130 Mr. 0. Heaviside. [June 15, 



when similarly treated, gives 



= e v . (109) 



That is, e v l = 1, which is a case of Taylor's theorem, if we do not go 

 too close to the boundary where the operand begins. That is, 

 regarding the operand as I? 1 (a?), ^ i s turned to 



Application of Generalized Exponential to obtain other Generalized 

 Formula} involving the Logarithm. 



56. Now return to the fundamental exponential formula 



<? = 2 a>'/(r), (110) 



and derive from it some other logarithmic formulae. Differentiate to 

 r, then 



'/'(r). (Ill) 



A second differentiation to r gives 



= _ 6 *(log*) 2 + 2 aff"(r). (112) 



A third differentiation gives 



= e(loga0 8 +2 </""0), (H3) 



and so on. Or, all together, 



2 <f 0) _ 2 x-f(r) 2 <r (r) 



-sitfM- -2~TO- ~2i?rw = ..... (114) 



Now combine them to see if they fit. Thus, we have the elementary 

 formula 



and this, by the use of (114), becomes 



; . 2*(/-/+-<g'+....)(r), (116) 



which, by Taylor's theorem, is the same as 



as required. 



57. Again, differentiate (111) to x. We obtain 



= 



= -2 of/'W + e^-' + S ^(r+l)/'(r+l), (118) 



by using (111) again, and (110). So 



