paperclip

Comment: Division by zero doesn't necessarily give you an error. If you take lim(x->0)+ [1/x] you'll get infinity, which is an acceptable mathematical answer. lim(x->0)- [1/x] gives you negative infinity. It's not an error, per se, but more an indeterminable state, unless the limit is defined. But, as you said tormod, this doesn't have practical applications. Quick question: If travel at the speed of light necessitates a passage of time somewhere, does travel through a wormhole do likewise? Since light speed is thought to be the "barrier" for all speed possible, then travelling through a wormhole would be instantaneous and circumvent this speed limit. So would the universe compensate by forwarding (or reversing?) time elsewhere? -paperclip

According to the standard model (afaik) light exists as a wave-particle duality. Sometimes it's a wave, sometimes it's a particle. To get light to travel through a vacuum and eliminate aether, the particle theory became widely accepted. Light isn't matter, even as a particle (photon) because it carries energy and momentum, but has no mass. Being able to pass through glass is *sorta* tricky, because not all light passes through glass. The visible spectrum does, yes, but ultraviolet light doesn't. Gamma rays are absorbed by mostly everything that exists. So even though something's transparent, that doesn't mean all light passes through it, just some. Like x-rays passing through flesh but not bone. It's the source of the heat that makes the light. For deflection or reflection, it is easier to think of light as a wave. Any wave passing though a medium under certain conditions will deflect or reflect. I think that about does it. For more information I would suggest http://www.howstuffworks.com and http://www.particleadventure.org (not completely sure if the second link is correct). -paperclip

For angles of a triangle: Using the law of sines (a / sin(A) = b / sin(B) = c / sin©) you can derive something to the tune of: sin(A) = (a sin(B)) / b Using the law of cosines (a^2 = b^2 + c^2 - 2bc cos(A)) you can derive: cos(A) = (-a^2 + b^2 + c^2) / (2bc) And, finally, with a trig identity (tan(A) = sin(A) / cos(A)) you can substitute and get: tan(A) = [ (a sin(B)) / b ] / [ (-a^2 + b^2 + c^2) / (2bc) ] Or you could simply use a calculator. If you're just using the value of the angle, you can always fall back on the unit circle. Seems a much easier way than calculating a bunch of factorials. Kal-El: May I ask where you found these formulas and if there's any proof supplied? They seem intuitively flawed, the cosine one pops out first. If you take x^(2/2!) (or x^[2/(2!)]) you end up with x. 1 - x...if x is greater than one, you get a negative number. Then, if it's meant to be x^[n/(n!)], you can never get past zero again, thus a negative cosine and you might not even be out of the first quadrant. -paperclip