PROOF OF TAYLOR'S THEOREM. 83 



109. The following proof of Taylor's Theorem deserves 

 notice, as it depends only on the equation which is proved 

 geometrically in Art. 103. Let 



be called F(x], then F' (x) = - l^ <f> n+1 (x}. 



I n 



Now, by Art. 103, F(x}=F (z) + (x - z] F' {z + 6 (x - z)}. 

 Also F(z)=Q, 



and F'z + ex-z = - 6n(z ~ x:)n < 



z _ 

 therefore ' 



_ 

 ^) (x} (z x] <f>' (x} i ^l~ <f>" (x) 



Put Ji for z x, then 



I /! 



[n_ 



110. The result of the preceding Article gives us an 

 expression for the remainder after n+ 1 terms of the expansion 

 of (f> (x + A), differing in form from that we found before. If 

 we assume 6 = 1 r the remainder becomes 



111. In the proofs given of Taylor's Theorem, we have 

 supposed all the functions that occur to be continuous. If 

 the function we wish to expand, or any of its differential 

 coefficients up to the (n + l) th inclusive, be infinite for values 



02 



