the " Method of Least Squares" 327 



included between positive and negative infinity, we must have 



4.° 



*/ 



e mx2 d% = ] 



Hence m must be negative; if we call it — ft 2 , it is easy to 

 show that A is equal to — = ; so that finally 



/(**)= -4= «-***, 



V IT 



which is what may be called Gauss's function. 



But to this demonstration, though it leads to an intelligible 

 conclusion, the original objection still applies: the probability 

 that the stone drops on the elementary area dxdy is not, ge- 

 nerally speaking, equal to f{x^)f{y^)dxdy ; so that the equation 

 for determining the form of the function, namely, 



is not legitimately established. 



To illustrate this, let vr{xy)dxdy be the probability that the 

 stone falls on the elementary area in question ; then the con- 

 dition that the probability of a deviation of gjven magnitude 

 is constant will be expressed by 



*a{xy) = ia( </x 2 +y 2 . 0) (A.) 



Moreover, we shall plainly have 



A^)^J^_^m{xy)dy 

 and 



f(f)=Jl^{xy)dx; 



and in order that the demonstration may be valid, we must have 

 or 



/.foe /* + « 



^{xy)dyj _^{xy)dx = <n{xy). . . . (B.) 



If this be true, then, and then only, equation (A.) may be re- 

 placed by 



/(* 2 )/(j/ 2 )=/(0)/(^+/). 



But in order that (B.) may be true, m{xy) must evidently be 

 the product of two factors ; one of them a function of y only, 

 and the other of x 9 and the integral of each factor taken be- 

 tween infinite limits must be equal to unity. Combining this 



