[564001] Calculus II
Yuki CHINO
Department of Applied Mathematics, National Yang Ming Chiao Tung University
March 7, 2023
Review: Convergence of series
Summary: How to check the convergence of series
1 Check Absolute convergence.
2 Consider Integral test & Comparison test (dominated
convergent theorem)
3 Check Ratio & Root test
Review: Convergence of series
Ratio/Root test
For the sequence an , Let
Ratio test
Ratio test
(1) If r < 1, then the series
(2) If r > 1, then the series
an+1
=: r
n→∞
an
p
lim n |an | =: r
lim
n→∞
∞
X
n=1
∞
X
an is absolute convergent.
an is divergent.
n=1
Review: Alternative series
Convergence of alternative series
Let an be a sequence satisfying that for all n ∈ N,
an ≥ an+1
and
lim an = 0,
n→∞
then the alternative series
∞
X
(−1)n an converges.
n=1
Contents and Goal
Today’s contents
• Power series and Taylor’s theorem
I Power series
I Taylor’s theorem
I Taylor series/expansion
Today’s Goal (Aim)
Be able to compute the radius of convergence
Be able to compute the Taylor approximation
Power series
Power series
∞
X
an xn = a0 + a1 x + a2 x2 + · · · + an−1 xn−1 + an xn + · · ·
n=0
More generally, the center of power series can be shifted
∞
X
an (x − a)n
n=0
= a0 + a1 (x − a) + a2 (x − a)2 + · · ·
+ an−1 (x − a)n−1 + an (x − a)n + · · · ,
which is called a power series centered at a.
Radius of convergence for power series
Radius of convergence
The threshold value distinguishing the series is convergent or divergent
is called radius of convergence.
Example 4.1.
Compute the radius of convergence of the following series
∞
X
xn
n=1
n!
.
Radius of convergence
Theorem 4.1. (d’Alembert test for the radius of convergence)
For the sequence an , Let
lim
n→∞
an+1
=: r
an
If r exists, then the radius of convergence for the power series
∞
X
1
an xn is , where
r
n=1
1
=∞
0
1
= 0.
∞
Radius of convergence
Theorem 4.2. (Cauchy-Hadamard test for the r.o.c.)
For the sequence an , Let
lim
n→∞
p
n
|an | =: r
If r exists, then the radius of convergence for the power series
∞
X
1
an xn is , where
r
n=1
1
=∞
0
1
= 0.
∞
Radius of convergence
Example 4.2.
Compute the radius of convergence of the following series:
(1).
∞
X
n2 xn
(2).
n=0
(3).
∞
X
(5).
n!
n=0
2
r n xn
(4).
n=0
∞
X
∞
X
(n + 1)n
(n!)2 2n
x
(2n)!
n=0
∞
X
xn
xn
2
n=0
(6).
∞
X
(3 + (−1)n )n xn
n=0
Radius of convergence
Question 4.3.
Let
A(x) =
∞
X
an xn ,
n=1
B(x) =
∞
X
bn x n
n=1
be power series with the radius of convergence r1 and r2 . We consider
the power series
∞
X
P (x) =
(an + bn ) xn .
n=1
1. If r1 < r2 , then prove that the radius of convergence of P (x) is
r1 .
2. If r1 = r2 , then the radius of convergence of P (x) is not r1 .
Show the counterexample.
Taylor’s theorem: Step 1
f (x) =
∞
X
an (x − a)n
n=0
= a0 + a1 (x − a) + a2 (x − a)2 + · · · + an (x − a)n + · · · .
I f 0 (x) = a1 + 2 · a2 (x − a) + 3 · a3 (x − a)2 + · · ·
I f 00 (x) = 2! · a2 + 3 · 2 · a3 (x − a) + 4 · 3 · a4 (x − a)2 + · · ·
I f 000 (x) = 3! · a3 + 4 · 3 · 2 · a4 (x − a) + · · ·
.
I ..
I f (n) (x) = n! · an + (n + 1) · · · 3 · 2 · an+1 (x − a) + · · ·
a0 = f (a), a1 = f 0 (a), a2 =
f 000 (a)
f (n) (a)
f 00 (a)
, a3 =
, · · · , an =
,···
2!
3!
n!
Taylor’s theorem: Step 1
Proposition 4.3.
If the function f (x) can be expressed by a power series centered at
a, then its coefficients are given by
an =
f (n) (a)
n!
n∈N
that is, we have the expression for the function
f (x) =
∞
X
f (n) (a)
n=0
n!
(x − a)n .
* In general, we don’t know whether the function can be expressed by a
power series or not. In this sense, the proposition is not so useful.
Taylor’s theorem: Step 2
The mean value theorem (Review)
Let f be continuous on a closed interval [a, b] and differentiable on
the open interval (a, b). Then, there exists a constant c ∈ (a, b) such
that
f (b) − f (a)
f 0 (c) =
.
b−a
•
b=a+h
and
c = a + θh
(h > 0, θ ∈ (0, 1))
The mean value theorem
Let f be continuous and differentiable on an interval I. Then, for
h > 0 and a ∈ I, there exists a constant θ ∈ (0, 1) such that
f (a + h) = f (a) + f 0 (a + θh) · h.
Taylor’s theorem: Step 2
1.
f (a + h) = f (a) + f 0 (a + θ1 h) · h
2.
f 0 (a + θ1 h) = f 0 (a) + f 00 (a + θ2 (θ1 h)) · (θ1 h)
⇒
3.
f (a + h) = f (a) + f 0 (a) · h + f 00 (a + θ1 θ2 h) · θ1 h2
f 00 (a + θ1 θ2 h) = f 00 (a) + f 000 (a + θ3 (θ1 θ2 h)) · (θ1 θ2 h)
⇒
f (a+h) = f (a)+f 0 (a)·h+f 00 (a)·θ1 h2 +f 000 (a+θ1 θ2 θ3 h)·θ12 θ2 h3
Repeat (n + 1) times
Taylor’s theorem: Step 2
x = a + h,
b1 = 1,
b2 = θ1 ,
b3 = θ12 θ2 , · · ·
f (x) = f (a) + f 0 (a)b1 (x − a) + f 00 (a)b2 (x − a)2 + f 000 (a)b3 (x − a)3
+ · · · + f (n) (a) · bn · (x − a)n + f (n+1) (a + θh) · bn+1 · (x − a)n+1 ,
We can apply Proposition 4.3. now !!
1
1
1
1
, b3 = , b4 = , · · · , bn = ,
2
3!
4!
n!
f (n+1) (a)
1
bn+1 = (n+1)
·
f
(a + θh) (n + 1)!
b1 = 1, b2 =
Taylor’s theorem
Taylor’s formula
Let n be a natural number and f be a function differentiable (n + 1)
times in its domain I. Then, for any x, a ∈ I, there exists a constant
θ ∈ (0, 1) such that
f (x) = f (a) +
f 00 (a)
f 0 (a)
· (x − a) +
· (x − a)2
1!
2!
+
f 000 (a)
· (x − a)3 + · · ·
3!
+
f (n+1) (ξ)
f (n) (a)
· (x − a)n +
· (x − a)n+1 ,
n!
(n + 1)!
where ξ = a + θ(x − a) ∈ (x, a) or (a, x).
Taylor’s theorem
f (x) =
n
X
f (k) (a)
k=0
k!
· (x − a)k +
f (n+1) (ξ)
· (x − a)n+1 ,
(n + 1)!
where f (0) (a) = f (a) and 0! = 1.
n
X
f (k) (a)
I P (x) :=
· (x − a)k : n-th order Taylor polynomial
k!
k=0
I Rn+1 :=
f (n+1) (ξ)
· (x − a)n+1 : Lagrange form of the remainder
(n + 1)!
I If lim Rn = 0,
n→∞
f (x) =
n
X
f (k) (a)
k=0
k!
* a = 0: Maclaurin series
· (x − a)k : Taylor series
Example
ex =
∞
X
xn
n=0
n!
=1+
x
x2
xn
+
+ ··· +
+ ··· .
1!
2!
n!
Exercises
Question 4.4.
Compute Maclaurin series for the following function with showing the
Lagrange form of remainder goes to zero.
(1) f (x) = sin x
(2) f (x) = cos x
Announcement
Summary
1 Computations for radius of convergence
2 Computation for range of convergence
3 Taylor/Maclaurin series expansion
I Next lecture: 3/10 (Fri.) 13:20-15:10, SC106.
I Office Hour: Thursday 16:00-17:30
Of course, you can make an appointment
(email: y.chino@math.nctu.edu.tw)