Recent content by Paragon

Alright, I quess you double-integrated some quadratic, but I don't mind you to explain a bit more. I think that I got the solution by now, but I'm interested in this, too. In particular, look at this page: http://mathworld.wolfram.com/GoldenRectangle.html Note that this is only my personal...

... ... I'm sorry, but I don't speak English...:smile:

This is nice. But... If that is what I think it is, then you proved the desired golden ratio for a general quartic with any two distinct points of inflection (that's that I did). As the ratio have to involve two distict points of inflexion, there is, I think, nothing more to extent. Hence...

Nope PS: ... and that is not mah meth =)

I formed the following statement: A "W"-shaped quartic function f(x) has two points of inflection B and C. A line through the points B, C passes through f(x) again at A and D. The ratio AB:BC:CD simplifies to 1 : \phi : 1. So, AB = CD and \phi = 1.61803399... , also known as the golden...

Indeed, the fact that the polynomial had rational zeros was a presumption. And p, q are factors of a_0 and a_n. The statement: "A finite sum of integers is clearly an integer itself." hit the nail =) Thanks guys, this helped a lot!

I want to find a nice and elegant proof of the Rational Root theorem, but I get stuck. I read some stuff on the Internet, but I have not found a complete proof of the theorem. Here's my try: Say we have a polynomial: F(x) = \sum ^{n}_{r = 0} a_{n}x^{n} = a_{n}x^{n} + a_{n - 1}x^{n-1} +...

Hi,:-p so, I can't solve the embarissing: x^{2} = -\frac{1}{2}ln(x) , where x \in ]0, 2] or (0 < x \geq 2 ) any hep would be nice... thanx for your pacience!

One's parents and (especially) grandparents don't know when to stop lying to you...

54. The truth is to lie in choir, untruth is to lie alone.

41. We will die

29. The great question of youth, is that you question the time spent by a thousand fools. 30. As you grow older, you'll find the only things you regret are the things you didn't do. 31. Be happy while you're living, for you're a long time dead. 32. It is not in You, but in...

amazing - no one even mentioned SlipKnot...

hey, Motai? here we go: http://www.imaginarygirlfriends.com/ Cheers! :wink:

Our knowledge is as inadequate as the source whence it is coming from.