| lokieee | I have S is the part of the cone z^2 = x^2 + y^2; that lies between the planes z= 1 and z = 3. And they say z = sqrt(x^2+y^2); over the region. D = {(x,y)| 1 <= x^2+y^2 <= 9 } how did they get those bounds? did they just plug in 3 for z and 1 for z? thats what it looks like |
| Kobbler | j0 to all and j0 to none |
| lokieee | yah that has to be it! |
| diseaser | what is the exponent rule for taking the square root of a cube root of x? ie: sqrt(x^(1/3)) ? |
| Kasadkad | sqrt(x^(1/3)) = (x^(1/3))^(1/2) = x^(1/3*1/2) = x^(1/6) |
| diseaser | excellent, thank you Kasadkad |
| Kasadkad | sure |
| diseaser | Kasadkad: what about squaring?: (x^(1/3))^2 = x^((1/3)*(2)) = x^(2/3) ? |
| Kobbler | Kasadkad, where did the ^(1/2) come from that? |
| diseaser | Kobbler: sqrt(x) = x^(1/2) |
| Kobbler | i read it and inderstand the consept a little but trying to figure it out |
| diseaser | x^(1/3) = cube root of x (dunno how to type that on irc heh) |
| Kobbler | ok i think that makes sence to me |
| diseaser | Kobbler: exponent rules: x^(a/b) = b-root(x^a), thus: sqrt(x) = sqrt(x^1) = x^(1/2) im sure others here can explain it much better than I just attempted hehe |
| Kobbler | for some reason when ever you guys put in sqrt (i dont know what it is here cause its comming up as an emotocon) |
| lokieee | waht chat client are u usig |
| Kobbler | ice chat |
| diseaser | oh I did ( x ) like sqrt ( x ) |
| Steve|Office | Kasadkad: I had an idea. Assume that some rational function is a root, and then derive a contradiction. |
| Kobbler | thanks |
| Steve|Office | In particular, assume that (f(a)/g(a))^p = a. Since this is in char p, I can use the Freshman's dream to expand f(a)^p and g(a)^p. Then multiplying both sides by g(a)^p, I have two equal polynomials that do not share any powers of a so f = g = 0 but g cannot be 0. |
| Kobbler | wait how do you multiply that? (im working off of intergrated 3 highschool math which i slept through for the most part) |
| Kasadkad | Steve|Office: Not having any roots doesn't imply it's irreducible, though |
| NotFist | I was wondering is there a site that teaches beginner linear algebra? |
| WILDSTYLE | NotFist, sosmath.com I think. |
| Steve|Office | Oh right, damn. |
| Kobbler | what is linear algebra, i reconize the term but i dont know for sure |
| NotFist | WILDSTYLE, Matrix Algebra? |
| WILDSTYLE | ... yes? |
| NotFist | oh ok ty |