Let's discuss Mathematics

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

1. 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

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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

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4. 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)

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That sounds right 7. 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

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Lohrenswald likes this.
9. Lohrenswald老仁森林

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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

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Is omega max a constant or does it vary ?

11. CKSChieftain

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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.

12. 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. CKSChieftain

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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老仁森林

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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 