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JT
Posts:
1,040
Registered:
4/7/12


I do not understand result from wolfram 0.49999999999=((n/2)1)/n n=1 (grease monkey arithmetic?)
Posted:
Jun 3, 2013 5:55 AM


On 6 Maj, 11:50, Dan <dan.ms.ch...@gmail.com> wrote: > On May 6, 2:57 am, JT <jonas.thornv...@gmail.com> wrote: > > > Compare this onehttp://www.wolframalpha.com/input/?i=0.49999999%3D%28%28n%2F2%291%29... > > n = 1.x10^8 > > > and this onehttp://www.wolframalpha.com/input/?i=0.499999999999999999999999999999... > > > n=1? > > > Is there something wrong with wolfram alpha, or can it not handle big > > numb libraries? > > k = (n/2  1) / n => > n k = (n/2  1) => > n( 1/2  k) = 1 => > > n = 1 / (0.5  k) > > k = 0.49 => 0.5  0.49 = 0.01 => n = 1/ 0.01 = 100 > k = 0.4999 => 0.5  0.4999 = 0.0001 => n = 1/ 0.0001 = 10000 > > Both wolfram results are correct .Small variations in a parameter can > produce huge variations in the result, if you're on the other side of > the fraction : > > http://library.thinkquest.org/2647/media/odd1ox.gif > > For the record, I think you meant to input 0.4999 =  (n/2 + 1) / n > This has solution n = 1 as the parameter goes to 0.5 > > 0.4999 = (n/2  1) / n > has an indefinite solution of n going to +/ infinity as the parameter > goes to 0.5
What make you beleive the result could be 1 Dan?



