THE METHOD OF LEAST SQUARES. 59 



and that the others might, for the sake of simplicity, be dropt 

 out. But whatever his views may have been, his conclusion 

 is unintelligible. 



The demonstration may, however, be amended so as to 

 avoid this difficulty, and we will suppose that the reviewer 

 meant something different from what he has expressed. Let 

 / (x 2 ) dx be the probability of a deviation parallel to the axis 

 of abscissae, of which the magnitude lies between x and x + dx. 

 Then/(?/ 2 ) dy is similarly the probability of a deviation parallel 

 to the axis of ordinates, and lying between y and y 4- dy. Thus 

 the probability that the stone drops on the elementary area 

 dxdy, of which the corner next the origin has for its co-ordi- 

 nates x and ?/, seems to be f(& 2 )f(y 2 ) dxdy\ and as all devia- 

 tions of equal magnitude are equally probable, this probabi- 

 lity must remain unchanged as long as the sum of the squares 

 of x and y remains the same ; so that we have for determining 

 the unknown function the equation 



=/(o) />'+/), . 



of which the solution is 



and as the deviation must of necessity have some magnitude 

 included between positive and negative infinity, we must have 



f 



a 



Hence m must be negative; if we call it - #*, it is easy to 



show that A is equal to p? ; so that finally 



VTT 



; 



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, gene- 



