Treadstone

panic has it right, its simply convention that maintains continuity in notation. THe eastiest example is this... x^3/x^2 = x^(3-2) = x^1= x x^3/x^4 = x^(3-4) = x^(-1) so what is x^(-1)? well if you write it out on paper is cross out fators its a little easier to see, but this should give you the idea. x^3/x^4 = (x*x*x)/(x*x*x*x) notice, three of the x's on top cancel out with 3 on bottom so that means that x^3/x^4 = (x*x*x)/(x*x*x*x) = 1/x So, because of this, we adopt the notation x^(-n) = 1/x^n in order to maintain uniform notation.

I have to do a presentation on a math article for a senior topics class in a couple weeks and i was wondering if anyone had a good article on hyperbolic space that has a proof in it that is fairly graspable. Or if anyone has any interesting article so long as it has a proof in it and is not super long. Thanks guys Nathan

it has to do with maintaining continuity of notation. Same with numbers to the zero power equaling 1.

can someone solve this system for me...its killing me to do it by hand and i dont have my hands on any software x'' - x' + y' + 2y = t y'' + x' - 2x - y = cos2t where x and y are functions of t i just need the general solutions for x and y.....thanks fellas

thanks bro, that was really helpful

thanks bro, that was really helpful

thanks a lot man, that realy helped me out, good to see i'm on the right track....did you happen to maple the 2nd one? Also, i'm getting screwed up on something else, same section, but i dont remember if the derivative of x'/2 is x''/2 or x''...any help there?

thanks a lot man, that realy helped me out, good to see i'm on the right track....did you happen to maple the 2nd one? Also, i'm getting screwed up on something else, same section, but i dont remember if the derivative of x'/2 is x''/2 or x''...any help there?

if anyone is on and able to check my work on these couple diff eqs i would really appreciate it. The prof doesnt assign problems with solutions so i cant ever check my work . That is frustrating....heres a couple x' = 3x - 2y + sin(t) y' = 4x - y - cos(t) i used the differential operator and the method of undetermined coeffcients to get this answer..dont think its right because its so ugly x = [e^(t)] * [Acos(2t) + Bsin(2t)] - (1/10)sin(t) + (7/10)cos(t) y = [Ce^(t)] * [Dcos(2t) + Esin(2t)] + (11/10)sin(t) + (3/10)cos(t) And, if you are up to it, heres another one...D is the differential operator (D-3)[x] + (D-1)[y] = t (D+1)[x] + (D+4)[y] = 1 got answers useing the same method as above... y = Ae^(11t) + (1/10)t + (3/11) x = Be^(3t) - (9/20)t + (19/330) Thanks for anyone who responds...anyone going to IUP should avoid Dr. Kuo, lol, check her rating on http://www.ratemyproffesor.com and i've had her for 4 classes, she is terrible...my only Cs in math

if anyone is on and able to check my work on these couple diff eqs i would really appreciate it. The prof doesnt assign problems with solutions so i cant ever check my work . That is frustrating....heres a couple x' = 3x - 2y + sin(t) y' = 4x - y - cos(t) i used the differential operator and the method of undetermined coeffcients to get this answer..dont think its right because its so ugly x = [e^(t)] * [Acos(2t) + Bsin(2t)] - (1/10)sin(t) + (7/10)cos(t) y = [Ce^(t)] * [Dcos(2t) + Esin(2t)] + (11/10)sin(t) + (3/10)cos(t) And, if you are up to it, heres another one...D is the differential operator (D-3)[x] + (D-1)[y] = t (D+1)[x] + (D+4)[y] = 1 got answers useing the same method as above... y = Ae^(11t) + (1/10)t + (3/11) x = Be^(3t) - (9/20)t + (19/330) Thanks for anyone who responds...anyone going to IUP should avoid Dr. Kuo, lol, check her rating on http://www.ratemyproffesor.com and i've had her for 4 classes, she is terrible...my only Cs in math

ah, i know a little about homomorphisms, so i think i know a little about what it is now...at least enough to tell people when they ask

well it sounds pretty cool, though i only have a bit more of an understanding than i do now, lol....i'll post my findings while i'm taking the class

gonna take a class next semester on toplogy and i was wondering what it was. The course description was...less than descripted, damn mathmeticians

i am proficient in linear algebra...though forigve me if this post is short, as i just finsihed doing 6 sections worth of diff eq and proofs homework and my brain in a little fried. We just started doing rings, could you give me something to look for in the lectures that might help me figure out what set product modulo is?

due tomorrow, oh how i hate procrastination.. need to solve y''=2y+2tan^3x anywho, i know about the parameter thing, but how do i solve y''=2y....just use the charactoristic equation r^2=0? seemed to easy... edit**figured it out

i am familar with unions of sets and we just started talking about rings so i know a little about the + and * used in rings, though they dont have to be the same + and * as in interger stuff. Basicly, i know the def'n of rings, the 5 ring axioms, and a little about morphisms (homo and iso)...but thats about it. thinking about rings and such, if i define a ring as (AxB, +,*,z) such that the + and * put items from AxB into eq classes and still satisfy the ring axioms, would that be more along the right track of figureing out set product modulo? Oh, i write the z for 0 so i make sure not to mix up the 'zero' item in the ring with the integer 0....and you still havent told me your math background

alright, let me take a wack at it...though it could be more questions and thoughts than actual answers..edit**as i write this i become less and less sure about it, lol, please be kind... the product of integers can be defined as... na=a+a+a+a...n times so 5a=a+a+a+a+a So can we define a set product similarly? AxB would be...B+B+B+B...A times? what is B+B? I dont think there is a direct relation such as this because i dont think B+B makes any sense. So...AxB = {(a1,b) , (a2,b) , ... , (ak,b)} where k is the cardinality of A (thinking that that the cardinality is how many things are in A) and b is an element of B, though in this case none of the b's equal each other (i didnt want to associate a1 and b1 but rather the first element in A, a1, with some element in B, b) So...to relate that to modulo operations...let (*d) = modulo multiplication with the base of d, and let (x*d) be modulo set product with a base of d. m(*d)n = n+n+n+n+...+n m times then put into its equivalence class that coresponds to d which would be... (n+n+n+n+...+n m times)/d + q = mn | m,n,d,q are intergers and q is the eq class A(x*d)B = {(a1,b) , (a2,b) , ... , (ak,b)} where k is the cardinality of A, then put into eq classes that corespond with d...but what would that mean? would d need to be a set, D? so the eq classes would be based on another set product of the (AxB)xD? Now i'm kinda stuck and i have to get going to my OR class... Thanks for helping me with this, i really appreciate it. Just out of curiousity, what is your math background, you remind me of a prof..which isnt a bad thing Nathan

this is similar to a lot of proofs that Euclid used before calculus and other rigorous type math was found. so then...a set which contains all the elements of the product of two sets would have a cardinality equal to the procuct of the original two sets. So that kinda makes sense...graphicly a set product has the dimensions equal to that of the lengths of the sets that made it up (well in really loose terms anyway). So the overall contents of that 'set product shape' would have an area, which we could probably call its contents, equal to that of the product of the lengths of the sets that made it up.....damn and now i have to go to work, i'll look over the rest when i get back tonight. Thanks!

let me think about that...i'm glad you told me there is one...i wasnt sure if you were trying to make me think of one, knowing there wasnt, just to shoot me down, lol

Very nicely put, i think this is the best disarming statement i've heard in a long time....very nice

i think maybe if you took out all the bias in the article it wouldnt be so terribly written...all the "supposed"s just detract from what they are trying to say....however then you would actually have to read what they are saying and this it would be painfully obvious how clueless creation scientists are....

i dont think you're understanding my purpose for the post lol....simply put after being exposed to modular addtion, such as the example above, and modular multiplication...i was wondering if there is modular set product and if so what it did.

with the (x,y) dependency thing, i might be mixing up notations...for example, in the ordered pair (x,y) as it pertains the function f(x)=y, y would be dependent on x and through the relation f you would get the pair (x,y). Perhaps this is not the same as set product. But there is modulo arithmetic, such as addition of equivalence classes. If we define the operation (+3) as modular addition of eq classes charactorized by the remainder after division by 3 or a multple of 3 then [1] (+3) [2] = [1 + 2] = [3] = [0] because [1] = {...,-5,-2,1,4,2...} [2] = {...,-4,-1,2,5,...} [0] = {...,-6,-3,0,3,6,...} notice that [0] = [3], [6], [9] or any other class [3n] you can construct similar equality lists for [2] and [1]. At any rate, i was trying to figure out that since there are modular arithmetic operation like the mod+, and mod* if there was a mod set product....did i clearify my question better?

i dont know, lol, i was just thinking...i'll think some more and see what i come up with...but for now... modular operation create equivalence classes, which are partitians of larger sets. Set products are, in terms of Cartesian products (and this could be wrong), are relations on sets where by 1 set is made dependent on another set, ie XxY yeilds elements that are (x,y) like in a function where y is dependent on x. So a modular set production would take a large set, put it into partitians and then relate the partitians (set producting the mod)? Or would it partitian relations (moding the set product)? This is where i dont know if what i thinking is interesting and novel or useless and nonsensical .....i'll post more when i get to thinking about it more, i gotta head to proofs II class