sparknotes
The Taylor Series
The Remainder Term
We now use integration by parts to determine just how good of an approximation is given by the Taylor polynomial of degree n , p n(x) . By the fundamental theorem of calculus,
f (b) - f (a) = 
f'(t)dt
|
Integrating by parts, choosing - (b - t) as the antiderivative of 1 , we have
|
f'(t)dt
|
= | - f'(t)(b - t)|a
b +
f
(2)(t)(b - t)dt
|
|
| = |
f'(a)(b - a) +
f
(2)(t)(b - t)dt
|
Again, integrating by parts yields
|
f
(2)(t)(b - t)dt
|
= |
 - f
(2)(t)
 +
f
(3)(t)
dt
|
|
| = |
f
(2)(a) +
f
(3)(t)
dt
|
Putting these equations together, we have
|
f (b) = f (a) + f'(a)(b - a) + f
(2)(a) +
f
(3)(t)
dt
|
Continuing the process, we arrive at
f (b) = f (a) + f'(a)(b - a) + ... + f
(n-1)(a) +
f
(n)(t)
dt
|
Substituting x for b , we have an expression for f (x) , called Taylor's formula at x = a , involving the familiar Taylor polynomial of degree n - 1 for f and an integral called the remainder term and denoted by r n(x) :
| f (x) | = |
f (a) + f'(a)(x - a) + ... + f
(n-1)(a) + 
f
(n)(t)
dt
|
|
| = | p n-1(x) + r n(x) |
Therefore, in order to compute how close p n(x) is to f (x) , we need to find the magnitude of the remainder term. Fortunately, there is a simpler way to express r n(x) .
Letting x be fixed for a moment, choose numbers m and M in the interval [a, x] so that f (m) is the minimum value of f on the interval and f (M) the maximum value. Then for any t in [a, x] ,
|
f
(n)(m)≤f
(n)(t)≤f
(n)(M)
|
The corresponding integrals must satisfy similar inequalities:
|
f
(n)(m)
dt≤r
n(x)≤f
(n)(M)
dt
|
or
f
(n)(m) ≤r
n(x)≤f
(n)(M)
|
By the intermediate value theorem,
|
f
(n)(t)
|
takes on all values between its minimum and maximum in the interval [a, x] , so there exists some c in [a, x] such that
r
n(x) =  (x - a)n
|
The expression on the right looks very much like the n th term of the Taylor polynomial--the only difference is that the derivative is evaluated at some number c in the interval [a, x] rather than at a .
Now that the remainder term is in a more manageable form, we can try to bound it. Suppose we have a bound, B n , for the absolute value of the n th derivative of f on the interval [a, x] . That is,
| | f (n)(c)|≤B n |
for all c in [a, b] . Then we have the bound
| r
n(x)|≤ | x - a|n
|
To conclude, we restate the results of this section in the case a = 0 . The Taylor formula in this case reads
f (x) = f (0) + f'(0)x + ... + f
(n-1)(0) + 
f
(n)(t)
dt
|
with remainder term bounded by
|
| r
n(x)|≤| x|n
|
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