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Full text of "Scientific Papers - Vi"

origin, the area is triangular and increases as o-2; afterwards it becomes hexagonal, and after passing through the form of a regular hexagon, when its area is a maximum, returns backwards through the same phases.
The calculations by the sequence formula present no difficulty of principle.
When n  4, I find
(0r<a),    fa(<r) = <r*IQaa;
(ar< 2a),    fa (a-} = {as - 4 (a- - a}3}/6as;
when 2a < a- < 4a, the above values are repeated symmetrically. In this case there is no discontinuity either in fa, or fa, or fa'. When <r = 2a, that is in the middle of the range,
fa  f 3    fa' = 0-
The calculations might be pursued to higher values of n without much trouble. In all cases there is symmetry with respect to the middle of the range. The functions <j>n are algebraic and rise in degree by a unit at each step. At the beginning of the range 0Jl+1 (a) = (ff/a)nfnl, so that the contact at both ends of the representative curves with the line of abscissae becomes of high order.
Again, since cr must lie somewhere between 0 and na, we must have
(a) da-1 a == 1;
from the above expressions we may test this in the cases of n = 2, 3, 4.
A plot of the curves for these cases is given in Fig. 1. The ordinate represents (j> (<r) and the abscissa represents a itself with a taken as unity, so that the area of each curve is unity.
Fig. 1.
In order to pass from these curves in which a is the sum of the distances from one end to the representative curves for the mean distance, which must lie between 0 and a, we have merely to reduce the scale of the abscissae in the ratio n : 1, and to increase the scale of the ordinates in the same ratio, so that the area is preserved. For instance, when n = 4, the middle ordinate will be increased from f to f.f the line within the square which is drawn perpendicular to the diagonal through the origin, a itself corresponding to the position of the line- as measured along the diagonal *.