Recent content by kjartan

For part two, all we need to do is believe what we are given and apply the triangle inequality. Suppose (a,b) are in C, then 2<a2+b2<4 notice that: |x-a|= sqrt[(x-a)2] <= sqrt[(x-a)2 + (y-b)2] Similarly for y. Observing that C is an annulus with an inner radius of sqrt[2] and an outer radius...

Update: after squaring, the image obtained is what is called a "cardioid shape" not "ellipse-ish." Now that I have the shape, I am working "backwards" to obtain a description. This confirms the shape: http://en.wikipedia.org/wiki/Cardioid#Cardioids_in_complex_analysis The rest is algebraic...

I don't have an equation at this point. Basically, I just plotted some points. If theta is taken with respect to the origin, not the center of this circle, I identified the points on the original circle (before squaring) corresponding to +/- (0, pi/12, pi/8, pi/6, pi/4, pi/3) and then...

Homework Statement sketch the curve in the z-plane and sketch its image under w=z^2 |z-1|=1 Homework Equations z=|z|e^(iArgz) argw=2argz The Attempt at a Solution At first I simply sketched the solution for a circle centered at (1,0) in the z-plane and then mapped that to...

Whoops! Thanks! Ok, now finally . . . integrating e^(-(y1+y2)) with our new limits of integration and then taking d/du, to obtain f(u)=4ue^(-2u) Presto! Thanks! Ok, so I should have paid more attention to their use of "average" in the question, I kind of read over that. I appreciate...

Ok, almost there . . . So, since we have (y1+y2)/2, we switch the upper limits of integration as such: \int^{2u}_{0}\int^{2u-y2}_{0}e^(-(y1+y2)/2)dy1dy2= 4(1-e^(-u)-ue^(-u)). Then take d/du, obtaining: 4ue^(-u)I must be tired, I'll have to take another look at this to figure out what I'm doing...

Thanks for taking a look! What I did was this. f(y1) = e^(-y1), and f(y2) = e^(-y2), since the general form for an exponentially distributed random variable is f(y) = (1/\beta)*e^(-y/\beta). Where the mean is beta, and we are given that the mean is 1. Then, since y1 & y2 are independent...

1. Suppose that two electronic components in the guidance system for a missile operate independently and that each has a length of life governed by the exponential distribution with mean 1 (with measurements in hundreds of hours). (a) Find the probability density function for the average length...

Well, I think I see how the first example should be read. <[12],[20]> = <[4]> in ℤ_40, since <[12],[20]> ⊆ <[4]> since [12] = [4]⊙[3] and [20] = [4]⊙[5] <[12],[20]> ⊇ <[4]> since [4] = [12]⊙[2] ⊖ [20]⊙[1] Hopefully my thinking is correct here. Then, given <[x],...,[y]>, we find the...

If we call <a> the "span" of a, then I need some clarification on the concept of span. def. if G is a group and a∈G, then <a> denotes the set of all integral powers of a. Thus, <a> = {a^n : n∈ℤ} thm. Let S be any subset of a group G, and let <S> denote the intersection of all of the...

Thanks for clearing this up. I was also thinking that (-1,1) - K meant the set of all x-1/n between -1 and 1 (basically, the open interval (-2,1) in R). Hopefully this will help becu: We're looking at a basis element in the K-topology, B = {x in R: -1 < x < 1}\{1/n: n is a natural...

Relevant equation: y(x,t) = Acos(kx - wt) where k = 2pi/wavelength and w = 2pi*f (d^2y(x,t)/dt^2)/(d^2y(x,t)/dx^2) = w^2/k^2 = v^2 So, v = |w|/|k| = (2730s^-1)/(172m^-1) = 15.9m/s 1.50m/15.9m/s = 0.0945s to go up the string Next, m = w/g = 1.25N/9.81m/s^2 = 0.127kg u = m/L =...

I was just working on this same problem and couldn't find any help online, but I figured it out myself; so, although this is rather late, perhaps it will help someone else. Anyway, at first I wanted to solve for wavelength the same way that ku1005 did, but, in fact, this is not necessary...

Thanks much, Tom, your reply was most helpful and I appreciate it! Thanks for the tip about LaTex, too.

Homework Statement This is my first post here. I'm particularly unsure about b(ii) Thanks for any and all replies! I apologize in advance if I haven't used the correct conventions, but I hope that this is legible. I will learn the correct conventions for future posts but was pressed...