Mathematics

Jake Rose

1:56 AM

Anybody here?

Jake Rose

2:11 AM

If S is real symmetri, T is real anti-symmetric show that det(T+iS−1) is non zero

Anybody got an idea as to how to 'deduce' this?

Adam

2:55 AM

anyway I too have an issue with the way in which the supreme fascist says things ought to be, I feel as if I am missing something axiomatic here that will help this sit better with me $$\prod _{j=n+1}^{n-k-1}a_{{j}}=\prod _{j=n-k}^{n}\frac{1}{a_{{j}}}$$

Prashin Jeevaganth

4:31 AM

Does anyone have any knowledge of game theory here?

Have some questions at an elementary level not involving much calculations, just some theoretical concepts to grasp for me

JBis

4:46 AM

@PrashinJeevaganth I would suggest just asking. If someone can (and wants to) answer then they will.

Prashin Jeevaganth

In a zero-sum game, just considering pure strategy, we can find a saddle point due to the existence of a dominant strategy A. Is there any motivation for either party to choose to pursue a mix strategy to try to milk out more gains from the other player?

Secret

a mixed strategy zero sum game always have an equilibrium. It might be useful to check whether the equilibrium is tending towards the pure strategy limit to assess whether extra gain can be obtained by going from pure to mixed

Prashin Jeevaganth

@Secret Is it true by Nash's Theorem, any mixed strategy(zero-sum or variable sum) definitely has an equilibrium, regardless of which in the pure strategy there is a Nash equilibrium

Can't understand why sometimes there's one from each case and sometimes we only have to depend on the mixed strategy one

Nobody recognizeable

6:40 AM

Hey guys i was thinking from definition a harmonic function $\phi$ satisfies $\nabla^2 \phi $=0 so does that indicate sinx is not a harmonic function.

Secret

@PrashinJeevaganth yes, every finite game has a Nash equilibrium, but not necessary a pure strategy Nash equlibrium

1

Q: Why Are Sine and Cosine Called Harmonic Functions

I thought that the definition of a harmonic function f such that $\nabla^2f=0$ In one dimension, doesn't this mean a function who's second derivative is zero? IE $\frac{d^2f}{dy^2}=0$ However, the sine nor cosine function's second derivative does not equal zero, but in many textbooks they are...

because historical reasons

Unknown MathMan

Which introductory book on logic will be best for an Undergraduate? I mean discussed everything clearly

Nobody recognizeable

@Secret thanks.

Secret

7:07 AM

logic questions are best to be asked in the logic room, as 21820 and co are the experts there

Prashin Jeevaganth

7:29 AM

Does anyone understand how x.x gets me x+1?

I dont get how the derivation of x is the multiplicative inverse of x+1 gets me the other result

Secret

basically trial and error. There are only 4 elements in this field and 0 and 1 cannot be the multiplicative inverse of x

if x.x=1, it leads to the contradiction that x=1, thus by elimination, only x+1 remains

Prashin Jeevaganth

@Secret So am I right to conclude that the result has nothing to do with the above derivation that x is the multiplicative inverse of x+1?

Basically when making some weird finite field like this, I have to trial and error out for all cases

Secret

well, for any finite field, there are cases that always holds: x.x cannot be x else inverting gives x=1, and x.x cannot be 0 else inverting gives x=0. Otherwise, unless there are further contradictions, x.x can be anything else

For this finite field, x.x=1 lead to x=1 thus it is also ruled out.

I don't know if there is some general way to test what the possible result when a given field element is multiplied to another

x.(x+1) = 1 as deduced above
Also x.(x+1)=x.x+x by distributivity. This latter equation can be used to find further restrictions

One possible way is to consider the contrapositive. I.e. we knew that if an element has a multiplicative inverse, you cannot be absorbing i.e. ax=a cannot hold.

Thus we can phrase the problem this way in trying to find the inverse of x:

Akiva Weinberger

7:54 AM

Why does the crochet hyperbolic plane not contradict Hilbert's theorem

Secret

Pick any a in the field, you do not want ax=a. Now we an already ignore 0 and 1 since they can never be multiplicative inverses of other elements. That leaves x and x+1. Now suppose x(x+1)=x+1. Expanding we get x.x+x=x+1 and by adding the additive inverse, we get x.x=1, but that is not allowed. We obviously don't want x.x=x either, because then x=1. This means x(x+1)=/=x+1. Now because this is a field, there are no zero divisors, meaning x(x+1)=0 is out. we can also easily ruled out x(x+1)=x+1 and x(x+1)=x as they will then be absorbing elements, which does not exists in fields.

Prashin Jeevaganth

@Secret I saw a loophole in this question ... we could define 2 = 1+1 which is actually 0 in the field

Consider x·(x+1) = 1, and from its LHS, x.(x+1)+x = x.(x+(1+1)) = x.(x+0)= x.x

Secret

In fields, the following cannot hold for any elements ab=ba=a, ab=0

We are interested in finding c such that xc=1. Now c cannot be x,0 or 1 because then it will obey any of the above equations (and zero is never a multiplicative inverse in any field, ring etc.). That leaves x+1 as the answer to the inverse of x

Responding to the new paragraph next:

Prashin Jeevaganth

RHS would be 1+x (when we add x to both sides)

hence we have found another multiplicative inverse

Secret

x.(x+1)=1 => x.x+x=1
x+1 = x + x.(x+1) = x.(1+x+1)=x.x
x+1=x.x
x+x+1=x.x+x
1=x.x
x=1
contradiction.
This means 1+1 cannot be 0

if any of the pathways lead to a contradiction, that assignment is forbidden because equality is transitive

wait typo

Prashin Jeevaganth

8:08 AM

@Secret but the question said addition of coefficients are done in the Z/2Z world

Secret

x+1 = x + x.(x+1) = x.(1+x+1)=x.x
x+1=x.x
x+x+1=x+x.x
x(1+1)+1=x(1+x)
0+1=x(1+x)
1=x(1+x)
No contradiction

so no there isn't another inverse of x

Prashin Jeevaganth

@Secret ah ok badly phrased...

oh cool

actually since there are only 3 non-zero elements in this field

and 1 doesn't need to have an inverse(because 1.1 = 1)

we could say informally that x is the multiplicative inverse of x+1 right?

Secret

yes, all two sided inverses are unique. This is why fields are so nice

Prashin Jeevaganth

How big of a field do people usually use?

this is a 4 element one lul

Secret

depends on the domain of study. Polynomials over finite fields are important in algebraic geometry (details I understood nothing of except I am aware of it when mathien and co talk about it in this chat, as well leaky talking about ideals and stuff) and number theory (related to factorisation stuff). Infinite fields such as the reals and complexes made up most of analysis

Prashin Jeevaganth

8:20 AM

Me learning stuff that I don't understand and questioning 1+1=2

-Math

Secret

The field with four elements as an example of finite fields is important in the theory of irreducible polynomials, it is known as GF(4)

Prashin Jeevaganth

The moment I saw "irreducible" XD

Secret

In mathematics, the following is guarentee to give me a BSOD:

1. Having more than 3 inequality signs appearing in the same line of a proof (bonus point if there is one or more absolute value of the form |f(x)=g(a)|

2. Anything that has to do with the word "prime"

3. Irreducible polynomials

I have ZERO intuition on any of these

Prashin Jeevaganth

what's a BSOD?

Secret

knowyourmeme.com/memes/blue-screen-of-death-bsod

basically you computer ceased to respond and hangs

so in this context, my brains hangs when I am exposed to the above 3 things

(Number theory as a whole used to be the 4th, but after I understood it is basically the algebra of factorising numeric expression, I started to get some intuition on it)

My current understanding is that number theory governs the rules of how numbers are partitioned

from fermat last theorem, to sum of two primes problem, to collatz etc., it is the science on figuring how how to split numbers apart in certain ways

I also like general topology more because it will replace almost every $>$ with $\supset$ which strangely, will not cause me BSOD

Prashin Jeevaganth

8:30 AM

Splitting numbers ...

How many ways to express 1 and 0

lol

Secret

and most importantly, the lack of a metric consideration means the absolute sign can rarely appear in topology, which is often give me a lot of headache in mental arithmetic

Akiva Weinberger

5

Q: Galois Field GF(4)

Question: Why is the table of GF(4) look like the one below? I know it has to do with the fact that 4 is composite Let GF(4) = {0,1,B,D} Addition: $$ \begin{array}{c|cccc} + & 0& 1& B & D \\ \hline 0& 0 & 1 & B & D \\ 1 & 1 & 0 & D & B \\ B & B & D & 0 & 1 \\ D & D &...

field-theory finite-fields

Seems related to what you were talking about

Prashin Jeevaganth

@AkivaWeinberger that's the higher level topic XD

Akiva Weinberger

Strange notation^

Prashin Jeevaganth

I hope they dont test Galois fields in my exams XD

haha

Anyone know why the reason why 9,10 are False,True respectively? I just plugged in the formula for random walks to my understanding

got a 10^-5 number for Q9, and 500- 0.04 for Q10

Trying to find my sanity

Secret

8:41 AM

I cannot visualise probabilistic dyanamics very well without reams of paper

and it is at least 8 years since I last did a random walk problem

Prashin Jeevaganth

@Secret omg, it's either your a genius or you are very old

I can't imagine my age -8

Secret

I am a PhD in computational chemistry

I have all the age you need to not give the negative number

lol

Also:

Nov 15 at 11:03, by Secret

Prashin Jeevaganth

I didn't understood that either

LOL

Secret

Nov 15 at 11:15, by Secret

another detail is that I first learnt about axioms, definitions, elementary derivations, general cases, and then examples and counterexamples, before converging to the theorems themselves

My learning is very lopsided. I do not follow the usual step by step and my logic make erratic jumps alot from one random math field to another

The only thing which I can say to have a rather solid core on is linear algebra

anything else is basically picked up throughout this chat and some random reading

Prashin Jeevaganth

@Secret would you happen to know why the answer for the following questions is F,T,F?

Secret

8:50 AM

I have not studied anything about geometric partitions, thus I don't know

Prashin Jeevaganth

does this have anything to do with that?

Secret

(and I am surprised this is actually a field of study)

Prashin Jeevaganth

OMG, I don't even understand what I'm getting into

how come all my questions come from a different math field

fields, random walks and now this

Secret

Envy-free cake-cutting

An envy-free cake-cutting is a kind of fair cake-cutting. It is a division of a heterogeneous resource ("cake") that satisfies the envy-free criterion, namely, that every partner feels that their allocated share is at least as good as any other share, according to their own subjective valuation. When there are only 2 partners, the problem is easy and has been solved in Biblical times by the divide and choose protocol. When there are 3 or more partners, the problem becomes much more challenging. Two major variants of the problem have been studied: Connected pieces, e.g. if the cake is a 1-...

wtf

what is in your exam syllabus?

Prashin Jeevaganth

Some pretty insane shit, taught on a very skimpish level, but the questions escalate quite quickly

Secret

9:06 AM

pretty applied stuff. Is this commerce mathematics?

Elsa

I'm stuck at the Second Isomorphism Theorem , precisely I'm confused with the meaning of HN/N , where H is a subgroup of G and N is normal subgroup of G. Shouldn't HN/N be equal to H/N ?