TABLE OF CONTENTS The Peano form of the remainder in the Taylor formula says that the remainder is smaller than the last term by an order of magnitude. Sometimes a more precise estimate is needed. It is provided by the next theorem.
If f, f', ,f ( n ) are continuous on an interval I with end-points a, b and f ( n +1) exists on I except possibly a finite subset of I, then the function t 
f ( n +1)( t)( b ? t) n/ n! is integrable on I and
| (15.50) |  |
where
| (15.51) |  |
If one is prepared to use Exercise 15.7.6 then the exceptional set in the above theorem can be countable.
By induction. For n=0 the result is just the Fundamental Theorem. [12] For n=1 we have
Employing integration by parts (15.44) with u= t ? b and v'= f ? ? we have
Turning to the induction hypothesis, by Equation (15.44) with u= f ( n +1) and v( t)=( b ? t) ( n +1)/( n+1)! we have (the existence of the first integral is guaranteed by the induction hypothesis)
This means
Equation (15.50) is often given a different form
These ways of writing the formula have the advantage that they do not suggest that h is positive or that x>
