Power Series 1 1.1

JIA-MING (FRANK) LIOU

Abstract.

Contents

1. Power Series 1

1.1. Polynomials and Formal Power Series 1

1.2. Radius of Convergence 2

1.3. Derivative and Antiderivative of Power Series 4

1.4. Power Series Expansion defined by Geometric Series 6

1.5. Power Series Defined by Differential Equations 9

1.6. The Convergence of Taylor Series 10

1.7. Definition of e^z 12

1.8. More Eamples 12

1. Power Series

1.1. Polynomials and Formal Power Series. Letxbe a variable anda0,· · · , anbe real numbers. A polynomial in x with coefficientsa₀,· · ·, a_n is an expression

P(x) =a₀+a₁x+· · ·+a_nxⁿ,

Ifan6= 0, nis called the degree ofP. The space of polynomials in x with coefficients inR is denoted by R[x]. For each real number c∈R, defineP(c) =a₀+a₁c+· · ·+a_ncⁿ. Then P :R→Rdefines a function. Hence a polynomial inx can be thought of as a function on R. Evey polynomial function is a smooth function on R which means that P^(k)(x) exists for each k≥0. Moreover,P(0) =a0,

P⁰(x) = a₁+ 2a₂x+ 3a₃x²+· · ·+na_nxⁿ⁻¹

P⁰⁰(x) = 1·2a₂+ 2·3a₃x+ 3·4a₄x²· · ·+ (n−1)na_nxⁿ⁻²

· · · = · · · P⁽ⁿ⁾(x) = n!an.

Hence we haveP(0) =a0, P⁰(0) =a1, P⁰⁰(0) = 2!a2,· · · , P⁽ⁿ⁾(0) =n!an. Suppose that we have two polynomialsP(x) =Pn

i=0a_ixⁱ and Q(x) =Pm

j=0b_jx^j. Then P(x)Q(x) =

n

X

i=0 m

X

j=0

aibjx^i+j =

m+n

X

k=0 k

X

i=0

aibk−i

! x^k.

Hence the product of two polynomials is again a polynomial.

1

Let (an)n≥0 be a sequence of real numbers. The formal power series inxwith coefficients a_n is an expression

(1.1) a₀+a₁x+a₂x²+· · ·+a_nxⁿ+· · ·=

∞

X

n=0

a_nxⁿ.

The space of formal power series with coefficients inR is denoted by R[[x]]. Suppose that A(x) = P∞

n=0a_nxⁿ and B(x) = P∞

n=0b_nxⁿ are two formal power series in R[[x]]. Define their sum and their product by

A(x) +B(x) =

∞

X

n=0

(a_n+b_n)xⁿ A(x)B(x) =

∞

X

n=0

c_nxⁿ,

where c_n = Pn

k=0a_kbn−k. Hence we know that R[x] is a subset of R[[x]], i.e. every poly- nomial is a formal power series. Here comes a question: When does a formal power series define a function on an interval ofR?

1.2. Radius of Convergence. The following theorem gives us an example when a formal power series converges.

Theorem 1.1. (Abel) Suppose that r is a real number so that the series

∞

X

n=0

anrⁿ is con-

vergent. For each |c| < |r|,

∞

X

n=0

ancⁿ is absolutely convergent. Conversely if

∞

X

n=0

anrⁿ is

divergent, then for|c|>|r|,

∞

X

n=0

ancⁿ is also divergent.

Proof. Since the seriesP∞

n=0a_nrⁿ is convergent, lim

n→∞a_nrⁿ= 0. Hence there exists N >0 so that for anyn≥N,|a_nrⁿ|<1. Thus |a_n|< r⁻ⁿfor n≥N. Therefore for any n≥N,

|a_ncⁿ|<

c r

n

. Since|c/r|<1, the geometric seriesP∞

n=N+1(c/r)ⁿ is convergent. By the comparison test, P∞

n=N+1|a_ncⁿ|is convergent and thusP∞

n=0a_ncⁿ is convergent.

Definition 1.1. R >0 is called the radius convergenceof (1.1) if (1) for any|c|< R,P∞

n=0a_ncⁿ is absolutely convergent;

(2) for any|c|> R,P∞

n=0ancⁿ is divergent.

Theorem 1.2. Suppose that lim

n→∞

pn

|a_n|=ρ. Then the radius of convergence of (1.1) is R=ρ⁻¹. Similarly, if lim

n→∞|a_n+1

a_n |=ρ, thenR=ρ⁻¹.

Proof. Letρ be the limit of (a_n+1/a_n).Let a_n(x) =a_nxⁿ.Then a_n+1(x)

a_n(x) = a_n+1 a_n x.

Hence

n→∞lim

a_n+1(x) a_n(x)

=ρ|x|.

We know that ifρ|x|<1,by ratio test, the infinite series

∞

X

n=0

an(x) =

∞

X

n=0

anxⁿis absolutely convergent. In other words, the power series is convergent if |x|< 1/ρ. Take R = 1/ρ. If

|x|> R, then ρ|x|> 1. By ratio test, the power series is divergent. Hence we prove that R= 1/ρis the radius of convergence of the power series.

Let us take a look at some examples.

Example 1.1. Find the radius of convergence of the following three formal power series (1) A₁(x) =

∞

X

n=1

(−1)ⁿxⁿ n ; (2) A₂(x) =

∞

X

n=1

xⁿ n²; (3) A₃(x) =

∞

X

n=1

nxⁿ.

Sol. By ratio test, the radii of convergence of the above three formal power series are 1.

Note that P

n=1(−1)ⁿ/n is convergent by the Leibnitz test. A(x) is also convergent at x = 1 but when x = −1, the series P

n=11/n is divergent by the p-test. Hence A1(x) is convergent on −1 < x ≤ 1 or I₁ = (−1,1]. Similarly, A₂(x) is convergent on both x=±1. Therefore A2(x) is convergent onI2 = [−1,1]. We also find thatA3 is convergent on I₃ = (−1,1) and divergent atx=±1. I₁, I₂, I₃ are called the interval of convergence of A₁, A₂, A₃ respectively.

Definition 1.2. An intervalIonRis of the form [a, b] or [a, b) or (a, b] or (a, b). An interval I is called the interval of convergence if for every c∈ I,P∞

n=0ancⁿ is convergent and for any d6∈I,P∞

n=0a_ndⁿ is divergent.

Corollary 1.1. I can only be [−R, R], or (−R, R], or [−R, R) or (−R, R).

IfI is the interval of convergence of the formal power seriesA(x) =P∞

n=0anxⁿ, then for each c∈I,P∞

n=0a_ncⁿ is convergent. For each c∈I, define A(c) =

∞

X

n=0

a_ncⁿ

Then the formal power seriesA(x) defines a function onI.

Example 1.2. The radius of convergence of

∞

X

n=0

xⁿ n!.

Sol. Since an= _n!¹,an+1/an = 1/(n+ 1) for all n. Hence limn→∞(an+1/an) = 0. Thus the radius of convergence is R=∞.

Remark. Later we will show that this power series converges to e^x.

Theorem 1.3. (Multiplication Theorem) Suppose that two power series A(x), B(x) are convergent on |x|< R. Then A(x)±B(x), A(x)B(x) are all convergent on|x|< R.

1.3. Derivative and Antiderivative of Power Series. Suppose thatp(x) =a0+a1x+

· · ·+anxⁿ=Pn

k=0akx^k be a polynomial. Then

p⁰(x) =a₁+ 2a₂x+· · ·+na_nxⁿ⁻¹ =

n

X

k=0

ka_kx^k−1

Given a power series A(x) =P∞

n=0anxⁿ, one might think that if A is differentiable, then A⁰(x) =

∞

X

n=0

na_nxⁿ⁻¹.

On the other hand, an antiderivative of p is given by Z

p(x)dx=C+a0x+a1

2 x²+· · ·+ an

n+ 1aⁿ⁺¹=C=

n

X

k=0

ak

k+ 1x^k+1. Hence one might guess

Z

A(x)dx=c+

∞

X

n=0

an

n+ 1xⁿ⁺¹.

Theorem 1.4. Suppose that the radius of convergence of (1.1) isR. Then the power series of

B(x) =

∞

X

n=0

na_nxⁿ⁻¹, C(x) =c+

∞

X

n=0

a_n n+ 1xⁿ⁺¹ are alsoR.

Suppose the radius of convergence of (1.1) is R. Suppose that |x| < R and |x₀| < R.

Then

A(x)−A(x0) =

∞

X

n=0

an(xⁿ−xⁿ₀).

In order to calculate the difference quotient ofA, we need:

Lemma 1.1. Let x, ybe real numbers. Then

xⁿ−yⁿ= (x−y)(xⁿ⁻¹+xⁿ⁻²y+· · ·+xyⁿ⁻²+yⁿ).

Hence the difference quotient ofA atx0 is A(x)−A(x₀)

(x−x0) =

∞

X

n=0

a_n(xⁿ⁻¹+xⁿ⁻²x₀+· · ·+xxⁿ⁻²₀ +xⁿ⁻¹₀ ).

We know that

x→xlim₀(xⁿ⁻¹+xⁿ⁻²x₀+· · ·+xxⁿ⁻²₀ +xⁿ⁻¹₀ ) =nxⁿ⁻¹₀ .

Hence one finds that (not so rigorous)A⁰(x₀) =B(x₀). ThusA⁰(x) =B(x) for any|x|< R.

One could also see that C⁰(x) =A(x). Hence C(x) =c+R

A(x)dx.

Theorem 1.5. Suppose that the radius of convergence of (1.1) is R. Then for all|x|< R, A⁰(x) =B(x),

Z

A(x)dx=C(x).

We have already see that the radius of convergence of f(x) = P∞

n=0xⁿ/n! is R = ∞.

Hence the function f(x) is smooth on R. Then by the differentiation theorem for power series, we find that

f⁰(x)−f(x) = 0, f(0) = 1.

One can show that the differential equation has a unique solution and the solution is defined to bee^x. Hencef(x) =e^x for all x∈R, i.e.

e^x=

∞

X

n=0

xⁿ n!.

Corollary 1.2. Let A(x) =

∞

X

n=0

anxⁿ be a convergent power series on |x|< R. Then A is k-times differentiable for allk atx= 0 anda_n= A⁽ⁿ⁾(0)

n! for all n≥0. As a consequence, A(x) =

∞

X

n=0

A⁽ⁿ⁾(0)

n! xⁿ, |x|< R.

Theorem 1.6. (Taylor’s Theorem) Suppose that A(x) =

∞

X

n=0

anxⁿ converges on |x|< R.

Let −R < a < R. Then A is k-times differentiable at x =aand for |x−a|< R− |a|, we have

A(x) =

∞

X

n=0

A⁽ⁿ⁾(a)

n! (x−a)ⁿ.

By the theorem, we find thatA is k-times differentiable on|x|< R, i.e. A is a smooth function on|x|< R. A functionf : (a, b)→Ris called smooth iff isk-times differentiable at every point of (a, b) for allk≥0. A convergent power series on|x|< Rdefines a smooth function on |x| < R by the Taylor’s theorem. Conversely, given a smooth function f on

|x|< R, can we find a power series which is convergent tof on |x|< R?

Definition 1.3. Letf be a C^∞-function on |x|< R. The formal power series defined by

∞

X

n=0

f⁽ⁿ⁾(0) n! xⁿ

is called the Maclaurin series generated byf atx= 0. In general, the Taylor series generated by f atx=ais the following formal power series

(1.2)

∞

X

n=0

f⁽ⁿ⁾(a)

n! (x−a)ⁿ.

If the Taylor series (1.2) converges to f, we say that the Taylor series (1.2) is the Taylor expansion of f.

The Taylor series generated by f may not be convergent to the original function. For example,

Example 1.3. Let f(x) be the following function onR: f(x) =

e^−1/x2, forx6= 0;

0, x= 0.

Then its Taylor series is 0 but f(x) is not a zero function.

Remark. Note that if the function f is defined by a convergent power series, the Taylor series generated byf is convergent to f by Taylor’s theorem.

1.3.1. Uniqueness of Power Series.

Theorem 1.7. If|x|< R, the power series A(x) =

∞

X

n=0

anxⁿ= 0.

Then an= 0 for all n≥0.

Proof. SinceA(x) on 0 on|x|< R, thenA^(k)(x) = 0 on|x|< R. Hencean=A^(k)(0)/n! = 0

for all n.

Corollary 1.3. (Uniqueness of Power Series) Suppose that A(x) =

∞

X

n=0

anxⁿ and B(x) =

∞

X

n=0

b_nxⁿ are both convergent on|x|< R. Then A(x) =B(x) if and only if a_n=b_n for all n≥0.

Proof. For each|x|< R,P∞

n=0a_nxⁿ andP∞

n=0b_nxⁿ are both convergent. Hence

∞

X

n=0

(a_n−b_n)xⁿ=

∞

X

n=0

a_nxⁿ−

∞

X

n=0

b_nxⁿ= 0.

By the previous theorem,a_n−b_n= 0 for all n≥0, i.e. a_n=b_n for all n≥0.

Now, we have already shown that the convergence power series on |x| < R defines a function on |x|< R. We might ask ourself the following questions. To which function does the power series converge?

1.4. Power Series Expansion defined by Geometric Series.

Example 1.4. Find a function f(x) so that (1) f(x) =

∞

X

n=0

xⁿ.

(2) g(x) =

∞

X

n=1

nxⁿ⁻¹. (3) h(x) =

∞

X

n=1

xⁿ⁺¹ n+ 1.

Sol. The radius of convergence of these three functions are 1. These three functions are all convergent on |x|<1. The power series f is a geometric series andf(x) = 1/(1−x). One also finds thatg(x) =f⁰(x) = 1/(1−x)² for|x|<1. Moreover,h⁰(x) =f(x) andh(0) = 0.

Hence h(x) =−ln(1−x) for |x|<1.

Example 1.5. Find a function f(x) so that (1) f(x) =

∞

X

n=1

nxⁿ.

(2) f(x) =

∞

X

n=1

xⁿ n . (3) f(x) =

∞

X

n=1

xⁿ n².

Sol. For x 6= 0, f(x)/x = P∞

n=1nxⁿ⁻¹. Let g(x) = P∞

n=0xⁿ. Then g⁰(x) = f(x)/x. We know thatg(x) = 1/(1−x). Henceg⁰(x) = 1/(1−x)². Thusf(x) =x/(1−x)² on |x|<1.

1.4.1. Taylor Expansion of ln(1 +x). Letf(x) = ln(1 +x),x >0. Thenf⁰(x) = 1/(1 +x).

For|x|<1, we know that

(1.3) 1

1 +x = 1−x+x²−x³+· · ·+ (−1)ⁿxⁿ+· · ·=

∞

X

n=0

(−1)ⁿxⁿ. By the fundamental theorem of calculus, for anyx >−1,

ln(1 +x) = Z x

0

dt 1 +t. For|x|<1, we can integrate (1.3) term by term and obtain

ln(1 +x) =

∞

X

n=0

(−1)ⁿ Z x

0

tⁿdt=

∞

X

n=0

(−1)ⁿxⁿ⁺¹ n+ 1. We can compute the following series by the Taylor expansion of f(x) = lnx.

(1.4)

∞

X

n=0

(−1)ⁿ

n+ 1 = 1−1 2 +1

3 −1

4 +· · ·+(−1)ⁿ

n+ 1 +· · ·. By (1.3), for|x|<1,

1 1 +x =

n

X

k=0

(−1)ⁿxⁿ+(−1)ⁿxⁿ⁺¹ 1 +x . Hence for |x|<1,

ln(1 +x) =

n

X

k=0

(−1)ⁿxⁿ⁺¹ n+ 1+

Z x 0

(−1)ⁿtⁿ⁺¹ 1 +tdt.

For eachn∈N, the partial sum of (1.4) is sn= 1−1

2 +1 3− 1

4+· · ·+(−1)ⁿ

n =

n

X

k=0

(−1)^k k+ 1. Then we obtain the following identity:

ln 2 =sn+ Z 1

0

(−1)ⁿtⁿ⁺¹ 1 +tdt.

This shows that

|s_n−ln 2| ≤ Z 1

0

tⁿ⁺¹dt= 1 n+ 2. We conclude that lim

n→∞sn= ln 2.

1.4.2. Taylor Expansion of tan⁻¹x. Let f(x) = tan⁻¹x,x ∈ R. Then f⁰(x) = 1/(1 +x²) for all x∈R. For|x|<1, f⁰(x) has the following series expansion

(1.5) 1

1 +x² =

∞

X

n=0

(−1)ⁿx²ⁿ. Integrating (1.5), we obtain

tan⁻¹x=

∞

X

n=0

(−1)ⁿx²ⁿ⁺¹ 2n+ 1. Similarly, one can show that

π

4 = 1−1 3+ 1

5−1

7 +· · ·+ (−1)ⁿ

2n+ 1+· · ·=

∞

X

n=0

(−1)ⁿ 1 2n+ 1. We leave it to the reader as an exercise.

Theorem 1.8. (Abel Theorem) Suppose that

∞

X

n=0

a_nis convergent. Then we have already

know thatf(x) =

∞

X

n=0

anxⁿ converges uniformly on |x|<1. Then

x→1−lim f(x) =

∞

X

n=0

an.

Proof. Letsnbe the n-th partial sum of (an). Then an=sn−sn−1. Hence

n

X

k=0

(s_k−sk−1)x^k = (1−x)

n−1

X

k=0

s_kx^k+s_nxⁿ. Hence for |x|<1, we obtain

f(x) = (1−x)X

n=0

s_nxⁿ.

Denotes=P

n→∞s_n. Then for >0, there existsN >0 so that forn≥N,|s_n−s|< /2.

By P∞

n=0xⁿ= 1/(1−x) for |x|<1, we find

|f(x)−s| ≤(1−x)

N

X

n=0

|s_n−s||x|ⁿ+ 2.

Since (s_n) is convergent, so is (s_n−s). Thus there exists M >0 so that |s_n−s| ≤ M for all n. Therefore

|f(x)−s| ≤(1−x)N M+ 2.

For the same > 0, choose δ = /(2M N) > 0. Then whenever |x−1| < δ, we have

|1−x|M N < /2. In this case we find that for |1−x|< δ,|f(x)−s|< . Thus we prove

that limx→1−f(x) =s

1.5. Power Series Defined by Differential Equations.

1.5.1. Taylor Expansion of sinx and cosx. Let us consider the following second order dif- ferential equation

(1.6) y⁰⁰+y= 0.

Suppose that y=P∞

n=0anxⁿ is a solution. Then we find the following recursive relation a_n+2 =− 1

(n+ 1)(n+ 2)a_n, n≥0.

Hence one finds that

a_2n= (−1)ⁿ

(2n)!a₀, a_2n+1 = (−1)ⁿ (2n+ 1)!a₁. Then y is formally given by

y=a₀

∞

X

n=0

(−1)ⁿ

(2n)!x²ⁿ+a₁

∞

X

n=0

(−1)ⁿ

(2n+ 1)!x²ⁿ⁺¹. LetC(x) and S(x) be the following formal power series

C(x) =

∞

X

n=0

(−1)ⁿ

(2n)!x²ⁿ, S(x) =

∞

X

n=0

(−1)ⁿ

(2n+ 1)!x²ⁿ⁺¹. Then y=a₀C(x) +a₁S(x).

Proposition 1.1. The radii of convergence of C and S are both ∞ and they are both solution to (1.6). Moreover, C(0) = 1, C⁰(0) = 0 andS(0) = 0 andS⁰(0) = 1.

One can also check that c(x) = cosx and s(x) are both solutions to (1.6); c(0) = 1, c⁰(0) = 0 and s(0) = 0, s⁰(0) = 1. Now, we can show thatC(x) =c(x) andS(x) =s(x). To show thatC(x) =c(x) is equivalent to show thatC(x)−c(x) = 0. Letg(x) =C(x)−c(x).

Then g(0) = 0 and g⁰(0) = 0 and g also satisfies (1.6). Define h(x) = (g(x))²+ (g⁰(x))². Then for allx,

h⁰(x) = 2g(x)g⁰(x) + 2g⁰(x)g⁰⁰(x) = 2g⁰(x)

g(x) +g⁰⁰(x)

= 0.

Hence h is a constant function by the mean value theorem. Thus h(x) = 0 for all x ∈ R which shows thatg(x) = 0 for all x∈R. ThereforeC(x) =c(x) for allx∈R.

1.5.2. Taylor Expansion of (1 +x)^α. Letf(x) = (1 +x)^α. Then

(1.7) (1 +x)f⁰(x) =αf(x),

f(0) = 1. Assume that A(x) = P∞

n=0a_nxⁿ is a solution to (1.7). By (1.7), one can show that

a_n+1= α−n

n+ 1a_n, n≥1 and a₁ =αa₀. One can check that

a₂= α(α−1)

2 a₀, a₃ = α(α−1)(α−2)

3! a₀,· · ·a_n= α(α−1)· · ·(α−n+ 1) n! a₀· · ·. Since f(0) = 1, a₀ = 1. One can show that the radius of convergence of the formal power series is 1. Hence the power series is convergent on |x|<1. We also denote

α n

= α(α−1)· · ·(α−n+ 1)

n! .

Since the power series is convergent on |x| < 1, the power series solve (1.7) on |x| < 1.

Hence by the differentiation theorem for power series, A(x) =

∞

X

n=0

α n

xⁿ

satisfies (1.7). Now, we want to show that f(x) = A(x) for |x| < 1. Define h(x) = f(x)−A(x). Thenh(0) = 0 andhalso satisfies (1.7). Multiplying (1.7) by (1 +x)^−α−1, we find

(1 +x)^−αh⁰(x)−α(1 +x)^−α−1h(x) = d

dx(1 +x)^−αh(x) = 0.

Hence (1 +x)^−αh(x) = C for all x ∈R. Thus h(x) =C(1 +x)^α, x ∈R. Since h(0) = 0, C= 0. We find that h(x) = 0 for |x|<1. We conclude that

f(x) =

∞

X

n=0

α n

xⁿ.

We can also compute the Taylor series generated by f at x = 0 by direct computing the derivatives:

f⁽ⁿ⁾(x) =α(α−1)· · ·(α−n+ 1)(1 +x)^α−n, n≥0.

Hencef⁽ⁿ⁾(0) =α(α−1)· · ·(α−n+ 1). Thus the Taylor series generated by f atx= 0 is given by

∞

X

n=0

α(α−1)· · ·(α−n+ 1)

n! xⁿ

and its radius of convergence is 1. How do we know that the Taylor series generated by f atx = 0 is convergent to f? By using differential equations, we can find a power series which converges tof and thus the power series is its Taylor series by the Taylor’s theorem.

Although the computation of the Taylor series generated byf is simpler, we didn’t know if the Taylor series is convergent tof. Therefore, we need to study the convergence of Taylor series.

1.6. The Convergence of Taylor Series. Suppose thatf is a differentiable function on an interval I. Then for each a < x, the slope of the secant line is given by

s= f(x)−f(a) x−a .

The mean value theorem states that we can find a pointc lies betweenx andy so that the slope of the tangent line to x=cis s, i.e.

f(x)−f(a)

x−a =f⁰(c).

Thusf(x)−f(a) =f⁰(c)(x−a), i.e. we have

f(x) =f(a) +f⁰(c)(x−a).

Suppose that f is n+ 1-times differentiable. Similarly, for a < x, we can always find a < c < xso that

(1.8) f(x) =f(a) +f⁰(a)

1! (x−a) +· · ·+f⁽ⁿ⁾(a)

n! (x−a)ⁿ+f⁽ⁿ⁺¹⁾(c)

(n+ 1)! (x−a)ⁿ⁺¹.

Definition 1.4. The polynomialPn(x) defined by Pn(x) =

n

X

k=0

f^(k)(a)

k! (x−a)^k

is called the n-th Taylor polynomial of f at x = a. The last term in (1.8) is called the remainder off atx=a and denoted byR_n(x, a).

Theorem 1.9. (Taylor’s Theorem) Letf be aC^∞-function onR. Let a < xbe points in R. Then there existsa < c < x Then (1.8) holds.

Proof. By the fundamental theorem of calculus, we find that f(x) =f(a) +

Z x a

f⁰(t)dt.

Using integration by parts,

f(x) =f(a) +f⁰(a)(x−a)− Z x

a

(t−x)f⁰⁰(t)dt.

Inductively,

f(x) =

n

X

k=0

f^(k)(a)

k! (x−a)^k+ Z x

a

(x−t)ⁿ

n! f⁽ⁿ⁺¹⁾(t)dt.

By the mean value theorem, show that there existsa≤c_x,a≤xso that f(x) =

n

X

k=0

f^(k)(a)

k! (x−a)^k+f⁽ⁿ⁺¹⁾(cx,a)

(n+ 1)! (x−a)ⁿ⁺¹. Thereforef(x) =P_n(x) +R_n(x, a), where

Rn(x, a) =f⁽ⁿ⁺¹⁾(cx,a)

(n+ 1)! (x−a)ⁿ⁺¹

Theorem 1.10. Suppose that |f⁽ⁿ⁺¹(t)| ≤ M for all a ≤t ≤ x for all n ∈ N. Then the Taylor series generated by f is convergent tof for|x−a|< R. for R >0.

Proof. LetPn(x) be then-th Taylor polynomial of f atx=a. By the Taylor’s theorem,

|P_n(x)−f(x)|=|R_n(x, a)| ≤ M Rⁿ⁺¹ (n+ 1)!. Since limn→∞M Rⁿ⁺¹/(n+ 1)! = 0, by the Sandwich principle,

n→∞lim Pn(x) =f(x).

Hence the series

∞

X

n=0

f⁽ⁿ⁾(a)

n! (x−a)ⁿ is convergent tof(x) for |x−a|< R.

Example 1.6. Calculate the Taylor series of C(x) = cosx and S(x) = sinx at x = 0 directly from the definition and show that their Taylor series converge toC andS.

Sol. We leave it to the reader as an exercise.

1.7. Definition of e^z. For a complex number z∈C, define e^z=

∞

X

n=0

zⁿ n!.

Theorem 1.11. Given a real number θ, letz=iθ. Then e^iθ = cosθ+isinθ, θ∈R.

Proof. Forn= 2k,iⁿ= (−1)^k and for n= 2k+ 1,iⁿ= (−1)^ki. Hence e^iθ =

∞

X

n=0

iⁿθⁿ n! =

∞

X

k=0

(−1)^k

(2k)!θ^2k+i

∞

X

k=0

(−1)^k (2k+ 1)!θ^2k+1

= cosθ+isinθ.

Theorem 1.12. For real numbersθ,ϕ, we have

(1) e^i(θ+ϕ) =e^iθe^iϕ;

(2) forn∈Z,e^inθ = (e^iθ)ⁿ. 1.8. More Eamples.

Example 1.7. Find the Taylor expansion of the following functions (1) f₁(x) = 1

x atx= 1 and x= 2.

(2) f2(x) = 1

x² atx= 1.

(3) f₃(x) = 1

x³ atx= 1.

(4) f4(x) = x

1−x atx= 0.

Sol. Since x=x−1 + 1, f1(x) = 1

1 + (x−1) =

∞

X

n=0

(−1)ⁿ(x−1)ⁿ=

∞

X

n=0

(−1)ⁿn!

n! (x−1)ⁿ, |x−1|<1.

We also find thatf⁽ⁿ⁾(0) = (−1)ⁿn!. Sincex=x−2 + 2, f1(x) = 1

2 + (x−2) = 1 2

1 1 + (x−2)/2

= 1 2

∞

X

n=0

(−1)ⁿ

x−2 2

n

=

∞

X

n=0

(−1)ⁿ

2ⁿ⁺¹ (x−2)ⁿ, |x−2|< 1 2.

We also find thatf⁽ⁿ⁾(2) = (−1)ⁿn!/2ⁿ⁺¹. Note thatf₁⁰(x) =−f₂(x) andf₂⁰(x) =−2f₃(x).

As long as you find f₂ by taking the derivative off₁, you can findf₂ and thus f₃. For the last problem, consider

1 1−x =

∞

X

n=0

xⁿ, |x|<1.

Then by the multiplication theorem for power series, on|x|<1, f₄(x) =x· 1

1−x =

∞

X

n=0

xⁿ⁺¹.

Example 1.8. Find the Taylor expansion of the following functions:

(1) f1(x) = cos 2x atx= 0.

(2) f₂(x) =x²sinx atx= 0.

(3) f3(x) =x²cosx² atx= 0.

(4) f4(x) =e^x atx= 1.

(5) f₅(x) =xe^x atx= 0.

Sol. Since cost=

∞

X

n=0

(−1)ⁿ t²ⁿ

(2n)!, cos 2x =

∞

X

n=0

(−1)ⁿ2²ⁿ x²ⁿ

(2n)!. Similarly, one can find the Taylor expansion off₂, f₃, f₅ atx= 0. Sincee^x=ee^x−1, we know that

f4(x) =e^x =e·

∞

X

n=0

(x−1)ⁿ

n! =

∞

X

n=0

e

n!(x−1)ⁿ.

Department of Mathematics, University of California, Davis, CA 95616, U.S.A.

E-mail address:[email protected]