Panini*Hobbit@mp3 Posted April 22, 2012 Posted April 22, 2012 Love the first pic. I would definitely go in there after that drawing Quote
speedfact Posted April 23, 2012 Posted April 23, 2012 So what's the answer to the second pic? 42 1 Quote
GI-JOE Posted April 23, 2012 Posted April 23, 2012 42 OMFG did I just see you reference something from popular culture!? The end is nigh. Quote
rolf Posted April 23, 2012 Posted April 23, 2012 The second pic doesn't has an answer. Actually, it are only three equations The first equation is something weird, because you cant integrate e^(x^2) and i thought e^(-x^2) neither. (similar to int(-1 to 1)(sqrt(1-x^2) dx), but that answer is easier if you draw it ) The second looks like a fourier transformation. See here what you can do with that. The third is the abc formula, given ax^2 + bx + c=0, what the solutions are for x to make this function valid. Note that if b^2-4ac < 0, then there is no solution. I only had that face for the first equation tho, and I still don't get it 1 Quote
speedfact Posted April 23, 2012 Posted April 23, 2012 OMFG did I just see you reference something from popular culture!? The end is nigh. yes i read that book.....thanks jerk 1 Quote
Latino555 Posted April 23, 2012 Posted April 23, 2012 (edited) The second pic doesn't has an answer. Actually, it are only three equations The first equation is something weird, because you cant integrate e^(x^2) and i thought e^(-x^2) neither. (similar to int(-1 to 1)(sqrt(1-x^2) dx), but that answer is easier if you draw it ) The second looks like a fourier transformation. See here what you can do with that. The third is the abc formula, given ax^2 + bx + c=0, what the solutions are for x to make this function valid. Note that if b^2-4ac < 0, then there is no solution. I only had that face for the first equation tho, and I still don't get it I knew you would reply! I have a feeling that you are a highschool trig or algebra teacher Edited April 23, 2012 by Latino555 Quote
rolf Posted April 23, 2012 Posted April 23, 2012 I knew you would reply! I have a feeling that you are a highschool trig or algebra teacher LOL. But I'm sorry, I'm just a student, currently for over 6.5 years already and I need almost another year to finish my education, which should originally only take 5 years. My grades at highschool were so disappointing they wouldn't even hire me I guess. On the opposite, my education took a bit longer due to a weird detour (and will take at least 6.5 years, so the overshoot isn't that much), my current education is ranked 34th worldwide in its field, which means not a free graduation . My main interest is in Algorithms, although I appreciate Formal Logic and Visualization as well every once in a while. My addiction to weird nerd jokes (mainly in maths and computer science) and wikipedia gives me the other information. For the third formula, I know how to derive it (which, to be honest, is way easier than using the formula). The second I learned/memorized during one year that I spend on Electrical Engineering (5 years ago). The first is trivial to see what they mean, since its answer is absolutely unrelated and that if you'd try to derive it yourself you'd crash and burn. So, in that case, don't derive it I know that by taylor expansion you can rewrite the formula, and by --- I guess --- some series by Leibniz or Euler you'd be able to rewrite it to sqrt(pi), but how exactly I do not know. It's slightly different than the other formula I gave, since that one you'd have to draw and immediately see the answer, but it's similar in the sense that you'd crash and burn if you'd try it a different way. Seriously, read wikipedia _a_ _lot_ and you'd understand it easier, or recognize it easier. Math isn't hard, it's just logic Quote
Gubbi Posted April 23, 2012 Posted April 23, 2012 Did not know this topic was going to end up in a nerd fest but oh well. You cannot derive e^x^2 because it is no bound function and the area below the function is evolving to infinity. You do not have this problem for e^-x^2 because it is basically a Gaussian function which has, of course, a finite area below the function. However you can't integrate it just like this, but there is a little trick: Int_{-inf,inf} e^-x^2 dx = W Multiply by another integral like: Int_{-inf,inf} Int_{-inf,inf} e^-x^2 e^-y^2 dx dy Since e^-y^2 is independent of dx, you can move it out of the 2nd integral: Int_{-inf,inf} e^-y^2 Int_{-inf,inf} e^-x^2 dx dy W Int_{-inf,inf} e^-y^2 dy = W^2 Int_{-inf,inf} Int_{-inf,inf} e^(-x^2-y^2) dx dy = W^2 Change to polar coordinates: Int_{0,inf}Int_{0,2pi} e^-r^2 r dtheta dr (r^2 = x^2 + y^2) Solve theta integral: 2pi Int_{0,inf} e^-r^2 r dr Now you can subsitute: 2pi (-e^-r^2)/2 and insert boundaries: 0 - 2pi (-1)/2 = pi Since W^2 = pi -> W = sqrt{pi) which is what we were looking for. No need for any approximations. --> This was my exam question, so I remembered it hehe However the purpose of the first equation is to make it easy to integrate Gaussian functions because you simply substitute your function to this form, know the solution and substitute back. Edit: This is a mess to read, if you like, I can write it down for you and scan it lol but the trick is only to get polar coordinates to get the r into it 4 Quote
Kat0n Posted April 23, 2012 Posted April 23, 2012 Did not know this topic was going to end up in a nerd fest but oh well. You cannot derive e^x^2 because it is no bound function and the area below the function is evolving to infinity. You do not have this problem for e^-x^2 because it is basically a Gaussian function which has, of course, a finite area below the function. However you can't integrate it just like this, but there is a little trick: Int_{-inf,inf} e^-x^2 dx = W Multiply by another integral like: Int_{-inf,inf} Int_{-inf,inf} e^-x^2 e^-y^2 dx dy Since e^-y^2 is independent of dx, you can move it out of the 2nd integral: Int_{-inf,inf} e^-y^2 Int_{-inf,inf} e^-x^2 dx dy W Int_{-inf,inf} e^-y^2 dy = W^2 Int_{-inf,inf} Int_{-inf,inf} e^(-x^2-y^2) dx dy = W^2 Change to polar coordinates: Int_{0,inf}Int_{0,2pi} e^-r^2 r dtheta dr (r^2 = x^2 + y^2) Solve theta integral: 2pi Int_{0,inf} e^-r^2 r dr Now you can subsitute: 2pi (-e^-r^2)/2 and insert boundaries: 0 - 2pi (-1)/2 = pi Since W^2 = pi -> W = sqrt{pi) which is what we were looking for. No need for any approximations. --> This was my exam question, so I remembered it hehe However the purpose of the first equation is to make it easy to integrate Gaussian functions because you simply substitute your function to this form, know the solution and substitute back. Edit: This is a mess to read, if you like, I can write it down for you and scan it lol but the trick is only to get polar coordinates to get the r into it This topic finally ends on a nerd answer Quote
rolf Posted April 24, 2012 Posted April 24, 2012 Nice answer Gubbi to those math nerds: If a=b, then a^2-ab=0. Hence, you divide by 0, so WTF Quote
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.