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} |