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Topic: Is it me or is it Wolfram?
Replies: 16   Last Post: May 13, 2013 4:51 PM

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LudovicoVan

Posts: 3,201
From: London
Registered: 2/8/08
Re: Is it me or is it Wolfram?
Posted: May 10, 2013 2:04 PM
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"JT" <jonas.thornvall@gmail.com> wrote in message
news:82f721d1-c691-47fa-8428-913e49966f62@m7g2000vbf.googlegroups.com...
> http://www.wolframalpha.com/input/?i=0.49999999999999999999999999999999999999999%3D%28n%2F2-1%29%2Fn
>
> n = -1.
> 0.49999999999999999999999999999999999999999 = (n/2-1)/n
>
> http://www.wolframalpha.com/input/?i=%3D%28100000000000000000000000000000000000000000%2F2-1%29%2F100000000000000000000000000000000000000000
>
> 0.49999999999999999999999999999999999999999=(100000000000000000000000000000000000000000/2-1)/
> 100000000000000000000000000000000000000000
>
> I do not understand to, can please someone explain why and how wolfram
> get -1 for the upper calculation, it is obvious using the one below
> what the solution is?
>
> And if there was two solutions should not Wolfram give them both? What
> is going on here, i am total newb to math calculators so tell me what
> is going on?


Take the equation k = (n/2-1)/n, and consider that your k is not fitting
into a float (most probably they are using doubles, i.e. the 64-bit floats,
but I haven't checked), so k is (apparently) rounded to 0.5. Then,
depending on how you transform the equation and the exact step at which you
substitute your value for k, you either get -Infinity or -1 (exercise left
to the reader, or I guess you could just check the step-by-step solution,
but I haven't).

That is how floating point works: you'd rather ideally use
arbitrary-precision rationals, otherwise, as mentioned already in the
thread, increase the precision of your floating point numbers. But I do not
think you can do any of these with Wolfram Alpha.

Julio





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