## 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 x_{0}that
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 x_{0}that
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, x_{0}+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, x_{0}+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, x_{0}+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, x_{0}+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}

^{(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, x_{0}+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}

^{(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, x_{0}+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}

^{(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, x_{0}+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}

^{(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, x_{0}+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}

^{(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

^{k} is
positive and (c, d ) ⊆ (x0− R, x_{0}+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

^{k} is
positive and (c, d ) ⊆ (x0− R, x_{0}+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}

^{(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

^{k} is
positive and (c, d ) ⊆ (x0− R, x_{0}+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

^{k} is
positive and (c, d ) ⊆ (x0− R, x_{0}+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}

^{(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:

[Cauchy]The function

f (x ) = e^{−1/x}^{2} x 6= 0

0 x = 0

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

### Remark:

[Cauchy]The function

f (x ) = e^{−1/x}^{2} 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:

^{∞}(−∞, ∞)
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 ,x}^{0}(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 ,x}^{0}(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 x_{0}
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 x_{0}
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 ,x_{0}

(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 ,x_{0}

(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 ,x_{0}

(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) F^{0}(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) G^{0}(t) = − f^{(n)}(t)

(n − 1)!(x − t)^{n−1}
for t ∈ (a, b).

Notice by the Chain rule that

(11) F^{0}(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) G^{0}(t) = − f^{(n)}(t)

(n − 1)!(x − t)^{n−1}
for t ∈ (a, b).

Notice by the Chain rule that

(11) F^{0}(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) G^{0}(t) = − f^{(n)}(t)

(n − 1)!(x − t)^{n−1}
for t ∈ (a, b).

Notice by the Chain rule that

(11) F^{0}(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) G^{0}(t) = − f^{(n)}(t)

(n − 1)!(x − t)^{n−1}
for t ∈ (a, b).

Notice by the Chain rule that

(11) F^{0}(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) G^{0}(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)G^{0}(c) = (F (x ) − F (x0))G^{0}(c)

= (G(x ) − G(x0))F^{0}(c)

= −G(x0)F^{0}(c)

= −G(x0)F^{0}(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)G^{0}(c) = (F (x ) − F (x0))G^{0}(c)

= (G(x ) − G(x0))F^{0}(c)

= −G(x0)F^{0}(c)

= −G(x0)F^{0}(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)G^{0}(c) = (F (x ) − F (x0))G^{0}(c)

= (G(x ) − G(x0))F^{0}(c)

= −G(x0)F^{0}(c)

= −G(x0)F^{0}(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)G^{0}(c) = (F (x ) − F (x0))G^{0}(c)

= (G(x ) − G(x0))F^{0}(c)

= −G(x0)F^{0}(c)

= −G(x0)F^{0}(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)G^{0}(c) = (F (x ) − F (x0))G^{0}(c)

= (G(x ) − G(x0))F^{0}(c)

= −G(x0)F^{0}(c)

= −G(x0)F^{0}(c).

c ∈ (x0,x ) such that

−F (x0)G^{0}(c) = (F (x ) − F (x0))G^{0}(c)

= (G(x ) − G(x0))F^{0}(c)

= −G(x0)F^{0}(c)

= −G(x0)F^{0}(c).

c ∈ (x0,x ) such that

−F (x0)G^{0}(c) = (F (x ) − F (x0))G^{0}(c)

= (G(x ) − G(x0))F^{0}(c)

= −G(x0)F^{0}(c)

= −G(x0)F^{0}(c).

c ∈ (x0,x ) such that

−F (x0)G^{0}(c) = (F (x ) − F (x0))G^{0}(c)

= (G(x ) − G(x0))F^{0}(c)

= −G(x0)F^{0}(c)

= −G(x0)F^{0}(c).

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

n!

f^{(n)}(c)(x − c)^{n−1}
(n − 1)!

=Rnf ,x_{0}

(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 ,x_{0}

(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 ,x_{0}

(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 ,x_{0}

(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 )| ≤ M^{n}

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 )| ≤ M^{n}

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 on**R and have**
Maclaurin expansions

sin x =

∞

X

k =0

(−1)^{k}x^{2k +1}

(2k + 1)! , cos x =

∞

X

k =0

(−1)^{k}x^{2k}
(2k )!

### Example:

Prove that sin x and cos x are analytic on**R and have**
Maclaurin expansions

sin x =

∞

X

k =0

(−1)^{k}x^{2k +1}

(2k + 1)! , cos x =

∞

X

k =0

(−1)^{k}x^{2k}
(2k )!

### Example:

Prove that e^{x} is analytic on**R and has Maclaurin**
expansion

e^{x} =

∞

X

k =0

x^{k}
k !.

### Example:

Prove that e^{x} is analytic on**R 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 (x_{0}− r , x_{0}+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 (x_{0}− r , x_{0}+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 , x_{0}+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 , x_{0}+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 , x_{0}+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 , x_{0}+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 , x_{0}+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

^{k}, for all
x ∈ (−R, R). Suppose that (x0− r , x0+r ) ⊆ (−R, R) and
fix x ∈ (x0− r , x_{0}+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

^{k}, for all
x ∈ (−R, R). Suppose that (x0− r , x0+r ) ⊆ (−R, R) and
fix x ∈ (x0− r , x_{0}+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:

∞

P

k =0

^{k}, for all
x ∈ (−R, R). Suppose that (x0− r , x0+r ) ⊆ (−R, R) and
fix x ∈ (x0− r , x_{0}+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:

∞

P

k =0

^{k}, for all
x ∈ (−R, R). Suppose that (x0− r , x0+r ) ⊆ (−R, R) and
fix x ∈ (x0− r , x_{0}+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| + |x_{0}| < 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 − x_{0}| + |x_{0}|)^{k}

< ∞.

Since

∞

P

k =0

aky^{k} converges absolutely at
y := |x − x0| + |x_{0}| < 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 − x_{0}| + |x_{0}|)^{k}

< ∞.

Since

∞

P

k =0

aky^{k} converges absolutely at
y := |x − x0| + |x_{0}| < 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 − x_{0}| + |x_{0}|)^{k}

< ∞.

Since

∞

P

k =0

aky^{k} converges absolutely at
y := |x − x0| + |x_{0}| < 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 − x_{0}| + |x_{0}|)^{k}

< ∞.

Since

∞

P

k =0

aky^{k} converges absolutely at
y := |x − x0| + |x_{0}| < 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 − x_{0}| + |x_{0}|)^{k}

< ∞.

Since

∞

P

k =0

aky^{k} converges absolutely at
y := |x − x0| + |x_{0}| < 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 − x_{0}| + |x_{0}|)^{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 − x_{0})^{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 − x_{0})^{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 − x_{0})^{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 − x_{0})^{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 − x_{0})^{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 − x_{0})^{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)^{k}x^{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)^{k}x^{2k +1}

2k + 1 x ∈ (−1, 1).

Theorem (Lagrange)

Let n ∈**N. If f ∈ C**^{n}(a, b), then

Rn(x ) := Rnf ,x0(x ) = 1 (n − 1)!

Z x
x_{0}

(x − t)^{n−1}f^{(n)}(t)dt

for all x , x0 ∈ (a, b).

Theorem (Lagrange)

Let n ∈**N. If f ∈ C**^{n}(a, b), then
Rn(x ) := Rnf ,x0(x ) = 1

(n − 1)!

Z x
x_{0}

(x − t)^{n−1}f^{(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−1}dt
it follows that

Rn+1(x ) = − 1 (n − 1)!

Z x
x_{0}

(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−1}dt
it follows that

Rn+1(x ) = − 1 (n − 1)!

Z x
x_{0}

(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−1}dt
it follows that

Rn+1(x ) = − 1 (n − 1)!

Z x
x_{0}

(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−1}dt
it follows that

Rn+1(x ) = − 1 (n − 1)!

Z x
x_{0}

(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−1}dt
it follows that

Rn+1(x ) = − 1 (n − 1)!

Z x
x_{0}

(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−1}dt
it follows that

Rn+1(x ) = − 1 (n − 1)!

Z x
x_{0}

(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−1}dt
it follows that

Rn+1(x ) = − 1 (n − 1)!

Z x
x_{0}

(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

u^{0}(t)v (t)dt = 1
n!

Z x x0

(x −t)^{n}f^{(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

u^{0}(t)v (t)dt = 1
n!

Z x x0

(x −t)^{n}f^{(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

u^{0}(t)v (t)dt = 1
n!

Z x x0

(x −t)^{n}f^{(n+1)}(t)dt.

Hence, the formula holds for n + 1. 2