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Let's discuss Mathematics

Discussion in 'Science & Technology' started by ParadigmShifter, Mar 16, 2009.

  1. Lohrenswald

    Lohrenswald 老仁森林

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    I also have that b and c are eachothers complex conjugate, so identical if they're real, so those solutions you suggested can't fit.

    The problem is I can't just assume that b and c are real, and I still can't figure out how to decide their values when assuming a+d=0
     
  2. AdrienIer

    AdrienIer Deity

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    a=0, b=i, c=-i, d=0 works right ? Which means b and c aren't necessarily real ?

    Edit : if you limit yourself to b and c being purely imaginary thenfor 0<k<1 b=k*i, c=-k*i and a=sqrt(1-k²) works in general
     
  3. Lohrenswald

    Lohrenswald 老仁森林

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  4. Lohrenswald

    Lohrenswald 老仁森林

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    especially since assuming a+d=0, I'm essentially left with

    a^2+bc=1

    d^2+bc=1

    and since a^2=d^2 that's essentially just one equation with two unknowns

    I mean I still have b=c*, but like

    I could try like b=b(real)+ib(imag) I guess
     
  5. Lohrenswald

    Lohrenswald 老仁森林

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    asked for help, so in case people are curious, the answer is a and d in cosinus form and b and c in sinus form (either real or imaginary for the latter)
     
  6. AdrienIer

    AdrienIer Deity

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    That sounds right :)
     
  7. Lohrenswald

    Lohrenswald 老仁森林

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    does the like infinite sum or whatever of 1/n! converge to something?

    that is like 1+1/2!+1/3!+1/4!+...

    I know it's kinda important to like figure out how and why these kinds of things are such and such but right now I kinda just need to know if it does and what it is

    edit: 2 seems like a reasonable guess, but I'm not sure

    edit2: or maybe it's more 1.75, but that's just from adding up until n=10
     
  8. Petek

    Petek Alpha Centaurian Administrator Supporter

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    Lohrenswald likes this.
  9. Lohrenswald

    Lohrenswald 老仁森林

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    Given that one in 2.4, I'm supposed to get 2.5, but I can't nomatter what I try
    Stirling's approximation here is N!= N^N times e^-N times square root (2 pi N)

    thing is the N^N for N+ or - N seems uncrackable, and the e-factors dissapear, even though they're supposed to be there at the final answer
     
  10. AdrienIer

    AdrienIer Deity

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    Is omega max a constant or does it vary ?
     
  11. CKS

    CKS Deity

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    I think omega max is N! divided by ((N/2)!(N/2)!) for even N and N! divided by (((N+1)/2)! ((N-1)/2)!) for odd N. (Assuming spin is either +1 or -1, giving S = N, N-2, N-4, etc.) This should help keep the e factors in the final answer.
     
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  12. Lohrenswald

    Lohrenswald 老仁森林

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    It shouldn't depend on S, but N can probably be in there

    CKS's answer might make sense in some way, but I've asked for help at university today. Got a lot of info but still stuck

    so:

    the e factors given by stirling's approximation should go away, and you can put the square roots into one factor, then there's the like N and S pyramid

    I was told to like divide by the square root factor (looking at like an equation with omega on one side) and then do ln on both sides, which with like all the exponentials and factors and such you can do all kinds of things to like simplify that with those logarithm rules

    but then my achilles heel: apparantly we're supposed to taylor expand some of those things to get it easier
    problem now is I can't do that

    there's a trick, so we can simplify the two taylor polynomials I need to

    ln(1+x) and ln(1-x) (x=S/N, but my problem right now is more general)

    I think ln(1+x) shold become x+(x^2)/2 (and stop it there because x is relatively small)

    problem is I can't get it on that form

    I know being stuck at that stage is like super inept of me, but I don't know how to do it
     
  13. Lohrenswald

    Lohrenswald 老仁森林

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    Okay I got it now

    It was kinda hard to like work with expanding at x=0 for a function with arguement 1+x

    I'll update if I find omega max

    EDIT:

    decided to do this for some reason, so this is omega max



    (made a mistake in the N+S squares thing and fixed it lazily)

    gotta be honest I don't like that S, but I don't see how I can get rid of it
     
    Last edited: Sep 17, 2019
  14. Lohrenswald

    Lohrenswald 老仁森林

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    someone said it'd probably not be that bad to estimate that N^2+S^2 is close to N^2

    It feels kinda cheap, but most of the time S is fairly small compared to N (although sometimes it's a bit big), and the taylor expansion earlier kinda had to assume S/N was a fairly small number

    N here is actually ten thousand, so it's kind of at the edge of those like judgements being valid, but I guess I'll have to live with it

    On to the next thing though

    say you want to rewrite a variable x to ly (l times y) (It's physics so it's to get rid of dimensions but whatever)

    or maybe more precisely y=x/l (so l should have like opposite dimensions to x so that y is dimensionless)

    now the derivate, d/dx

    the problem I'm trying to solve like makes it seem that d/dx should be equal to (1/l)d/dy, but to me it seems more obvious it would be ld/dy

    because like in the other notation f'(x)=f'(ly)=lf'(y) no?

    Am I missing something?
     
  15. CKS

    CKS Deity

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    df/dx = df/dy * dy/dx. Since dy/dx = 1/l, df/dx = (1/l) df/dy. So d/dx = (1/l) d/dy.

    In the prime notation, you are taking the derivative with respect to something else (like t), which is why you get the opposite result.
     
  16. Lohrenswald

    Lohrenswald 老仁森林

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    It's quantum mechanics but it's fairly math shifted

    I've thought about expressing the total S in the sum according to the Clebsch-Gordan table, but otherwise I've not really figured out much

    any ideas?

    I'm a bit scarce in saying what I've tried because this is a kinda desperate place to ask help, I mainly do just to not squander the oppurtunity, but I can answer questions if needed
     
  17. Lohrenswald

    Lohrenswald 老仁森林

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    The answer I think is 9/8 J with degeneracy 4 and -3/8 J with degeneracy 6

    it only took me several days to figure this out
     
  18. Mouthwash

    Mouthwash Escaped Lunatic

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    Posts like that make me thankful I've forgotten my math beyond basic multiplication. I mean, I'll have to relearn some to get a degree, but I assure you there won't be quantum physics involved.
     
  19. Lohrenswald

    Lohrenswald 老仁森林

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    For the record I was wrong, and that exam went pretty bad
    there were 8 eigenstates because there were 8 basis vectors for example

    anyway I have another question

    I've got something in the form of 1/((x^2)*(exp(1/x)-1)^2) (other constants too, of course)

    I'm suppossed to evaluate this in the limit of x goes to zero and infinity (or very large), but I'm not sure how to
     
  20. AdrienIer

    AdrienIer Deity

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    Write the exp(x) as 1+x+x^2+x^3+... etc for the limit on 0.
    For infinity there is no tension between the exp and the x^2 so just use your usual limit rules
     

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