Mathematical methods in the physical sciences 3rd edition Mary L. Boas
Chapter 1 Infinite series, Power series
(무한급수, 멱급수)
Lecture 1 Infinite series, convergence
1
1. Geometric series (기하급수)
- Geometric sequence (기하수열)
a, ar, ar2 , ar3 ,, arn 1 ,
2,4,8,16,,22 
n 1
 2 n ,, 
2 4 8
2
1, , , ,,1   
3 9 27
3
n 1
2
 
3
r  2 1
n 1
,  ,0
r
2
1
3
- Series (급수): an indicated sum of a given sequence.
cf. Infinite series, Geometric series (무한급수, 기하급수)
2  4  8  16    22
n 1
2 4 8
2
1  
  
3 9 27
3

n 1

2
- Summation of the geometric series, S_n
Geometric sequence : an  arn 1
Sn  a  ar  ar2  ar3    arn 1
rS n 
ar  ar2  ar3    arn 1  arn
Sn  rS n  a  arn

a 1 rn
 (1  r ) Sn  a 1  r  Sn 
1 r

n


- Summation of the infinite geometric series
For r  1,
lim Sn  S 
n 
a
1 r
 lim r  0
n
n 
"convergent"
For r  1, "divergent or oscillating"
3
Example) Traveling distance of bouncing ball
v1  evo
v2  e2vo
v3  e3vo
vn  envo
v4  e4vo
vo


vn  evn 1 , e 2  1
2
2
vn
e 2 vn 1
1
2
m ghn  m vn  hn 

 e 2 hn 1
2
2g
2g

htotal
v02
(ev0 ) 2 1
v02 e 2 v02 1
 h0  2 hn 
2


2
2g
2g 1 e
2g
g 1  e2
n 1
4
3. Application of series (급수의 응용)
- It is possible for the sum of an infinite series to be nearly the same as the sum
of a fairly small number of terms at the beginning of the series. Many applied
problems can not be solved exactly, but we may be able to find an answer in
terms of an infinite series, and then use only as many terms as necessary to obtain
the needed accuracy.
x 2 x3
ex. e  1  x      e x  1  x for x  1
2! 3!
x
- There is more to the subject of infinite series than making approximations.
We will see how we can use power series (that is, series whose terms are
powers of x) to give meaning to functions of complex numbers.
- Other infinite series: Fourier series (sines and cosines), Legendre series,
Bessel series, and so on.
5
4. Convergent and divergent series (수렴급수, 발산급수)
lim S n  S (finite number).
n 
a. If the partial sum S_n of an infinite series tend to a limit S, the series is
called convergent. Otherwise, it is called divergent.
b. The limiting value S is called the sum of the series.
c. The difference R_n=S-S_n is called the remainder (or the remainder after n
terms).
lim Rn  lim S  S n   S  S  0, if the series is convergent .
n
n
cf. Convergent and Divergent series (Summation)
a  ar  ar2  ar3    arn 1  arn  
to get summation when converged
to check if it converges or diverges
7
5. Testing series for convergence; the preliminary test
(수렴에 대한 검사; 예비검사)
- Preliminary test (divergence condition)
a. If the terms of an infinite series do not tend to zero, the series diverges.
b. If they do, we must test further.
lim an  0,
n 
ex.

a
n
diverges.
1 2 3 4
   
2 3 4 5
8
6. Convergence tests for series of positive terms; absolute convergence
(양의 항으로 이루어진 급수에 대한 수렴검사 ; 절대수렴)
If some of the terms of a series are negative, we may still want to consider the
related series which we get by making all the terms positive; that is, we may
consider the series whose terms are the absolute values of the terms of our
original series.
If the new series converges, we call the original series absolutely convergent. It
can be proved that if a series converges absolutely, then, it converges.
9
A. The comparison test (비교검사)
1) smaller than the convergent series
m1  m2  m3   : convergent
a1  a2  a3   should be absolutelyconvergent, if an  mn .
2) larger than the divergent series
d1  d 2  d3   : divergent
a1  a2  a3   should diverges, if an  d n .
10
B. The integral test (적분검사)
0  an 1  an
a. If
b. If

an dn is finite,  an converges.

an dn is infinite,  an diverges.



Example.

1 1 1
1     .
2 3 4


1

dn  ln n  .
n
11
C. The ratio test (비율검사)
: in cases that we cannot evaluate the integral.
an 1
n 
,
an
  lim  n .
n 
A a series with   1 converges.
A a series with   1 diverges.
If   1, the ratio test does not tellus anything.
Example
1
1 1
1
  
2! 3!
n!
n 
1
1
1
 
,
n  1! n! n  1
1
0
n  n  1
  lim n  lim
n 
So, the above series should be convergent.
12
D. A special comparison test (may skip this.)
(특별 비교검사)
 n 1 bn : convergent series of positive terms, an  0,

and an / bn  a finite limit
 n 1 an converges.

 n 1 d n : divergentseries of positive terms, an  0,

and an / bn  a limit greater than 0 (or )
 n 1 an diverges.

13
7. Alternating series (교대급수)
- Alternating series: a series whose terms are alternatively plus and minus.
- Test for alternating series: An alternating series converges if the absolute value
of the terms decreases steadily to zero.
an1  an , limn an  0
 1
1 1 1
1   
2 3 4
n
1
1
1

 , lim  0
n  1 n n  n
n 1
Example.
  : converges,
14
8. Conditionally convergent series (제한적인 수렴급수)
- A series converges, but does not converge absolutely. In this case, it is a
conditionally convergent series.
9. Useful facts about series (급수에 대한 유용한 사실)
1. The convergence or divergence of a series is not affected by multiplying
every term of the series by the same nonzero constant. Neither is it affected by
changing a finite number of terms (for example, omitting the first few terms.)
2. Two convergent series may be added (or substracted) term by term. The
resulting series is convergent, and its sum is obtained by adding the sums of the
two given series.
3. The terms of an absolutely convergent series may be rearranged in any order
without affecting either the convergence or the sum.
15
Mathematical methods in the physical sciences 3rd edition Mary L. Boas
Chapter 1 Infinite series, Power series
Lecture 2 Power series (Taylor expansion)
10. Power series (멱급수)

n
2
3
a
x

a

a
x

a
x

a
x

 n
0
1
2
3
n0

or
 an x  a   a0  a1 x  a x  a2 x  a   a3 x  a   
n
n 0

S  x    an x n
n 0
for convergent an x n
cf. interval of convergence (수렴 구간)
2
3
11. Theorems about power series (멱급수에 대한 정리)
Theorem 1. A power series may be differentiated or integrated term by term; the
resulting series converges to the derivative or integral of the function represented
by the original series within the same interval of convergence as the original series.
Theorem 2. Two power series may be added, subtracted, or multiplied; the
resolution series converges at least in the common interval of convergence.
Theorem 3. One series may be substituted in another provided that the values of
the substituted series are in the interval of convergence of the other series.
Theorem 4. The power series of a function is unique, that is, there is just one
power series which converges to a given function.
12. Expanding function in Power series (함수의 멱급수 전개)
(using the differentiation…)
First set like this.
sin x  a0  a1 x  a2 x 2  a3 x 3  a4 x 4    an x n  
i) At x  0, sin x  0  a0  0
ii) First derivative (첫번째 미분계수)
cos x  a1  2a2 x  3a3 x 2  4a4 x 3  , at x  0,
a1  1
iii) Second derivative (두번째 미분계수)
 sin x  2a2  2  3a3 x  3  4a4 x 2  , at x  0,
a2 
0
0
1 2
iv) T hird derivative (세번째 미분계수)
 cos x  2  3a3  2  3  4a4 x  , at x  0,
a3 
1
1

1 2  3 3!
In thisway, we can determinethecoefficients of thepower series.
x3 x5 x7
sin x  x     
3! 5! 7!
x2 x4 x6
cf. cos x  1     
2! 4! 6!
(Maclaurin series : T alyor series about the origin)
- General Talyor series for f(x)
f x   a0  a1 x  a   a2 x  a   a3 x  a   a4 x  a     an x  a   
2
3
4
f x   a1  2a2 x  a   3a3 x  a     nan x  a 
n 1
2
f x   2a2  2  3a3 x  a     n  1nx  a 
n2
f 3 x   2  3a3    n  2n  1nx  a 
n 3
n



f n  x   1  2  3n  2n  1nan  terms containing powers of x  a 
For x  a,
f a   a0 , f a   a1 , f a   2a2  2!a2 , f 3 a   2  3a3  3!a3 , f n  a   n!an
 f x   f a 
 x  a  f a  
For x  0,
1
1
1
2
3 3 


x  a  f a   x  a  f a     x  a n f n  a   
2!
3!
n!
x2
x 3 3 
x n n 
f x   f 0  xf 0 
f 0 
f 0   
f 0  
2!
3!
n!
13. Techniques for obtaining power series expansions
(멱급수 전개를 얻는 방법)
There are often simpler ways for finding the power series of a function than the
successive differentiation process in Section 12. Theorem 4 in Section 11 tells
us that for a given function there is just one power series. Therefore we can
obtain it by any correct method and be sure that it is the same Maclaurin series
we would get by using the method of Section 12. We shall illustrate a variety of
methods for obtaining power series.
First, please memorize these basic series for your timesaving.

sin x  
n0

cos x  
 1n x 2 n 1  x  x3  x5  x 7  
(2n  1)!
3!
5!
7!
 1n x 2 n
x2 x4 x6
1 
 
(2n)!
2! 4! 6!
n 0

xn
x 2 x3 x 4
e   1 x 
 

2! 3! 4!
n  0 n!
x

 1n 1 x n
n0
n
ln1  x   
1  x 
p
x 2 x3 x 4
 x    
2
3
4
1  x  1
 p n
p p  1 2 p p  1 p  2 3
    x  1  px 
x 
x 
2!
3!
n 0  n 

x 1
A. Multiplication (곱하기)
Ex.1


x3 x5
x  1sin x  x  1 x    
3! 5!


3
4
x
x
 x  x2    
3! 3!
Ex. 2



x 2 x3 x 4
x2 x4
e cos x  1  x 
 
 1 

 
2! 3! 4!
2! 4!



x
x3 x 4
1 x  

3
6
 
  Ree  
cf. eix  cos x  i sin x, Re eix  cos x

e x cos x  Re e x eix
1 i x
B. Division (나누기)
1
Ex.1 ln1  x  
x
n 1

 1   1n 1 x n

1
x 2 x3 x 4
 1 x n 1
 x       

x
2
3
4
x
n
n
n0
n0

x x 2 x3
1 
 
2 3
4
Ex.2
1
1 x
1  x  x 2  x3  
1 x 1
1 x
x
 x  x2
x2
x 2  x3
 x3
Ex. 3
x3 2 5
x   x 
3! 15
sin x
x2 x4
x3 x5
tan x 
1 
x   
cos x
2! 4!
3! 5!
C. Binomial Series (이항 급수)
 p
   1,
0 
 
 p
   p,
1 
 
 p  p  p  1
 
2 
2!
 
 p  p p  1 p  2  p  n  1
 
n 
n!
 
ex. .1
 1 2 x 2   1 2 3 x3  
1
1
 1  x   1  x 
1 x
2!
3!
 1  x  x 2  x3  
D. Substitution of a polynomial or a Series for the variable in another series
(다항식이나 급수를 다른 급수의 변수로 바꾸어 넣기)
ex. 1
e
x2
X2 X3 X4
 e 1 X 



2!
3!
4!
X
  x2!  x3!  x4!
 1  x2 
2 2
2 3
2 4

x 4 x 6 x8
1 x 

 
2! 3! 4!
2
ex.2
e
tan x
X2 X3 X4
 e 1 X 



2!
3!
4!
X
2
3




x3 2 5
x3 2 5




x


x


x


x





3
3 15
3 15

 
x
2 5


 


 1   x   x   

3 15
2!
3!


x 2 x3 3 4
1 x 
  x 
2
2 8
E. Combination of methods (방법들의 결합)
ex. How to expand arctan x in a Taylor series
dt
t

arctan
t
 arctan x.
0 1  t 2
0
x
1  t 
2 1
 1 t2  t4  t6  


x
dt
x3 x5 x7
2
4
6
0 1  t 2  0 1  t  t  t   dt  x  3  5  7  
x
x3 x5 x7
arctanx  x     .
3
5
7
F. Using the basic Maclaurin series (x=0) (기본적인 Maclaurin 급수의 사용)
ex. 1
ex. 2
ln x  ln1  x  1  x  1 
 3 
3
cos x  cos   x 
2
 2 
3

x
2

1
x  12  1 x  13  1 x  14  
2
3
4
3 


  sin  x 

2 


3  1 
3 
 1

x


x





  .
2  5! 
2 
 3! 
3
5
G. Using a computer (컴퓨터 사용)
x3 x 4 x5
e cos x  1  x   

3
6 30
x
x3
x3 x 4
x3 x 4 x5
S1  1  x, S3  1  x  , S 4  1  x   , S5  1  x   
3
3
6
3
6 30
Original
S1
S3
S4
S5
40
20
0
-5
-4
-3
-2
-1
0
-20
-40
1
2
3
4
5
Mathematical methods in the physical sciences 3rd edition Mary L. Boas
Chapter 1 Infinite series, Power series
Lecture 3 Application of Power series
33
15. Some uses of series (급수의 활용)
1) Numerical computation (수치 계산)
: With computers and calculators so available, you may wonder why we would
ever want to use series for numerical computation. Here is an example to warn
you of the pitfalls of blind computation.
ex.1
ln
1 x
 tan x
1 x
x  0.0015

 

x3 x5 x7
x 3 2 5 17 7
  x   
    x   x 
x  
3
5
7
3 15
315

 

x5 4 7


x 
15 45
~ 5.06  1016
x  0.0015
error order, x^7~10^-21
34
ex.2

d4  1
d 4  1  2 x 6 x10
2
sin x 
 4   x  
   
4 
dx  x
3! 5!
 x  0.1 dx  x 

35
2) Summing series (급수 더하기)
1
1 1 1
  
2 3 4
x 2 x3 x 4
 ln1  x   x     
2
3 4
x 1
3) Evaluation of definite integral (구간 적분의 계산)
ex.
 2 x 6 x10


0 sin x dx  0  x  3!  5!  dx
1
2
1
1
1
1



3 7  3! 11 5!
 0.333  0.02381 0.00076  

 0.31028
36
4) Evaluation of indeterminate forms (부정 형태의 계산)
1  ex
lim
x 0
x


x 2 x3
1  1  x    
2! 3!
  lim  1  x    1
 lim 


x 0
x 0
x
2
!


cf. L’Hopital’s rule
f x 
f x 
lim
 lim
, if
x  a g x 
x  a g  x 
1 e
1 e
lim
 lim
x 0
x 0
x
x 
x
ex.

f a   g a   0, g a   0.
x


 ex
 lim
 1.
x 0 1
cf: confer (= compare)
e.g.: exempli gratia (= for example)
i.e.: id est (= that is)
37
proof)
f x 
f 0  f 0x  1 2! f 0x 2  1 3! f 0x 3  
lim
 lim
x 0 g x 
x  0 g 0   g 0 x  1 2! g 0 x 2  1 3! g 0 x 3  
f 0x  1 2! f 0x 2  1 3! f 0x 3  
 lim
x  0 g 0 x  1 2! g 0 x 2  1 3! g 0 x 3  
f 0  1 2! f 0x  1 3! f 0x 2   f 0
f x 
 lim


lim
x  0 g 0   1 2! g 0 x  1 3! g 0 x 2  
g 0 x 0 g x 
38
5) Series approximation (급수의 어림)
ex.1 Equation of motion of a simple pendulum
F   kx
F  kx (storing force)
d2
F  m a  m 2  mx
dt
 kx  mx
x  A sin t   
k  2 
 kAsin t      Am 2 sin t      2   

m  T 
2
39

l
m gcos
m gsin 
mg
F  m a  mx  m l
F   m gsin 
 m gsin   m l
 
g
sin  0
l
' difficult t o solve'
g
     0 ( sin    for   1)
l
: simple harmonic oscillator
40
15-28 Special relativity (특수 상대성)

2
v
E  m0c 1 
2
1  v c 
1
2
2
For v
2
c2
2
2
1 2
2
1
1  v c 
2

c
1
2
2

c
1
2
2
 
X 3 2
1 2
 X  1 v 2 
2 4
2 c
1
2
 1,


2
v
cf . m  m0 1 
2
 1  X 
E  m0c 1  v
2

c
1
2
1
 
1 v2
2
2 c
 
1
 1 2 
 m0c 2 1  v 2   m0c 2  m0v 2
2
 2 c 
kinetic energy
rest mass energy
41
Homework
Chapter 1
1-1
6-8, 6-22
13-7, 13-27
(15-27, 28, 29, 31, and 33: You do not have to do these as a homework. But, it
is likely that one of them will show up in the midterm exam.)
42