(navigation image)
Home American Libraries | Canadian Libraries | Universal Library | Community Texts | Project Gutenberg | Children's Library | Biodiversity Heritage Library | Additional Collections
Search: Advanced Search
Anonymous User (login or join us)
See other formats

Full text of "Scientific Papers - Vi"

One Dimension.
In this case the required information for any finite n is afforded by Bernoulli's theorem. There are n + 1 possible resultants, and if we suppose the component amplitudes, or stretches, to be unity, they proceed by intervals of tu>o from + n to — n, values which are the largest possible. The probabilities of the various resultants are expressed by the corresponding terms in the expansion of (£ •+• £)n. For instance the probabilities of the extreme values ħ n are (1/2)'1. And the probability of a combination of a positive and b negative components is
albl'    ...........................
in which a + b = n, making the resultant a — b. The largest values of (6) occur in the middle of the series, and here a distinction arises according as n is even or odd. In the former alternative there is a unique middle term when a — b =^H; but in the latter a and b cannot be equated, and there are two equal -middle terms corresponding to a = %n + %, b ~ \n — \, and to a = \n — \, b =^n 4- £. The values of the second fraction in (6) are the series of integers in what is known as the " arithmetical triangle." We have now to consider the values of
n I albl
to be found in the neighbourhood of the middle of the series.    If n be even, the v o-l ue of the term counted s onwards from the unique maximum is
( '
If n "be odd, we have to choose between the two middle terms.    Taking for instance, a = \n -I- £, b = ^n — $, the 5th term onwards is
n I
The expressions (8) and (9) are brought into the same form when we replace s by the resultant amplitude a\ When n is even, x = — 2s; when n is odd, x {$ — 2 (s — •£), so that in both cases we have on restoration of the factor (i)'1
The difference is that when n is even, so has the (n +1) values
0,       ħ2,       ħ4,       ħ6, ... ħn; and when n is odd, the (n + 1) values
ħ1,       ħ3,        ħ5, ... ħw.
The expression (10) may be regarded as affording the complete solution of the problem proposed; it expresses the probability of any one of the possible, of which I had already given the solution for the case of n infinite J, the identity depending of course upon the vector character of the components.