Artboard Created with Sketch. Close Search Dialog
! Error Created with Sketch.

The Taylor Series

Math
Summary

The Remainder Term

Summary 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    

Integrating by parts, choosing - (b - t) as the antiderivative of 1, we have


f'(t)dt=- f'(t)(b - t)|ab + 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 rn(x):


f (x)=f (a) + f'(a)(x - a) + ... + f(n-1)(a) + f(n)(t)dt  
 =pn-1(x) + rn(x)  

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],

f(n)(m)f(n)(t)f(n)(M)    

The corresponding integrals must satisfy similar inequalities:

f(n)(m)dtrn(x)≤f(n)(M)dt    

or

f(n)(m)rn(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

rn(x) = (x - a)n    

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,

| f(n)(c)|≤Bn    

for all c in [a, b]. Then we have the bound

| rn(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

| rn(x)|≤| x|n