Recent content by jav

G finite abelian group WTS: There exist sequence of subgroups {e} = Hr c ... c H1 c G such that Hi/Hi+1 is cyclic of prime order for all i. My original thought was to create Hi+1 by reducing the power of one of the generators of Hi by a prime p. Then the order of Hi/Hi+1 would be p, but...

Check your an+1 term. The [(n+1)!] should be squared.

Homework Statement Know: modulus(z) < 3 WTS: |Im(z2 - zbar + 6)| <12 where zbar is the complex conjugate Homework Equations z = x + iy The Attempt at a Solution |Im(z2 - zbar + 6)| = |Im(x2 + 2i*x*y - y2 - x + iy + 6)| = |2xy + y| So I want to show |2xy + y|< 12 I already proved it...

I have just taken real analysis I and II (single and multivariable respectively) as an undergrad. I am planning on taking the grad analysis course next year. From what I gather Rudin seems to be the time-tested bible of analysis, but I don't know at what level the book is written. Can...

This expression can be written as -1/9 * d/dx (1/(1-x/9)). Then think geometric series. Can you see it from there?

It looks like you are integrating z^2 instead of z. Also shouldn't there be an extra r in the integral? ie. \int\int z*r dr d\vartheta whereas it looks like you are doing \int\int z^2 dr d\vartheta

Ah of course, thanks.

Sorry, I was out all weekend, but I should have given what I started with. I took the natural log, and used l'hospital's rule, but I ran into a problem when I was trying to prove that lim(ln(an)) = A -> ln(lim(an)) = A ie. continuity. Clearly, it is true and the function is continuous, but I...

I know lim n^(1/n) = 1 n->infininity Does anyone have ideas on how to prove this? I feel like its something simple I am missing. Thanks

This is true if the function is even. Try not using this when you solve for the integral. ie. use 1 and 3 as your new bounds for your u integral

Try alternating signs, and it becomes straightforward.

In this case, you are trying to solve for the probability that card X is an ace (without knowing any of the other cards). If you solve for the P(a|b) you are solving for the probability of event a occurring given that you know b has occurred. ie you want to solve for P(X is an ace and...

Think he means groups of (1/2)^n terms ie 1 + 1/2 + (1/3 + 1/4) + (1/5 + 1/6 + 1/7 + 1/8) + (1/9 + ...+ 1/16) + (1/17+...+ 1/32) ... > 1 + 1/2 + (1/4 + 1/4) + (1/8 + 1/8 + 1/8 + 1/8) + (1/16 + ...+ 1/16) + (1/32+...+ 1/32) ... And note that s2n-1 = n+1 for the second sequence of...

If they have the same pdf, then they by definition have the same mean and variance.

That function would be discontinuous at x=2 and x=-2