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§11.9 Representations of Functions as Power Series

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1      國立交通大學應用數學系 莊重教授 

§11.9 Representations of Functions as Power Series

 源頭函數:

1 ... , | | 1.

1 1

0

2

x x x

x x n

n

(i). 直系函數:

Example 1:

     

1 , 1 | | 1. 1

1 1

1 2 2

0 0

2 2

2

 

x x

x x x

x

n n

n n

n

Example 2:

2.

|

| 2 , 2

2 1 1 2

1 2 1 2

1

0 1 0

 

x x x

x x n n

n

n n

Example 3:

  , | | 2 . 1 2

2 2

1 2 1 2

2

0 1

3

0 3 3

3

   

 

 

 

 

 

   

x x x

x x x

x x

n

n n n n

n

(ii). 旁系函數:(related to 直系函數 by 微分和積分)

*註記:

1. Power series 在絕對收斂的範圍(即收斂區間)可作逐項微分或積分

的動作。

2. 有時須微分或積分不只一次才到直系。

(2)

2      國立交通大學應用數學系 莊重教授 

Example 4:

(i). Find the power series representation of ln ( 1- x ).

(ii). Express ln 2 as a convergent infinite series with all positive terms.

Solution:

 

. 1

|

| ..., 1 1

1

... 1 3 1 2

ln

2 0

0 1 3

2

x x

x x x

n c x x

x x c x

n n

n n

dx

d



Let x = 0 ⇒ ln 1 = 0 = c.

 

. 1

|

| ,

1

|

| 1 , 1

ln

1 0

1

 

n x x n x x x

n n n

n

Example 5:

(i). Find the power series representation of

f(x)tan1x.

(ii). Let  

0 0

tan 1

) (

n n n

x tdt a x

x

f

. Then

a8 ?

Solution:

 

  1 , | | 1 . 1

1

. 1

|

| 1 tan

0

2 2

0

2 1

 

x x x

x x c

x

n

n n n

n n

dx

d

       

Let x = 0 ⇒ tan

-1

0 = 0 = c + 0 ⇒

c

0.

 

1.

|

| 1 , 2 tan 1

0

1 2

1

 

 

n x x x

n

n n

Example 6:

Find the power series representation of

1

.

) 1

( 2

x x

f

(3)

3      國立交通大學應用數學系 莊重教授 

Solution:

 

. 1

|

| 1 ,

1

. 1

|

| , 1

1

0 1

1 2

x x x

x nx x

n n dx d n

n



Example 7:

Evaluate  1 1

x7 dx

as a power series.

Solution:

 

 

  | | 1 . 1

1 7 1 1 1

1

0

1 7 0

7 7 0

7

 

 

 





x n c

x dx x

dx x x dx

n

n n n

n n n n

n

Example 8:

1.

?, | | 1.

1

1

x

nx

n n

2. (i)

?, | | 1.

1

x nx

n n

(ii)

?

12

n

n

n

3. (i) 

1

?, | | 1.

2

x x

n n

n

n

(ii)

?

2 2

2

n

n

n n

(iii)

?

12

2

n

n

n

(4)

4      國立交通大學應用數學系 莊重教授 

Solution:

1. For |x| < 1, we have

  .

1 1 1

1

2 1

1 1

1

x x dx x d

dx x d dx nx d

n n n

n n

n

 

 

 

 

 

 

 

  

2. (i) For |x|<1, we have

 

2

1 1

1 1 x

nx x x nx

n n n

n

(此例子為旁系+直系)

(ii) 2 .

2 1 1

2 1

2

2

1

 

 

  

n

n

n

3. (i) For |x|<1, we have

 

2

1

n

xn

n n

      .

1

1 2

3

2

2 1 2

2

1 2

2

2 2

x nx x

dx x d dx nx

x d x n n x

n n n

n n

n

 

 

 

 

  

(ii) 4 .

2 1 2 2 1

2

3

2

2

2

 

 

 

 

 

n

n

n n

(iii)

4 2 6.

2 2

2 2

2 2 1

2

1 1

2

1

2

n

n n

n n

n n

n n

n

n n

n n

n n n

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