Chapter 7 Infinite Series (無窮級數)

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Chapter 7 Infinite Series

(無窮級數)

Hung-Yuan Fan (范洪源)

Department of Mathematics, National Taiwan Normal University, Taiwan

Spring 2019

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本章預定授課範圍

7.1 Sequences

7.2 Series and Convergence

7.3 The Integral and Comparisons Tests 7.4 Other Convergence Tests

7.5 Taylor Polynomials and Approximations 7.6 Power Series

7.7 Representation of Functions by Power Series

7.8 Taylor and Maclaurin Series

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Section 7.1 Sequences

(序列)

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Sequences (序列)

An infinite sequence of real numbers is denoted by {an} = {an}n=1={a1, a2,· · · , an,· · · },

where anis the nth term (第 n 項) of the sequence for n∈ N.

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Def. (序列的收斂性)

A sequence{an} converges to a limit L, denoted by

nlim→∞an= L, if∀ ε > 0, ∃ M > 0 s.t.

n > M =⇒ |an− L| < ε.

Otherwise, we say that{an} diverges if the limit does not exist.

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示意圖 (承上頁)

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Thm 7.1 (函數在 ∞ 處的極限 ⇒ 序列的收斂性)

Let f be a real-valued function having the limit

xlim→∞f(x) = L.

Ifan= f(n) ∀ n ∈ N, then

nlim→∞an= L.

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pf: Let ε > 0 be given arbitrarily. Since lim

x→∞f(x) = L,∃ M > 0 s.t.

x > M =⇒ |f(x) − L| < ε.

Withan= f(n) for all n∈ N, ifn > M, then

|an− L| = |f(n) − L| < ε.

Thus, it follows from the Def. that

nlim→∞an= lim

n→∞f(n) = L.

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Example (Thm 7.1 的反敘述不成立)

Consider f(x) = sin(πx) for all x > 0 and let an= f(n) for n∈ N.

Then

nlim→∞an= lim

n→∞f(n) = lim

n→∞sin(nπ) = lim

n→∞0 = 0, but we know that

xlim→∞f(x) = lim

x→∞sin(πx) does not exist!

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Thm 7.2 (Limit Laws for Sequences)

Suppose that lim

n→∞an= L and lim

n→∞bn= K. Then

1 lim

n→∞(an± bn) = L± K.

2 lim

n→∞(c· an) = c· L for all c ∈ R.

3 lim

n→∞(an· bn) = L· K.

4 lim

n→∞

an bn = L

K ifbn̸= 0 ∀ n ∈ N andK̸= 0.

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Thm 7.3 (Squeeze Theorem for Sequences)

If∃ M > 0 s.t.an≤ cn≤ bn ∀ n > M, and

nlim→∞an= L = lim

n→∞bn,

then the sequence{cn} converges to the same limit L, i.e.,

nlim→∞cn= L.

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How to find the bounds for {c

n

}?

Sincen!≥ 2n for all n≥ 4, it follows that

|cn| = (−1)n n!

= 1 n! 1

2n, and hence we immediately obtain

−1

2n ≤ cn 1 2n for alln≥ 4.

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Thm 7.4 (Absolute Value Thoerem)

Let{an} be a sequence of real numbers. Then

nlim→∞|an| = 0 ⇐⇒ limn

→∞an= 0.

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Monotonic and Bounded Sequences (1/2)

Def. (單調序列的定義)

A sequence{an} is said to be monotonic (單調的) if its terms are nondecreasing (遞增的)

an≤ an+1 ∀ n ∈ N, or its terms arenonincreasing (遞減的)

an≥ an+1 ∀ n ∈ N.

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Monotonic and Bounded Sequences (2/2)

Def. (有界序列的定義)

(1) The sequence {an} isbounded above (於上有界) if∃ M ∈ R s.t. an≤ M ∀ n ∈ N.

(2) The sequence {an} isbounded below (於下有界) if∃ N ∈ R s.t. an≥ N ∀ n ∈ N.

(3) The sequence{an} is bounded (有界的)if it isbounded above and bounded below, i.e. ∃ M > 0 s.t. |an| ≤ M ∀ n ∈ N.

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Thm 7.5 (Bounded Monotonic Sequences)

If the sequence{an} isbounded and monotonic, then it converges, i.e.,∃! L ∈ R s.t. limn

→∞an= L.

Example 9 (Thm 7.5 的例子)

(a) The sequence {an} = {1/n} is bounded and nonincreasing, since

|an| ≤ 1 and an= 1 n 1

n + 1 =an+1 ∀ n ∈ N. So, it must converge by Thm 7.5 with lim

n→∞an= 0.

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Thm 7.5 (Bounded Monotonic Sequences)

If the sequence{an} isbounded and monotonic, then it converges, i.e.,∃! L ∈ R s.t. limn

→∞an= L.

Example 9 (Thm 7.5 的例子)

(a) The sequence {an} = {1/n} is bounded and nonincreasing, since

|an| ≤ 1 and an= 1 n 1

n + 1 =an+1 ∀ n ∈ N.

So, it must converge by Thm 7.5 with lim

n→∞an= 0.

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Section 7.2

Series and Convergence

(級數與收斂)

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Def. (Partial Sums of a Series)

(a) An infinite series (無窮級數) of real numbers is of the form Xan=

X n=1

an= a1+ a2+ a3+· · · .

(b) For each n∈ N, the nth partial sum (第 n 個部分和) ofP an is defined by

Sn= Xn

i=1

ai= a1+ a2+· · · + an.

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Def. (Convergence of a Series)

(a) We say thatP

an converges if the sequence{Sn} converges with lim

n→∞Sn= S. In this case, S is called the sum of the series and write S =P

an. (b) We say thatP

an diverges if the sequence{Sn} diverges.

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Two Questions for Series

Two Questions

1 Does a given series converge or diverge?

2 What is the sum of a convergent series?

Note: these questions are not always easy to answer,

especially the second one. (通常需要藉助數值方法求得近似和!)

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Geometric Series (幾何級數)

For a̸= 0 and ratio (公比) r ̸= 0, the geometric series is given by X

n=0

arn= a + ar + ar2+ ar3+· · · .

Thm 7.6 (幾何級數的收斂性)

X n=0

arn=



 a

1− r if |r| < 1, diverges if|r| ≥ 1.

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Example 3 (Thm 7.6 的例子)

(a)

X n=0

3 2n =

X n=0

3

1 2

n

= 3

1− (1/2) = 6 by Thm 7.6.

(b) X n=0

3 2

n

diverges because the ratio|r| = |3/2| = 3/2 ≥ 1.

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Thm 7.7 (Properties of Infinite Series)

Suppose thatP

an= A and P

bn= B are convergent series.

1 P

(c· an) = c· P an

= c· A for all c ∈ R.

2 P

(an± bn) = P an

± P bn

= A± B.

3 In general, we know that X(anbn)̸= X

an X bn

, X an bn

̸=

Pan Pbn.

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Thm 7.8

IfP

an converges, then lim

n→∞an= 0.

(收斂級數第 n 項所形成的序列必定趨近 0!)

pf: If the series

P

an= S converges, then we know that

nlim→∞Sn= S = lim

n→∞Sn−1, where Sn=

Pn i=1

ai is the nth partial sum ofP

an. So, we immediately obtain

nlim→∞an= lim

n→∞(Sn− Sn−1) = S− S= 0.

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Thm 7.9 (The nth Term Test; 第 n 項測試法)

If lim

n→∞an̸= 0, the series P

an diverges.

(若第 n 項不趨近到 0, 則級數必定發散!)

Note: 此定理為 Thm 7.8 的反敘述,常用於判斷級數的發散性。

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Divergence of Harmonic Series (示意圖)

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Section 7.3

The Integral and Comparisons Tests

(積分與比較測試法)

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Thm 7.10 (The Integral Test; 積分測試法)

If f ispositive, conti. and↘ on [1, ∞), andan= f(n) ∀ n ∈ N, then

X n=1

an and Z

1

f(x) dxboth converge or both diverge.

Note: 上述定理只說明級數與瑕積分是同收同發,但並沒有說明

兩者相等喔!

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Def. (p-級數的定義)

(a)

X n=1

1

np is called a p-series (p-級數) with p > 0.

(b) If p = 1, then X n=1

1

n is called a harmonic series (調和級數).

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Thm 7.11 (p-級數的收斂與發散)

The p-series

X n=1

1

np converges forp > 1, and diverges for 0 < p≤ 1.

Note: 此定理只探討 p-級數的收斂性,但無法由此求出該級數

的和!

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計算 p-級數的和 (檔名: p_series.m)

p = 2.7; data = [];

for k = 0:8 N = 10^k;

n = 1:N;

S_N = sum(1./(n.^p));

data = [data;N S_N];

end

semilogx(data(:,1),data(:,2),'bo-');

title('Partial sums of a p-series with p = 2.7');

xlabel('\bf n');

ylabel('\bf S_n');

P

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程式執行結果 (承上例)

下列圖形顯示 P

n=1

n1p 的收斂與發散,其中 p = 2.7 和 p = 0.7:

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Thm 7.12 (Direct Comparison Test; 直接比較法)

If0 < an≤ bn for all n∈ N, then

1 P

bnconverges =P

anconverges, and

2 P

an diverges =P

bn diverges.

口訣:

(1) 大的級數收斂保證小的級數也收斂!

(2) 小的級數發散保證大的級數也發散!

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Thm 7.13 (Limit Comparison Test; 極限比較法)

Letan, bn> 0 ∀ n ∈ Nwith lim

n→∞

an bn = L.

1 If0 < L <∞, then P

anand P

bn both converge or both diverge.

2 IfL = 0, then P

bn converges =P

an converges, and Pan diverges =P

bn diverges.

3 IfL =∞, then P

anconverges =P

bn converges, and Pbndiverges =P

an diverges.

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Section 7.4

Other Convergence Tests

(其他收斂測試法)

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Alternating Series (交錯級數)

An alternating series is of the form

S = X n=1

(−1)n+1an= X n=1

(−1)n−1an= a1− a2+ a3− a4+· · · ,

where the nth terman> 0 for all n∈ N.

Questions

Does the alternating series always converge?

How to estimate the sum S of a convergent alternating series?

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Thm 7.14 (Alternating Series Test; 交錯級數測試法)

If the sequence ofpositive terms{an} satisfies

1 ∃ M > 0 s.t. an≥ an+1 for all n > M, and

2 lim

n→∞an= 0,

then the alternating series S =P

(−1)n+1an converges with the property

|Sn− S| ≤ an+1 ∀ n ∈ N.

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Absolute and Conditional Convergence

Def. (Types of Convergence for a Series)

(1) P

an is absolutely convergent (絕對收斂) if P

|an| converges.

(2) P

an is conditionally convergent (條件收斂) if P an converges, but P

|an| diverges.

Thm 7.16 (絕對收斂保證原級數收斂)

P|an| converges =⇒P

anconverges.

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Absolute and Conditional Convergence

Def. (Types of Convergence for a Series)

(1) P

an is absolutely convergent (絕對收斂) if P

|an| converges.

(2) P

an is conditionally convergent (條件收斂) if P an converges, but P

|an| diverges.

Thm 7.16 (絕對收斂保證原級數收斂)

P|an| converges =⇒P

anconverges.

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Example (Thm 7.16 的反例)

The series P

n=1 (−1)n

n converges by the Alternating Series Test, but P

n=1|an| = P

n=1 1

n is a divergent p-series with p = 1!

In fact, any conditionally convergent series gives a counterexample of Thm 7.16.

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Thm 7.17 (The Ratio Test; 比值法)

Letan̸= 0 ∀ n ∈ Nwith ρ = lim

n→∞

an+1 an

.

1 ρ < 1 =P

an converges absolutely.

2 ρ > 1 or ρ =∞ =P

an diverges.

3 ρ = 1 =⇒ the test is inconclusive.

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Thm 7.18 (The Root Test; 根式法)

Letan̸= 0 ∀ n ∈ Nwith ρ = lim

n→∞

pn

|an| = limn

→∞

|an|1n .

1 ρ < 1 =P

an converges absolutely.

2 ρ > 1 or ρ =∞ =P

an diverges.

3 ρ = 1 =⇒ the test is inconclusive.

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Section 7.5

Taylor Polynomials and

Approximations

(泰勒多項式與近似)

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Def. (Taylor and Maclaurin Polynomials)

Suppose that f has n derivatives at c∈ dom(f).

(1) A polynomial of the form

Pn(x) = Xn k=0

f(k)(c)

k! (x− c)k

= f(c) + f(c)(x− c) + · · · + f(n)(c)

n! (x− c)n is called the nth Taylor poly. (n 階泰勒多項式) for f at c.

(2) Ifc = 0, then Pn(x) = Xn k=0

f(k)(0)

k! xk is called the nth Maclaurin poly. (n 階馬克勞林多項式) for f.

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多項式 P

1

(x) 和 P

2

(x) 的示意圖 (承上例)

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多項式 P

3

(x) 和 P

4

(x) 的示意圖 (承上例)

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多項式 P

6

(x) 的示意圖 (承上例)

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Def. (The Remainder of P

n(x))

Let f have (n + 1) derivatives on an interval I containing c.

(1) Rn(x)≡ f(x) − Pn(x) is called the remainder (剩餘項) associated with Pn(x).

(2) |Rn(x)| = |f(x) − Pn(x)| is the error associated with Pn(x).

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Thm 7.19 (Taylor’s Theorem; 泰勒定理)

If f has (n + 1) derivatives on an interval I containing c, then

∀ x ∈ I,∃ z between x and c s.t.

f(x) = Xn k=0

f(k)(c)

k! (x− c)k+ Rn(x) = Pn(x) + Rn(x), where the Lagrange form of the remainder is given by

Rn(x) = f(n+1)(z)

(n + 1)!(x− c)n+1.

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Section 7.6 Power Series

(冪級數)

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Def. (以 c 點為中心的冪級數)

An infinite series of the form

X n=0

an(x− c)n= a0+ a1(x− c) + a2(x− c)2+· · ·

is called a power series (冪級數) centered at c∈ R.

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冪級數的收斂行為

Three Types of Convergence for Power Series

For a power series P

n=0

an(x− c)n, you will see that

1 it converges only at x = c or

2 it converges only for|x − c| < R with R > 0 or

3 it converges for all x∈ R.

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Radius and Interval of Convergence

Denote

R = Radius of Convergence (收斂半徑), I = Interval of Convergence (收斂區間).

Type I of Convergence

X n=0

an(x− c)n converges only at x = c.

=⇒R = 0 and I = {c}.

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Type II of Convergence

X n=0

an(x− c)nconverges absolutely for |x − c| < R, and diverges for |x − c| > R.

=⇒R > 0 and I = (c − R, c + R).

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Type III of Convergence

X n=0

an(x− c)nconverges for all x∈ R.

=⇒R = ∞ and I = (−∞, ∞).

Example 4 (Type III 的例子)

Find the radius of convergence for the power series X

n=0

(−1)n

(2n + 1)!x2n+1.

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Note

In Section 7.8, we will further show that sin x =

X n=0

(−1)n

(2n + 1)!x2n+1 = x−x3 3! +x5

5! x7

7! +· · · ∀ x ∈ R.

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Endpoint Convergence (端點收斂性)

Besides the interval of convergence I = (c− R, c + R) in Type II, we also have

I = [c− R, c + R), (c − R, c + R] or [c − R, c + R]

for different endpoint convergence of a power series.

Example 5 (端點收斂的例子)

Find the interval of convergence of the power series X xn

= x +x2 +x3

+ x4

+· · · .

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Differentiation and Integration of Power Series

For the cases of Type II or Type III, we consider a real-valued function f defined by

f(x) = X n=0

an(x− c)n ∀ x ∈ I,

where I = (c− R, c + R) with R > 0, or I = (−∞, ∞).

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Two Questions

Is f differentiable and integrable on the open interval I?

If yes, what are f (x) and Z

f(x) dx?

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Thm 7.21 (冪級數的微分與積分公式)

Let f(x) =

X n=0

an(x− c)n be well-defied on I = (c− R, c + R).

1 Term-by-term differentiation (逐項微分):

f(x) = X n=0

d

dx[an(x− c)n] = X n=1

nan(x− c)n−1 ∀ x ∈ I.

2 Term-by-term integration (逐項積分):

Z

f(x) dx = X n=0

h Z an(x− c)ndxi

=C+ X n=0

an

n + 1(x− c)n+1 ∀ x ∈ I,

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Remarks

(1) The radii of convergence of f (x) and Z

f(x) dx are the same as that of f(x) =P

an(x− c)n.

(2) But, their intervals of convergence may differ from f.

(微分和積分後的收斂半徑與 f 相同,但收斂區間略有不同!)

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Section 7.7

Representation of Functions by Power Series

(以冪級數作為函數的表示式)

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Geometric Power Series (幾何冪級數)

The function f(x) = 1

1− x is well-defined for x̸= 1.

f has a geometric power series centered at x = 0, i.e.,

f(x) = 1 1− x =

X n=0

xn= 1 + x + x2+ x3+· · · for |x| < 1.

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示意圖 (承上頁)

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Operations with Power Series

Let f(x) =

X n=0

anxn and g(x) = X n=0

bnxn be well-defined.

1 For any k∈ R, f(kx) =X

n=0

an(kx)n= X n=0

(ankn)xn.

2 For any N∈ N, f(xN) = X n=0

an(xN)n= X n=0

anxnN.

3 f(x)± g(x) =X

n=0

anxn

±X

n=0

bnxn

= X n=0

(an± bn)xn.

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Section 7.8

Taylor and Maclaurin Series

(泰勒級數與馬克勞林級數)

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Thm 7.22 (The Form of a Convergent Power Series)

If f is a real-valued function defined on I = (c− R, c + R) by

f(x) = X n=0

an(x− c)n ∀ x ∈ I,

then f has derivatives of all orders on I, and moreover, we have an= f(n)(c)

n! ∀ n ∈ N, with 0! = 1 and f(0) = f.

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Def. (泰勒級數的定義)

Suppose that f has derivatives of all orders at c.

(1) A power series of the form X

n=0

f(n)(c)

n! (x− c)n= f(c) + f(c)(x− c) +f ′′(c)

2! (x− c)2+· · · is called the Taylor series (泰勒級數) for f at c.

(2) Ifc = 0, a power series of the form X n=0

f(n)(0)

n! xnis called the Maclaurin series (馬克勞林級數) for f.

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Convergence of Taylor Series

Suppose that f has derivatives of all orders on an open interval I containing c. It follows from Taylor’s Thm (Thm 7.19) that

∀ x ∈ I,∃ z between x and c s.t.

f(x) = Xn k=0

f(k)(c)

k! (x− c)k+ Rn(x) = Pn(x) + Rn(x), where the Lagrange form of the remainder is given by

Rn(x) = f(n+1)(z)

(n + 1)!(x− c)n+1.

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Question

When does the Taylor series for f at c always converge to f on the open interval I?

Thm 7.23 (泰勒級數的收斂性)

f(x) =

X n=0

f(n)(c)

n! (x− c)n conv. on I⇐⇒ lim

n→∞Rn(x) = 0 ∀ x ∈ I.

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Question

When does the Taylor series for f at c always converge to f on the open interval I?

Thm 7.23 (泰勒級數的收斂性)

f(x) =

X n=0

f(n)(c)

n! (x− c)n conv. on I⇐⇒ lim

n→∞Rn(x) = 0 ∀ x ∈ I.

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Remark

In order to prove that

nlim→∞Rn(x) = lim

n→∞

f(n+1)(z)

(n + 1)!(x− c)n+1= 0 ∀ x ∈ I, we often use the Squeeze Theorem and the following fact

nlim→∞

|x − c|n+1

(n + 1)! = 0 ∀ x ∈ R.

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Useful Taylor or Maclaurin Series (1/2)

(1) 1

1− x = X n=0

xn for −1 < x < 1.

(2) ln x = X n=1

(−1)n−1

n (x− 1)nfor 0 < x≤ 2.

(3) ex= X n=0

xn

n! for −∞ < x < ∞.

(4) sin x =

X (−1)n x2n+1 for −∞ < x < ∞.

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Useful Taylor or Maclaurin Series (2/2)

(5) cos x =

X n=0

(−1)n

(2n)!x2n for −∞ < x < ∞.

(6) sin−1x = X n=0

(2n)!

(2nn!)2(2n + 1)x2n+1 for −1 ≤ x ≤ 1.

(7) tan−1x = X n=0

(−1)n

2n + 1x2n+1 for−1 ≤ x ≤ 1. (8) Binomial Series (二項級數) withk∈ R:

(1+x)k = 1+

X n=1

k(k− 1) · · · (k− n + 1)

n! xn for −1 < x < 1.

(138)

Thank you for your attention!

Figure

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