Posted on 16 December 2016 - 05:01 PM
Purpkey wrote
I will get around to it eventually. That's a long one.
Posted on 16 December 2016 - 05:01 PM
Chocqlate wrote
Thank you, Chocqlate.
Posted on 17 December 2016 - 01:27 AM
1. Explain the differences between dipole-dipole attraction, hydrogen bonding, and London dispersion forces, as well as the reasoning behind why each force exists.2. Explain how to find the element described by a photoelectron spectroscopy graph.
3. Explain delta heat of vaporization and vapor pressure, and ways to find them.
4. Explain the difference between a face-centered cubic cell and a body-centered cubic cell, as well as the reasoning behind using these cubic cells.
5. Explain the meaning of the Clausius-Clapeyron equation, as well as the concept of how it was derived.
(I hope your chemistry is as good as your math and physics)
Posted on 17 December 2016 - 01:51 AM
why is a building called a building when it is already built?Posted on 17 December 2016 - 05:39 AM
SporkHandles wrote
Because when they were building it, they just decided to call it a building. The name just stuck after that point.
Posted on 17 December 2016 - 06:50 AM
Can you help me prove that if A commutes with a nilpotent matrix N, det(A+N)=det(A)Last edited on 01 January 2017 - 05:19 PM by AlbertEinstein
Bestaquiio wrote
Thankfully, I've gotten a lot of experience with linear algebra doing quantum mechanics.
Lemma 1: The determiniant of a nilpotent matrix is zero.
Proof: Let N be k-nilpoent, i.e., N^k=0. Then, det(N^k) = det(0) = 0 = det(N*N*…*N) = det^k(N). Thus, det(N) = 0.
Consider two cases.
Case 1: Suppose det(A) = 0. Since A,N commute, these must be simultaneously diagonalizable. Since N is nilpotent, its upper-triangular form is strictly upper triangular. Hence, A+N has the same diagonal entries as A. Since the determinant of an upper triangular matrix is the product of its diagonal entries, we must have det(A+N) = 0, since the determinant is basis independent.
Case 2: Suppose det(A) is not zero. Again, since A,N commute, they are simultaneously diagonalizable. We write this transformation as A = U(D_A)U^(-1), N = U(D_N)U^(-1), where D_A, D_N are the respective upper triangular forms of A,N and U is the matrix of eigenvectors. Then, det(A+N) = det(U(D_A+D_N)U^(-1)) = det(UU^(-1))det(D_A+D_N) = det(D_A+D_N). Using the same reasoning as in the first case, D_A + D_N must have the same diagonal entries as A. Hence, det(A+N) = det(A).
QED
Edit: I will do those chemistry questions later.
Edit 2: It should read "simultaneously upper triangularizable" instead of "simultaneously diagonalizable." Commuting Hermitian operators are simultaneously diagonalizable.
Posted on 17 December 2016 - 01:20 PM
A number consists of three different digits. The sum of the five other numberswith three different digits that can be formed with these digits is 2003.
Determine which number
I will tell you the solution if you can't find it but you're einstein ;)))
Posted on 17 December 2016 - 02:23 PM
Hexus wrote
with three different digits that can be formed with these digits is 2003.
Determine which number
I will tell you the solution if you can't find it but you're einstein ;)))
Suppose the digits of this number are a, b, and c. Then, this number, say A, is A = 100a+10b+c.
The other numbers that can be formed by this are
100a+10c+b
100b+10a+c
100b+10c+a
100c+10a+b
100c+10b+c.
The sum of these numbers is 2003 = 122a+212b+221c.
This is a diophantine equation that you can either solve by guess and check or some algorithms from number theory (Euclidean).
I did some guess and check on c, seeing that c must be odd since the other two are even and 2003-{even} gives an odd number.
Thus, I found c = 7 works and gives a solution that works.
Hence, our number is 217.
Last edited on 30 December 2016 - 06:44 AM by euree
Posted on 30 December 2016 - 01:21 PM
Vykz wrote
Jinxful wrote...
Posted on 01 January 2017 - 04:04 AM
SporkHandles wrote
In English terms (or at least what I think) a building should be called a builting, but since that isn't correct use of the English language they used build instead and applied it with ing, which forms building. Now normally if a word ends with ing it means someone, or something, is performing an action (doing something). It somewhat does make sense that it's called a building since people are inside it (most buildings have people in them at some points) and atoms which move around, so technically the building is occupied. Whilst the building technically isn't doing anything itself and just stays there, there are things moving inside of it. If there's nothing inside a building for a long time, it's normally considered abandoned, depending on what type of building it is. This is just my opinion on a building though, surely there's a lot more opinions on the word building around that explains why it's called that.
@AlbertEinstein out of curiosity, how old are you? You seem to know a lot about mathematics.
Last edited on 01 January 2017 - 03:07 PM by SuperSandwish
1=42=8
3=12
4=?
Very simple, yet most get it wrong
Posted on 01 January 2017 - 05:04 PM
SuperSandwish wrote
2=8
3=12
4=?
Very simple, yet most get it wrong
1
Posted on 01 January 2017 - 05:09 PM
SuperNeil64 wrote
SuperSandwish wrote...
1
Posted on 01 January 2017 - 06:22 PM
Are you familiar with Beal's conjecture? It is in relation to Fermat's last theorem. I once read over a proof of it, but honestly I was lost. Is there a simple way for you to put it into layman's terms for me, or do I lack an understanding that is more complicated than one that can be shortened to fit into a forum post?Posted on 02 January 2017 - 12:39 PM
HIVLIK wrote
I can't speak much on Beal's conjecture, but I can say that since it is a conjecture, there is no proof for it yet. It is some sense a generalization of Fermat's last theorem, which, of course, was proven by Andrew Wiles in the mid-90s.
I can speak a bit on Fermat's last theorem and its proof. FLT was a mess to prove, and quite frankly, you'd have to be a very well-rounded mathematician to even UNDERSTAND the proof of it.
The proof focuses on algebra, but not algebra that most people are familiar with. Higher math deals with algebra broadly described as "abstract algebra." This field describes various algebraic structures like groups, rings, fields (which are rings), and vector spaces (specifically, linear algebra for vector spaces). Often as undergraduates, math majors take three semesters of abstract algebra: linear algebra, algebra I (groups and maybe group representations), and algebra II (rings, fields, Galois theory).
But those classes are just building blocks for the stuff that Andrew Wiles had to use. Other relevant algebraic fields are commutative algebra (Here, "commutative" is a*b = b*a) and representation theory (taking your algebraic structure and putting it into its respective "linear" form to make life easier, basically). Wiles used algebraic geometry (algebra and curves) and, naturally, (algebraic) number theory. He also had to work with elliptic curves and modular forms.
The proof went along the lines of (very loosely–maybe even wrong since I'm no expert on this):
1.) Show that the solutions to FLT form an elliptic curve.
2.) Prove the Modularity Theorem (Taniyama-Shimura conjecture at the time), which shows that all rational elliptic curves are modular.
3.) A solution to FLT would not be modular, so it would violate the Modularity Theorem. Hence, FLT is proven.
I've looked at the proof of this a few times, and I recognize some notation, but other than that, I don't really know what's going on. I'd need a lot more math.
Posted on 02 January 2017 - 03:53 PM
AlbertEinstein wrote
ThatOneCombo wrote...
We begin by creating a function with a vertical asymptote of 2x+1 and y-intercept 1.
f(x) = 2x + 1 + g(x), such that f(0) = 1 = 1 + g(0) —> g(0) = 0. (1)
We also want large x to give f(x) ~ 2x + 1, so lim(x->infinity) g(x) = 0. (2)
Let's define g(x) = p(x)/q(x). From (2), we require deg q(x) > deg p(x) (3). Combining f(x) into one fraction:
f(x) = (2xq(x) + q(x) + p(x))/q(x) (4). From (3) and (4), we clearly have that the numerator of f(x) has degree deg q(x) + 1.
Now, assuming you meant three distinct zeros, one with multiplicity two, we must have that the degree of the numerator of f(x) is 2 + 1 + 1 = 4 = deg q(x) + 1 —> deg q(x) = 3.
Since g(0) = 0, we must have p(0) = 0. Thus, p(x) can be written as p(x) = x r(x), deg r(x) = 1 or 0, from (3). For simplicity, choose deg r(x) = 0 and r(x) = 1. Hence, p(x) = x.
So, f(x) = ((2x+1)q(x)+x)/q(x). Let q(x) = ax^3 + bx^2+ cx + d. Then, the numerator of f(x) is 2ax^4+ (2b+a)x^3 + (2c+b)x^2 + (2d+c+1)x + d. Let's choose the three distinct roots as j,k,l. Then, the numerator must factor as n(x-j)^2 (x-k)(x-l).
This gives a system of equations for a, b, c, d , when we choose our roots. Precisely, 2a = n, -n(2j+k+l) = 2b+a, n(kl+2jk +2jl+j^2) = 2c+b, -n(2jkl+j^2 k +j^2 l) = 2d+c+1, j^2 k l n = d.
Choosing j = 1, k = -1, l = 2, n = 2, we have a=1, b= -7/2, c= 11/4, d = -4.
So just one example would be f(x) = (2x^4-6x^3-7x^2+2x-4)/(x^3-(7/2)x^2+(11/4)x-4).
Posted on 02 January 2017 - 03:58 PM
Mom: Son what are you doing you're supposed to be doing your homeworkMe: I'm on the badlion forums
Mom: How is that helping you with your homework
Me: *Shows her this thread*
I love this man
Last edited on 02 January 2017 - 07:55 PM by *deleted-96553
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