Let's discuss Mathematics

[AdrienIer]

Third degree isn't really a subject we study (except if there's an obvious root, which means it's just the study of a second degree equation), although sometimes when we start talking about complex numbers it's mentioned that "i" was first invented for third degree equations.

 

[Kyriakos]

Hm, isn't it a bit underwhelming to just ask you if you know what logarithms are and how to do power multiplication?

Maybe it was just 1 out of 5 million questions or something.

 

[Samson]

Hm, isn't it a bit underwhelming to just ask you if you know what logarithms are and how to do power multiplication?

Maybe it was just 1 out of 5 million questions or something.

The past oxford admissions exams are all online, the maths ones here. Does not look trivial to me.

 

[Kyriakos]

The past oxford admissions exams are all online, the maths ones here. Does not look trivial to me.

I suppose it is directed to people who just graduated from highschool (?) and it has a large number of questions.
But how is the log base2 of 3 upper/lower bound question not trivial for that age?
I am not saying it should have math olympiad-type questions, just that one question seemed a bit strange

 

[Kyriakos]

Maybe this should have been in the Oxford entrance exam

 

[Samson]

Is the statement "Covid cases are exponential" the same as saying "R is constant", or more exactly "R and generation interval are constant, or in an inverse relationship"?

An exponential relationship is one such that y = a + b * R ^ x? So would a constant rate of infection be exponential (R = 1)?

 

[CKS]

Is the statement "Covid cases are exponential" the same as saying "R is constant", or more exactly "R and generation interval are constant, or in an inverse relationship"?

In math, yes. Or, "R is approximately constant."
In English, no. In English, "Covid cases are exponential," or "Covid cases are growing exponentially," just means that the number of cases is growing fast. Some people may use exponential correctly in math terms, but not very many people (at least in the USA).

An exponential relationship is one such that y = a + b * R ^ x? So would a constant rate of infection be exponential (R = 1)?

It would be exponential, but we would not say that. Usually we say that something is growing exponentially or falling exponentially, rather than just saying it is exponential, since whether it is growing or falling is kind of important. If it were neither rising or falling, we'd describe it more simply as being constant, even through it can be described as exponential.

 

[AdrienIer]

I wouldn't say that it just means "growing fast", more that the speed of the spread is rising as the number of cases rises.

 

[Lohrenswald]

beta=ln(x^2+2ixy-y^2-a^2)+Uix-Uy

anyone know how this can be written in a form like ci+d?

 

[monikernemo]

beta=ln(x^2+2ixy-y^2-a^2)+Uix-Uy

anyone know how this can be written in a form like ci+d?

The logarithm part can be refactored to:
ln(x+iy+a)+ln(x+iy-a)

Now suppose that a is fixed, and you take some suitable branch cut for ln (such that x+iy -a, x+iy-a is not in the domain)so ln(x+iy+a)+ln(x+iy-a) is holomorphic for this domain. In particular, if I am not wrong, you can find (real) harmonic functions, A, B such that

A(x,y) +iB(x,y) = ln(x+iy+a) + ln(x+iy-a).

Hope this helps.

 

[Kyriakos]

I have a question, about a property of powersets. Namely:
Can you prove that any element of a set will exist in half of the set's powerset, without using the method in the proof that the number of subsets is always 2^(number of elements in the original set)?
Cause obviously it is trivial to just go "you change one of the 2s to a 1, so you have half".

 

[monikernemo]

I am interested in there possibly being some method different enough so that it may reveal more about the subsets and their relations to the original set.

Not really, after all sets have very little to no structure.

 

[Kyriakos]

Not really, after all sets have very little to no structure.

I don't agree with this (but it's not the point, and certainly not the scope).

The question is just if there is any different method at all - forget about the reasons why it could be of interest.

I have a question, about a property of powersets. Namely:
Can you prove that any element of a set will exist in half of the set's powerset, without using the method in the proof that the number of subsets is always 2^(number of elements in the original set)?
Cause obviously it is trivial to just go "you change one of the 2s to a 1, so you have half".

Basically the question is if there is some non-identical method to go about proving this, cause if there was then it could be used in synergy to the other method to show other stuff for tied subjects (and that's what interests me).

 

[monikernemo]

I don't agree with this (but it's not the point).

The question is just if there is any different method at all - forget about the reasons why it could be of interest.

Sets are a blank canvas, and I don't think there is anything more one can derive from structure of sets.

A method could be: Let A be subsets containing an element x and B be subsets not containing the element x. So there is an obvious bijection from A to B by f: A -> B with f(S) = S\{x} and g: B -> A with g(S) = S U {x}. In particular if X finite one has |A| +|B| = |P(X)| with |A| = |B| by the earlier bijection.

 

[Kyriakos]

Sets are a blank canvas, and I don't think there is anything more one can derive from structure of sets.

A method could be: Let A be subsets containing an element x and B be subsets not containing the element x. So there is an obvious bijection from A to B by f: A -> B with f(S) = S\{x} and g: B -> A with g(S) = S U {x}. In particular if X finite one has |A| +|B| = |P(X)| with |A| = |B| by the earlier bijection.

Let me translate this to english:

There is a 1 to 1 correspondence between A and B (from both sides). A contain x, B do not. One goes from a subset A to a B by just excluding x, and from B to A by containing it. The cardinality of the (finite) sets A,B is equal to the poweset only including those with element x cardinality.
(erased here something dumb I had written ^_^ ) But far more importantly I did not get how in your proof it is an "obvious" bijection (the question itself is to show that x exists in half of the subsets; you seem to start your proof by using that as true).

In other words: intuitively anyone can say "obviously any element of the original set will exist the same number of times in the powerset as any other" - what I asked is for a proof which is not using crucially stuff from the proof 2^n
Presented in a different way: what's intuitive for you is not for a machine.

 

[monikernemo]

you didn't account for subsets that have neither x or x (the empty set) nor for those that have both (the tautology of the original set). But far more importantly I did not get how in your proof it is an "obvious" bijection (the question itself is to show that x exists in half of the subsets; you seem to start your proof by using that as true).

(1) we assume law of excluded middle, so sets either contain or do not contain x.

(2) we also assume standard ZFC axioms, so in particular, A is well-formed formula:

A = {S in P(X): a in S} and B= {S in P(X): a not in S}
And it's left as an exercise to figure out why A U B = P(X) and A, B are disjoint.

(3): finally, it is an exercise (for you) to show that the maps are indeed a bijection

(4) I did not assume that half of the subsets already contain x, I merely asserted the existence A and B, which are well-formed if one assumes ZFC.

 

[Kyriakos]

I will examine your post, but it is a bit disappointing*
That obviously doesn't mean it is of no use to me.
I do question, however, if what you presented is really non-tautological with the 2^n proof. Recall that I have no use for any proof which is essentially the same. So while I will for the time being take your word this isn't the same, I do really doubt it is so ^_^

*perhaps only due to the different pacing, cause I am not yet into generalizations using the ZFC and assorted types.

 

[monikernemo]

I will examine your post, but it is a bit disappointing
That obviously doesn't mean it is of no use to me.
I do question, however, if what you presented is really non-tautological with the 2^n proof. Recall that I have no use for any proof which is essentially the same. So while I will for the time being take your word this isn't the same, I do really doubt it is so ^_^

Maybe if you can qualify what is a non-equivalent proof, and the original proof you had in mind?

I find it disappointing as well that the asker is unwilling to fill in the logical gaps, after all maths is not a spectator sport.

 

[Kyriakos]

Maybe if you can qualify what is a non-equivalent proof, and the original proof you had in mind?

I find it disappointing as well that the asker is unwilling to fill in the logical gaps, after all maths is not a spectator sport.

You are somewhat right, but not re the proof. I mentioned (ok, technically I alluded to...) a very specific proof
Here it is:
Given that the number of subsets (including empty and tautology to the original) of any original set with x number of elements is 2^x (this is true because there are only two cases for element x1 to be in the subsets or not, and this gets multiplied by the co-existence or lack of the x2,x3...xn, so it is 2(2)(2)... n times = 2^n
And since if you take one element out, this leads to one of those 2s being transformed to a 1 (cause there is only one way to not be there), it follows the sets which contain (or do not contain) any specific element are 2^n divided by 2.

Also, do excuse my tone - it's just me and I am generally playful and megalomanic. That said, in light of the above proof, do you still think that your own suggestion is not crucially the same?

Edit: Lastly, I think the equivalent question would be: can you show (without using the proof in this post as your proof or part of it) that any element of the original set will be in the powerset as many times as any of the other elements.

 

[Kyriakos]

On a different forum I read an idea which while still isn't that different, can be of use:
Someone suggested that to prove all elements of the original set appear in equal number in the powerset (which you can do formally using P(X)=2^x proof and the tied to it proof that each is in half of the subsets), you could just rename x1 to x2 etc, given obviously the position does not matter (thus you get a decently different proof of the second sentence, about all elements of X being equally in P(X)).
I like this idea, not just due to it being stealthy, but because it allows you to (for other things) tie this to probability where position does matter.
So something good came out of my questions ^_^ (here too! Thanks for your replies, @monikernemo )