Sequences and Series of Functions

Home , Series (mathematics)

Li Jiayi, Joanna

14 July, 2014

1 Contents

1 Review: Sequences and Series of Real Numbers 4

1.1 Sequences ...... 4

1.1.1 Sequences ...... 4

1.1.2 Some properties of a convergent sequence ...... 4

1.1.3 Monotone Sequences ...... 5

1.1.4 Subsequences and the Bolzano-Weiertrass Theorem ...... 6

1.1.5 Cauchy’s Convergence Criterion ...... 7

1.2 Series ...... 7

1.2.1 Series ...... 7

1.2.2 Some tests for convergence ...... 8

1.2.3 Absolute and Conditional Convergence ...... 8

2 Sequences and Series of Functions 9

2.1 Pointwise convergence of sequences of functions ...... 9

2.2 Uniform convergence of sequences of functions ...... 11

2.2.1 Uniform convergence and boundedness ...... 13

2.2.2 Uniform convergence and continuity ...... 14

2.2.3 Uniform convergence and Riemann integration ...... 16

2.2.4 Uniform convergence and differentiation ...... 17

2 2.3 Uniform convergence of series of functions ...... 20

2.3.1 Some tests for uniform convergence of series of functions ...... 20

2.3.2 the Power Series ...... 20

3 1 Review: Sequences and Series of Real Numbers

1.1 Sequences

1.1.1 Sequences

Deﬁnition 1.1. A sequence of real numbers is a real-valued function deﬁned on the set of natural numbers, i.e. a function f : N −→ R.

In other words, a sequence can be written as f(1), f(2), f(3), ... We shall denote by an such a sequence where an = f(n) ∈ R. The number an is called the n-th term of the sequence.

Deﬁnition 1.2. A sequence {an} is said to converge to a real number L if ∀ > 0, ∃N ∈ N such that if n > N, then |an − L| < . n Example 1.1. Prove that the sequence converges to 1, that is, prove that n + 1 n lim = 1 x→∞ n + 1

Proof. We are required to show that for any given > 0, we can ﬁnd a natural number N n −1 such that if n > N, then | − 1| = | | < . n + 1 n + 1 1 1 We ﬁrst note that < for any N ∈ . Now that for any given , by the Archimedean n + 1 n N 1 Property, there exists N ∈ such that < . It follows that if n > N, then N N 1 1 1 < < < n + 1 n N and so we are done.

1.1.2 Some properties of a convergent sequence

Theorem 1.1 (Uniqueness of Limits). Let {an} be a convergent sequence. Then the limit of the sequence is unique.

4 Deﬁnition 1.3. A sequence {an} is bounded above if there exists M ∈ R such that an < M for all n ∈ N. Similarly, a sequence {an} is bounded below if there exists m ∈ R such that an > m for all n ∈ N. A sequence {an} is bounded if it is bounded above and below.

Theorem 1.2 (Boundedness of Convergent Sequences). If {an} is a convergent sequence, then it is bounded.

Theorem 1.3 (Arithmetic Properties of Convergent Sequences)). Let {an} and {bn} be two sequences. Suppose lim an = A and lim bn = B for some A, B ∈ . Then n→∞ n→∞ R

(a) lim (an + bn) = lim an + lim bn = A + B n→∞ n→∞ n→∞

(b) lim (anbn) = ( lim an)( lim bn) = AB n→∞ n→∞ n→∞

an limn→∞ an A (d) lim = = , provided that bn 6= 0 for all n ∈ N and B 6= 0. n→∞ bn limn→∞ bn B

Theorem 1.4 (Sandwich Theorem). Let {an}, {bn} and {cn} be sequences. Suppose that

an ≤ bn ≤ cn

for all n ∈ , and that lim an = lim cn = L. Then lim bn = L. N n→∞ n→∞ n→∞

1.1.3 Monotone Sequences

Deﬁnition 1.4. Let {an} be a sequence. We say that {an} is increasing if an ≤ an+1 for all n ∈ N and is decreasing if an ≥ an+1 for all n ∈ N. We say that {an} is a monotone if it is either increasing or decreasing.

Theorem 1.5. (a) If {an} is increasing and bounded above, then {an} converges.

(b) If {an} is decreasing and bounded below, then {an} converges.

5 1.1.4 Subsequences and the Bolzano-Weiertrass Theorem

Deﬁnition 1.5. Let {an} be a sequence of real numbers, and let n1 < n2 < n3 < ··· be a strictly increasing sequence of natural numbers. Then the sequence

an1 , an2 , an3 ···

is called a subsequence of {an} and is denoted by {ank }, where k ∈ N indexes the subsequence.

Theorem 1.6. If {an} is a sequence converging to L, then every subsequence {ank } also converges to L.

Theorem 1.7. Let {an} be a sequence. The {an} has a monotone subsequence. , Proof. To prove this theorem, we call the k-th term dominant if it is greater than or equal to all the following terms. In other words, the term ak is dominant if ak ≥ am for all m ≥ k. There are two cases to consider:

Case 1: There are inﬁnitely many dominant terms in the sequence {an}. Then we have a inﬁnite subsequence an1 , an2 , an3 ,... Now ani is dominant for all i ∈ N and ni < ni+1, so we must haveani ≥ ani+1 . Similarly, we have

an1 ≥ an2 ≥ an3 ≥ · · ·

Hence we have a decreasing subsequence ank and we are done.

Case 2: There are only ﬁnitely many dominant terms in the sequence {an}. (including the case where there is no dominant term). Suppose aN is the last dominant term, and let n1 = N + 1. Now an1 is not dominant and so there must exist n2 > n1 such that an1 ≤ an2 .

Continuing this way, we obtain an increasing subsequence an1 , an2 , an3 , ··· as desired.

Theorem 1.8 (the Bolzano-Weiertrass Theorem). Let {an} be a bounded sequence. Then

{an} has a convergent subsequence.

6 1.1.5 Cauchy’s Convergence Criterion

Deﬁnition 1.6. A sequence {an} is called a Cauchy sequence if ∀ > 0, ∃N ∈ N such that if n, m > N, then |an − am| < .

Theorem 1.9. Let {an} be a convergent sequence. Then {an} is a Cauchy sequence.

Theorem 1.10. Let {an} be a Cauchy sequence. Then {an} is a convergent sequence. ,

1.2 Series

1.2.1 Series

Deﬁnition 1.7. Let {an} be a sequence of real numbers. The sequence {sk} can be deﬁned by

s1 = a1

s2 = a1 + a2

s3 = a1 + a2 + a3 . .

k X sk = a1 + a2 + a3 + ··· + ak = an n=1

∞ X is called the sequence of partial sums of the series an. n=1 The term an is called the n-th term of the series and the term sk is called the k-th partial ∞ X sum of the series. The series an is said to converge to some number L if the associated n=1 sequence {sk} of the partial sums converges to L. We write ∞ X an = L n=1

7 and we refer to L as the sum of the series. If the sequence {sk} of partial sums diverges, then ∞ X we say that the series an diverges. n=1

1.2.2 Some tests for convergence

∞ X Theorem 1.11 (Cauchy Convergence Criterion for Series). The series an converges if n=1 and only if for any > 0 there exists N ∈ N such that if n > m > N, then

|am+1 + am+2 + ··· + an| < .

Theorem 1.12 (Comparison Test). Suppose that {an} and {bn} are sequences satisfying

0 ≤ an ≤ bn for all n ∈ N.

∞ ∞ X X (a) If bn, then anconverges n=1 n=1 ∞ ∞ X X (b) If an diverges, then bn diverges. n=1 n=1

Theorem 1.13 (Ratio Test). Let {an} be a sequence with an > 0 for all n ∈ N. Suppose that a lim n n→∞ an+1 ∞ X exists and equals r. The the series an converges if r < 1 and diverges if r > 1. n=1

1.2.3 Absolute and Conditional Convergence

∞ X Deﬁnition 1.8. Let {an} be a sequence of real numbers. The series an is said to be n=1 ∞ ∞ X X absolutely convergent if the series |an| is convergent. The series an is said to be n=1 n=1 ∞ X conditionally convergent if it converges but does not converge absolutely, that is, an con- n=1 ∞ X verges, but |an| diverges. n=1

8 2 Sequences and Series of Functions

We will now discuss the convergence of sequences and series of real-valued functions de-

ﬁned on a set X. In most cases, X would be a subset of R. Since we are dealing with the convergence problems of sequences and series, we will naturally be interested in knowing if the limit functions, whenever they exist, will preserve properties such as boundedness, continuity, uniform continuity, differentiability and integrability. In other words, if the sequence of functions under consideration is bounded, continuous, differentiable or integrable, and if the limit function of the sequence exists, does it carry these properties?

2.1 Pointwise convergence of sequences of functions

Deﬁnition 2.1. Suppose {fn}, n = 1, 2, 3 ··· is a sequence of functions deﬁned on a set

E, and suppose that the sequence of numbers {fn(x)} converges for every x ∈ E. We can deﬁne a function f by

f(x) = lim fn(x), x ∈ E. n→∞

Under these circumstances, we say that {fn} converges on E and that f is the limit, or the limitfunction, of {fn}. Sometimes we shall use a more descriptive terminology and shall say that ”{fn} converges for every x ∈ E”, and if we deﬁne

n X f(x) = lim fi(x), x ∈ E, n→∞ i=1 X then the function f is called the sum of the series fn.

Note 2.1. The main problem which arises is to determine whether important properties of functions are preserved under the limit operations mentioned above. For instance, if the functions {fn} are bounded, continuous, differentiable, or integrable, is the same true of the

9 0 0 limit function? What are the relations between fn and f , say, or between the integrals of fn and that of f ? To say that f is continuous at a limit point x means

lim f(t) = f(x). t→x

Hence, to ask whether the limit of a sequence of continuous functions is continuous is the same as to ask whether

lim lim fn(t) = lim lim fn(t), t→x n→∞ n→∞ t→x i.e., whether the order in which limit processes are carried out is immaterial. We shall now show by means of several examples that limit processes cannot in general be interchanged without affecting the result. Afterward, we shall prove that under certain conditions the order in which limit operations are carried out is immaterial.

Example 2.1. For m = 1, 2, 3, ··· , n = 1, 2, 3, ··· , let

m s = . m,n m + n

Then, for every ﬁxed n,

lim sm,n = 1, m→∞ so that

lim lim sm,n = 1. n→∞ m→∞ On the other hand, for every ﬁxed m,

lim sm,n = 0, n→∞ so that

lim lim sm,n = 0. m→∞ n→∞

Example 2.2. Let x2 f (x) = , x ∈ ; n = 0, 1, 2, ··· , n (1 + x2)n R

10 and consider ∞ ∞ X X x2 f(x) = f (x) = . n (1 + x2)n n=0 n=0

Since fn(0) = 0, we have f(0) = 0. For x 6= 0, the last series in the above equation is a convergent geometric series with sum 1 + x2. Hence   0 if x = 0, f(x) =  2 1 + x if x 6= 0, so that a convergent series of continuous functions may have a discontinuous sum.

Note 2.2. After these examples, which show what can go wrong if the limit processes are interchanged carelessly, we will deﬁne a new mode of convergence, stronger than point wise convergence as deﬁned in Definition2.1, which will enable us to arrive at positive results.

2.2 Uniform convergence of sequences of functions

Deﬁnition 2.2. We say that a sequence of functions {fn}, n = 1, 2, 3, ··· , converges uniformly on E to a function f if for every > 0 there exists N ∈ N such that n ≥ N implies

|fn(x) − f(x)| ≤ for all x ∈ E.

Theorem 2.1 (Cauchy Critirion ). The sequence of functions {fn}, deﬁned on E, converges uniformly on E, if and only if for, every > 0 there exists an integer N such that m ≥ N, n ≥ N, x ∈ E implies

|fm(x) − fn(x)| ≤ .

Proof. Assume that fn → f uniformly on E. Then, given > 0, we can ﬁnd N such that n > N implies |f (x) − f(x)| < for all x ∈ E. n 2

11 Taking m > N, we also have |f (x)−f(x)| < . Thus it follows from the triangular inequality m 2 that

|fn(x) − fm(x)| < |fn(x)| − f(x)| + |f(x) − fm(x)| < .

Conversely, suppose that the Cauchy Criterion is satisﬁed. Then, for each x ∈ E, the sequence {fn(x)} converges. Let f(x) = lim fn(x) for all x ∈ E. We have to prove that the n→∞ convergence is uniform. If > 0 is given, we can choose an N such that the Cauchy Criterion implies that ∀n > N, k = 1, 2, 3,..., |f (x) − f (x)| < n n+k 2 for all x ∈ E. Therefore lim |fn(x) − fn+k(x)| = |fn(x) − f(x)| ≤ k→∞ 2 for all x ∈ E. Hence, n > N implies |fn(x) − f(x)| < for all x ∈ E. This proves that fn → f uniformly in E.

Theorem 2.2. Suppose

lim fn(x) = f(x)(x ∈ E). n→∞ Put

Mn = sup |fn(x) − f(x)|. x∈E

Then fn → f uniformly on E if and only if Mn → 0 as n → ∞. x2n Example 2.3. Let {f } be a sequence of functions deﬁned on such that f (x) = n R n 1 + x2n for all x ∈ R, n ∈ N. Prove that {fn} converges pointwise but not uniformly on R.

Proof. Since lim fn(x) exists for all x ∈ , according to pointwise convergence, the limit n→∞ R function f is given by   0 if |x| < 1  1 f(x) = if |x| = 1 2  1 if |x| > 1

12 Put

Mn = sup |fn(x) − f(x)|. x∈E 1 In this case, we have M = . Thus, it follows from Theorem 2.2 that {f } does not converge n 2 n to f uniformly on R.

2.2.1 Uniform convergence and boundedness

Theorem 2.3. If {fn} is a sequence of bounded functions on E, and if fn → f uniformly on E, then f is bounded on E.

Proof. Let = 1. Since fn → f uniformly on E, ∃ n ∈ N such that n > N implies

|fn(x) − f(x)| < 1 for all x ∈ E. Particularly, consider n = N + 1. It follows that

|fN+1(x) − f(x)| < 1 for all x ∈ E, that is

|f(x)| < |fN+1(x)| + 1 for all x ∈ E. Since fN+1 is bounded on E, there exists B > 0 such that

|fN+1(x)| < B for all x ∈ E. Thus, |f(x)| < B + 1 for all x ∈ E. This proves that f is bounded on E.

13 Example 2.4. A sequence of bounded functions that are pointwise convergent, but the limit function is not bounded.   1 n if x ∈ (0, ] Consider f (x) = n n 1 1  if x ∈ ( , 1]. x n

The limit function f is given by

1 f(x) = lim fn(x) = n→∞ x for all x ∈ (0, 1].

Since ∀n ∈ N, |fn(x)| < n + 1 for all x ∈ (0, 1], all the functions in the sequence are bounded. However, f(x) → ∞ as x → 0+, the limit function f is not bounded on (0, 1].

2.2.2 Uniform convergence and continuity

Theorem 2.4 ( ). Suppose fn → f uniformly on a set E. Let x be a limit point of E, and suppose that ,

lim fn(t) = An (n = 1, 2, 3, ··· ). t→x

Then {An} converges, and

lim f(t) = lim An. t→x n→∞ In other words, the conclusion is that

lim lim fn(t) = lim lim fn(t) t→ x n→∞ n→∞ t→ x

Proof. Let > 0 be given. By the uniform convergence of {fn}, there exists N such that n ≥ N, m ≥ N, t ∈ E imply

|fn(t) − fm(t)| ≤ .

Letting t → x, we obtain

|An − Am| ≤

14 for n ≥ N, m ≥ N, so that {An} is a Cauchy sequence and therefore converges, say to A. Next,

|f(t) − A| ≤ |f(t) − fn(t)| + |fn(t) − An| + |An − A|.

|f(t) − A| ≤ ,

provided that t ∈ V ∩ E, t 6= x. This is equivalent to lim f(t) = lim An. t→x n→∞

Theorem 2.5. If {fn} is a sequence of continuous functions on E, and if fn → f uniformly on E, then f is continuous on E.

Note 2.3. If x0 is an accumulation point of E, then f is continuous at x = x0 if and only if

lim lim fn(x) = lim lim fn(x). x→x0 n→∞ n→∞ x→x0

If x0 is an isolated point of E, then f is automatically continuous at x = x0. Thus, proof of Theorem 2.5 follows directly from Theorem 2.4.

Note 2.4. Is the converse of Theorem 2.5 true? The converse is not true in general, i.e. uniform convergence is a sufﬁcient but not necessary condition to transmit continuity from individual terms to the limit function. In Example 2.5, we will see a non-uniformly convergent sequence of continuous functions with a continuous limit function.

15 2 n Example 2.5. Consider {fn} deﬁned on [0, 1] such that fn(x) = n x(1 − x) for all n ∈

N, x ∈ [0, 1]. The limit function f exists and

2 n f(x) = lim fn(x) = lim n x(1 − x) = 0 n→∞ n→∞ for all x ∈ [0, 1]. Obviously, each term in the sequence, as well as the limit function, is continuous on [0, 1].

However, Let Mn = sup |fn(x) − f(x)|. Since lim Mn > 0, it follows from Theorem 2.2 x∈[0,1] n→∞ that {fn} does not converge to f uniformly on [0, 1].

2.2.3 Uniform convergence and Riemann integration

Theorem 2.6 ( ). Suppose fn ∈ R([a, b]) for n = 1, 2, 3,... and suppose fn → f uniformly , on [a, b]. Then f ∈ R([a, b]), and Z b Z b f = lim fn. a n→∞ a

(The existence of the limit is part of the conclusion.)

Proof. It sufﬁces to prove this for real fn. Put

n = sup |fn(x) − f(x)|, the supremum being taken over a ≤ x ≤ b. Then

fn − n ≤ f ≤ fn + n, so that the upper and lower integrals of f satisfy Z b Z Z Z b (fn − n) ≤ f ≤ f ≤ (fn + n). a a Hence Z Z 0 ≤ f − f ≤ 2n(b − a).

16 Since n → 0 as n → 0, the upper and lower integrals of f are equal. Thus f ∈ R([a, b]). Another application of the inequality above now yields Z b Z b | f − fn | ≤ n(b − a). a a Z b Z b This implies f = lim fn. a n→∞ a Note 2.5 (Riemann integrability). By f ∈ R([a, b]), we mean that the function f is Riemann integrable on [a,b]. The deﬁnition is as follows:

Let f :[a, b] −→ R be a bounded (not necessary continuous) function on [a, b]. Then f is said to be Riemann integrable on [a, b] if U(f) = L(f), where

U(f) = inf{U(f, P ): P is a partition of [a, b]}

L(f) = sup{L(f, P ): P is a partition of [a, b]}.

In this case, the common value of U(f) and L(f) is called the Riemann integral of f on [a, b], Z b and is denoted by f(x)dx. a

2.2.4 Uniform convergence and differentiation

By analogy with Theorem 2.3, 2.4, and 2.6, one might expect the following result to hold:

Theorem 2.7 ( ). Suppose that {fn}is a sequence of functions, differentiable on [a, b] and - such that {fn(xo)} converges for some point x0 on [a, b]. If {fn} converges uniformly on

0 0 [a, b], then {fn} converges uniformly on [a, b], to a function f , and

0 0 f (x) = lim fn(x)(a ≤ x ≤ b). n→∞

Actually, in Example 2.6, we will see that this cannot be true.

0 Example 2.6. A sequence of differentiable functions {fn} with limit 0 for which {fn} diverges.

17 sin nx Let f (x) = √ if x ∈ , n = 1, 2, 3,... n n R Then lim fn(x) = 0 for all x ∈ . n→∞ R 0 √ 0 But fn(x) = n cos nx, so lim fn(x) does not exist for any x ∈ . n→∞ R

Theorem 2.8 ( ). Assume that each term of {fn} is a real-valued function having a ﬁnite , derivative at each point of an open interval (a, b). Assume that for at least one point x0 in

(a, b) the sequence {fn(x0)} converges. Assume further that there exists a function g such

0 that fn → g uniformly on (a, b). Then:

a) There exists a function f such that fn → f uniformly on (a, b).

b) For each x in (a, b) the derivative f 0(x) exists and equals g(x).

Proof. Assume that c ∈ (a, b) and deﬁne a new sequence {gn} as follows:  f (x) − f (c)  n n if x 6= c,  x − c gn(x) =  0 fn(c) if x = c.

The sequence {gn} so formed depends on the choice of c. Convergence of {gn(c)} follows

0 from the hypothesis, since gn(c) = fn(c). We will prove next that {gn} converges uniformly on (a, b). If x 6= c, we have

h (x) − h (x) g (x) − g (x) = n m , n m x − c

0 0 0 where h(x) = fn(x)−fm(x). Now h (x) exists for each x ∈ (a, b) and equals fn(x)−fm(x). Applying the Mean Value Theorem, we get

0 0 gn(x) − gm(x) = fn(x1) − fm(x1)

0 where x1 lies between x and c. Since {fn} converges uniformly on (a, b), we can use the

Cauchy condition to deduce that {gn} converges uniformly on (a, b).

Now we can show that {fn} converges uniformly on (a, b). Let us form a particular sequence

18 {gn} corresponding to the special point c = x0 for which {fn(x0)} is assumed to converge. We can write

fn(x) = gn(x)(x − x0) + fn(x0) an equation which holds for every x ∈ (a, b). Hence we have

fn(x) − fm(x) = fn(x0) − fm(xo) + [gn(x) − gm(x)](x − x0).

This equation, with the help of the Cauchy condition, establishes the convergence of {fn} on (a, b). This proves (a).

To prove (b), return to the sequence {gn} deﬁned for an arbitrary point c in (a, b) and let

0 G(x) = lim gn(x). The hypothesis that fn exists means that lim gn(x) = gn(c). In other n→∞ x→c words, each gn is continuous at c. Since gn → G uniformly on (a, b), the limit function G is also continuous at c. This means that

G(c) = lim G(x), x→c the existence of the limit being part of the conclusion. But, for x 6= c, we have

fn(x) − fn(c) f(x) − f(c) G(x) = lim gn(x) = lim = . n→∞ n→∞ x − c x − c

Hence, the derivative f 0(c) exists and equals G(c). But

0 G(c) = lim gn(c) = lim fn(c) = g(c); n→∞ n→∞ hence f 0(c) = g(x). Since c is an arbitrary point of (a, b), this proves (b).

Note 2.6 (Mean Value Theorem). Let f :[a, b] −→ R be a function which is continuous on [a, b] and differentiable on (a, b). There exists c ∈ (a, b) such that

f(b) − f(a) f 0(c) = . b − a

19 2.3 Uniform convergence of series of functions

Deﬁnition 2.3. Given a sequence {fn} of functions deﬁned on a set S. For each x in S, let

n X sn(x) = fk(x)(n = 1, 2, 3, ··· ). k=1 X If there exists a function f such that sn → f uniformly on S, we say the series fn(x) converges uniformly on S and we write

∞ X fn(x) = f(x)(uniformly on S). n=1

2.3.1 Some tests for uniform convergence of series of functions

Theorem 2.9 (Cauchy Condition for uniform convergence of series). The inﬁnite series X fn(x) converges uniformly if, and only if, for any > 0 there is an N such that n > N implies

n+p X | fn(x)| < for each p = 1, 2, 3, . . . , and for every x in S. k=n+1

Theorem 2.10 (Weierstrass M-test). Let {Mn} be a sequence of non-negative numbers such that

0 ≤ |fn(x)| ≤ Mn, for n = 1, 2, 3, . . . and for every x in S. X X Then fn(x) converges uniformly on S if Mn converges.

2.3.2 the Power Series

In this section, we shall derive some properties of functions which are represented by power series, i.e., functions of the form ∞ X n f(x) = cnx n=0

20 or, more generally, ∞ X n f(x) = cn(x − a) . (∗) n=0 There are called analytic functions. We shall restrict ourselves to real values of x. Instead of circles of convergence(complex values of x), we shall therefore encounter intervals of convergence. If (*) converges for |x − a|,R, f is said to be expanded in a power series about the point x = a. As a matter of convenience, we often take a = 0 without any loss of generality.

1. Properties of functions represented by real power series

Theorem 2.11. Suppose the series

∞ X n cnx (∗) n=0 converges for |x| < R, and deﬁne

∞ X n f(x) = cnx (|x| < R). x=0 Then (*) converges uniformly on [−R +, R −], no matter which > 0 is chosen. The function f is continuous and differentiable in (−R,R), and

∞ 0 X n−1 f (x) = ncnx (|x| < R). n=0 X Theorem 2.12. Suppose cn converges. Put

∞ X n f(x) = cnx (−1 < x < 1). n=0 Then ∞ X lim f(x) = cn. x→1 n=0 Theorem 2.13. suppose ∞ X n f(x) = cnx , n=0

21 the series converging in |x| < R. If −R < a < R, then f can be expanded in a power series about the point x = a which converges in |x − a| < R − |a|, and

∞ X f (n)(a) f(x) = (x − a)n (|x − a| < R − |a|). n! n=0

Note 2.7. This is known as T aylor0s T heorem.

2. Power-series expansions for the exponential and trigonometric functions