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04-17-2013 , 05:54 AM
Quote:
Originally Posted by Cangurino
BTW the cross product is a vector...
Fair point. Hmmm, I wonder if I copied the question down correctly.
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04-17-2013 , 06:38 AM
Spoiler:
(b*c^3 - b^3*c) - (a*c^3 - a^3*c) + (a*b^3 -
a^3*b) =(a-b)*(b-c)*(c-a)*(a+b+c)


How to get it here (but i am afraid still not a general method for other complex cases unless they are kind of similar in structure);

Spoiler:
The logic to derive this without Mathematica is something like this;

First notice that b-a term is evident to be a factor if you group the original terms properly , the motivation being the ab^3-a^3b term or the b^3-a^3 times c term etc. Basically the b-a comes out easily as differences of cubes or squares or linear terms all having the linear difference term as common factor.

At the same time the same can be said for the term c-b (again notice the c^3-b^3 or c-b or c^2-b^2 terms that all factor c-b out.

And still the same can be said about a-c term! LOL (same reordering of terms to generate it).

So the initial expression we have is at the same time multiple of (b-a), (c-b) and (a-c). That motivates to think its is actually something like;

(a1*a+b1*b+c1*c) (<- general expression of 3 terms a,b,c ,cant be more than first power in each term to be 4th order in the end product and cant have a constant term either) times (b-a)* (c-b)*(a-c)

Obviously we then see than a1=b1=c1=1 to match the terms easily when expanded.


So i used a kind of number theory semi-bs argument (lol) that the original expression has always factors (b-a), (c-b), (a-c) which we easily see by regrouping, hence it had to be proportional to their product in general as one can imagine all kinds of numbers eg primes for the differences (b-a), (c-b), (a-c) and the expression having to have all at the same time as factors so it has to be coming out in general also as sensible hint that the combined product is right as factor in general (b-a)*(c-b)*(a-c)). Then after that all that was left is to consider the general term left allowed to get the 4th order terms so the a+b+c was generated.

Last edited by masque de Z; 04-17-2013 at 07:01 AM.
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04-17-2013 , 07:36 AM
Quote:
Originally Posted by five4suited
is this where I should post asking someone to explain how the physics of explosions in contained spaces works? does it magnify or mitigate the explosion?
Simple blast waves in an open space create a rapid rise in air pressure usually lasting less than 10 milliseconds. In enclosed environments the reflection of blast waves from walls and other surfaces creates complex waves of longer duration. This allows greater transfer of energy to the body, increasing the risk of primary blast injuries such as tympanic perforation and blast lung and increasing displacement of the body wall, which may cause a shearing effect on larger organs, especially abdominal viscera.

-Blast injury in enclosed spaces


High-order explosives produce a supersonic overpressure shock wave, while low order explosives deflagrate (subsonic combustion) and do not produce an overpressure wave. A blast wave generated by an explosion starts with a single pulse of increased air pressure, lasting a few milliseconds. The negative pressure (suction) of the blast wave follows immediately after the positive wave. The duration of the blast wave, i.e., the time an object in the path of the shock wave is subjected to the pressure effects, depends on the type of explosive material and the distance from the point of detonation. The blast wave progresses from the source of explosion as a sphere of compressed and rapidly expanding gases, which displaces an equal volume of air at a very high velocity. The velocity of the blast wave in air may be extremely high, depending on the type and amount of the explosive used. Indeed, while a hurricane-force wind (approximately 200 km/h) exerts only 0.25 PSI overpressure (i.e. 1.72 kPa), a lethal blast-induced overpressure of 100 PSI (i.e. 690 kPa) travels with a velocity of approximately 1500 mph (i.e. 2414 km/h). An individual in the path of an explosion will be subjected not only to excess barometric pressure, but to pressure from the high-velocity wind traveling directly behind the shock front of the blast wave. The magnitude of damage due the blast wave is dependent on: 1) the peak of the initial positive pressure wave (bearing in mind that an overpressure of 60-80 PSI or 414-552 kPa is considered potentially lethal); 2) the duration of the overpressure; 3) the medium in which it explodes; 4) the distance from the incident blast wave; and 5) the degree of focusing due to a confined area or walls. For example, explosions near or within hard solid surfaces become amplified two to nine times due to shock wave reflection. As a result, individuals between the blast and a building generally suffer two to three times the degree of injury compared to those in open spaces.[6]

-Blast injury mechanism
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04-17-2013 , 11:26 AM
thank you bruce.
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04-17-2013 , 02:22 PM
Quote:
Originally Posted by masque de Z
(b*c^3 - b^3*c) - (a*c^3 - a^3*c) + (a*b^3 - a^3*b) = (a-b)*(b-c)*(c-a)*(a+b+c)

First notice that b-a term is evident to be a factor...

At the same time the same can be said for the term c-b...

And still the same can be said about a-c term!
TY masque. Much obliged.

In the solution you say that (a - b) is a factor, but in the explanation you say that (b - a) is a factor, etc. Was this a trivial mistake, or am I missing something deeper?
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04-17-2013 , 03:36 PM
Quote:
Originally Posted by lastcardcharlie
TY masque. Much obliged.

In the solution you say that (a - b) is a factor, but in the explanation you say that (b - a) is a factor, etc. Was this a trivial mistake, or am I missing something deeper?
Its the same it doesnt matter (as long as in the end one keeps the signs right), the expression given initially is the correct one i think as i checked it. The b-a thing later is given as such because of the original way i found it, you know with b*c^3-a*c^3 subterms as motivation (c^3 common). However in the end i tried to produce an expression that made symmetry sense ie ;

(a-b)*(b-c)*(c-a)*(a+b+c) implies you get always the difference of one variable minus the next one in the order a,b,c (you know cyclically).
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04-17-2013 , 10:55 PM
Have what's probably a simple probability question:

Prob of A given B

And it comes with a graph.

I understand that the equation is prob (AB)/prob B, but I always get confused on which part of the graph is which, and if I should be adding parts of the graph or not.

So if I had, for instance (forgive the crudeness, on an ipad):

Genderlikes beerhates beertotal
Man .42.28 .70
Woman.075.225 .30
Total.495 .5051

If I wanted to find prob that someone liked chocolate, given the were a man, I would take .42/.7?

I think just coming up with that question and making the chart make sense answered the question for me, but I'll ask just to be sure.

Again, apologies for the ****ty chart, hope it makes sense.
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04-17-2013 , 11:15 PM
Correct. Assuming you're talking about beer rather than chocolate.
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04-17-2013 , 11:33 PM
Whoops, forgot to edit out all the chocolate!

Thank you, sir.

Last question, I swear, and only cuz I took bad notes.

Trying to find the slope of a prediction line. In my notes I have:

b = r ( std dev of y / std dev of x)

And off to the side I have "r = ratio?"

Guess I didn't hear what the professor said. But it seems important to this problem I'm doing. The google suggests that it's correlation, not ratio.

Would this then be the proper formula for calculating the r value I need?

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04-18-2013 , 04:25 AM
Hey, guys, I'm learning Linear Algebra: http://ocw.mit.edu/courses/mathemati...bra-fall-2011/

While the material that I've seen so far has been well-explained and pretty awesome, I am disappointed in the homework assignments: http://ocw.mit.edu/courses/mathemati...Ses1.1prob.pdf

Is there a place where I can get a grip-load of homeworks to try out, or will I just have to create a bunch of my own stuff and learn it that way? I don't mind creating my own hw's since apparently many systems aren't solvable and it would probably give me some intuition on the properties of said systems, but... rather work through stuff others created.
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04-18-2013 , 03:01 PM
Quote:
Originally Posted by daveT
Is there a place where I can get a grip-load of homeworks to try out
If you're willing to make an investment I would recommend the Elementary Linear Algebra book by Howard Anton and Chris Rorres (I'm assuming you're completely new to linear algebra, sorry if you're not). It is a massive book and covers all areas in a lot of detail, with loads of practise questions. I have found it has given me a really good base of knowledge 3 years on from using it, especially because it helps to establish loads of equivalent properties. For example, whilst A invertible IFF det(A) = 0 is the a pretty well-known "equivalent" statement in linear algebra, there are loads of other equivalent statements to this that I found came in very handy for me over the years.
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04-18-2013 , 03:12 PM
There are a lot of problems in Strang's book as well, which goes along with that course. It's fairly reasonably priced as well.
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04-18-2013 , 03:18 PM
I was hoping someone could help me understand the equivalence between the definitions for functions to be continuous between topological spaces, ie:

For X and Y topological spaces, and f:X-->Y a function, my notes don't prove why these definitions are equivalent (possibly because I'm missing something pretty obvious!):

1. f continuous IFF for U open in Y f^-1(U) is open in X
2. f continuous IFF for C closed in Y, f^-1(C) is closed in X

I can see why this is true when f is surjective, because then f^-1(Y) = X, so for F closed in Y, U=Y\F is open in Y and f^-1(Y) = f^-1(U u F) = f^-1(U) u f^-1(F) (because U and F are disjoint) so X = f^-1(U) u f^-1(F) implies f^-1(U) = X\f^-1(F) open in X (and then it is also easy to see the converse here). But if f is not surjective, then all that follows is that f^-1(U) is in X\f^-1(F) (open in X), so why does it follow in this case that f^-1(U) is open in X, given definition 2?
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04-19-2013 , 01:04 AM
Quote:
Originally Posted by TaliskerBA
I was hoping someone could help me understand the equivalence between the definitions for functions to be continuous between topological spaces, ie:

For X and Y topological spaces, and f:X-->Y a function, my notes don't prove why these definitions are equivalent (possibly because I'm missing something pretty obvious!):

1. f continuous IFF for U open in Y f^-1(U) is open in X
2. f continuous IFF for C closed in Y, f^-1(C) is closed in X

I can see why this is true when f is surjective, because then f^-1(Y) = X, so for F closed in Y, U=Y\F is open in Y and f^-1(Y) = f^-1(U u F) = f^-1(U) u f^-1(F) (because U and F are disjoint) so X = f^-1(U) u f^-1(F) implies f^-1(U) = X\f^-1(F) open in X (and then it is also easy to see the converse here). But if f is not surjective, then all that follows is that f^-1(U) is in X\f^-1(F) (open in X), so why does it follow in this case that f^-1(U) is open in X, given definition 2?
Isn't the preimage of a complement always the complement of the preimage? I don't we need surjectivity for that. So we have that f^-1(U)=X\f^-1(F). For assume that f(x)\notin F, that means that f(x)\in U, so x\in f^-1(U), and vice versa.
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04-19-2013 , 01:47 AM
Let C be any closed set with complement U. Then F(x) in U OR F(x) in C forall x in X. Then F^-1(U) u F^-1(C) = X. So if F^-1(U) open then F^-1(C) closed
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04-19-2013 , 03:05 AM
If you replace "or" with "or but not both" and the union by a disjoint union I agree.
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04-19-2013 , 06:36 AM
Anyone good with logic? Ive been traveling alot, so haven't been up to date. Got an assignment recently. Feel free to pm. I am willing to pay for help. I dont want you to solve it for me, i want you to teach me how it works. Here's my first task. How does it look?

Young smokers either identify with their future selves or fail to identify with their future selves. If young smokers identify with their future selves, then they are irrational if they know smoking causes cancer.
If young smokers fail to identify with their future selves, then they act without due regard for another person (namely, their future self), assuming that they know smoking causes cancer.
And given that young smokers act without due regard for another person, they are immoral.
But while young smokers do know that smoking causes cancer, they are not immoral. Therefore, young smokers are irrational, and they identify with their future selves.



identify with their future selfs=P
To be irrational=Q
To know smoking causes cancer=S
Acting without regard to another person=R
To be immoral=T
Young smokers=U


(P ∧ U) ∨¬(P ∧U ), ((P ∧ U) ∧ S )==> Q, ¬((P ∧ U) ∧ S)==>R, (U ∧ R) ==> T, (U ∧ ¬S) ==>¬T ⊧ (U ∧ Q ∧ P)
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04-19-2013 , 07:25 AM
Quote:
Originally Posted by TaliskerBA
If you're willing to make an investment I would recommend the Elementary Linear Algebra book by Howard Anton and Chris Rorres (I'm assuming you're completely new to linear algebra, sorry if you're not). It is a massive book and covers all areas in a lot of detail, with loads of practise questions. I have found it has given me a really good base of knowledge 3 years on from using it, especially because it helps to establish loads of equivalent properties. For example, whilst A invertible IFF det(A) = 0 is the a pretty well-known "equivalent" statement in linear algebra, there are loads of other equivalent statements to this that I found came in very handy for me over the years.
Quote:
Originally Posted by SenorKeeed
There are a lot of problems in Strang's book as well, which goes along with that course. It's fairly reasonably priced as well.
Ah, bummer. I was hoping for some free stuff. I'm sort of a "cheap" student. Yeah, just starting to learn this stuff. I'm planning to take some programming courses in the future and they all require Linear Algebra knowledge. Probably wouldn't make a huge difference since, as long as I have a decent base, I will see what specific areas I need to brush up on when I am faced with it.

Reading the Amazon reviews on each of those books is confusing. Can't satisfy anyone these days. I'm guessing Anton / Rorres is more "classical" math, a 'la proofs and rigor whereas Strang is more focused on the "engineering" side, is that a decent, albeit insulting and generic, summary?

Regardless, it is definitely an interesting class so far. Strang's lectures are pretty solid. It's a different perspective on things to say the least.
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04-19-2013 , 10:48 AM
Quote:
Originally Posted by SenorKeeed
There are a lot of problems in Strang's book as well, which goes along with that course. It's fairly reasonably priced as well.
Quote:
Originally Posted by daveT
Ah, bummer. I was hoping for some free stuff. I'm sort of a "cheap" student. Yeah, just starting to learn this stuff. I'm planning to take some programming courses in the future and they all require Linear Algebra knowledge. Probably wouldn't make a huge difference since, as long as I have a decent base, I will see what specific areas I need to brush up on when I am faced with it.

Reading the Amazon reviews on each of those books is confusing. Can't satisfy anyone these days. I'm guessing Anton / Rorres is more "classical" math, a 'la proofs and rigor whereas Strang is more focused on the "engineering" side, is that a decent, albeit insulting and generic, summary?

Regardless, it is definitely an interesting class so far. Strang's lectures are pretty solid. It's a different perspective on things to say the least.
You mention programming. What's your goal in learning linear algebra? If you're trying to learn the abstract side of things (more relevant to math), I'd push you one way. If you're trying to learn the engineering side of things, I'd push you another.

In any case, my recs won't be free books, but as with all books there are (legal) east Asian copies around as well as pdfs of often more questionable origin.

I didn't love Strang's book (2nd ed of Intro. to Linear Algebra), but it was definitely example- and problem-heavy, and it would probably be a nice book for you to have. I also have the 4th ed of Anton's "Elementary Linear Algebra," which looks similar.

If you want a book in front of you and are really averse to spending, you can get old versions of Strang's "Linear Algebra and its Applications" for like $3-5 on half.com, so cost shouldn't be prohibitive.

If you want a different viewpoint -- which imo really helps solidify what's really going on, but is a little more mathy -- when you've gotten through this OCR course, pick up Axler's "Linear Algebra Done Right" and go through it. You'll see things in a new light.
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04-19-2013 , 11:36 AM
Quote:
Originally Posted by Wyman
You mention programming. What's your goal in learning linear algebra? If you're trying to learn the abstract side of things (more relevant to math), I'd push you one way. If you're trying to learn the engineering side of things, I'd push you another.

In any case, my recs won't be free books, but as with all books there are (legal) east Asian copies around as well as pdfs of often more questionable origin.

I didn't love Strang's book (2nd ed of Intro. to Linear Algebra), but it was definitely example- and problem-heavy, and it would probably be a nice book for you to have. I also have the 4th ed of Anton's "Elementary Linear Algebra," which looks similar.

If you want a book in front of you and are really averse to spending, you can get old versions of Strang's "Linear Algebra and its Applications" for like $3-5 on half.com, so cost shouldn't be prohibitive.

If you want a different viewpoint -- which imo really helps solidify what's really going on, but is a little more mathy -- when you've gotten through this OCR course, pick up Axler's "Linear Algebra Done Right" and go through it. You'll see things in a new light.
Good question. There is no legitimately stated goal, but I'll say that courses that require LA are artificial intelligence, quantum computing, probabilistic graphical models, and 3d visualization. In these areas, I will have to understand the relationships between vectors, matrix multiplications, trigonometry, and the easy calculus stuff, at least from what I've seen of it.

I don't mind diving into esoterica either. I'm not that good at "real math," but I find the insights I gain from grinding out proofs and seeing relationships between apparently not-related ideas to be mind expanding and healthy. That Axler book suggestion looks intriguing and I've got that one book-marked now.

Despite the above, please don't assume that is my "goal." I've been keeping an open mind throughout this learning adventure and I wouldn't want to detract myself from that path.
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04-19-2013 , 12:12 PM
Understood.

Let me say this. IMNSHO, linear algebra may well be the most important math class for someone going into any sort of engineering that is computer related (or computer science, or...). It's worth really learning this stuff.
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04-19-2013 , 12:25 PM
Then I'll do this: I'll work through the current course and then work through the Axler book. I think I can create enough examples to do gain a decent base knowledge to get through the more difficult stuff.

Thanks for the feedback, everyone.
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04-20-2013 , 11:40 AM
I don't understand why the tension T in the strings would be mg/2 (according to the solutions manual), and I'm not sure how the hint indicates that this would be the case. Would someone mind explaining the concept to me?

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04-20-2013 , 12:19 PM
Quote:
Originally Posted by Acemanhattan
I don't understand why the tension T in the strings would be mg/2 (according to the solutions manual), and I'm not sure how the hint indicates that this would be the case. Would someone mind explaining the concept to me?

The block has mass m, so it exerts a force of Mg on the pulley. Two pieces of string lead to the pulley, and each exerts a force equal to the tension T in the string. Since the pulley doesn't move the forces have to be in an equilibrium, so 2T=Mg.
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04-20-2013 , 12:55 PM
Quote:
Originally Posted by Cangurino
The block has mass m, so it exerts a force of Mg on the pulley. Two pieces of string lead to the pulley, and each exerts a force equal to the tension T in the string. Since the pulley doesn't move the forces have to be in an equilibrium, so 2T=Mg.
Thanks for the explanation
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