Recent content by SomeRandomGuy

nevermind, I believe I solved it

Prove that a number a is a zero divisor mod n iff gcd(a,n)>1. I have the proof assuming a is a zero divisor mod in => gcd(a,n)>1. However, going in the other direction has me stumped. Basically, here is all I have so far that I feel is correct: Assume gcd(a,n)=d>1 => da' = a and dn' = n...

First, i'd like to thank you for your response. I understand everything you said except what I left in quotes. Why did you integrate 1 from -2pi to 2pi? Other than that, I already had the support being [-1,1] and I know how to find mean and variance when given some P(X).

Let X be the sin of the angle theta in radians chosen uniformly from (-pi/2,pi/2). Find the mean and variance of X. HINT: X = sin(theta). Specify the support of X and check to see if your result describes a p.d.f. Anyone got any idea's? I managed to solve the majority of other problems and...

I have y(t) = (t-C)^2 when y>=0. I get the same thing when y<0 as well, by separation of variables. I use t-C rather than t+C thanks to a hint from my professor from yesterday's lecture. So, is this the solution I am looking for?

dy/dt = 2*sqrt(abs(y)) = 2*sqrt(y) y>=0, 2*sqrt(-y) y<0 isn't 2*sqrt(-y) when y<0 = 2*sqrt(y)

Consider dy/dt = 2*(abs(sqrt(y))) 1.Show that y(t)=0 is a solution for all t. I did this part 2.Find all solutions (hint, give solution like y(t)=... for t>=0, y(t)=... t<0). He told us in class that t=0 isn't necessarily the point we should be concerned with 3.Why doesn't this...

Yes, MOLS is mutually orthogonal latin squares. the notation (4, 7^2, 3) referes to an [n, M, d] code where n is the length of each vector in the code, M is the number of vectors, and d is the minimum distance between them. Anyway, I figured it our, so it's all good. Thanks for the responses.

Construct a [4, 7^2, 3] code. I know it exists because 7 is prime, so there are 6 MOLS. However, I am not quite sure how to go about constructing this code.

Thanks for the help guys, I made some pretty dumb mistakes in a few of my posts that I caught tonight after looking things over. I think I have the answers now. when f(x) = |x^2-1|, it's not differentiable at x = 1, -1. when f(x) = x|x|, it's differentiable at 0. Is this right? If needed, ill...

Ok, I did what you guys said and took the right and left hand derivatives for this function. I got the right hand derivative to be infiinity, while the left is -infinity. So, it's not differentiable at x = 0. I understand this, and I used this technique on the 4 other problems. I have one...

I don't understand. Are which two limits the same? The definition of the derivative on |x| and x|x|?

How do you determine when f is differentiably from a real analysis standpoint (no graphs and calculus)? Would I simply look for a point of discontinuity? We have 4 problems on our homework assignment involving this issue and I don't see one example in my notes or the book adressing it. Here is...

Yes, limx->c+ f(x) = L if limf(x_n)=L for all x_n > c where lim(x_n)=c. Similar definition for for limx->c- f(x). Since f was defined as an increasing function, then x_n is either increasing or decreasing depending on what limit we are taking. How can we conclude that limf(x_n) = L...

Been bangin my head against the wall all weekend thinkin about this question: Let f:R->R be an increasing function. Proce that lim x->c+f(x) and lim x-c-f(x) (right and left hand limits) must each exist at every point c in R. There's more to the question, but if I can get this part solved...