tag:blogger.com,1999:blog-3722233.post4426115031519844843..comments2020-11-24T09:46:09.481-06:00Comments on Computational Complexity: e to the pi vs pi to the eLance Fortnowhttp://www.blogger.com/profile/06752030912874378610noreply@blogger.comBlogger11125tag:blogger.com,1999:blog-3722233.post-8763038525815865212013-09-25T14:37:51.148-05:002013-09-25T14:37:51.148-05:00e^x>x+1 for all x<>0 . Choose x to be (pi...e^x>x+1 for all x<>0 . Choose x to be (pi/e)-1. The rest is left to the reader as an exercise ;-)Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-3722233.post-21835956363788714732013-05-05T09:04:53.308-05:002013-05-05T09:04:53.308-05:00Here's how I approached it - is there a functi...Here's how I approached it - is there a function that has both outputs? Start by taking logs of each. Then factor out what is common to both and what one is left with is the logarithm of something divided by that something - that is, (log e)/e and (log Pi)/Pi. This suggests the function (log x)/x. The further analysis and graph of this function shows that e raised to the Pi is THE maximum and every other output (Pi raised to the e being just one) is less than that. Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-3722233.post-63603313690502198042010-02-22T07:31:21.236-06:002010-02-22T07:31:21.236-06:00Is there an element of "elegance" regard...Is there an element of "elegance" regarding the use of log(x)/x instead of going brute force with x^e/e^x ?<br />I know that log(x) - 1 = 0 is easier to solve than x^e*e^x-e*x^(e-1)*e^x = 0. But it's not that much easier. <br />(OK, the second derivative is a real drag.)<br /><br />RickAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-3722233.post-17094967136089110762010-02-17T11:36:44.322-06:002010-02-17T11:36:44.322-06:00To the last Anon- good point, not I doubt I knew t...To the last Anon- good point, not I doubt I knew then, or even now, a <br />``good reason'' why<br />pi\ne e. Also, I should have<br />clarified the original post by saying that my thoughts BACK THEN were NOT so clearly thought out.<br />I am putting my muddled thoughts from back then into clearer language then I could ever have used then.GASARCHhttps://www.blogger.com/profile/06134382469361359081noreply@blogger.comtag:blogger.com,1999:blog-3722233.post-64839859039808151932010-02-17T08:43:51.053-06:002010-02-17T08:43:51.053-06:00"I realized that even if I found out the answ..."I realized that even if I found out the answer it would not tell me a reason for the answer"<br /><br />I am curious whether at that age you knew any "reason" behind the fact that $\pi \neq e".Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-3722233.post-68368529106402584892010-02-17T03:10:13.744-06:002010-02-17T03:10:13.744-06:00You can just look at the partial derivatives of f(...You can just look at the partial derivatives of f(x,y) = x^y. You can use these to see that f grows faster by changing y when x and y are both at least e. <br /><br />More formally, f_x = x^y ln x and f_y = y x^(y-1). If a>e, then ln a < a/e, so f_x(a,e) < f_y(e,a). So by integrating over a on both sides from e to pi, you get f(e,pi)>f(pi,e). A similar calculation should give you something like Boris's claim.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-3722233.post-27145190079648380372010-02-16T20:14:48.569-06:002010-02-16T20:14:48.569-06:00If you're going to prove it, you may as well g...If you're going to prove it, you may as well give a complete proof...<br /><br />Raising each side to 1/(\pi e), we see that the question is equivalent to determining which of \pi^{1/\pi} or e^{1/e} is larger. Taking the derivative of the function x^{1/x} shows that this function is maximized at e. <br /><br />Looking at this now, I'm pleasantly amazed to recall that I solved this in high school. (Given the trick the proof is easy, but finding the trick is not immediate.)Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-3722233.post-52330303696054593812010-02-16T14:09:39.142-06:002010-02-16T14:09:39.142-06:00Rutherford B. Hayes, the man who in the far future...Rutherford B. Hayes, the man who in the far future will be thought to have invented the modem.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-3722233.post-19264047359070702142010-02-16T12:20:13.825-06:002010-02-16T12:20:13.825-06:00@Anonymous: Sorry, I meant "closer to e"...@Anonymous: Sorry, I meant "closer to e" in a vague sense. I meant to put it in quotes, but forgot. As Gasarch mentions, the actual relevant quantity is log(x)/x. This is unimodal with a maximum at e. Its value at 2 and 4 are equal, so in my vague sense, 3.9 is closer to e than 2 is.<br /><br />I guess that's what I get for making imprecise claims.Unknownhttps://www.blogger.com/profile/06229416831400925212noreply@blogger.comtag:blogger.com,1999:blog-3722233.post-10865760471118346932010-02-16T12:10:38.029-06:002010-02-16T12:10:38.029-06:00Boris, are your sure this is correct? Take a=2 and...Boris, are your sure this is correct? Take a=2 and b=3.9Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-3722233.post-16817623211402510132010-02-16T10:40:28.769-06:002010-02-16T10:40:28.769-06:00You may enjoy the following related facts:
Of the...You may enjoy the following related facts:<br /><br />Of the two expressions a^b and b^a, in general, the larger one is the one whose base is closer to e. For example, with integers, this is almost always the smaller one, except for 3^2 is bigger than 2^3. The "proof" is the one you mentioned. (Note: 1 is infinitely far from e for these purposes.)<br /><br />e^π is known to be transcendental, but π^e is not.Unknownhttps://www.blogger.com/profile/06229416831400925212noreply@blogger.com