Taylor series via division

Taylor series via division

Let the real (

http://planetmath.org/RealFunction

) (or

complex

(

http://planetmath.org/ComplexFunction

))

functions

f

and

g

have the

Taylor series

f

⁢

(

x

)

=

a

0

+

a

1

⁢

(

x

-

a

)

+

a

2

⁢

(

x

-

a

)

2

+

…

g

⁢

(

x

)

=

b

0

+

b

1

⁢

(

x

-

a

)

+

b

2

⁢

(

x

-

a

)

2

+

…

on an

interval

I

(or a circle in

ℂ

) centered at

x

=

a

.  If

b

0

≠

0

, then also the

quotient

f

⁢

(

x

)

g

⁢

(

x

)

apparently has the

derivatives

of all orders (

http://planetmath.org/HigherOrderDerivatives

) on

I

.  It is not hard to justify that if one divides (

http://planetmath.org/Division

) the series of

f

by the series of

g

, the obtained series

f

⁢

(

x

)

g

⁢

(

x

)

=

c

0

+

c

1

⁢

(

x

-

a

)

+

c

2

⁢

(

x

-

a

)

2

+

…

(1)

is identically same as the Taylor series of

f

⁢

(

x

)

/

g

⁢

(

x

)

on

I

.

We consider the

coefficients

c

n

of (1) as undetermined .  They can be determined by first multiplying, using

Cauchy multiplication rule

, the series (1)
and the series of

g

and then by comparing the gotten coefficients of powers (

http://planetmath.org/GeneralAssociativity

) of

x

-

a

with the corresponding coefficients of the series of

f

.  Accordingly, we have the conditions

a

0

=

b

0

⁢

c

0

,

a

1

=

b

0

⁢

c

1

+

b

1

⁢

c

0

,

a

2

=

b

0

⁢

c

2

+

b

1

⁢

c

1

+

b

2

⁢

c

0

,

…

(2)

Since for every

n

, the equation

a

n

=

b

0

⁢

c

n

+

b

1

⁢

c

n

-

1

+

b

2

⁢

c

n

-

2

+

…

+

b

n

⁢

c

0

holds and

b

0

≠

0

,  we get the

recurrence relation

c

n

=

-

b

1

b

0

c

n

-

1

-

b

2

b

0

c

n

-

2

-

…

-

b

n

b

0

c

0

+

a

n

b

0

(

n

=

0

,

1

,

2

,

…

)

.

(3)

Example.

We will calculate the

Bernoulli numbers

, which are the numbers

B

n

appearing in the Taylor series of

x

e

x

-

1

expanded with the powers of

x

:

x

e

x

-

1

=

∑

n

=

1

∞

B

n

n

!

⁢

x

n

(4)

This function has really all derivatives in the point

x

=

0

,  since in this point the

inverse

(

http://planetmath.org/InverseNumber

)

e

x

-

1

x

=

1

+

x

2

!

+

x

2

3

!

+

…

naturally has the derivatives and the value 1 distinct from zero.  Let us think the division of

x

by the Taylor series of

e

x

-

1

.

Corresponding to (1), we denote the of (4) as

c

0

+

c

1

⁢

x

+

c

2

⁢

x

2

+

…

.  When we now think this series and the series

x

+

x

2

2

!

+

…

+

x

n

n

!

+

…

of the

denominator

of

x

e

x

-

1

to be multiplied, the result must be

x

, i.e. the coefficients of all powers of

x

except the first power are 0.  So the two first conditions corresponding to (2) are

c

0

=

1

,

c

1

+

1

2

⁢

c

0

=

0

;  thus

c

0

=

B

0

=

1

,

c

1

=

B

1

=

-

1

2

.

Setting the coefficient of

x

n

equal to zero gives the formula

c

0

n

!

+

c

1

(

n

-

1

)

!

+

…

+

c

n

-

2

2

!

+

c

n

-

1

=

0

(5)

for

n

≥

2

.  Putting here

c

i

=

B

i

i

!

to (5) we obtain

B

0

0

!

⁢

n

!

+

B

1

1

!

⁢

(

n

-

1

)

!

+

B

2

2

!

⁢

(

n

-

2

)

!

+

…

+

B

n

-

2

(

n

-

2

)

!

⁢

2

!

+

c

n

-

1

(

n

-

1

)

!

=

0

,

and multiplying this by

n

!

,

(

n

n

)

⁢

B

0

+

(

n

n

-

1

)

⁢

B

1

+

(

n

n

-

2

)

⁢

B

2

+

…

+

(

n

2

)

⁢

B

n

-

2

+

(

n

1

)

⁢

c

n

-

1

=

0

.

This yields, by substituting the values of

B

0

and

B

1

and recalling that the

odd Bernoulli numbers are zero

(

n

>

2

), the recursion formula

1

-

2

⁢

k

2

+

(

2

⁢

k

+

1

2

)

⁢

B

2

+

(

2

⁢

k

+

1

4

)

⁢

B

4

+

…

+

(

2

⁢

k

+

1

2

⁢

k

-

2

)

⁢

B

2

⁢

k

-

2

+

(

2

⁢

k

+

1

2

⁢

k

)

⁢

B

2

⁢

k

=

0

for the even Bernoulli numbers

B

2

⁢

k

(

k

=

1

,

2

,

…

).  It gives successively

-

1

2

+

3

⁢

B

2

=

0

,

-

3

2

+

10

⁢

B

2

+

5

⁢

B

4

=

0

,

-

5

2

+

21

⁢

B

2

+

35

⁢

B

4

+

7

⁢

B

6

=

0

,

…

From here we obtain

B

2

=

1

6

,

B

4

=

-

1

30

,

B

6

=

1

42

,  and so on.

Remark.

The method of using undetermined coefficients in division of

power series

is especially simple in the case that the denominator in (1) is a

polynomial

, because the number of the terms in the recursion formula (3) is, independently on

n

, below a finite bound. Thus the method is applicable for expanding the

rational functions

to power series. For example, if we want to expand

1

1

+

x

2

with the powers of

x

-

1

, we write

1

+

x

2

=

2

+

2

⁢

(

x

-

1

)

+

(

x

-

1

)

2

.  The two first conditions corresponding to (2) are

2

⁢

c

0

=

1

and

2

⁢

c

1

+

2

⁢

c

0

=

0

,  whence

c

0

=

1

2

and

c

1

=

-

1

2

.  The coefficient of

(

x

-

1

)

n

gives the condition

2

⁢

c

n

+

2

⁢

c

n

-

1

+

c

n

-

2

=

0

,  whence the simple recursion formula

c

n

=

-

c

n

-

1

-

1

2

⁢

c

n

-

2

; the use of this is much more comfortable than the

long division

1

:

(

2

+

2

⁢

(

x

-

1

)

+

(

x

-

1

)

2

)

.

References

1

Ernst Lindelöf

:

Differentiali- ja integralilasku
ja sen sovellutukset I

. Second edition.  WSOY, Helsinki (1950).

Title

Taylor series via division

Canonical

name

TaylorSeriesViaDivision

Date of creation

2014-12-02 17:46:48

Last modified on

2014-12-02 17:46:48

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

20

Author

pahio (2872)

Entry type

Topic

Classification

msc 30B10

Classification

msc 26A24

Classification

msc 41A58

Synonym

quotient of Taylor series

Synonym

calculating Bernoulli numbers

Related topic

BinomialCoefficient

Related topic

BernoulliNumber

Related topic

BernoulliPolynomialsAndNumbers

Related topic

ErnstLindelof

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