e_to_the_pi_i sykoze Yet another example of what I have said in the_god_debate_continues: e^(pi*i)+1=0 namely that e (2.7141..., not ecstasy) raised to the power of pi (3.1415...) multiplied by i (square root of -1) plus 1 is equal to zero. Its so amazing - here we have five to the most fundamental quantities to our understanding of nature and science, three of them seemingly unrelated, combining to make a whole, working unity. It blows the mind to think of how this works, but at least it does, and thats enough for me. 010902 ... kx21 e = emotion, i = ignorance and pi = dream? 010902 ... kx21 pi = perfect illusion, instead... 010902 ... kx21 Perfect Illusion = Supreme_enlightenment? 010903 ... kx21 It should be i = illusion, pi= perfect ignorance, and perfect ignorance = Supreme_Enlightenment... 010903 ... TalviFatin Hmm. Lets see. e=ecstacy, i=indecision, pi=blackberry pie. So, thats Ecstacy into the pie, which was made out of indecision because do I really want indigestion? Plus one..who is me...because I would be the only one to eat the pie with ecstacy in it..equals zero. Yep. If I eat that pie, i'm a zero. Loser. Non-hip individual. So this equation is true. 010904 ... blown cherry argh! Why is there never a 'proper' calculator around when I want one? I gotta think about this some more and then I'll come back. Still can't see how e^(i*anything) can work out as a nice finite negative number. 020218 ... unhinged that's kinda what i was thinking...mmmm, my days of calculus have long since been past 020218 ... phil it's really pretty simple to grasp 020219 ... MisterFunkadelic Okay, nobody tell Jay-Z about this one 020219 ... unhinged well i'm not grasping it so could you please explain it to me phil? 020219 ... screwing for virginity this is the most mind-bogeling thing i have ever seen! this is so amazing! and no, im not be sarcastic 021125 ... nonlucid i've always thought that was indescribably amazing - the reason i'd like to be a mathemetician one day but a lot of math is like that - it's so beautiful... though i guess that doesn't help much when you're in grade 8 and don't understand algebra and have a test the next day you haven't studied for but if you're an arrogant kid who's never had to work in the class and likes to ramble about stuff she hardly understands and knows nothing about, occasionally goes on flights of half-understanding but shies away from anything requiring work harder than fun it's cool 040628 ... REAListic optimIST inscrutinizable order and harmony within inscrutinizable chaos. 050821 ... e) ... 050821 ... " an_extraordinary_and_complicated_coincidence" 050821 ... anne-girl i learned the proof of this in cal class the other month woot 050822 ... -1 ... 050822 ... neesh for unhinged, many months late: (or really for a chance to delve through my old maths notes) there's such a thing as a taylor's expansion, that goes thus (i don't know why): f(x)= f(a)+ (x-a)f'(a)+ (((x-a)^2)/2!)f''(a)+ ...+ (((x-a)^n)/n!)f^n(a)+ ... from which the maclaurin series is derived, a taylor's expansion about the point zero (ie, a=0): f(x)= f(a)+ xf'(a)+ ((x^2)/2!)f''(a) etc now if you do maclaurin's expansion for sinx, you get: sinx= sin0 + xcos0 - ((x^2)/2!)sin0 - ((x^3)/3!)cos0 and so on. sin0=0, so we can ignore those terms, and cos0=1, so you get: sinx= x - (x^3)/3! + (x^5)/5! - (x^7)/7! etc similarly, cosx= 1 - (x^2)/2! + (x^4)/4! - (x^6)/6! and next let's put e^ix into the expansion: e^ix = 1 + (ix) + ((ix)^2)/2! + ((ix)^3)/3! + ((ix)^4)/4! + ((ix)^5)/5! + ... = 1 + ix - (x^2)/2! - i(x^3)/3! + (x^4)/4! + i(x^5)/5! + ... = [1 - (x^2)/2! + (x^4)/4! - ...] + i[x - (x^3)/3! + (x^5)/5! - ...] the first bracket is the same as the expansion of cosx and the second is the same as that of sinx, so: e^ix = cosx + isinx finally, let x=pi, and you've got: e^ipi = cospi + isinpi = -1 + 0 = -1 therefore, e^ipi + 1 = 0 phew! (sorry i was too lazy to fully go through every step, and sorry i don't know why taylor's theorem works) 060214 ... anne-girl the taylor series - f(x) = f(a) + (x-a)f'(a) + (x-a)^2f''(a)/2! + (x-a)^3*f'''(a)/3! ... basically, if you substitute a, all the terms cancel except the first, and you get f(a) = f(a) == true take the derivative & substitute a, you get f'(a) = f'(a) == true derivative again, f''(a) = f''(a) * 2 / 2! == true f'''(a) = f'''(a) * 2*3 /3! == true f''''(a) = f''''(a) *4*3*2 / 4! == true etc. nth derivative of f at a: f(n)(a) = f(n)(a) * n!/n!, check. not a good proof, but it's the basic idea 060215 ... ... yeah and if you stick a sausage in a roll you'll get a hot dog. perfect sense. 060215 ... pie eyed math hating e makes me pie eyed too! 060215 ... neesh oh i get it, cheers :) 060215 what's it to you? who go blather from