Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

Topic: I do not understand result from wolfram
Replies: 9   Last Post: Jun 3, 2013 5:55 AM

 Messages: [ Previous | Next ]
 JT Posts: 1,448 Registered: 4/7/12
Re: I do not understand result from wolfram
Posted: Jun 2, 2013 11:15 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%29-1%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

Evidently i was wrong found the post.

Date Subject Author
5/26/13 JT
6/2/13 JT
6/3/13 JT
5/26/13 RGVickson@shaw.ca
5/26/13 David Bernier
6/2/13 JT
6/2/13 JT
6/2/13 JT