If this is true, then the proof is no longer magical. You expect to find a unique real positive lambda; yes, we did. You expect that this is the largest root in magnitude [assuming my guess is right]; yes, it is. You expect that if lambda is the largest root, then the sequence is bounded by O(lambda^n); even that's true.
@robjohn Well, I will put it this way. My method will fail under these two cases: (1.) If the characteristic equation has negative coefficients in the RHS when you write it as x^d = something. (2.) x is not a positive real.
@JacobSchlather :) Right. So we should settle for every root is at most lambda* in magnitude... =)
Sorry for changing the goal post after seeing the counterexample. =)
@JM Nope, I am not. That will be useful, generally speaking. But I am assuming that God gave me the unique positive root. How big it is -- I don't care for this question.
Well, it's the same principle behind that cleverness. The characteristic equation for a_n = 2a_{n-2} is x^2=2, so the general solution is like c sqrt(2)^n + d (-sqrt(2))^{n}.
@JM If I think about it, I know which are not defined in TeX, but it is just easier to type \Beta. Also, I can fix up the font if needed (I forget if the plain Roman matches the Greek font face).
It was JM and it was just earlier this month. It felt longer ago.
oh, let f(t1,t2) = e^(i*omg*t1) e^(j*omg*t2). f(t1,t2) does trace out a circle as you vary t1 (or t2). I thought you were asking what effect has on a quaternion q: like f(t1,t2)*q
why it does not rotate............by multiplying e^(i*omg*t1) with e^(j*omg*t2) what is actually happening..............I want to interpret the variation of e^(i*omg*t1) e^(j*omg*t2) with t1 or t2 or both varying
no i wasn't intending to ask about multiplication by q
For eample consider a point on unit 4-sphere, and the line joining it to origin makes angles phi, theta, psi with i,j,k axes respectively.....then what is the quaternionic representation of that point....is it e^{i*phi} e^{j*heta} e^{k*psi} ? Is this correct
@QED e^(i*omg*t1) e^(j*omg*t2) always lies on the unit 4-sphere. I've checked its magnitude
Just curious, is it clear to anyone why x^8-24x^6+144x^4-288x^2+144 is irreducible over Q? Wolframalpha tells me it's the minimal polynomial of sqrt[(2+sqrt(2))(3+sqrt(3))] over Q, so it's surely irreducible, but I can't think of reason or test that would prove why.
@JM Oops, I just realised that there are over 40 questions with that tag. (I was initially mistaken that there are only around 5 questions.) Not sure about abolishing the tag anymore, but I still feel that the tag is pointless.
- if you propose it in meta, I will upvote you :=)
I suppose it will be much more efficient that way. How much time will it take to retag tens of questions removing the [natural-numbers] or [number] tag? In the current system, it's taking too much attention from a few people involved.
@JM If you are afraid that there will be too many bumps, if we start eradicating both number and algebra, there are two possibilities to avoid this: A. I can wait until number is done; B. We can do it on different days. (I retag regularly 10 questions from algebra tag each Wdnesday and each Saturday - I plan to go on with this timetable.)
@Srivatsan I see that this is not how this tag is being used, but I was just wondering: Do we have a tag for constructions of rational, real, natural numbers etc.? Like this question math.stackexchange.com/questions/85672/…
(and if we do not have such a tag, would it be useful in your opinion?)
I won't be at the computer until afternoon, but I will stay logged in. Feel free to ping me if you have something related to tagging debate; I'll check the messages later.
@MartinSleziak I did some cleanups for the [natural-numbers] and [number] tags today. I won't be touching them soon. [We could agree on a time-table...]
I would say, since you're doing [algebra] already, let's not stall that and take up the [numbers] tag. Either we do them in parallel, or if for some reason we should choose one, we can go ahead with [algebra].
BTW is it weird that when I came to my computer this morning, I check my new responses at math.SE first and my emails (some of them work-related) only later?
@RajeshD: [If in a game] each player has chosen a strategy and no player can benefit by changing his or her strategy while the other players keep theirs unchanged, then the current set of strategy choices and the corresponding payoffs constitute a Nash equilibrium.
@MartinSleziak I imagine it would be tagged under the resp. field: like, construction of reals = analysis, construction of rationals is elementary abstract algebra, but without the elementary qualification. Construction of natural numbers = set theory or logic or some such.
@Srivatsan: Von Neumann did actually contribute to it. It seems that the first two people studying strategies in games were Borel (surprise!?) and von Neumann. At least according to Wikipedia.
@RajeshD I am not sure what you mean. The space of strategies could even be finite, like {coöperate, defect}.
But it is not guaranteed that every game has a "pure" Nash equilibrium. But you always have a "mixed" NE. In that sense real numbers come in, probability comes in.
@AsafKaragila Of course, I agree with that. The question was whether some tag related to construction of varoius number systems (N,Z,Q,R,C) would be useful or not.
(Probably not, but I asked since number is about to be cleaned.)
I am not sure that such tag is relevant. There are too little questions, they would fit just fine into other sort of tags (each to its own) and we already have too many tags.
Can any of you guys explain to me what is a pullback?
@AsafKaragila: I used to think that every set had a cardinality. Apparently, only sets that can be well-ordered have a cardinality. But then again, why shouldn't there exist a bijection between sets A,B even only one of them can be well-ordered?
So maybe all sets do have a cardinality...
If A cannot be well-ordered find a bijection to a set B that can be well-ordered -> problem solved.
Yes, I think I figured it out. All sets are well-ordered in ZFC. So the cardinality of a set is the least ordinal such that there is a bijection to it.
