Advanced Calculus (I)

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Advanced Calculus (I)

WEN-CHINGLIEN

Department of Mathematics National Cheng Kung University

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7.4 Analytic function

Definition

A real-valued function is said to be (real) analytic on a nonempty, open interval (a,b) if and only if given x0 ∈ (a, b) there is a power series centered at x0that converges to f near x0; i.e., there exist coefficients {ak}k =0 and points c, d ∈ (a, b) such that c < x0<d and

f (x ) =

X

k =0

ak(x − x0)k

for all x ∈ (c, d )

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7.4 Analytic function

Definition

A real-valued function is said to be (real) analytic on a nonempty, open interval (a,b) if and only if given x0 ∈ (a, b) there is a power series centered at x0that converges to f near x0; i.e., there exist coefficients {ak}k =0 and points c, d ∈ (a, b) such that c < x0<d and

f (x ) =

X

k =0

ak(x − x0)k

for all x ∈ (c, d )

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Theorem (Uniqueness)

Let c,d be extended real numbers with c < d , let x0 ∈ (c, d) and suppose that f : (c, d) → R. If f (x ) =

P

k =0

ak(x − x0)k for each x ∈ (c, d ), then f ∈ C(c, d ) and

ak = f(k )(x0)

k ! , k = 0, 1, · · ·

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Theorem (Uniqueness)

Let c,d be extended real numbers with c < d , let x0 ∈ (c, d) and suppose that f : (c, d) → R. If f (x ) =

P

k =0

ak(x − x0)k for each x ∈ (c, d ), then f ∈ C(c, d ) and

ak = f(k )(x0)

k ! , k = 0, 1, · · ·

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Proof:

Clearly, f (x0) = a0. Fix k ∈N. By hypothesis, the radius of convergence R of the power series

P

k =0

ak(x − x0)k is positive and (c, d ) ⊆ (x0− R, x0+R). Hence, by Corollary 7.31, f ∈ C(c, d ) and

f(k )(x ) =

X

n=k

n!

(n − k )!an(x − x0)n−k

for x ∈ (c, d ). Apply, this to x = x0. The terms on the right side of (10) are zero when n > k and k !ak when n = k . Hence, f(k )(x0) = k !ak for each k ∈N. 2

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Proof:

Clearly, f (x0) = a0. Fix k ∈N. By hypothesis,the radius of convergence R of the power series

P

k =0

ak(x − x0)k is positive and (c, d ) ⊆ (x0− R, x0+R). Hence, by Corollary 7.31, f ∈ C(c, d ) and

f(k )(x ) =

X

n=k

n!

(n − k )!an(x − x0)n−k

for x ∈ (c, d ). Apply, this to x = x0. The terms on the right side of (10) are zero when n > k and k !ak when n = k . Hence, f(k )(x0) = k !ak for each k ∈N. 2

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Proof:

Clearly, f (x0) = a0. Fix k ∈N. By hypothesis, the radius of convergence R of the power series

P

k =0

ak(x − x0)k is positive and (c, d ) ⊆ (x0− R, x0+R).Hence, by Corollary 7.31, f ∈ C(c, d ) and

f(k )(x ) =

X

n=k

n!

(n − k )!an(x − x0)n−k

for x ∈ (c, d ). Apply, this to x = x0. The terms on the right side of (10) are zero when n > k and k !ak when n = k . Hence, f(k )(x0) = k !ak for each k ∈N. 2

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Proof:

Clearly, f (x0) = a0. Fix k ∈N. By hypothesis,the radius of convergence R of the power series

P

k =0

ak(x − x0)k is positive and (c, d ) ⊆ (x0− R, x0+R). Hence,by Corollary 7.31, f ∈ C(c, d ) and

f(k )(x ) =

X

n=k

n!

(n − k )!an(x − x0)n−k

for x ∈ (c, d ). Apply, this to x = x0. The terms on the right side of (10) are zero when n > k and k !ak when n = k . Hence, f(k )(x0) = k !ak for each k ∈N. 2

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Proof:

Clearly, f (x0) = a0. Fix k ∈N. By hypothesis, the radius of convergence R of the power series

P

k =0

ak(x − x0)k is positive and (c, d ) ⊆ (x0− R, x0+R).Hence, by Corollary 7.31,f ∈ C(c, d ) and

f(k )(x ) =

X

n=k

n!

(n − k )!an(x − x0)n−k

for x ∈ (c, d ). Apply, this to x = x0. The terms on the right side of (10) are zero when n > k and k !ak when n = k . Hence, f(k )(x0) = k !ak for each k ∈N. 2

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Proof:

Clearly, f (x0) = a0. Fix k ∈N. By hypothesis, the radius of convergence R of the power series

P

k =0

ak(x − x0)k is positive and (c, d ) ⊆ (x0− R, x0+R). Hence,by Corollary 7.31, f ∈ C(c, d ) and

f(k )(x ) =

X

n=k

n!

(n − k )!an(x − x0)n−k

for x ∈ (c, d ). Apply, this to x = x0. The terms on the right side of (10) are zero when n > k and k !ak when n = k . Hence, f(k )(x0) = k !ak for each k ∈N. 2

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Proof:

Clearly, f (x0) = a0. Fix k ∈N. By hypothesis, the radius of convergence R of the power series

P

k =0

ak(x − x0)k is positive and (c, d ) ⊆ (x0− R, x0+R). Hence, by Corollary 7.31,f ∈ C(c, d ) and

f(k )(x ) =

X

n=k

n!

(n − k )!an(x − x0)n−k

for x ∈ (c, d ). Apply, this to x = x0. The terms on the right side of (10) are zero when n > k and k !ak when n = k . Hence, f(k )(x0) = k !ak for each k ∈N. 2

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Proof:

Clearly, f (x0) = a0. Fix k ∈N. By hypothesis, the radius of convergence R of the power series

P

k =0

ak(x − x0)k is positive and (c, d ) ⊆ (x0− R, x0+R). Hence, by Corollary 7.31, f ∈ C(c, d ) and

f(k )(x ) =

X

n=k

n!

(n − k )!an(x − x0)n−k

for x ∈ (c, d ). Apply, this to x = x0. The terms on the right side of (10) are zero when n > k and k !ak when n = k . Hence, f(k )(x0) = k !ak for each k ∈N. 2

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Proof:

Clearly, f (x0) = a0. Fix k ∈N. By hypothesis, the radius of convergence R of the power series

P

k =0

ak(x − x0)k is positive and (c, d ) ⊆ (x0− R, x0+R). Hence, by Corollary 7.31, f ∈ C(c, d ) and

f(k )(x ) =

X

n=k

n!

(n − k )!an(x − x0)n−k

for x ∈ (c, d ). Apply, this to x = x0. The terms on the right side of (10) are zero when n > k and k !ak when n = k . Hence,f(k )(x0) = k !ak for each k ∈N. 2

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Proof:

Clearly, f (x0) = a0. Fix k ∈N. By hypothesis, the radius of convergence R of the power series

P

k =0

ak(x − x0)k is positive and (c, d ) ⊆ (x0− R, x0+R). Hence, by Corollary 7.31, f ∈ C(c, d ) and

f(k )(x ) =

X

n=k

n!

(n − k )!an(x − x0)n−k

for x ∈ (c, d ). Apply, this to x = x0. The terms on the right side of (10) are zero when n > k and k !ak when n = k . Hence, f(k )(x0) = k !ak for each k ∈N. 2

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Proof:

Clearly, f (x0) = a0. Fix k ∈N. By hypothesis, the radius of convergence R of the power series

P

k =0

ak(x − x0)k is positive and (c, d ) ⊆ (x0− R, x0+R). Hence, by Corollary 7.31, f ∈ C(c, d ) and

f(k )(x ) =

X

n=k

n!

(n − k )!an(x − x0)n−k

for x ∈ (c, d ). Apply, this to x = x0. The terms on the right side of (10) are zero when n > k and k !ak when n = k . Hence,f(k )(x0) = k !ak for each k ∈N. 2

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Proof:

Clearly, f (x0) = a0. Fix k ∈N. By hypothesis, the radius of convergence R of the power series

P

k =0

ak(x − x0)k is positive and (c, d ) ⊆ (x0− R, x0+R). Hence, by Corollary 7.31, f ∈ C(c, d ) and

f(k )(x ) =

X

n=k

n!

(n − k )!an(x − x0)n−k

for x ∈ (c, d ). Apply, this to x = x0. The terms on the right side of (10) are zero when n > k and k !ak when n = k . Hence, f(k )(x0) = k !ak for each k ∈N. 2

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Definition

Let f ∈ C(a, b) and let x0∈ (a, b). The Taylor expansion (or Taylor series) of f centered at x0 is the series

X

k =0

f(k )(x0)

k ! (x − x0)k.

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Definition

Let f ∈ C(a, b) and let x0∈ (a, b). The Taylor expansion (or Taylor series) of f centered at x0 is the series

X

k =0

f(k )(x0)

k ! (x − x0)k.

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Remark:

[Cauchy]

The function

f (x ) =  e−1/x2 x 6= 0

0 x = 0

belongs to C(−∞, ∞) but is not analytic on any interval that contains x = 0.

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Remark:

[Cauchy]

The function

f (x ) =  e−1/x2 x 6= 0

0 x = 0

belongs to C(−∞, ∞) but is not analytic on any interval that contains x = 0.

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Proof:

It is easy to see (Exercise 3, p.101) that f ∈ C(−∞, ∞) and f(k )(0) = 0 for all k ∈N. Thus the Taylor expansion of f about the point x0=0 is identically zero, but f (x ) = 0 only when x = 0. 2

(23)

Proof:

It is easy to see (Exercise 3, p.101) that f ∈ C(−∞, ∞) and f(k )(0) = 0 for all k ∈N. Thus the Taylor expansion of f about the point x0=0 is identically zero, but f (x ) = 0 only when x = 0. 2

(24)

Proof:

It is easy to see (Exercise 3, p.101) that f ∈ C(−∞, ∞) and f(k )(0) = 0 for all k ∈N. Thus the Taylor expansion of f about the point x0=0 is identically zero, but f (x ) = 0 only when x = 0. 2

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Proof:

It is easy to see (Exercise 3, p.101) that f ∈ C(−∞, ∞) and f(k )(0) = 0 for all k ∈N. Thus the Taylor expansion of f about the point x0=0 is identically zero, but f (x ) = 0 only when x = 0. 2

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Definition

Let f ∈ C(a, b) and x0∈ (a, b). The remainder term of order n of the Taylor expansion of f centered at x0is the function

Rn(x ) = Rnf ,x0(x ) := f (x ) −

n−1

X

k =0

f(k )(x0)

k ! (x − x0)k.

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Definition

Let f ∈ C(a, b) and x0∈ (a, b). The remainder term of order n of the Taylor expansion of f centered at x0is the function

Rn(x ) = Rnf ,x0(x ) := f (x ) −

n−1

X

k =0

f(k )(x0)

k ! (x − x0)k.

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Theorem

A function f ∈ C(a, b) is analytic on (a,b) if and only if given x0∈ (a, b) there is an interval (c,d) containing x0 such that the remainder term Rnf ,x0(x ) converges to zero for all x ∈ (c, d ).

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Theorem

A function f ∈ C(a, b) is analytic on (a,b) if and only if given x0∈ (a, b) there is an interval (c,d) containing x0 such that the remainder term Rnf ,x0(x ) converges to zero for all x ∈ (c, d ).

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Theorem (Taylor’s Formula)

Let n ∈N, let a,b be distinct extended real numbers, let f : (a, b) →R, and suppose that f(n)exists on (a,b). Then for each pair of points x , x0∈ (a, b) there is a number c between x and x0such that

Rnf ,x0

(x ) = f(n)(c)

n! (x − x0)n. In particular,

f (x ) =

n−1

X

k =0

f(k )(x0)

k ! (x − x0)k +f(n)(c)

n! (x − x0)n for some number c between x and x0.

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Theorem (Taylor’s Formula)

Let n ∈N, let a,b be distinct extended real numbers, let f : (a, b) →R, and suppose that f(n)exists on (a,b). Then for each pair of points x , x0∈ (a, b) there is a number c between x and x0such that

Rnf ,x0

(x ) = f(n)(c)

n! (x − x0)n. In particular,

f (x ) =

n−1

X

k =0

f(k )(x0)

k ! (x − x0)k +f(n)(c)

n! (x − x0)n for some number c between x and x0.

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Theorem (Taylor’s Formula)

Let n ∈N, let a,b be distinct extended real numbers, let f : (a, b) →R, and suppose that f(n)exists on (a,b). Then for each pair of points x , x0∈ (a, b) there is a number c between x and x0such that

Rnf ,x0

(x ) = f(n)(c)

n! (x − x0)n. In particular,

f (x ) =

n−1

X

k =0

f(k )(x0)

k ! (x − x0)k +f(n)(c)

n! (x − x0)n for some number c between x and x0.

(33)

Proof:

Without loss of generality, suppose that x0 <x . Define

F (t) := (x − t)n

n! and G(t) := Rnf ,t

(x ) = f (x )−

n−1

X

k =0

f(k )(t)

k ! (x −t)k for each t ∈ (a, b). In order to apply the Generalized

Mean Value Theorem to F and G, we need to be sure the hypotheses of that result hold.

(34)

Proof:

Without loss of generality,suppose that x0 <x . Define

F (t) := (x − t)n

n! and G(t) := Rnf ,t

(x ) = f (x )−

n−1

X

k =0

f(k )(t)

k ! (x −t)k for each t ∈ (a, b).In order to apply the Generalized

Mean Value Theorem to F and G, we need to be sure the hypotheses of that result hold.

(35)

Proof:

Without loss of generality, suppose that x0 <x . Define

F (t) := (x − t)n

n! and G(t) := Rnf ,t

(x ) = f (x )−

n−1

X

k =0

f(k )(t)

k ! (x −t)k for each t ∈ (a, b). In order to apply the Generalized

Mean Value Theorem to F and G,we need to be sure the hypotheses of that result hold.

(36)

Proof:

Without loss of generality, suppose that x0 <x . Define

F (t) := (x − t)n

n! and G(t) := Rnf ,t

(x ) = f (x )−

n−1

X

k =0

f(k )(t)

k ! (x −t)k for each t ∈ (a, b).In order to apply the Generalized

Mean Value Theorem to F and G, we need to be sure the hypotheses of that result hold.

(37)

Proof:

Without loss of generality, suppose that x0 <x . Define

F (t) := (x − t)n

n! and G(t) := Rnf ,t

(x ) = f (x )−

n−1

X

k =0

f(k )(t)

k ! (x −t)k for each t ∈ (a, b). In order to apply the Generalized

Mean Value Theorem to F and G,we need to be sure the hypotheses of that result hold.

(38)

Proof:

Without loss of generality, suppose that x0 <x . Define

F (t) := (x − t)n

n! and G(t) := Rnf ,t

(x ) = f (x )−

n−1

X

k =0

f(k )(t)

k ! (x −t)k for each t ∈ (a, b). In order to apply the Generalized

Mean Value Theorem to F and G, we need to be sure the hypotheses of that result hold.

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Notice by the Chain rule that

(11) F0(t) = −(x − t)n−1 (n − 1)!

for t ∈R. On the other hand,since d

dt

 f(k )(t)

k ! (x − t)k



= f(k +1)(t)

k ! (x − t)k− f(k )(t)

(k − 1)!(x − t)k −1 for t ∈ (a, b). and k ∈N, we can telescope to obtain

(12) G0(t) = − f(n)(t)

(n − 1)!(x − t)n−1 for t ∈ (a, b).

(40)

Notice by the Chain rule that

(11) F0(t) = −(x − t)n−1 (n − 1)!

for t ∈R. On the other hand, since d

dt

 f(k )(t)

k ! (x − t)k



= f(k +1)(t)

k ! (x − t)k− f(k )(t)

(k − 1)!(x − t)k −1 for t ∈ (a, b). and k ∈N,we can telescope to obtain

(12) G0(t) = − f(n)(t)

(n − 1)!(x − t)n−1 for t ∈ (a, b).

(41)

Notice by the Chain rule that

(11) F0(t) = −(x − t)n−1 (n − 1)!

for t ∈R. On the other hand,since d

dt

 f(k )(t)

k ! (x − t)k



= f(k +1)(t)

k ! (x − t)k− f(k )(t)

(k − 1)!(x − t)k −1 for t ∈ (a, b). and k ∈N, we can telescope to obtain

(12) G0(t) = − f(n)(t)

(n − 1)!(x − t)n−1 for t ∈ (a, b).

(42)

Notice by the Chain rule that

(11) F0(t) = −(x − t)n−1 (n − 1)!

for t ∈R. On the other hand, since d

dt

 f(k )(t)

k ! (x − t)k



= f(k +1)(t)

k ! (x − t)k− f(k )(t)

(k − 1)!(x − t)k −1 for t ∈ (a, b). and k ∈N,we can telescope to obtain

(12) G0(t) = − f(n)(t)

(n − 1)!(x − t)n−1 for t ∈ (a, b).

(43)

Notice by the Chain rule that

(11) F0(t) = −(x − t)n−1 (n − 1)!

for t ∈R. On the other hand, since d

dt

 f(k )(t)

k ! (x − t)k



= f(k +1)(t)

k ! (x − t)k− f(k )(t)

(k − 1)!(x − t)k −1 for t ∈ (a, b). and k ∈N, we can telescope to obtain

(12) G0(t) = − f(n)(t)

(n − 1)!(x − t)n−1 for t ∈ (a, b).

(44)

Thus, F and G are differentiable on (x0,x ) and continuous on [x0,x ]. By the Generalized Mean Value Theorem and the fact that F (x ) = G(x ) = 0, there is a number

c ∈ (x0,x ) such that

−F (x0)G0(c) = (F (x ) − F (x0))G0(c)

= (G(x ) − G(x0))F0(c)

= −G(x0)F0(c)

= −G(x0)F0(c).

(45)

Thus,F and G are differentiable on (x0,x ) and continuous on [x0,x ]. By the Generalized Mean Value Theorem and the fact that F (x ) = G(x ) = 0,there is a number

c ∈ (x0,x ) such that

−F (x0)G0(c) = (F (x ) − F (x0))G0(c)

= (G(x ) − G(x0))F0(c)

= −G(x0)F0(c)

= −G(x0)F0(c).

(46)

Thus, F and G are differentiable on (x0,x ) and continuous on [x0,x ]. By the Generalized Mean Value Theorem and the fact that F (x ) = G(x ) = 0, there is a number

c ∈ (x0,x ) such that

−F (x0)G0(c) = (F (x ) − F (x0))G0(c)

= (G(x ) − G(x0))F0(c)

= −G(x0)F0(c)

= −G(x0)F0(c).

(47)

Thus, F and G are differentiable on (x0,x ) and continuous on [x0,x ]. By the Generalized Mean Value Theorem and the fact that F (x ) = G(x ) = 0,there is a number

c ∈ (x0,x ) such that

−F (x0)G0(c) = (F (x ) − F (x0))G0(c)

= (G(x ) − G(x0))F0(c)

= −G(x0)F0(c)

= −G(x0)F0(c).

(48)

Thus, F and G are differentiable on (x0,x ) and continuous on [x0,x ]. By the Generalized Mean Value Theorem and the fact that F (x ) = G(x ) = 0, there is a number

c ∈ (x0,x ) such that

−F (x0)G0(c) = (F (x ) − F (x0))G0(c)

= (G(x ) − G(x0))F0(c)

= −G(x0)F0(c)

= −G(x0)F0(c).

(49)

Thus, F and G are differentiable on (x0,x ) and continuous on [x0,x ]. By the Generalized Mean Value Theorem and the fact that F (x ) = G(x ) = 0, there is a number

c ∈ (x0,x ) such that

−F (x0)G0(c) = (F (x ) − F (x0))G0(c)

= (G(x ) − G(x0))F0(c)

= −G(x0)F0(c)

= −G(x0)F0(c).

(50)

Thus, F and G are differentiable on (x0,x ) and continuous on [x0,x ]. By the Generalized Mean Value Theorem and the fact that F (x ) = G(x ) = 0, there is a number

c ∈ (x0,x ) such that

−F (x0)G0(c) = (F (x ) − F (x0))G0(c)

= (G(x ) − G(x0))F0(c)

= −G(x0)F0(c)

= −G(x0)F0(c).

(51)

Thus, F and G are differentiable on (x0,x ) and continuous on [x0,x ]. By the Generalized Mean Value Theorem and the fact that F (x ) = G(x ) = 0, there is a number

c ∈ (x0,x ) such that

−F (x0)G0(c) = (F (x ) − F (x0))G0(c)

= (G(x ) − G(x0))F0(c)

= −G(x0)F0(c)

= −G(x0)F0(c).

(52)

Hence, it follows from (11) and (12) that (x − x0)n

n!

 f(n)(c)(x − c)n−1 (n − 1)!



=Rnf ,x0

(x ) = (x − c)n−1 (n − 1)! . Solving this equation for Rnf ,x0 completes the proof. 2

(53)

Hence,it follows from (11) and (12) that (x − x0)n

n!

 f(n)(c)(x − c)n−1 (n − 1)!



=Rnf ,x0

(x ) = (x − c)n−1 (n − 1)! . Solving this equation for Rnf ,x0 completes the proof. 2

(54)

Hence, it follows from (11) and (12) that (x − x0)n

n!

 f(n)(c)(x − c)n−1 (n − 1)!



=Rnf ,x0

(x ) = (x − c)n−1 (n − 1)! . Solving this equation for Rnf ,x0 completes the proof. 2

(55)

Hence, it follows from (11) and (12) that (x − x0)n

n!

 f(n)(c)(x − c)n−1 (n − 1)!



=Rnf ,x0

(x ) = (x − c)n−1 (n − 1)! . Solving this equation for Rnf ,x0 completes the proof. 2

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Theorem

Let f ∈ C(a, b). If there is an M > 0 such that

|f(n)(x )| ≤ Mn

for all x ∈ (a, b) and n ∈N, then f is analytic on (a,b). In fact, for each x0∈ (a, b),

f (x ) =

X

k =0

f(k )(x0)

k ! (x − x0)k holds for all x ∈ (a, b)

(57)

Theorem

Let f ∈ C(a, b). If there is an M > 0 such that

|f(n)(x )| ≤ Mn

for all x ∈ (a, b) and n ∈N, then f is analytic on (a,b). In fact, for each x0∈ (a, b),

f (x ) =

X

k =0

f(k )(x0)

k ! (x − x0)k holds for all x ∈ (a, b)

(58)

Example:

Prove that sin x and cos x are analytic onR and have Maclaurin expansions

sin x =

X

k =0

(−1)kx2k +1

(2k + 1)! , cos x =

X

k =0

(−1)kx2k (2k )!

(59)

Example:

Prove that sin x and cos x are analytic onR and have Maclaurin expansions

sin x =

X

k =0

(−1)kx2k +1

(2k + 1)! , cos x =

X

k =0

(−1)kx2k (2k )!

(60)

Example:

Prove that ex is analytic onR and has Maclaurin expansion

ex =

X

k =0

xk k !.

(61)

Example:

Prove that ex is analytic onR and has Maclaurin expansion

ex =

X

k =0

xk k !.

(62)

Theorem

Suppose that I is an open interval centered at c and

f (x ) =

X

k =0

ak(x − c)k, x ∈ I.

If x0 ∈ I and r > 0 satisfy (x0− r , x0+r ) ⊆ I, then

f (x ) =

X

k =0

f(k )(x0)

k ! (x − x0)k

for all x ∈ (x0− r , x0+r ). In particular, if f is a C function whose Taylor series expansion converges to f on some open interval J, then f is analytic on J.

(63)

Theorem

Suppose that I is an open interval centered at c and

f (x ) =

X

k =0

ak(x − c)k, x ∈ I.

If x0 ∈ I and r > 0 satisfy (x0− r , x0+r ) ⊆ I, then

f (x ) =

X

k =0

f(k )(x0)

k ! (x − x0)k

for all x ∈ (x0− r , x0+r ). In particular, if f is a C function whose Taylor series expansion converges to f on some open interval J, then f is analytic on J.

(64)

Proof:

It suffices to prove the first statement. By making the change of variables w = x − c,we may suppose that c = 0 and I = (−R, R) i.e., that f (x ) =

P

k =0

akxk, for all x ∈ (−R, R). Suppose that (x0− r , x0+r ) ⊆ (−R, R) and fix x ∈ (x0− r , x0+r ). By hypothesis and the Binomial Formula,

(13)

f (x ) =

P

k =0

akxk

=

P

k =0

ak((x − x0) +x0)k

=

P

k =0

akPk j=0

 k j



x0k −J(x − x0)j.

(65)

Proof:

It suffices to prove the first statement. By making the change of variables w = x − c, we may suppose that c = 0 and I = (−R, R)i.e., that f (x ) =

P

k =0

akxk, for all x ∈ (−R, R). Suppose that (x0− r , x0+r ) ⊆ (−R, R) and fix x ∈ (x0− r , x0+r ). By hypothesis and the Binomial Formula,

(13)

f (x ) =

P

k =0

akxk

=

P

k =0

ak((x − x0) +x0)k

=

P

k =0

akPk j=0

 k j



x0k −J(x − x0)j.

(66)

Proof:

It suffices to prove the first statement. By making the change of variables w = x − c,we may suppose that c = 0 and I = (−R, R) i.e., that f (x ) =

P

k =0

akxk, for all x ∈ (−R, R).Suppose that (x0− r , x0+r ) ⊆ (−R, R) and fix x ∈ (x0− r , x0+r ). By hypothesis and the Binomial Formula,

(13)

f (x ) =

P

k =0

akxk

=

P

k =0

ak((x − x0) +x0)k

=

P

k =0

akPk j=0

 k j



x0k −J(x − x0)j.

(67)

Proof:

It suffices to prove the first statement. By making the change of variables w = x − c, we may suppose that c = 0 and I = (−R, R)i.e., that f (x ) =

P

k =0

akxk, for all x ∈ (−R, R). Suppose that (x0− r , x0+r ) ⊆ (−R, R) and fix x ∈ (x0− r , x0+r ). By hypothesis and the Binomial Formula,

(13)

f (x ) =

P

k =0

akxk

=

P

k =0

ak((x − x0) +x0)k

=

P

k =0

akPk j=0

 k j



x0k −J(x − x0)j.

(68)

Proof:

It suffices to prove the first statement. By making the change of variables w = x − c, we may suppose that c = 0 and I = (−R, R) i.e., that f (x ) =

P

k =0

akxk, for all x ∈ (−R, R).Suppose that (x0− r , x0+r ) ⊆ (−R, R) and fix x ∈ (x0− r , x0+r ). By hypothesis and the Binomial Formula,

(13)

f (x ) =

P

k =0

akxk

=

P

k =0

ak((x − x0) +x0)k

=

P

k =0

akPk j=0

 k j



x0k −J(x − x0)j.

(69)

Proof:

It suffices to prove the first statement. By making the change of variables w = x − c, we may suppose that c = 0 and I = (−R, R) i.e., that f (x ) =

P

k =0

akxk, for all x ∈ (−R, R). Suppose that (x0− r , x0+r ) ⊆ (−R, R) and fix x ∈ (x0− r , x0+r ). By hypothesis and the Binomial Formula,

(13)

f (x ) =

P

k =0

akxk

=

P

k =0

ak((x − x0) +x0)k

=

P

k =0

akPk j=0

 k j



x0k −J(x − x0)j.

(70)

Proof:

It suffices to prove the first statement. By making the change of variables w = x − c, we may suppose that c = 0 and I = (−R, R) i.e., that f (x ) =

P

k =0

akxk, for all x ∈ (−R, R). Suppose that (x0− r , x0+r ) ⊆ (−R, R) and fix x ∈ (x0− r , x0+r ). By hypothesis and the Binomial Formula,

(13)

f (x ) =

P

k =0

akxk

=

P

k =0

ak((x − x0) +x0)k

=

P

k =0

akPk j=0

 k j



x0k −J(x − x0)j.

(71)

Proof:

It suffices to prove the first statement. By making the change of variables w = x − c, we may suppose that c = 0 and I = (−R, R) i.e., that f (x ) =

P

k =0

akxk, for all x ∈ (−R, R). Suppose that (x0− r , x0+r ) ⊆ (−R, R) and fix x ∈ (x0− r , x0+r ). By hypothesis and the Binomial Formula,

(13)

f (x ) =

P

k =0

akxk

=

P

k =0

ak((x − x0) +x0)k

=

P

k =0

akPk j=0

 k j



x0k −J(x − x0)j.

(72)

Proof:

It suffices to prove the first statement. By making the change of variables w = x − c, we may suppose that c = 0 and I = (−R, R) i.e., that f (x ) =

P

k =0

akxk, for all x ∈ (−R, R). Suppose that (x0− r , x0+r ) ⊆ (−R, R) and fix x ∈ (x0− r , x0+r ). By hypothesis and the Binomial Formula,

(13)

f (x ) =

P

k =0

akxk

=

P

k =0

ak((x − x0) +x0)k

=

P

k =0

akPk j=0

 k j



x0k −J(x − x0)j.

(73)

Since

P

k =0

akyk converges absolutely at y := |x − x0| + |x0| < R, we have

P

k =0

ak k

P

j=0

 k j



x0k −J(x − x0)j

P

k =0

|ak|

k

P

j=0

 k j



|x0|k −J(|x − x0|)j

=

P

k =0

|ak|(|x − x0| + |x0|)k

< ∞.

(74)

Since

P

k =0

akyk converges absolutely at y := |x − x0| + |x0| < R,we have

P

k =0

ak k

P

j=0

 k j



x0k −J(x − x0)j

P

k =0

|ak|

k

P

j=0

 k j



|x0|k −J(|x − x0|)j

=

P

k =0

|ak|(|x − x0| + |x0|)k

< ∞.

(75)

Since

P

k =0

akyk converges absolutely at y := |x − x0| + |x0| < R, we have

P

k =0

ak k

P

j=0

 k j



x0k −J(x − x0)j

P

k =0

|ak|

k

P

j=0

 k j



|x0|k −J(|x − x0|)j

=

P

k =0

|ak|(|x − x0| + |x0|)k

< ∞.

(76)

Since

P

k =0

akyk converges absolutely at y := |x − x0| + |x0| < R, we have

P

k =0

ak k

P

j=0

 k j



x0k −J(x − x0)j

P

k =0

|ak|

k

P

j=0

 k j



|x0|k −J(|x − x0|)j

=

P

k =0

|ak|(|x − x0| + |x0|)k

< ∞.

(77)

Since

P

k =0

akyk converges absolutely at y := |x − x0| + |x0| < R, we have

P

k =0

ak k

P

j=0

 k j



x0k −J(x − x0)j

P

k =0

|ak|

k

P

j=0

 k j



|x0|k −J(|x − x0|)j

=

P

k =0

|ak|(|x − x0| + |x0|)k

< ∞.

(78)

Since

P

k =0

akyk converges absolutely at y := |x − x0| + |x0| < R, we have

P

k =0

ak k

P

j=0

 k j



x0k −J(x − x0)j

P

k =0

|ak|

k

P

j=0

 k j



|x0|k −J(|x − x0|)j

=

P

k =0

|ak|(|x − x0| + |x0|)k

< ∞.

(79)

Hence, by(13), Theorem 7.18, and Corollary 7.31, f (x ) =P

k =0akPk j=0

 k j



x0k −j(x − x0)j

=P j=0

 P

k =j

 k j



akx0k −j



(x − x0)j

=P j=0

 P

k =j

k !

(k − j)!ak(x0− 0)k −j (x − x0)j j!

=P j=0

f(j)(x0

j! (x − x0)j.2

(80)

Hence,by(13), Theorem 7.18, and Corollary 7.31, f (x ) =P

k =0akPk j=0

 k j



x0k −j(x − x0)j

=P j=0

 P

k =j

 k j



akx0k −j



(x − x0)j

=P j=0

 P

k =j

k !

(k − j)!ak(x0− 0)k −j (x − x0)j j!

=P j=0

f(j)(x0

j! (x − x0)j.2

(81)

Hence, by(13), Theorem 7.18, and Corollary 7.31, f (x ) =P

k =0akPk j=0

 k j



x0k −j(x − x0)j

=P j=0

 P

k =j

 k j



akx0k −j



(x − x0)j

=P j=0

 P

k =j

k !

(k − j)!ak(x0− 0)k −j (x − x0)j j!

=P j=0

f(j)(x0

j! (x − x0)j.2

(82)

Hence, by(13), Theorem 7.18, and Corollary 7.31, f (x ) =P

k =0akPk j=0

 k j



x0k −j(x − x0)j

=P j=0

 P

k =j

 k j



akx0k −j



(x − x0)j

=P j=0

 P

k =j

k !

(k − j)!ak(x0− 0)k −j (x − x0)j j!

=P j=0

f(j)(x0

j! (x − x0)j.2

(83)

Hence, by(13), Theorem 7.18, and Corollary 7.31, f (x ) =P

k =0akPk j=0

 k j



x0k −j(x − x0)j

=P j=0

 P

k =j

 k j



akx0k −j



(x − x0)j

=P j=0

 P

k =j

k !

(k − j)!ak(x0− 0)k −j (x − x0)j j!

=P j=0

f(j)(x0

j! (x − x0)j.2

(84)

Hence, by(13), Theorem 7.18, and Corollary 7.31, f (x ) =P

k =0akPk j=0

 k j



x0k −j(x − x0)j

=P j=0

 P

k =j

 k j



akx0k −j



(x − x0)j

=P j=0

 P

k =j

k !

(k − j)!ak(x0− 0)k −j (x − x0)j j!

=P j=0

f(j)(x0

j! (x − x0)j.2

(85)

Example:

Prove that arctan x is analytic on (-1,1) and has Maclaurin expansion

arctan x =

X

k =0

(−1)kx2k +1

2k + 1 x ∈ (−1, 1).

(86)

Example:

Prove that arctan x is analytic on (-1,1) and has Maclaurin expansion

arctan x =

X

k =0

(−1)kx2k +1

2k + 1 x ∈ (−1, 1).

(87)

Theorem (Lagrange)

Let n ∈N. If f ∈ Cn(a, b), then

Rn(x ) := Rnf ,x0(x ) = 1 (n − 1)!

Z x x0

(x − t)n−1f(n)(t)dt

for all x , x0 ∈ (a, b).

(88)

Theorem (Lagrange)

Let n ∈N. If f ∈ Cn(a, b), then Rn(x ) := Rnf ,x0(x ) = 1

(n − 1)!

Z x x0

(x − t)n−1f(n)(t)dt for all x , x0 ∈ (a, b).

(89)

Proof:

The proof is by induction on n. If n = 1,the formula holds by the Fundamental Theorem of Calculus.

Suppose that the formula holds for some n ∈N. Since Rn+1(x ) = Rn(x ) − f(n)(x0)

n! (x − x0)n and (x − x0)n

n! = 1

(n − 1)!

Z x x0

(x − t)n−1dt it follows that

Rn+1(x ) = − 1 (n − 1)!

Z x x0

(x − t)n−1 f(n)(t) − f(n)(x0) dt.

(90)

Proof:

The proof is by induction on n. If n = 1, the formula holds by the Fundamental Theorem of Calculus.

Suppose that the formula holds for some n ∈N. Since Rn+1(x ) = Rn(x ) − f(n)(x0)

n! (x − x0)n and (x − x0)n

n! = 1

(n − 1)!

Z x x0

(x − t)n−1dt it follows that

Rn+1(x ) = − 1 (n − 1)!

Z x x0

(x − t)n−1 f(n)(t) − f(n)(x0) dt.

(91)

Proof:

The proof is by induction on n. If n = 1,the formula holds by the Fundamental Theorem of Calculus.

Suppose that the formula holds for some n ∈N. Since Rn+1(x ) = Rn(x ) − f(n)(x0)

n! (x − x0)n and (x − x0)n

n! = 1

(n − 1)!

Z x x0

(x − t)n−1dt it follows that

Rn+1(x ) = − 1 (n − 1)!

Z x x0

(x − t)n−1 f(n)(t) − f(n)(x0) dt.

(92)

Proof:

The proof is by induction on n. If n = 1, the formula holds by the Fundamental Theorem of Calculus.

Suppose that the formula holds for some n ∈N. Since Rn+1(x ) = Rn(x ) − f(n)(x0)

n! (x − x0)n and (x − x0)n

n! = 1

(n − 1)!

Z x x0

(x − t)n−1dt it follows that

Rn+1(x ) = − 1 (n − 1)!

Z x x0

(x − t)n−1 f(n)(t) − f(n)(x0) dt.

(93)

Proof:

The proof is by induction on n. If n = 1, the formula holds by the Fundamental Theorem of Calculus.

Suppose that the formula holds for some n ∈N. Since Rn+1(x ) = Rn(x ) − f(n)(x0)

n! (x − x0)n and (x − x0)n

n! = 1

(n − 1)!

Z x x0

(x − t)n−1dt it follows that

Rn+1(x ) = − 1 (n − 1)!

Z x x0

(x − t)n−1 f(n)(t) − f(n)(x0) dt.

(94)

Proof:

The proof is by induction on n. If n = 1, the formula holds by the Fundamental Theorem of Calculus.

Suppose that the formula holds for some n ∈N. Since Rn+1(x ) = Rn(x ) − f(n)(x0)

n! (x − x0)n and (x − x0)n

n! = 1

(n − 1)!

Z x x0

(x − t)n−1dt it follows that

Rn+1(x ) = − 1 (n − 1)!

Z x x0

(x − t)n−1 f(n)(t) − f(n)(x0) dt.

(95)

Proof:

The proof is by induction on n. If n = 1, the formula holds by the Fundamental Theorem of Calculus.

Suppose that the formula holds for some n ∈N. Since Rn+1(x ) = Rn(x ) − f(n)(x0)

n! (x − x0)n and (x − x0)n

n! = 1

(n − 1)!

Z x x0

(x − t)n−1dt it follows that

Rn+1(x ) = − 1 (n − 1)!

Z x x0

(x − t)n−1 f(n)(t) − f(n)(x0) dt.

(96)

Let u = f(n)(t) − f(n)(x0), dv = (x − t)n−1 and integrate the right side of the identity above by parts. Since u(x0) = 0 and v (x ) = 0,we have

Rn+1(x ) = − 1 (n − 1)!

Z x x0

u0(t)v (t)dt = 1 n!

Z x x0

(x −t)nf(n+1)(t)dt.

Hence, the formula holds for n + 1. 2

(97)

Let u = f(n)(t) − f(n)(x0), dv = (x − t)n−1 and integrate the right side of the identity above by parts.Since u(x0) = 0 and v (x ) = 0, we have

Rn+1(x ) = − 1 (n − 1)!

Z x x0

u0(t)v (t)dt = 1 n!

Z x x0

(x −t)nf(n+1)(t)dt.

Hence, the formula holds for n + 1. 2

(98)

Let u = f(n)(t) − f(n)(x0), dv = (x − t)n−1 and integrate the right side of the identity above by parts. Since u(x0) = 0 and v (x ) = 0,we have

Rn+1(x ) = − 1 (n − 1)!

Z x x0

u0(t)v (t)dt = 1 n!

Z x x0

(x −t)nf(n+1)(t)dt.

Hence, the formula holds for n + 1. 2

Figure

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