tag:blogger.com,1999:blog-2235241775294044812.post4939868040221538431..comments2023-08-07T08:14:47.470-07:00Comments on Inquest of wisdom and compassion: The apology of Burmakin IBurmakinhttp://www.blogger.com/profile/01334148208700519171noreply@blogger.comBlogger6125tag:blogger.com,1999:blog-2235241775294044812.post-78061451410421611832010-09-28T13:29:26.255-07:002010-09-28T13:29:26.255-07:00Asymptoic line of the curve: time(t) on X coordina...Asymptoic line of the curve: time(t) on X coordinate,approximation(d) on Y coordinate<br /> f(x)=1/x satisfies the problem. The curve will never meet the X-axis where y becomes zero. The X-axis is the asymptote of the y=1/x curve at the point x is approaching infinity.Caspernoreply@blogger.comtag:blogger.com,1999:blog-2235241775294044812.post-58026314403594406852008-08-11T05:08:00.000-07:002008-08-11T05:08:00.000-07:00hi,may be we should read this one first?http://en....hi,<BR/><BR/>may be we should read this one first?<BR/><BR/>http://en.wikipedia.org/wiki/Zeno's_paradoxes#Achilles_and_the_tortoiseAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-2235241775294044812.post-26490309392993816712008-08-09T20:10:00.000-07:002008-08-09T20:10:00.000-07:00I think the time taken by the rabbit to reach the ...I think the time taken by the rabbit to reach the tortoise will be halved in each next chase. The problem is because the fraction always has some value. So the value of time taken is never equivalent to zero whereas it is approaching v.vvvv.... close to zero. Only when you get " time taken = zero " in one next chase, the puzzle is solved.<BR/><BR/>Maybe this is a series like this if time taken byAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-2235241775294044812.post-22675752002458988822008-08-04T19:56:00.000-07:002008-08-04T19:56:00.000-07:00Hi Burmakin,Just my two cents.Sujata addressed you...Hi Burmakin,<BR/><BR/>Just my two cents.<BR/><BR/>Sujata addressed you as "grandson" but her husband who is apparently much older than her, addressed you as "my son".If you could review this inconsistency, that will be great.<BR/><BR/>Peace,<BR/>CNTAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-2235241775294044812.post-53154439403149236752008-08-03T03:25:00.000-07:002008-08-03T03:25:00.000-07:00Remind me about the classic Greek story saying " i...Remind me about the classic Greek story saying " if u keep on jumping half the distance of your previous jump , at some point , u can't go on further anymore" :)Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-2235241775294044812.post-38749101325114764072008-07-29T09:32:00.000-07:002008-07-29T09:32:00.000-07:00Interesting 'twist' in calculus understanding!The ...Interesting 'twist' in calculus understanding!<BR/><BR/>The 'problem' seem to stem from relative speed. S started first, then R. But speed of R was said to be 5 times the speed of S. This means that as S slow down, R also slow down. If S is at rest, then R will have to be at rest because 5 times zero is zero.<BR/><BR/>As long as S stop moving, R will always be a fixed distance away. Hmm... Alvin S. C. Leehttps://www.blogger.com/profile/05231016382221154215noreply@blogger.com