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You are not logged in. #1 20090903 22:35:51
Very hard question on asymptotics (maybe there is no answer yet)I have the following problem for which I cannot find an answer. I posted this same problem in "physicsforums" and "mathhelpforum" and still nobody was able to give me an anwser. I looked everywhere in the internet (journal, books, etc) and also found nothing. Maybe some of you guys can give me some hints. where C is a contour, f and g are analytic functions defined over C. What is the asymptotic approximation of the closedform solution when t>infinity? In general you cannot use the classical methods from complex analysis like steepest descent or saddle point methods because these methods require for some h analytic. I found that there are solutions for general f when t is multiplying f, i.e., this is the general steepest descent. There are also solutions when Working with f, you'll find out that you can't cheat, e.g., won't help you. It only works for small values of t. So, for general f it seems that there is no general solution. What do you think? For a good reading on asymptotics, you can look at these books from google books http://books.google.es/books?id=_tnwmvHmVwMC&printsec=frontcover#v=onepage&q=&f=false http://books.google.es/books?id=xooq99A9anMC&printsec=frontcover#v=onepage&q=&f=false http://books.google.es/books?id=KQHPHPZs8k4C&printsec=frontcover#v=onepage&q=&f=false #2 20090903 22:40:09
Re: Very hard question on asymptotics (maybe there is no answer yet)I've made a mistake in one part. Where I write "It only works for small values of t" should be "It only works for small values of z" #3 20090904 11:17:04
Re: Very hard question on asymptotics (maybe there is no answer yet)Hi dannyv; Last edited by bobbym (20090904 11:19:01) In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #4 20090904 22:36:05
Re: Very hard question on asymptotics (maybe there is no answer yet)Thanks for the link. First time I hear about Haar's method. #5 20090904 22:37:35
Re: Very hard question on asymptotics (maybe there is no answer yet)Well, I will keep looking for a method, and If I find one I'll post it here. 