Or just for fun run Newton's method, here 7 times
(i.&0@:(=/){.{.)(j./0 _2+2^n) ":"0 (-_2 0 1&p.%+:)^:(n+i.2) 1x[n=:7
1.414213562373095048801688724209698078569671875376948073176679737990732478462107038850387534327641
Zsbán Ambrus wrote:
> On Wed, Sep 23, 2009 at 5:43 PM, Nicholas Spies <[email protected]> wrote:
>> I summed the first 100 digits of %:2 and got 442 instead of the
>> expected 475.
>> This lead me to look at
>
> Try something like this,
>
> ]s2 =: <.@:%: 2*10x^200
> 14142135623730950488016887242096980785696718753769480731766797379907324784621070388503875343276415727
> NB. J computes <.@:%: exactly on bigints, which we can verify like this:
> *:s2
> 1999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999997921066900256462438170679779231122858093966573842870223186995499050449327265013245005275895702631184
> *:>:s2
> 2000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000749494025002652535774057227650519015233310324596766369540354975031914284189227322706050964357914329
> NB. thus s2 is indeed the exact floor of (2^99) times square root
> two. (If J didn't have this built in, you could implement it by eg.
> interval halving or even something that converges faster but for this
> application simple interval halving would be fast enough)
> NB. now just sum its decimals:
> +/(100$10)#:s2
> 475
>
> There are other solutions of course.
>
> Ambrus
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>
--
Clifford A. Reiter
Mathematics Department, Lafayette College
Easton, PA 18042 USA, 610-330-5277
http://www.lafayette.edu/~reiterc
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