Exercise 5.20

Exercise 20: Formulate and prove an inequality which follows from Taylor’s theorem and which remains valid for vector-valued functions.

Answers

(analambanomenos, with fixes suggested by Dan “kyp44” Whitman
As in Theorem 5.15, let f be a real function on [ a , b ] and n a positive integer such that f ( n − 1 ) is continuous on [ a , b ] and f ( n ) ( t ) exists for t ∈ ( a , b ) . Let α and β be distinct points of [ a , b ] and define

P f , α ( t ) = ∑ k = 0 n − 1 f ( k ) ( α ) k ! ( t − α ) k .

Then by Theorem 5.15 there is a point x between α and β such that

| f ( β ) − P f , α ( β ) | ≤ ( β − α ) n n ! | f ( n ) ( x ) | .

We can extend this result to vector-valued functions just as Theorem 5.19 extended the Mean-Value theorem. Let f be a continuous mapping of [ a , b ] into R k such that f ( n − 1 ) is continuous on [ a , b ] and f ( n ) exists for t ∈ ( a , b ) . Let α and β be distinct points in [ a , b ] and define

P f , α ( t ) = ∑ k = 0 n − 1 f ( k ) ( α ) k ! ( t − α ) k .

Put z = f ( β ) − P f , α ( β ) , and let φ ( t ) = z ⋅ f ( t ) . Then by Theorem 5.15 there is a point x between α and β such that

| φ ( β ) − P φ , α ( β ) | = ( β − α ) n n ! | φ ( n ) ( x ) | = ( β − α ) n n ! | z ⋅ f ( n ) ( x ) | .

We also have

| φ ( β ) − P φ , α ( β ) | = | z ⋅ f ( β ) − z ⋅ P f , α ( β ) | = | z ⋅ z | = | z | 2 .

By the Schwartz inequality, we have

| z | 2 = ( β − α ) n n ! | z ⋅ f ( n ) ( x ) | ≤ ( β − α ) n n ! | z | | f ( n ) ( x ) |

so that

| f ( β ) − P f , α ( β ) | = | z | ≤ ( β − α ) n n ! | f ( n ) ( x ) | .

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2023-08-07 00:00
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