TAYLOR’S DEVELOPMENT 
221 
The demonstration depends upon the fact that i2 n (A, #) is A 
times the n th term /„(«, A) of the development of f'(x) about the 
point a + a. In fact let h = a + k. Then by 158 
/'(a + A) =/'(a + « + A) =/'(« + «) + ••• + + a) + ••• 
whose n ih term is 
/ n (a, A) = k -~-f (n) (a+ «). 
n — 1 . 
Let « = ^A, then 
B n (h, a) = A/ n («, A) 
as stated. 
The image A 0 , of D 0 is the half of a square of side A 0 , below the 
diagonal. 
To show that R n converges uniformly to 0 in D 0 we have only 
to show that / n ( a , A)= 0 uniformly in A 0 . (2 
To this end we have from 1) for all t in 51 
/' (a + 0 =/' (a) + tf"{d) + ~ f" O) + • • • (3 
Its adjoint 
<?(<)= l/'O) i +<!/"(“) I + ••• ( 4 
also converges in 51. 
By 161, 4 we can develop 4) about t = a, which gives 
£(«, A)= £<» + A£'(a)+ +••• 
n — 11 
But obviously 6r(«, A) is continuous in A 0 , and evidently all its 
terms are also continuous there. Therefore by 149, 3, 
■ ■ ^ , - 6r (n ~ ]) (a) = 0 uniformly in A 0 . (5 
n — 1! 
But if we show that 
|/< n) (a + a) | <a in ~v(a) (6 
it follows from 5) that 2) is true. Our theorem is then 
established.