# Calculus 3 2 giải tích

Infinite Series and Differential Equations
Nguyen Thieu Huy
Hanoi University of Science and Technology

Nguyen Thieu Huy (HUST)

Infinite Series and Diff. Eq.

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Series of Functions
Definition
Let {un (x)}∞
n=1 be a sequence of functions defined ∀x ∈ D ⊂ R. Formal
sum

un (x) is called a series of functions.
n=1

Nguyen Thieu Huy (HUST)

Infinite Series and Diff. Eq.

2 / 15

Series of Functions
Definition
Let {un (x)}∞
n=1 be a sequence of functions defined ∀x ∈ D ⊂ R. Formal
sum

un (x) is called a series of functions.
n=1

Here, when x is taken a concrete real value x0 ∈ D then

un (x0 ) is a
n=1

series of numbers.

Nguyen Thieu Huy (HUST)

Infinite Series and Diff. Eq.

2 / 15

Series of Functions
Definition
Let {un (x)}∞
n=1 be a sequence of functions defined ∀x ∈ D ⊂ R. Formal
sum

un (x) is called a series of functions.

n=1

Here, when x is taken a concrete real value x0 ∈ D then

un (x0 ) is a
n=1

series of numbers.
∞ n
2 (x+1)n
Ex. 1:
, ∀x ∈ R, is a series of functions. Substituting x by 3,
n!
n=1

we obtain
n=1

8n
n!

is a series of numbers.

Nguyen Thieu Huy (HUST)

Infinite Series and Diff. Eq.

2 / 15

Series of Functions
Definition
Let {un (x)}∞
n=1 be a sequence of functions defined ∀x ∈ D ⊂ R. Formal
sum

un (x) is called a series of functions.
n=1

Here, when x is taken a concrete real value x0 ∈ D then

un (x0 ) is a
n=1

series of numbers.
∞ n
2 (x+1)n
Ex. 1:
, ∀x ∈ R, is a series of functions. Substituting x by 3,
n!
n=1

we obtain
n=1

8n
n!

is a series of numbers.

Ex. 2: Series of functions

n=1

n=1
√1
n

1
nx ,

∀x ∈ R. Substituting x by 12 , we have

is a series of numbers.

Nguyen Thieu Huy (HUST)

Infinite Series and Diff. Eq.

2 / 15

Domain of Convergence
Definition

un (x), the set X ⊂ R is called
∞

un (x) is convergent ∀x ∈ X

n=1
domain of convergence ⇐⇒ ∞

un (x) is divergent ∀x ∈
/ X.

For series of functions

n=1

n=1

Nguyen Thieu Huy (HUST)

Infinite Series and Diff. Eq.

3 / 15

Domain of Convergence
Definition

un (x), the set X ⊂ R is called
∞

un (x) is convergent ∀x ∈ X

n=1
domain of convergence ⇐⇒ ∞

un (x) is divergent ∀x ∈
/ X.

For series of functions

n=1

n=1

In that case, for each x ∈ X we put S(x) :=

un (x) ∈ R, then we
n=1

obtain a function x → S(x). The function S(x) is called the sum of the

series of functions

un (x), and we say
n=1

un (x) is convergent on X to
n=1

the function S(x).

Nguyen Thieu Huy (HUST)

Infinite Series and Diff. Eq.

3 / 15

Domain of Convergence
Definition

un (x), the set X ⊂ R is called
∞

un (x) is convergent ∀x ∈ X

n=1
domain of convergence ⇐⇒ ∞

un (x) is divergent ∀x ∈
/ X.

For series of functions

n=1

n=1

In that case, for each x ∈ X we put S(x) :=

un (x) ∈ R, then we
n=1

obtain a function x → S(x). The function S(x) is called the sum of the

series of functions

un (x), and we say
n=1

un (x) is convergent on X to
n=1

the function S(x).

Ex. 1: Domain of convergence for
1
1−x

=

x n is (−1, 1). Its sum is

n=0

x n;

∀x ∈ (−1, 1).

n=0

Nguyen Thieu Huy (HUST)

Infinite Series and Diff. Eq.

3 / 15

Ex. 2: Domain of convergence for

Its sum S(x) :=
n=1

Nguyen Thieu Huy (HUST)

n=1
1
nx

1
nx

is (1, ∞) (property of p-Series).

is called ζ-Riemann function.

Infinite Series and Diff. Eq.

4 / 15

Ex. 2: Domain of convergence for

Its sum S(x) :=
n=1

n=1
1
nx

1
nx

is (1, ∞) (property of p-Series).

is called ζ-Riemann function.

♣ The previous tests can be applied to find domains of convergence.

Nguyen Thieu Huy (HUST)

Infinite Series and Diff. Eq.

4 / 15

Ex. 2: Domain of convergence for

Its sum S(x) :=
n=1

n=1
1
nx

1
nx

is (1, ∞) (property of p-Series).

is called ζ-Riemann function.

♣ The previous tests can be applied to find domains of convergence.

Ex. 3: Find the Domain of Convergence for
n=1

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Infinite Series and Diff. Eq.

xn
n ;

un (x) =

xn
n .

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Ex. 2: Domain of convergence for

Its sum S(x) :=
n=1

n=1
1
nx

1
nx

is (1, ∞) (property of p-Series).

is called ζ-Riemann function.

♣ The previous tests can be applied to find domains of convergence.

Ex. 3: Find the Domain of Convergence for
n=1

We have lim

n→∞

un+1 (x)
un (x)

= lim

n→∞

x n+1
n+1
xn
n

xn
n ;

|x|n
n→∞ n+1

= lim

un (x) =

xn
n .

= |x|.

Therefore, according to Ratio Test
1

If |x| < 1 then series is conv.

2

If |x| > 1 then series is div.

3

Remain to consider |x| = 1 (here Ratio Test cannot be applied)

Nguyen Thieu Huy (HUST)

Infinite Series and Diff. Eq.

4 / 15

Ex. 2: Domain of convergence for

Its sum S(x) :=
n=1

n=1
1
nx

1
nx

is (1, ∞) (property of p-Series).

is called ζ-Riemann function.

♣ The previous tests can be applied to find domains of convergence.

Ex. 3: Find the Domain of Convergence for
n=1

We have lim

n→∞

un+1 (x)
un (x)

= lim

n→∞

x n+1
n+1
xn
n

xn
n ;

|x|n
n→∞ n+1

= lim

un (x) =

xn
n .

= |x|.

Therefore, according to Ratio Test
1

If |x| < 1 then series is conv.

2

If |x| > 1 then series is div.

3

Remain to consider |x| = 1 (here Ratio Test cannot be applied)
∞ 1
n=1 n ⇒ div. (p-Series with p = 1)
(−1)n
becomes ∞
n=1 n . This is alternating series
1
1
n ∀n and 2) limn→∞ n = 0. So, it is conv.

a) For x = 1: Series becomes
b) For x = −1: Series
1
satisfying 1) n+1

Altogether, Domain of convergence is [−1; 1).
Nguyen Thieu Huy (HUST)

Infinite Series and Diff. Eq.

4 / 15

Convergence: revisited

un (x) be convergent on D to function S(x). Then, using the

Let
n=1

language ( , N) we can rewrite the notion of convergence as follows:

un (x) is convergent on D to function S(x)

Series
n=1

k

⇔ ∀x ∈ D; ∀ > 0, ∃Nx; ∈ N such that |S(x) −

un (x)| <

∀k > Nx; .

n=1

Nguyen Thieu Huy (HUST)

Infinite Series and Diff. Eq.

5 / 15

Convergence: revisited

un (x) be convergent on D to function S(x). Then, using the

Let
n=1

language ( , N) we can rewrite the notion of convergence as follows:

un (x) is convergent on D to function S(x)

Series
n=1

k

⇔ ∀x ∈ D; ∀ > 0, ∃Nx; ∈ N such that |S(x) −

un (x)| <

∀k > Nx; .

n=1

Ex. :
1

Series

x n is convergent on (−1, 1) to function

n=0

2

Series
n=1

1
nx

S(x) =
n=1

1
1−x

convergent on (1, ∞) to ζ-Riemann function

1
nx .

Nguyen Thieu Huy (HUST)

Infinite Series and Diff. Eq.

5 / 15

Convergence: revisited

un (x) be convergent on D to function S(x). Then, using the

Let
n=1

language ( , N) we can rewrite the notion of convergence as follows:

un (x) is convergent on D to function S(x)

Series
n=1

k

⇔ ∀x ∈ D; ∀ > 0, ∃Nx; ∈ N such that |S(x) −

un (x)| <

∀k > Nx; .

n=1

Ex. :
1

Series

x n is convergent on (−1, 1) to function

n=0

2

Series
n=1

1
nx

S(x) =
n=1

1
1−x

convergent on (1, ∞) to ζ-Riemann function

1
nx .

Convergence according to the above definition is sometimes called
pointwise convergence. To do convenient calculi on a series of functions,
we need the concept of uniform convergence.
Nguyen Thieu Huy (HUST)

Infinite Series and Diff. Eq.

5 / 15

Uniform convergence
Definition

un (x) is called uniformly convergent on D to function S(x)

Series
n=1

⇔ ∀x ∈ D; ∀ > 0, ∃N ∈ N depending only on , not on x, such that
k

|S(x) −

un (x)| <

∀k > N .

n=1

Nguyen Thieu Huy (HUST)

Infinite Series and Diff. Eq.

6 / 15

Uniform convergence
Definition

un (x) is called uniformly convergent on D to function S(x)

Series
n=1

⇔ ∀x ∈ D; ∀ > 0, ∃N ∈ N depending only on , not on x, such that
k

|S(x) −

un (x)| <

∀k > N .

n=1

We introduce the following two tests to check the uniform convergence:
Cauchy test for uniformly convergent Series of functions

un (x) and set D ⊂ R satisfying

Consider series
n=1

m

∀x ∈ D; ∀ > 0 ∃N ∈ N such that |

series

un (x)| <

∀m, k > N . Then,

n=k+1

un (x) is uniformly convergent on D.
n=1

Nguyen Thieu Huy (HUST)

Infinite Series and Diff. Eq.

6 / 15

Weierstrass test

un (x) and set D ⊂ R satisfying ∃ non-negative

Consider series of
n=1

numbers an ; n = 1, 2, · · ·
1

|un (x)|

an ∀x ∈ D; ∀n = 1, 2, · · · .

2

Series of numbers

an is convergent.
n=1

Then, Series

un (x) is uniformly convergent on D.
n=1

Nguyen Thieu Huy (HUST)

Infinite Series and Diff. Eq.

7 / 15

Weierstrass test

un (x) and set D ⊂ R satisfying ∃ non-negative

Consider series of
n=1

numbers an ; n = 1, 2, · · ·
1

|un (x)|

an ∀x ∈ D; ∀n = 1, 2, · · · .

2

Series of numbers

an is convergent.
n=1

Then, Series

un (x) is uniformly convergent on D.
n=1
m

Proof. We have

m

m

|un (x)|

un (x)
n=k+1

an ∀x ∈ D.
n=k+1

n=k+1

Since

an is convergent, it follows that
n=1

m

∀ > 0 ∃N ∈ N such that

an <

∀m, k > N .

n=k+1

Nguyen Thieu Huy (HUST)

Infinite Series and Diff. Eq.

7 / 15

Weierstrass test

un (x) and set D ⊂ R satisfying ∃ non-negative

Consider series of
n=1

numbers an ; n = 1, 2, · · ·
1

|un (x)|

an ∀x ∈ D; ∀n = 1, 2, · · · .

2

Series of numbers

an is convergent.
n=1

Then, Series

un (x) is uniformly convergent on D.
n=1
m

Proof. We have

m

m

|un (x)|

un (x)
n=k+1

an ∀x ∈ D.
n=k+1

n=k+1

Since

an is convergent, it follows that
n=1

m

∀ > 0 ∃N ∈ N such that

∀m, k > N . Therefore,

an <
n=k+1
m

∀x ∈ D; ∀ > 0 ∃N ∈ N such that |

m

un (x)|
n=k+1

an <
n=k+1

∀m, k > N . Due to Cauchy test, the series is uniformly convergent on D.
Nguyen Thieu Huy (HUST)

Infinite Series and Diff. Eq.

7 / 15

Examples

Ex. 1: Prove that series
n=1

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sin nx
n2 −n+1

is uniformly convergent on R.

Infinite Series and Diff. Eq.

8 / 15

Examples

Ex. 1: Prove that series
We have: 1)

sin nx
n2 −n+1

Nguyen Thieu Huy (HUST)

sin nx
n2 −n+1
n=1
1
∀x
n2 −n+1

is uniformly convergent on R.
∈ R; ∀n = 1, 2, · · · .

Infinite Series and Diff. Eq.

8 / 15

Examples

Ex. 1: Prove that series
We have: 1)

sin nx
n2 −n+1

sin nx
n2 −n+1
n=1
1
∀x
n2 −n+1

2) Series of numbers
n=1

limn→∞

1
n2 −n+1
1
n2

is uniformly convergent on R.
∈ R; ∀n = 1, 2, · · · .

1
n2 −n+1

is conv. (Comparing with

= 1 and Series
n=1

n=1
1
n2

1
,
n2

where

is conv. (Riemann)).

Due to Weierstrass Test, Series is uniformly convergent on R.

Nguyen Thieu Huy (HUST)

Infinite Series and Diff. Eq.

8 / 15

Examples

Ex. 1: Prove that series
We have: 1)

sin nx
n2 −n+1

sin nx
n2 −n+1
n=1
1
∀x
n2 −n+1

n=1

limn→∞

1
n2 −n+1
1
n2

∈ R; ∀n = 1, 2, · · · .

1

2) Series of numbers

is uniformly convergent on R.

n2 −n+1

is conv. (Comparing with

= 1 and Series
n=1

n=1
1
n2

1
,
n2

where

is conv. (Riemann)).

Due to Weierstrass Test, Series is uniformly convergent on R.

Ex. 2: Prove that series
n=0

Nguyen Thieu Huy (HUST)

x n is uni. conv. on [− 12 , 12 ].

Infinite Series and Diff. Eq.

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