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, pn(x). By the fundamental theorem of
calculus,
f (b) - f (a) =  f'(t)dt |
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Integrating by parts, choosing - (b - t) as the antiderivative of 1, we have
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f'(t)dt
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= |
- f'(t)(b - t)|ab + f(2)(t)(b - t)dt
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= |
f'(a)(b - a) + f(2)(t)(b - t)dt
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Again, integrating by parts yields
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f(2)(t)(b - t)dt
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= |
 - f(2)(t)  + f(3)(t)dt
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= |
f(2)(a) + f(3)(t)dt
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Putting these equations together, we have
| f (b) = f (a) + f'(a)(b - a) + f(2)(a) + f(3)(t)dt |
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Continuing the process, we arrive at
f (b) = f (a) + f'(a)(b - a) + ... + f(n-1)(a) + f(n)(t) dt |
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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 rn(x):
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f (x) |
= |
f (a) + f'(a)(x - a) + ... + f(n-1)(a) +  f(n)(t) dt
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= |
pn-1(x) + rn(x) |
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Therefore, in order to compute how close pn(x) is to f (x), we need to find the
magnitude of the remainder term. Fortunately, there is a simpler way to express
rn(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],
The corresponding integrals must satisfy similar inequalities:
| f(n)(m)dt≤rn(x)≤f(n)(M)dt |
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or
f(n)(m) ≤rn(x)≤f(n)(M) |
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By the intermediate value theorem,
| f(n)(t) |
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takes on all values between its minimum and maximum in the interval [a, x], so there
exists some c in [a, x] such that
rn(x) =  (x - a)n |
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The expression on the right looks very much like the nth 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, Bn, for the absolute value of the nth derivative of f on
the interval [a, x]. That is,
for all c in [a, b]. Then we have the bound
| rn(x)|≤ | x - a|n |
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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 |
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with remainder term bounded by
| | rn(x)|≤| x|n |
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