A.D.
M R Richards
5 Power Series
Taylor Series
Consider a function f(x) where its value and derivatives are known at x 0. Try to
evaluate the function at a “nearby” point x0 + h.
Try a power series expansion in h about the point x0.
f ( x0 + h) =  0   1 h   2 h 2  ...


=
n 0
n
h n (1)
Where the coefficients of  n need to be determined.
Setting h=0   0  f ( x0 )
Next, differentiating (1) with respect to h
f’(x0+h) =  1  2 2 h  3 3 h 2  ... (2)
Setting h=0  1  f ' ( x0 )
Similarly  2 
f ' ' ( x0 )
2!
The general result  n 
Hence Taylor series is

f ( n ) x0 n
f(x0 + h) = 
h
n!
n 0
1
f
n!
(n)
( x0 )
Valid for small h
The Taylor series about the origin (setting x0=0 and x=h) is the MacLaurin Series
MacLaurin Expansion

xn n
f ( x)  
f (0)
n 0 n!
e.g.
1)
f(x) = ex
 f n ( x)  e x
 f n (0)  1

ex =
xn
x2

1

x

 ...

2!
n 0 n!
Page 1 of 10
A.D.
M R Richards
2) f(x) = sin x
f1(x) = cos x
f2(x) = -sin x
f3(x) = -cos x
f4(x) = sin x

sin x  
n 0
 f0(0) = 0
 f1(0) = 1
 f1(0) = 0
 f1(0) = -1
 f1(0) = 0
x 2 n 1 (1) n
(2n  1)!
Similarly
(1) n x 2 n
(2n)!
n 0

cos x  
cos x  1 
x2 x4

2! 4!
Hence, from 1 and 2
e i  cos x  i sin x
And e i  1  0
Notes
1
 1  x  x2  x3
1 x
ln( 1  x)  x 
x2 x3 x4


2
3
4
(1  x)   1  x 
 (  1)
2!
x2 
 (  1)(  2)
3!
x3
2
In order to have a MacLaurin expansion the derivative must exist, i.e. e  x has no
derivative and no MacLaurin expansion.
The expansion for ex, cos x and sin x is valid for all x
The expansions
Page 2 of 10
1
and (1  x) are only valid for x  1
1 x
A.D.
M R Richards
For a function of 2 variables
f(x,y) the Taylor Series is
f(x0+h, y0+k) = f ( x0 , y 0 )  (h
f
f
1
2 f
f
f
 k )  ( h 2 2  2h
 k 2 2 )  ...
x
y
2!
xy
x
y
Where all derivatives are evaluated at (x0,y0)

1 n
D f
n  0 n!
f ( x0  h, y 0  k )  
Where D = h


k
x
y
Convergence Of A Power Series
The MacLaurin is an infinite power expansion.

f ( x)    n x n
Where  n 
n 0
f n (0)
n!
We need to know whether the power series has meaning, i.e. if the infinite sum
converges to a limit. In general convergence only occurs for a range of x values.
The formal way to define this is the partial series.
N
f N   un
Where u n   n x n
n 0
If lim exists we say the power series converges.
N 
If lim does not exist we say the power series diverges.
N 
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A.D.
M R Richards

Is
u
n 0
n
meaningful?
N
Or better does lim
N 
u
n 0
n
exist.
If it does then the sum converges.
If it does not then the sum diverges.
Note
N
A series may diverge because f N (  u N ) increases without bound as N tends
n 0

towards infinity, or does not approach a limit e.g.
 (1)
n
is meaningless, as the
n 0
partial sum oscillates between 0 and 1.
1
f1   (1) n  (1) 0  (1)1  0
n 0
2
f 2   (1) n  (1) 0  (1)1  (1) 2  1
n 0
3
f 3   (1) n  (1) 0  (1)1  (1) 2  (1) 3  0
n 0
Examples
1) Geometric Series
N
S N   x n  1  x  x 2  x 3  ....  x N
n 0
N
xS N  x x n  x  x 2  x 3  x 4  ....  x N 1
n 0
S N (1  x)  1  x N 1
1  x N 1
SN 
1 x
Hence S = lim S N only exists if x  1
N 
For x  1 the limit does not exists and the series diverges.
Hence the MacLaurin series f ( x) 
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1
 1  x  x 2  ... is only valid for  1  x  1
1 x
A.D.
M R Richards

2) A well known example of a divergent series is
n 1
Proof

1
1 1 1
 1     ...

2 3 4
n 1 n

1
1
1
1
1
1
1
1
1
1
1
1
1
1
 n  1  2  ( 3  4 )  ( 5  ...  8 )
n 1

 n  1  2  ( 4  4 )  ( 8  ...  8 )
n 1

1
1
 n  1  2  2  2  ...
n 1
Hence, diverges.
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1
n
A.D.
M R Richards
Absolute And Conditional Convergence

An infinite series S   u n is convergent and said to be absolutely convergent
n 0

provided
u
n 0
n
is also convergent. Absolute convergence implies convergence. (Note
un could be complex).
e.g.
1 1 1 1
    ...
2 4 8 16
Is convergent and absolutely convergent because
1 1 1 1
 u n  2  4  8  16  ...
1
1 1 1
 u n  2 (1  2  4  8  ...)
1
1 1
1
 u n  2 (1  2  2 2  2 3  ...)
1 1
 u n  2 . 1  1  Convergent
1
2
The series
Notes
If a series only contains real positive terms and is convergent, convergence and
absolute convergence will be the same.
1 1 1 1
    ... is convergent (via the Leibnitz Test), but is only
2 3 4 5
conditionally convergent as it is convergent but its absolutely sum diverges (as
1 1 1
1     ...  
2 3 4
The series 1 
Below are 4 useful tests of convergence of

u
n
A necessary but not sufficient condition is u n  0 as n   . If this is not
fulfilled series diverges, if it is fulfilled does not mean the series defiantly
converges.
e.g.
 n 2 Diverges
1
n
May or may not converge.
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A.D.

M R Richards
Comparison. If
u
n
is given and we can find a converging series
n
a non negative bn such that u n  bn then
Similarly if we find a divergent
u n  bn then
u
b
u
n
is also convergent.
n
with a non negative bn such that
is divergent.
Leibnitz Text For Alternating Series

If a series is of the form
 (1)
n 0
n
an
With
1)
Positive an
2)
an forming a decreasing sequence i.e. an+1 < an
3)
a n  0 as n  
Then it converges
e.g. 1 

1 1 1 1
   Converges.
2 3 4 5
Ratio Test
Comes from comparison test applied with
Let us suppose u n  0 for any value of n
u n1
n  u
n
Define L  lim
If
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L < 1 Series is convergent
L > 1 Series is divergent
L = 0 Test fails
with
n
n

n
n
n
n
b
x
n
A.D.
M R Richards
Examples
Use the ratio test to determine the convergence of various MacLaurin series:
1)

ex  
n 0
xn
n!
With u n 

e x   un
xn
n!
Then u n1 
x ( n1)
(n  1)!
n 0
Then
u n 1
x n 1 n!

.
un
(n  1)! x n
Since (n + 1)! = (n + 1) n!
u n 1
x n 1 n!

.
un
(n  1)! x n
u n 1
x

un
n 1
x
L  lim
 0x
n  n  1
MacLaurin series convergent for all values of x
2)
x 2 n 1
sin x   (1)
(2n  1)!
n 0
u n 1
(1) n 1 x 2( n 1) 1

(2(n  1)  1)!
u n 1
 (1) n x 2 n 3

(2n  3)!

n

sin x   u n
(1) n x 2 n 1
With u n 
(2n  1)!
n 0
Then
Then
u n 1
 x 2 n 3 (2n  1)!

.
un
(2n  3)! x 2 n !
u n 1
x2

un
(2n  3)( 2n  2)
x2
 0x
n  ( 2n  3)( 2n  2)
L  lim
Page 8 of 10
MacLaurin series convergent for all values of x
A.D.
M R Richards
3)
f ( x) 
1
1 x
With u n  x n

f ( x)   x n
Then u n 1  x n 1
n 0
Then
u n 1 x n 1
 n
un
x
u n 1
 x
un
MacLaurin series convergent for x  1
L  lim x  x
n 
4)
f ( x)  ln( 1  x)
x2 x3 x4


 ...
2
3
4
(1) n 1 x n

With u n 
(1) n 1 x n
f ( x)  
n
n
n 1
f ( x)  x 

f ( x)   u n
n 1
Then
u n 1 (1) n  2 x n 1
n

. n
un
n 1
x (1) n 1
u n 1
 xn

un
n 1
nx
n
1
 x lim
 x
n


1
n 1
n 1
1
n
MacLaurin series convergent for x  1
L  lim
n 
Page 9 of 10
 x lim
n 
Then u n 1 
(1) n  2 x n 1
n 1
A.D.
M R Richards
5)
Find the full range of convergence of

xn
x 2 x3
 x

 ...

2
3
n 1 n
Ratio Test
xn
un 
n
x n 1
u n 1 
n 1
Then
xn
x1
u n 1


1
un
n 1
1
n
u n1
 x
n  u
n
 lim
Hence Series diverges if x  1
Series converges if x  1
The ratio test fails when x = 1, x = -1
For x = 1 Series becomes

1n
1 1 1
 1     ... Which diverges

2 3 4
n 1 n
For x = -1 Series becomes

(1) n
1 1 1
 1     ...

n
2 3 4
n 1
(1) n
1 1 1
 1(1     ...)

n
2 3 4
n 1

Which converges (Leibnitz test)
So full range of convergence is  1  x  1
Page 10 of 10