JT
Posts:
943
Registered:
4/7/12
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Re: Is it me or is it Wolfram?
Posted:
May 10, 2013 3:24 PM
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On 10 Maj, 20:04, "Julio Di Egidio" <ju...@diegidio.name> wrote:
> "JT" <jonas.thornv...@gmail.com> wrote in message
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> news:82f721d1-c691-47fa-8428-913e49966f62@m7g2000vbf.googlegroups.com...
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> >http://www.wolframalpha.com/input/?i=0.499999999999999999999999999999...
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> > n = -1.
> > 0.49999999999999999999999999999999999999999 = (n/2-1)/n
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> >http://www.wolframalpha.com/input/?i=%3D%2810000000000000000000000000...
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> > 0.49999999999999999999999999999999999999999=(100000000000000000000000000000000000000000/2-1)/
> > 100000000000000000000000000000000000000000
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> > 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?
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> > 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?
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> 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).
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> 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.
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> Julio
No that was not the answer given in any of the primitive math
calculations i did around 97-98, but this is the answer from
mathematica and wolfram.
And just one more thing, for all the prostitutes in mathematics go
fuck yourself.
http://www.youtube.com/watch?v=1SkUxknvRlc
Fuck your kebab
http://www.youtube.com/watch?v=ndN_5IrPOhc
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