[QUOTE=Krinkels;47273018]What does 'characteristic roots of systems' mean?[/QUOTE]
The best way I can explain it from my class is the function of the system. The characteristic roots are what allow you to get the function so you can graph it. In my case, it's through a time domain
[QUOTE=Krinkels;47273018]What does 'characteristic roots of systems' mean?[/QUOTE]
It means the roots of the characteristic polynomial, which is defined as p(λ) = det(A - λI) for any square matrix A.
As JohnnyMo1 already mentioned, eigenvectors v are all nonzero vectors that satisfy the equation Av = λv for some scalar values λ (these are called the eigenvalues). From this definition, you can use some [URL="http://en.wikipedia.org/wiki/Eigenvalues_and_eigenvectors#Characteristic_polynomial"]basic linear algebra[/URL] to arrive to the equation det(A - λI) = 0. This allows you to actually compute what the eigenvectors and eigenvalues are.
I decided my physics degree didn't have anywhere near enough maths subjects in it, so I've picked up a bunch of books on different topics to teach to myself slowly to broaden (and hopefully deepen) my skill set.
Top of the list are set theory, topology, differential geometry, and tensor calculus (doing a General Relativity honours project this year, so they make sense to know fairly well). Also picked up a book on graph theory because why the hell not. We also never formally learned linear algebra (it just sort of got developed through quantum mechanics classes), so I've gotten myself some books in that. Got some books on analysis, too (real and complex). Hopefully these should keep me busy for the foreseeable future. So much maths!
i need to brush up on logarithmic differentiation
good friend of mine is doing his phd in topology and sometimes he makes it look like art
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Ok, so I have this really basic question about permutations/combinations etc. which is something I never fully understood from Algebra 2. I'm Calc now but these kind of questions still stump me. I know this is miles below what you guys usually do but please bear with me :tinfoil:
"In how many distinguishable ways can the seven letters in the word MINIMUM be arranged, if all the letters are used each time?"
So, distinguishable means it's a permutation right? The answer says that since the word has 7! permutations and M has 3! perms and I has 2! perms you do: 1/((3!)(2!)) and then 7!/(12) to get 420 as the answer. I don't get what is happening when you multiply the permutations of letters with a 1 on top and then divide the total permutations by the resulting denominator.
You're right to use permutations, since you're rearranging all seven letters.
So you use 7! for the number of ways that seven letters can be arranged.
But not all the letters in MINIMUM are unique - if you were to switch the fifth and seventh letter you would still have MINIMUM, and they specified distinguishable ways. So you need to take these into account, and say that you're always going to arrange the Ms and Is in a particular order so that you don't count the same permutation twice. This is what you're doing when you divide the total number of permutations by the denominator.
For a simpler example, suppose you were to find all the distinguishable ways to write AAB. There are 3:
BAA
ABA
AAB
But there are six permutations of (A1)(A2)B:
B(A1)(A2)
B(A2)(A1)
(A1)B(A2)
(A2)B(A1)
(A1)(A2)B
(A2)(A1)B
Now, to get the number of permutations when (A1) and (A2) are not distinct, we can count just the ones that have (A1) before (A2). This is half of them, since we're fixing one permutation of (A1)(A2) out of 2!.
Shot in the dark here. I've been stumped by this for a while, and while I'm more than happy to take the orthogonality of eigenvectors as given, I get the feeling that this is something I'm likely to be tested on and need to figure out.
[img]http://puu.sh/gwNU1/47e9111e35.png[/img]
I don't see how rewriting the sines as exponentials helps at all. You end up with a rather ugly bunch of exponentials, and even if multiplied out, you still get a summation of exponentials, which I don't see how is equated to the Kronecker delta function. Any tips on how to go about this?
Have you not talked about adjoints? I don't know why you'd bother with any of this, it's almost trivial to prove with adjoints.
[editline]12th March 2015[/editline]
Wait you're talking about matrices and not linear operators in general so I guess not.
I think we might have; for reference this is in a differential equations class.
Are you referring to the proof that ends in
(ƛ1-ƛ2)(ᴪ1,ᴪ2) = 0, and since ƛ1=/=ƛ2 then (ᴪ1,ᴪ2) must be 0?
Yep.
[t]http://i.imgur.com/WmgCidu.png[/t]
?
You have to use the chain rule; find dy/dx when x=3, and multiply that for dx/dt
[editline]13th March 2015[/editline]
[sp]54[/sp]
[QUOTE=Cosa8888;47315468]You have to use the chain rule; find dy/dx when x=3, and multiply that for dx/dt
[editline]13th March 2015[/editline]
[sp]54[/sp][/QUOTE]
Actually i figured it out right after i posted, i didnt have to use the chain rule, it was simply the derivitive 3x squared then substitute in x so 27
Oops, I've been doing enough integrals to confuse them with derivatives, so I did a half derivate creating a horrible abomination.
It's 27, you are right.
I don't follow. You're looking for dy/dt, you need to use the chain rule.
y(t) = x(t)^3 + 3
dy/dt = (dy/dx)(dx/dt)
dy/dt = 3(dx/dt)(x(t)^2)
x = 3: dy/dt = 3(2)(3^2) = 54
Oh. This is why I have so many bad grades. Yes, it's 54.
It's gonna be pi day for Americans..
3/14/15 9:26:53sec
3.141592653 etc
[QUOTE=piranhamatt;47320555]It's gonna be pi day for Americans..
3/14/15 9:26:53sec
3.141592653 etc[/QUOTE]
meh
They gave us free berry pies yesterday at the Physics Department :dance:
Do you guys know where I can find a lot, and I mean A LOT, of maths excersices on all topics on the internet? I need to practice for an extremly important test at the end of this year and spending money on books it's not possible right now.
[QUOTE=Cosa8888;47323004]Do you guys know where I can find a lot, and I mean A LOT, of maths excersices on all topics on the internet? I need to practice for an extremly important test at the end of this year and spending money on books it's not possible right now.[/QUOTE]
What level are you looking for?
High school; Logarithm, algebra, complex numbers, euclidean geometry, probability and statistics. That kind of stuff. I'll need to be perfect by the end of the year.
Just saw a post on mathoverflow from a dude who isn't in academia asking for people to check his proof of the Riemann hypothesis. Took me (with no graduate education in math) until the second line after the abstract to find a crippling mistake.
Why do people insist on doing this stuff?
Is it still there? Can you post the link?
[QUOTE=JohnnyMo1;47324930]Just saw a post on mathoverflow from a dude who isn't in academia asking for people to check his proof of the Riemann hypothesis. Took me (with no graduate education in math) until the second line after the abstract to find a crippling mistake.
Why do people insist on doing this stuff?[/QUOTE]
Links?
[QUOTE=Krinkels;47325116]Is it still there? Can you post the link?[/QUOTE]
It got deleted before I could even read the single comment in response, lol.
The mistake I saw (and stopped reading after) was claiming that the Riemann zeta function of s was defined as sum from 1 to infinity of 1/n^s (whereas that only converges when Re(s) > 1 and you have to do an analytic continuation to extend that sum to the rest of the plane minus s = 1). Kind of a problem when you claim to have proved that all the non-trivial zeroes of the Riemann zeta function have real part 1/2 when your formula DOESN'T CONVERGE FOR POINTS WITH REAL PART 1/2.
[QUOTE=JohnnyMo1;47333484]It got deleted before I could even read the single comment in response, lol.
The mistake I saw (and stopped reading after) was claiming that the Riemann zeta function of s was defined as sum from 1 to infinity of 1/n^s (whereas that only converges when Re(s) > 1 and you have to do an analytic continuation to extend that sum to the rest of the plane minus s = 1). Kind of a problem when you claim to have proved that all the non-trivial zeroes of the Riemann zeta function have real part 1/2 when your formula DOESN'T CONVERGE FOR POINTS WITH REAL PART 1/2.[/QUOTE]
That mathematical babble is turning me on :3
[QUOTE=Adeptus;47333531]That mathematical babble is turning me on :3[/QUOTE]
[img_thumb]http://anticache.img0.joyreactor.com/pics/post/comics-SMBC-math-nsfw-629982.gif[/img_thumb]
So today I was presented with a little problem:
Find x in x^(x-5)=(1/8)^(8-x)
I tried solving it with logarithm but apprently x does not exist, is there an answer to this?
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