Recent content by JFo

Nevermind, it just hit me. If R is the region of integration then R = {(x,y)| g(x)>y and y>=0}, or more compactly R = {(x,y)| 0<=y< g(x)} Which of course is just the area below the graph of g(x) and above the x-axis (g is assumed non negative) eliminating the points where g(x)=0.

I'm reading through a proof (the full theorem statement is at the bottom of the post) in a book on probability and I'm having trouble following a line in the proof. The line reads as follows: \int_{0}^{\infty} \int_{x:g(x)>y} f(x) dx dy = \int_{x:g(x)>0} \int_{0}^{g(x)} dy f(x) dx Where...

So if we have a theorem, and are then able to show that we can drop one of the hypotheses (or replace it with a weaker hypothesis) and still obtain the same result, would that be considered "strengthening" or "weakening" Intuitively I would think strengthening, but then I don't know how that...

Thanks. I know it's mostly semantics, but I couldn't find a definition, or even a description, anywhere.

Could someone explain to me what it means for a theorem to be a strong(er) form or a weak(er) form of another theorem? I've heard these terms used over and over, but never bothered to ask. If I had to guess at a definition, I'd say that if q is a theorem then we say p is stronger if p...

Thank you muzza! I really need to get rid of this extra chromosome!

Remove the last digit from a number and subtract twice this digit from the new (shorter) number. Show that the original number is divisible by 7 iff this difference is divisible by 7. I have only the division algorithm and the fact that the integers are closed under...

When it's said that a group G acts "trivially" on a set X does that mean \forall g \in G, \forall x \in X, gx = x ??

for a given eigenvalue \lambda, the dimension of the corresponding eigenspace E_{\lambda} must satisfy 1 \leq dim(E_{\lambda}) \leq multiplicity(\lambda)

What is the rank of A? What is the rank of A^tA?

Oh, your right. I got two theorems mixed up. Thanks again for your help

Thanks, that's useful advice. Now that I think of it though, the proof above doesn't rely on the fact that S_3 and S_11 are Sylow p-subgroups. I could have just stated that there exist subgroups of orders 9 and 11 in G (since 9 and 11 are powers of a prime dividing |G|), and the same results...

There is only 1 Sylow 3-subgroup and 1 Sylow 11-subgroup in a group of order 99. Denote these as S_3 and S_{11}. |S_3| = 9 and |S_{11}| = 11. S_{11} is cyclic and and every nonzero element of S_{11} is of order 11. This implies that S_3 \cap S_{11} = \{e\}. Therfore |S_3S_{11}| = 99 so...

Prove all groups of order 99 are abelian: I'm stuck right now on this proof, here's what I have so far. proof: Let G be a group such that |G| = 99, and let Z(G) be the center of G. Z(G) is a normal subgroup of G and |Z(G)| must be 1,3,9,11,33, or 99. Throughout I will make repeated...

I suppose if \tau is a transpostion, then \tau^2 = i, thus \phi(i) = \phi(\tau^2) = \phi(\tau)^2 = 0. This implies \phi(\tau) = 0 since there are no nonzero elements of order 2 in Z_5. Ahhh... this then implies that that all of S_5 gets mapped to 0 since every permutation can be written as...