Don't have a theory to offer, but you can compare to next term and
keep exact result to almost the end.
10^.-/%/"1(1 3,:1 1)&(+/ .*)^:(40+i.2)]1 0x
_22.2352
Devon McCormick wrote:
> Members of the Forum -
>
> If I'm approximating, e.g. the square root of 3, with a matrix method which
> returns an extended precision numerator and denominator, when I work out the
> decimal equivalent of this, at what point do I run out of significant
> digits?
>
> For example,
>
> (1 3,:1 1)&(+/ .*)^:(5+i.5)],.1 0 NB. Successive approximations of %:3
> 76
> 44
>
> 208
> 120
>
> 568
> 328
>
> 1552
> 896
>
> 4240
> 2448
> (%:3)-%/(1 3,:1 1)&(+/ .*)^:10],.x: 1 0 NB. How far off are successive
> approximations?
> _6.6086991e_6
> (%:3)-%/(1 3,:1 1)&(+/ .*)^:20],.x: 1 0
> _1.2607915e_11
> (%:3)-%/(1 3,:1 1)&(+/ .*)^:30],.x: 1 0
> _2.220446e_16
> (%:3)-%/(1 3,:1 1)&(+/ .*)^:40],.x: 1 0 NB. More than 16 digits
> 0
> (1 3,:1 1)&(+/ .*)^:40],.x: 1 0 NB. How precise is this in decimal?
> 144052522725670912
> 83168762773110784
> 2^.(1 3,:1 1)&(+/ .*)^:40],.x: 1 0 NB. Bits/number - relevant?
> 56.999373
> 56.206891
>
>>From the progression of exponents in the cases for 10, 20, and 30
> iterations, I'm guessing the answer should be about 21 significant digits
> but is there a way to better quantify this?
>
> Thanks,
>
> Devon
--
Clifford A. Reiter
Mathematics Department, Lafayette College
Easton, PA 18042 USA, 610-330-5277
http://www.lafayette.edu/~reiterc
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