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The Taylor Series

The Remainder Term

Problems

Problems

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) dtr 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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