area=. (1 % 4 * ]) * 3 o. (2 %~ o.1) * 1 - %
Thus, the optimal areas with a perimeter of length 100, for some instances,
must be:
(*:100)&* @: area 2 3 5 100 1000 1000000 1000000000
1250 1443.37567 1538.84177 1591.41853 1591.54812 1591.54943 1591.54931
The semicircle optimal area must be:
(*:100)&* @: area _
0
No, unfortunately J does not interpret the above sentence in that sense.
If I'm not wrong then J is right about this one:
pi(n-1) pi 0
tg ------- tg ----
2 n 2 0
lim -------------- = ---------- = ---- = 0
n -> oo 4 n oo oo
But the accuracy gives us a much earlier decline to zero, because
3 o. 1r2p1*1-% 10x^ 16 +i.6
3.53014e15 1.63318e16 1.63318e16 1.63318e16 1.63318e16 1.63318e16
while 4 n gets bigger and bigger.
I guess your formula should be:
area17=: [: % (4*])*3 o. 1p1 %+:
(*: 100) * area17 2 3 5 7 100 10000 1e12 1e200
1250 1443.38 1538.84 1564.75 1591.42 1591.55 1591.55 1591.55
but
(*: 100) * area17 _
_
--
Met vriendelijke groet,
@@i = Arie Groeneveld
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