because we are only comparing
the horse with itself.

However it is nonsense in the second interpretation: If all pairs of sets of horses with n=1 have to be the same color for the basis
case to be true then the basis case is not true.

Are we trying to argue to get from n=1 to n=2 that given that a horse is the same color as itself then if we look at the set
containing both horses take away one horse it is the same as itself therefore n=2 works?

There is an inconsistent phrase in the theorem of "the" set of horses as compared to "a" set of horses. This is a big difference.
Either n=1 just has to represent one isolated set of one horse, or it has to represent every set including every pairwise comparison
between two sets to be valid, otherwise the inductive step will not be valid.

I think that the case for n=1 probably is supposed to be that each horse however chosen is the same color as itself.
However the n=2 case would be that any pair of horses however chosen are the same color.
We obviously cannot get from n=1 to n=2 so there has to be a flaw. If we have two horses {horse1, horse2} then
remove n=1 of them we get {horse1} or {horse2} either way the inductive step from n=1 to n=2 will not be proven.
The fact that the hypothesis holds for the sets individually does not tell us anything about the pairwise comparison between
the two horses. If we number the set from 1 to (2-1) then this is 1 to 1 ie horse1. If we number them from 2 to 2 then this
is horse2. Of course if we could prove the hypothesis for the case n=2 then we would have proven it for all n because if we
can arbitarily choose any pair and show that they have the same color then all horses are the same color.
So yes my revised answer would be that we cannot get from n=1 to n=2. This assumes the interpretation that we are
taking ANY pair of horses for n=2 and not just one isolated set. (If it is an isolated set then all of the n=k to n=k+1
steps are flawed, because we are always taking in a new horse that we don't know anything about.)
The inductive step could be said to work from n=2 to n=3 because each pair is covered by the n=2 case, but if you
go through the inductive step for n=1 to n=2 then of course it is nonsense, because there is no overlap between the
one size smaller subsets of {horse1, horse2}.

The image to illustrate the above post should be in this post. I have noticed that you cannot type in any Latex in the same
post as an image upload.

Note that the top left and top right illustrations both represent the fractions (1/3) = (2/6) = (6/18)
The middle two represent (5/6) = (15/18)
The lowest two represent the answer to the addition: (21/18) = (7/6)  [As a mixed number this is 1 and a sixth see Latex in above post.]

Question to Mandy: Have you understood this illustration ? Do you have any questions about it ?

(a) 11/36
(b) 2/5

I am absolutely sure about the first one, but have a slight nagging doubt about the second.
The thing that occurs to me is Bayes Theorem, and I haven't studied that for a long time.
I therefore decided to think about how many possible combinations added to 8:

6,2
5,3
4,4
3,5
2,6

Then noticed that two of them contained a six. So (2/5)

The thing that made me think of Bayes was that it was "something given something else", but I cannot think how
Bayes could be used in this example. Perhaps it is ((2/36)/(5/36))
Bayes formula would be:  P(K|C) = P(KC)/P(C)
So would it be valid to let  P(KC) be the chance of getting an 8 which also contains a six ?
With P(C) as the chance of getting a total of 8.
With P(K) as the chance of the role containing a six.
With P(K|C) as the chance of the role containing a six given that the total is 8.

(Someone needs to check all of that because I am not sure of it. Haven't done this for ages.)

You have got a good point there.

It seems that the real solution does work after all.

If you cube the real solution you get 16, then take the negative square root.

One of my graphics calculators does not even accept the input, and the other gives the first of my complex number answers.

So let us first work out (-4) ^ 2

Well this is 16 because (-4) ^ 2 = (-4) x (-4) = -(-(16)) = 16

Then we get (16 / 3)

So  (((-4) ^ 2)/3) = (16/3)

On the other hand: -((4^2)/3) = -(16/3) 

Now how about ((-4) ^ 2) / ((-4) ^ 3) = 16/(-64) = 1/(-4) = -(1/4)

If there are an odd number of negative terms in a multiplication then the result is negative.

If there are an even number of negative terms in a multiplication then the result is positive.