The radius is half a chord of the n-dimensional sphere at each stage. That chord is a vertical to a line segment equal in length to the old radius and is the height of an isosceles triangle lying on its side with that old radius as the lengths of its two equal sides. The base (not the side to which the height goes) of the isosceles triangle is always 1 as being the radius of the n-dimensional sphere.
As such, the cosine of the angle subtended by the half-chord is 1/(2*r) where r is the previous smaller radius. The new radius is just the sine of the same angle and therefore sqrt(1 - 1/(4*r^2)).
Tabulated:
dim r 1/r
n - 1 0.866025404 1.154700538
n - 2 0.816496581 1.224744871
n - 3 0.790569415 1.264911064
n - 4 0.774596669 1.290994449
n - 5 0.763762616 1.309307341
n - 6 0.755928946 1.322875656
n - 7 0.750000000 1.333333333
n - 8 0.745355992 1.341640786
n - 9 0.741619849 1.348399725
n - 10 0.738548946 1.354006401
n - 11 0.735980072 1.358732441
n - 12 0.733799386 1.362770288
n - 13 0.731925055 1.366260102
n - 14 0.730296743 1.369306394
n - 15 0.728868987 1.371988681
n - 16 0.727606875 1.374368542
n - 17 0.726483157 1.376494403
n - 18 0.725476250 1.378404875
n - 19 0.724568837 1.380131119
n - 20 0.723746864 1.381698559
n - 21 0.722998805 1.383128150
n - 22 0.722315119 1.384437310
n - 23 0.721687836 1.385640646
n - 24 0.721110255 1.386750491
n - 25 0.720576692 1.387777333
n - 26 0.720082300 1.388730150
n - 27 0.719622917 1.389616668
n - 28 0.719194952 1.390443574
n - 29 0.718795288 1.391216687
n - 30 0.718421208 1.391941091
n - 31 0.718070331 1.392621248
n - 32 0.717740563 1.393261092
n - 33 0.717430054 1.393864105
n - 34 0.717137166 1.394433378
n - 35 0.716860439 1.394971665
n - 36 0.716598572 1.395481430
n - 37 0.716350399 1.395964881
n - 38 0.716114874 1.396424004
n - 39 0.715891053 1.396860592
n - 40 0.715678085 1.397276262
I assume what's wanted is a closed form function that fits this.
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Posted by Charlie
on 2014-09-15 08:26:49 |
The sequence, if my trig is right
