Recent content by Fibo's Rabbit

Ah, and it all makes a lot more sense now.

Homework Statement Why does lim( (1+(1/n))^n ) = e?Homework Equations If a_n convergent to a, and b_n converges to b, then (a_n * b_n) converges to (a * b)The Attempt at a Solution The lim(1 + (1/n)) = 1. If you multiply (1 + (1/n)) by itself n-times, you get the equation (1 + (1/n))^n, so...

Here is the proof provided in my textbook that I don't really understand. Suppose that x' and x'' are both limits of (xn). For each ε > 0 there must exist K' such that | xn - x' | < ε/2 for all n ≥ K', and there exists K'' such that | xn - x'' | < ε/2 for all n ≥ K''. We let K be the larger...

Thank you so much for your help. I really appreciate it. I understood how to get to the solution before your final post, but for some reason I posted something that didn't make any sense...perhaps it was too late. You've helped me so that I can solve similar types of problems when I encounter...

\int\limits_0^1\int\limits_0^1\int\limits_0^{y^2/z}dxdydz doesn't give a meaningful answer, because the last antidifferentiation makes you take ln(0). I know 0 < y2 < x*z < 1, how do I represent that as the limits in my triple integral?

say we have (x, y, z), then xz > y^2. Then xz - y^2 > 0. So we want the area under the function f(x, y, z) = xz -y^2, but only when f is positive (right?) \iiint\limits_D xz - y^2 dzdydx = \frac{-1}{12} \hspace{1cm} D = [0,1]\times[0,1]\times[0,1] Okay...that gives me the signed volume, but I...

There are 9 distinct possibilities for relations between a, b, and d. a > d > b a > d < b a > d = b a < d > b a < d < b a < d = b a = d > b a = d < b a = d = b If a > d > b then ad > (b^2) If a > d = b then ad > (b^2) If a < d > b then ad > (b^2) If a = d > b then ad > (b^2)...

Homework Statement What proportion of 2x2 symmetric matrices with entries belonging to [0, 1] have a positive determinant? Homework Equations A^{T} = A If A = [[a, b], [c, d]] Then det(A) = ad - bc. But A is symmetric, so c = b. So det(A) = ad - b^2 So, in order for A to have a...