Recent content by c0dy

I think method 2 is how he wanted us to solve it. We were also using a similar method for solving partial differential equations. Is there a decent textbook which focuses solving ODE or PDEs this way?

Aha, that makes sense. Thanks a lot.

I'm thinking, dv/dt = (v_2, -v_1, 0) + (1, 0, 0) = (v_2 + 1, -v_1, 0) Then integrate? v(t) = (t*(v_2 + 1) + c1, -t*v_1 + c2, 0) => v(0) = (c1, c2, 0) = (1, -1, 0) => c1 = 1, c2 = -1

Homework Statement http://i.imgur.com/zmtJ64Z.jpgHomework Equations B is a column vector {{0},{0},{1}} E is a column vector {{1},{0},{0}} v(t) is also a column vector {{v_1},{v_2},{v_3}} The Attempt at a Solution I have calculated v(t) x B to get (v_2,-v_1,0) and I made a matrix, | 0 1 0|...

Ok that makes sense. Then from there I could say: Let E = \{s_k\} ^{\infty}_{k=n}. Since s^* is finite, then E is bounded from above, E \subset \{s_k\} and E is not empty, then a supremum exists in E. And then taking limit as n \rightarrow \infty, E would consist of only supE which is exactly s^*

Any hints?

Homework Statement Prove that, s^{*} = \lim_{n \rightarrow \infty} \sup_{k \geq n} s_k Assume that s^{*} is finite. Homework Equations Definition of s^{*} is here: http://i.imgur.com/AWfOW.png The Attempt at a Solution I started out writing what I know. By assuming s^{*} is...

If I have a lexicographic ordering on ℂ, and I define a subset, A = \{z \in ℂ: z = a+bi; a,b \in ℝ, a<0\}. I have an upper bound, say α = 0+di. My question is does only the real part, Re(α) = 0 define the upper bound? Or does the Im(α) = d have nothing to do with bounds in general? Since it...

Wow, you know how dumb I feel right now? Thanks a lot for your help + patience. I've had a history of over-complicating problems because I never look for obvious things. Be sure I'll have more problems as I'm taking Analysis, Abstract Algebra and a calc-based Statistics course, but I'm not...

Ok, so two people have shown me this for associativity: x*(y*z) = (z*x)*y = (x*z)*y = y*(z*x) = (y*x)*z <- How? = (x*y)*z Is that step assuming associativity? I do not know why 'x' can be brought into the parenthesis and taking out y', such as...

Sorry I forgot to mention that I have proven for commutativity. Let x = e, x*(y*z) = (z*x)*y e*(y*z) = (z*e)*y y*z = z*y Therefore, * is commutative. And the part you warned me about I did feel as if it was not right... but as of right now I don't know how I could introduce parenthesis to...

Oh that's right... could I do this: x*(y*z) = (z*x)*y = (z*e)*y = z*y = z*y*e = z*(y*e) = z*(e*y) = (e*y)*z = (x*y)*z

And for associativity: x*(y*z) = (z*x)*y = (z*e)*y = z*y = (z*y)*e = e*(y*z) = x*(y*z) Quite verbose I know, but is this legal?

Hm, that is a good point. I didn't know I could let x = e. I can let x = e because any arbitrary element in S could be the identity element?

Homework Statement Assume that * is an operation on S with identity element e and that, x * (y * z) = (x * z) * y for all x,y,z \in S. Prove that * is commutative and associative. 2. The attempt at a solution I want to prove commutativity first as that may make it easier to prove...