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JT
Posts:
1,306
Registered:
4/7/12


Re: I do not understand result from wolfram
Posted:
May 7, 2013 12:02 AM


On 6 Maj, 19:46, JT <jonas.thornv...@gmail.com> wrote: > On 6 Maj, 19:33, JT <jonas.thornv...@gmail.com> wrote: > > > > > > > > > > > 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 > > > I am not sure i understand how 1 could be correct, if n is the number > > of sides of a polygon (n/21)/n*360 is inner angle of a uniform > > polygon in degrees, and (n/21)/n is in revolutions. > > The series will of course converge into a straight line at 0.5 > > revolutions when reaching infinity as you noted, but for any natural n > > the series should just come closer to 0.5 . > > > 0.49=(100/21)/100 > > 0.499=(1000/21)/1000 > > 0.4999=(10000/21)/10000 > > 0.49999=(100000/21)/100000 > > 0.499999=(1000000/21)/1000000 > > 0.4999999=(10000000/21)/10000000 > > > Are you saying there is a max number number of sides of a polygon? > > Or are you saying that (n/21)360 is not correct formula for sum of > angles of any uniform polygon? > I feel a bit lost here.
Isn't (n/21)the sum of angles in turns of a polygon of n sides/ vertices?


Date

Subject

Author

5/7/13


JT

5/7/13


JT


