Recent content by srn

I am confused by the concept of stability and condition. As I understand it, condition is defined by how much the output changes when the input changes. But why is it linked to the problem and not the algorithm? What if I have two algorithms that calculate the same thing but in a completely...

Great, thanks both!

In the definition, 1) why must you find a n_0 \in N such that \forall N \geq n_0? You might as well say find a n_0 \in R such that \forall N > n_0. Just a matter of simplicity? 2) Why must |x_n - a| < \epsilon hold? I think |x_n - a| \leq \epsilon is fine as well, given that it must hold...

Uh, right. So the intersection is (0,0) but U + V \neq R^2. There's no scalars so that a\cdot (1,0) + b\cdot (1,1) = (0,1), for example. So U and V cannot form R^2 and the direct sums are hence not equal? edit: ooops, a = -1 and b = 1 :) so they do actually form R^2 Sidenote: If U=<(1,0) then...

If (R,S, +) is a vectorspace with U, W as subspaces, then U \oplus W = \{u + w | u \in U, w \in W\} and every s \in S can only be written in one possible way (as the sum of vectors of U and W). I.e. it's every possible combination of elements in (R, U, +) and (R, W, +). Suppose U=<(1,0)>, V =...

Thanks for the replies. And sorry, clearly posted this too late because I messed up the symbol in the question. :( Meant to say direct sum indeed... Hey. :) Yes they do. But here the direct sums are not equal though, right? I guess you meant the tensor product? Sorry :(

U⊕V = U⊕W; find U, V and W I need to give an example of different vectorspaces U, V, W so that U \oplus V = U \oplus W. Can anyone give a hint please? It's basically asking for V and W such that u_i + v_i = u_i + w_i yet V and W have to be different. How?

Got it. Thanks, guess I just needed a little push in the back. :)

I need to determine the affine transformations used to produce the following image: Been staring at it for an hour and it's frustrating me to no end because it's probably really easy. Clearly it gets scaled by 1/3 and there are 3 linear transformations that put it at (0,0), (1/3,1/3) and...

Hi, thanks for your reply. Yep, I realize that; though what I meant was that I would have suspected that the field for r > R2 would be weaker in comparison to a situation where there is no conducting sphere. I.e. it is weaked more than just from the distance from q. But I suppose from these...

Thank you both for replying and thank you for fixing my Latex, I'll be more accurate next time. For r < R1: that should have been in the denominator indeed, sorry, clumsy, I did know that. For R1 < r < R2: you're right, it should be zero. I was thorougly confused by an exercise in my book...

Homework Statement A conducting sphere of radius R2 has a central cavity of radius R1 that holds a charge q in its centre. Determine the electrical field for r > R2, r < R1 and R1 < r < R2 and determine the charge density induced by q. I'm not allowed to include a link to my figure, but I'm...

Awesome, that clarifies a lot. Thanks vela!

Ah, not so hard after all :( j := (y+.2)^2+(z+.2)^2 = 0.04 k := 0.07^2+(y+.15)^2+z^2 = .02255; > solve({j, k}, {z, y}); {y = -.2815230378, z = -0.01736924054}, {y = -0.4212402098e-1, z = -0.07721899475} That should be it then? Does anyone know how to draw a plane through 3 points in...

edit: should probably be in precalc sorry about that Homework Statement I have three points, two known and one partially known, only the x coordinate (3D) Point one (A): (-0,2; -0,2; -0,2) Point two (B): (-0.13; -0.15; 0) Point three (C): (-0,2; unknown; unknown) I also know that...