cosine

This is what I would have thought, but this assumes that a and b are rational, which is not neccesarily the case. They could be irrational.

I got confused with the proof right away when you said [math] x>y [/math] and then you said [math]x-y>0[/math] which implies by [math]x-y+y>0+y[/math] that [math]x>y[/math] which contradicts your first statement.

Mathematically speaking, laws are axioms and postulates, etc. In other words the assumed. Theorems are the results derived from laws. A "Theory" is a term to describe the large body of laws and theorems of a certain system. For example Euclid's Elements is a book containing a large part of Euclidean Geometric Theory. Names of famous theorems can be confusing and may not actually satisfy what is implied by their names. The law of cosines is not really a law, it is a theorem. (Unless for some reason you based your system of geometry by making it an axiom, but that would be lame.) Pythagreas's theorem, while believed to be first proved rigorously by Pythagreas, was known well before his time.

To exemplify my more-that-just-a-plain-ol'-infinite sadness!

Thanks! ... *tear* :-( :-( :-( :-( :-( :-( :-( :-( :-( :-( :-( times infinity plus infinty times A MILLION

Hey, I was just reading through really old posts, and if you look at the 100,000 posts thread, there are promises of an arcade... gahh where is it!?

How come I can't edit post 19? By the way. Follow that step in post 19, the last visible one I mean. It becomes a difference of perfect cubes on the top, aka x+h - x. that gets you pretty much where you need to be to solve the problem.

[math]f(x) = x^{1/3}[/math] [math]f'(x) = \lim_{h\to0}\frac{(x+h)^{1/3} - x^{1/3}}{h}[/math] [math]f'(x) = \lim_{h\to0}\frac{(x+h)^{\frac{1}{3}} - x^{\frac{1}{3}}}{h}\left(\frac{(x+h)^{\frac{2}{3}} + (x+h)^{\frac{1}{3}}(x)^{\frac{1}{3}} + x^{\frac{2}{3}}}{(x+h)^{\frac{2}{3}} + (x+h)^{\frac{1}{3}}(x)^{\frac{1}{3}} + x^{\frac{2}{3}}}\right)[/math] [math]f'(x) = \lim_{h\to0}\frac{x + h - x}{h\left((x+h)^{\frac{2}{3}} + (x+h)^{\frac{1}{3}}(x)^{\frac{1}{3}} + x^{\frac{2}{3}\right)}[/math] [math]f'(x) = \lim_{h\to0}\frac{h}{h\left((x+h)^{\frac{2}{3}} + (x+h)^{\frac{1}{3}}(x)^{\frac{1}{3}} + x^{\frac{2}{3}\right)}[/math] [math]f'(x) = \lim_{h\to0}\frac{1}{\left((x+h)^{\frac{2}{3}} + (x+h)^{\frac{1}{3}}(x)^{\frac{1}{3}} + x^{\frac{2}{3}\right)}[/math] [math]f'(x) = \lim_{h\to0}\frac{1}{\left((x+h)^{\frac{2}{3}} + (x^{2}+hx)^{\frac{1}{3}} + x^{\frac{2}{3}\right)}[/math]

haha yeah... I was first making it and I was like... how to manipulate it to get rid of the actual variables? then you would have whatever number came out, every time, then you would always know the answer.

Thanks!

[math]1234^{1234}=10^{x}[/math] [math]1234(\log(1234))=x[/math] [math]1234(\log(1234)) = 3814.6829070663730305378590465079...[/math] [math]1234^{1234} = (10^{3814})(10^{.6829070663730305378590465079...})[/math] [math]10^{.6829070663730305378590465079...} = 4.8184467781382543701667186035719...[/math] [math]1234^{1234} = 4.8184467781382543701667186035719... * 10^{3814}[/math] Sorry about the large amount of digits, I'm sort of paranoid about inaccurate decimal representations, but I was too tired to figure out a probably obvious exact form. But anyway, hope this helps.

I'm really excited to see that done here though. What can we do to get it implemented?

[math]\frac{karma}{dogma} = \frac{kar(ma)}{dog(ma)} = \frac{kar}{dog}[/math]

How easy would it be to implement something like that around here? Is there any possible way of testing something like that out? Also, as a suggestion to the score system. The "score" could be adjusted by how many posts the person has posted. That way, a bright newcomer could get rating comparable to someone who has been here a while. For example: Newby joins and posts 10 posts, and gets 50 points. So lets say his "karma rating" is 50pts/10psts = 5. Oldie has been around for a while, has posted 10 million posts, and has 50 million points. so 50 million pts/10million psts = 5. So you can see that perhaps these two have comparable user ratings. You may also want to argue though that Oldie should have a bit higher rating for being able to sustain his status, where the next 100 posts of Newby could give him 0 points. I think the "postive feedback" point system is a nice idea to work with.

Yeah thats what I was trying to say. Either I get an unliscenced copy or none at all. Does the latter benefit anyone as opposed to the former?

Hey, I have sort of an ethical problem here... I really want a math software toolkit such as Mathematica, Matlab, or Maple, but they are very expensive (drop at least 100 bucks easy, 130 for Mathematica). I hate to take things without a liscence, I appreciate the work of the creators of the projects and think they should be payed reasonably for their work. But those prices are insane (and yes I know other software reaches into hundreds, even thousands of dollars, and that is insane too). I just can't afford it. I checked what liscences my school has, and they can run 50 copies of Matlab at any given time on one of the computers in the department building. I would have to go through a long process and even after this the situation is extremely inconvienient. Is it ethical to download an unliscenced version of the software? Thanks very much, I look foward to hearing all points of view.

By the way, I feel like you guys may like to know just what it is this integral means. For various distance formulas of the form [math]x^{k} + y^{k} = z^{k}[/math] what is the value of pi? Thus in the unit circle of that k is: [math]y = \left(1 - x^{k}\right)^{\frac{1}{k}}[/math] So the pi is going to be the arclength of the unit circle from -1 to 1. Though since your distance formula changed, the arclength formula changed. [math]\left(dx^{2} + dy^{2}\right)^{\frac{1}{2}}[/math] becomes: [math]\left(dx^{k} + dy^{k}\right)^{\frac{1}{k}}[/math] Which leads to the general integral of this discussion. It should give the corresponding pi for every k.

Cool thanks, those are very interesting values. How did you get at them exactly? What does FT o'C mean? and I can tell you that for k = 2 the answer is pi.

Do you mean when k=0?

Could anyone help me find this integral? I've been going back to it occasionally on and off for a year and a half. I once had access to mathematica, which gave me a strange string for the integral. At the time I couldn't understand it, but I may be able to understand it now. Here is the integral I'm trying to find: [math]\int^{1}_{-1}\left[(-x)^{k}(1-x^{k})^{1-k} + 1\right]^{\frac{1}{k}}dx[/math] If anyone could help I would be very much obliged, it is an interesting integral.