Advanced Calculus (I)

Advanced Calculus (I)

WEN-CHINGLIEN

Department of Mathematics National Cheng Kung University

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 )

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 )

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, · · ·

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, · · ·

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

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

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

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

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

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

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

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

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

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

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

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

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.

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.

Remark:

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.

Remark:

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.

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

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

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

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

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 ) = R_n^{f ,x0}(x ) := f (x ) −

n−1

X

k =0

f^{(k )}(x0)

k ! (x − x0)^k.

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 ) = R_n^{f ,x0}(x ) := f (x ) −

n−1

X

k =0

f^{(k )}(x0)

k ! (x − x0)^k.

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 ).

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 ).

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.

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.

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.

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.

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.

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.

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.

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.

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.

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).

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).

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).

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).

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).

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).

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).

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).

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).

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).

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).

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).

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).

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

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

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

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

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)

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)

Example:

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

sin x =

∞

X

k =0

(−1)^kx^{2k +1}

(2k + 1)! , cos x =

∞

X

k =0

(−1)^kx^2k (2k )!

Example:

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

sin x =

∞

X

k =0

(−1)^kx^{2k +1}

(2k + 1)! , cos x =

∞

X

k =0

(−1)^kx^2k (2k )!

Example:

Prove that e^x is analytic onR and has Maclaurin expansion

e^x =

∞

X

k =0

x^k k !.

Example:

Prove that e^x is analytic onR and has Maclaurin expansion

e^x =

∞

X

k =0

x^k k !.

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.

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.

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

akx^k, 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

akx^k

=

∞

P

k =0

ak((x − x0) +x0)^k

=

∞

P

k =0

akPk j=0

 k j



x0k −J(x − x0)^j.

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

akx^k, 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

akx^k

=

∞

P

k =0

ak((x − x0) +x0)^k

=

∞

P

k =0

akPk j=0

 k j



x0k −J(x − x0)^j.

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

akx^k, 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

akx^k

=

∞

P

k =0

ak((x − x0) +x0)^k

=

∞

P

k =0

akPk j=0

 k j



x0k −J(x − x0)^j.

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

akx^k, 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

akx^k

=

∞

P

k =0

ak((x − x0) +x0)^k

=

∞

P

k =0

akPk j=0

 k j



x0k −J(x − x0)^j.

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

akx^k, 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

akx^k

=

∞

P

k =0

ak((x − x0) +x0)^k

=

∞

P

k =0

akPk j=0

 k j



x0k −J(x − x0)^j.

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

akx^k, 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

akx^k

=

∞

P

k =0

ak((x − x0) +x0)^k

=

∞

P

k =0

akPk j=0

 k j



x0k −J(x − x0)^j.

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

akx^k, 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

akx^k

=

∞

P

k =0

ak((x − x0) +x0)^k

=

∞

P

k =0

akPk j=0

 k j



x0k −J(x − x0)^j.

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

akx^k, 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

akx^k

=

∞

P

k =0

ak((x − x0) +x0)^k

=

∞

P

k =0

akPk j=0

 k j



x0k −J(x − x0)^j.

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

akx^k, 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

akx^k

=

∞

P

k =0

ak((x − x0) +x0)^k

=

∞

P

k =0

akPk j=0

 k j



x0k −J(x − x0)^j.

Since

∞

P

k =0

aky^k 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

|a_k|(|x − x₀| + |x₀|)^k

< ∞.

Since

∞

P

k =0

aky^k 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

|a_k|(|x − x₀| + |x₀|)^k

< ∞.

Since

∞

P

k =0

aky^k 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

|a_k|(|x − x₀| + |x₀|)^k

< ∞.

Since

∞

P

k =0

aky^k 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

|a_k|(|x − x₀| + |x₀|)^k

< ∞.

Since

∞

P

k =0

aky^k 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

|a_k|(|x − x₀| + |x₀|)^k

< ∞.

Since

∞

P

k =0

aky^k 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

|a_k|(|x − x₀| + |x₀|)^k

< ∞.

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

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

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

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

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

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

Example:

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

arctan x =

∞

X

k =0

(−1)^kx^{2k +1}

2k + 1 x ∈ (−1, 1).

Example:

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

arctan x =

∞

X

k =0

(−1)^kx^{2k +1}

2k + 1 x ∈ (−1, 1).

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).

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).

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.

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.

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.

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.

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.

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.

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.

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

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

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