Recent content by Carl140

Yes, sorry about that. It is from s=0 to s=1.

Hi! Define an integral operator K: L^2 (0,1) -> L^2(0,1) by: Kx(t) = Integral[ (1+ts)exp(ts)x(s) ds from t=0 to t=1]. Why is "obvious" that K is a one-to-one operator? I know K is one to one if Kx(t) = 0 implies x(t) = 0 but I don't see why this is true. Can you please explain why?

Hello! I'm applying to different universities in the US. Can you please give your opinion about the University of Missouri - Columbia? (http://www.math.missouri.edu) They have several professors with PhDs from prestigious universities so I think it's a good option, isn't it? what do you think?

Hi all! I'm applying to several universities in the US. Among them is the University of Utah, I'm applying for a PhD in pure mathematics. Do you know what's the reputation of this university? I was told it is good, but I'd prefer to read comments from Americans or from anyone who studied...

Homework Statement Hello. I want to study the stability of the origin of the following problem: dx/dt = -2y dy/dt = x + 2y dz/dt = -2z So the eigenvalues of this 3 x 3 matrix are -2, 1 + i, 1-i. The eigenvectors are (0,0,1) , (2,-1-i,0), (-2,-1+i,0)...

Hi Matt, thanks for your reply. Assume there exists a point q in B(f(x),b) such that q is not in f(B(x,c)). Now since c>a it follows that B(x,a) is contained in B(x,c). Hence f(B(x,a)) is contained in f(B(x,c)). Since q is not in f(B(x,c)) then q is not in f(B(x,a)). But f is continuous so...

Gotcha guys. |f(x)-f(y)| / |x-y|^a<= K |x-y|^(b-a). Since 0 < b-a< 1 then |x-y|^(b-a) < |x-y|. Actually I meant: then |f(x)-f(y)|/|x-y|^a <= K |x-y|^a |x-y|. But x,y are both points in [c,d] so |x-y| <= d. Thus we get |f(x)-f(y)| <= K |x-y|^a * d so the constant is C = k*d, correct?

Halls: Sorry, I do not follow your hint/suggestion, what do you mean?

OK, thanks again. My try: Since f is Lipschitz of order b then there exists a constant K >0 such that for all x, y in [a,b] we have: |f(x)-f(y)|<= K |x-y|^b. Observe K|x-y|^b = K|x-y|^a |x-y|^(b-a). Therefore |f(x)-f(y)| <= K |x-y|^a |x-y|^(b-a) and thus: |f(x)-f(y)|/|x-y|^a <= K...

Homework Statement Let (X,d) and (Y,d') be metric spaces and f: X-> Y a continuous map. Suppose that for each a>0 there exists b>0 such that for all x in X we have: B(f(x), b) is contained in closure( f(B(x,a))). Here B(f(x),b) represents the open ball with centre f(x) and radius b...

OK, thanks for your reply. So |f(x)-f(y)|<= K |x-y|^b implies |f(x)-f(y)|<= K |x-y|^a |x-y|^(b-a). Therefore: |f(x) - f(y)| /|x-y|^a <= |x-y|^(b-a) But x and y are both in [a,b] so |x-y| <= |x|+|y| = b + b = 2b. Therefore |f(x)-f(y)|/|x-y|^a <= (2b)^(b-a). So our constant C is then...

Homework Statement 1. Let 0 < a < b <= 1. Prove that the set of all Lipschitz functions of order b is contained in the set of all Lipschitz functions of order a. 2. Is the set of all Lipschitz functions of order b a closed subspace of those of order a? Homework Equations I know...

That's all I'm given, I forgot to state that also S is assumed to be of infinite extent, basically S represents a part of the ventricular tissue which is separated by two walls: the epicardium and the endocardium.

Sorry. Here it is: After /doc it should be /presentations/pdf/ScollanThesisDefense.pdf That is: /doc/presentations/pdf/ScollanThesisDefense.pdf It is working, just checked it.

Homework Statement When one calculates electric potentials, it involves integrating over the charge distribution, and for a surface with a uniform charge distribution, you encounter an integral of the form: \int_{\mathcal{S}} \frac{d^2x'}{|\vec{x}-\vec{x'}|} Where \vec{x} is the vector...