Recent content by a_Vatar

A continuous function on an interval(in R), should possesses an intermediate value property. That's why it's impossible

C'mon, guys. There must be something wrong with this naive construction, otherwise it would be used as an example(on planetmath for instance).

Instead of creating a new thread, I decided to just write in here. If we define a set X={p in (0, 1), p in R} and let E = X \ Q. Now, E non-empty(since Q is countable and X is not) and bounded, furthermore, no point of Q is in E. Now what remains to be shown is that if p in E, then p is a...

HallsofIvy, I'm studying with Rudin’s Principles of Mathematical Analysis. They define lim sup exactly as sup of a set of all subsequential limits. So let E be a set of all subsequential limits and let x = lim sup(an) = sup(E), then if an > x = lim sup(an) for all but finitely many an, it is...

Hi guys. I'm apprently stuck on the basics of the analysis. On the proof that Q lacks least upper boundary property to be precise. The example I have uses a set A (p in Q | p > 0, p^2 < 2) then q is defined as p - \frac{p^{2} - 2}{p + 2} . Then they show that if p is in A then q is in A too...

but where do this coefficient come from: \sqrt{a+1} Is there a general formula to expend a quartic? Or am I missing something obvious Thanks.

http://mathworld.wolfram.com/Lemniscate.html eq. 13 - 15

How did you get this expression? \frac{1+t^2}{(\sqrt{1+a}t^2+2\sqrt{a-b}t+\sqrt{1+a}) (\sqrt{1+a}t^2-2\sqrt{a-b}t+\sqrt{1+a})} when I factor {a(1-t^2)^2 + 4bt^2 + (1+t^2)^2} = (a+1)t^4 -2(a-2b-1)t^2 + (a+1) then say, u=t^2 and u1,2 = \frac{(a-2b-1)\pm 2\sqrt{(1+b)(b-1)}}{a+1}...

\int \frac{1}{a*cos^2(\phi) + b*sin^2(\phi) + 1} d\phi after letting t=\tan \frac{\phi}{2} I get \int \frac{1+t^2}{a(1-t^2)^2 + 4bt^2 + (1+t^2)^2} this does not look to facinating, so I let u=t^2 and this simplifies it a bit further, but introduces \sqrt{u} into denominator; I...

Well, you don't. What is the integral of y = ax ? ax^2 / 2, right? Can you integrate r^2 = 2a^2 (cos^2theta) then, integrate wrt to theta not a?

should be r^2 = 2a^2 (cos^2theta); sketch the curve to find the limits of intergration, I'd say for the right loop you integrate from -pi/4 to pi/4. Constant a controls the form of the curve. Try sketching it here http://graph.seriesmathstudy.com/

Hint: convert to Lemniscate equation into polar form and calculate area from there

Hi, I'm trying to evaluate this integral over a hemisphere: \int cos(\theta)^{a*cos^2(\phi) + b*sin^2(\phi)} dw where dw - solid angle measure,\phi is azimuthal angle and \theta is polar angle. Thus we have: \int cos(\theta)^{a*cos^2(\phi) + b*sin^2(\phi)} dw = \int \int...

This picture show what differential is for a function f(x) http://www.bymath.com/studyguide/ana/sec/ana4a.gif Basically, it is the change in the linear approximation for a function for a change in x, dx. dy/dx = f '(x) -> differential dy = f '(x) dx When dx is small dy is a good...

Thanks for help guys! Finally got the trick of dividing one equation the other :) This yields an expression for t = 1/t0 for A)