series inversion

series inversion

The method of

series inversion

allows us to derive the

power series

of an

inverse function

f

-

1

given the power series of

f

.

Clearly, since we are representing

function

f

and

f

-

1

by power series,
the function

f

must necessarily be

holomorphic

; that is, it is

differentiable

as a complex-valued function on an open subset of the

complex plane

.
(It follows that

f

-

1

must also be holomorphic.)

Holomorphic functions

include the

elementary functions

studied in

calculus

such as

sin

,

cos

,

tan

and

exp

.

For the method to work smoothly, it is best to assume that we want
to invert

f

at around the origin, and its value there is zero.
There is no loss of generality, since
if

f

⁢

(

b

)

=

a

, we can apply series inversion to the function

g

defined
by

f

⁢

(

z

)

=

a

+

g

⁢

(

z

-

b

)

.
Then

f

-

1

⁢

(

w

)

=

g

-

1

⁢

(

w

-

a

)

+

b

;
we obtain the power series for

f

-

1

centred at

a

.

Also, it must be true that

f

′

⁢

(

0

)

≠

0

, otherwise

f

will not even
be

invertible

around the origin.

An example

We explain the method by an example, for

f

⁢

(

z

)

=

arctan

⁡

z

.
In the following, we will consistently use the notation

O

⁢

(

z

n

)

to denote a holomorphic function

h

⁢

(

z

)

whose power series expansion begins
with

z

n

. And similarly when

z

is replaced with the variable

w

.

First, we start with the well-known power series expansion for

w

=

arctan

⁡

z

:

w

=

z

-

z

3

3

+

z

5

5

+

O

⁢

(

z

7

)

.

(1)

The number of explicit terms in the power series expansion determines
the number of terms that we will be able to obtain
in the power series expansion of

f

-

1

.
So in this case, we are going to seek an expansion
of

f

-

1

⁢

(

w

)

=

tan

⁡

w

up to (but excluding) the

w

7

term.

A simple rearrangement of (

1

) gives

z

=

w

+

z

3

3

-

z

5

5

+

O

⁢

(

z

7

)

(2)

Now we

substitute equation (

2

) into itself

.
Of course, usually when we substitute an equation into itself
we do not get anything, but here it works because we can

perform simplications using the

O

notation

.
So for instance, in the following,
in second term

z

3

/

3

on the right of equation (

2

), we put in
equation (

2

) simplified to

z

=

w

+

O

⁢

(

z

3

)

.
Why we should choose this simplication will be clear in a moment:

z

=

w

+

1

3

⁢

(

w

+

O

⁢

(

z

3

)

)

3

-

z

5

5

+

O

⁢

(

z

7

)

(3)

=

w

+

1

3

⁢

(

w

3

+

O

⁢

(

z

3

)

⁢

(

w

2

+

⋯

)

)

-

z

5

5

+

O

⁢

(

z

7

)

(4)

=

w

+

1

3

⁢

(

w

3

+

O

⁢

(

w

3

)

⋅

O

⁢

(

w

2

)

)

+

O

⁢

(

w

5

)

.

(5)

In equation (

5

) we used the fact that the expansion
for

z

=

f

-

1

⁢

(

w

)

must begin

with a

w

term, i.e.

f

-

1

⁢

(

0

)

=

0

. Also note that we are guaranteed that the

w

and

z

terms
have non-zero

coefficients

, because

f

′

⁢

(

0

)

≠

0

. (Otherwise
we would not be able to isolate

z

in equation (

2

).)

Now equation (

5

) simplifies to

z

=

w

+

w

3

3

+

O

⁢

(

w

5

)

,

(6)

which is already an achievement, because we have identified
exactly what the

w

3

term must be.

We substitute (

6

) into the

z

3

and

z

5

terms of (

2

),
and obtain:

z

=

w

+

1

3

⁢

(

w

+

w

3

3

+

O

⁢

(

w

5

)

)

3

-

1

5

⁢

(

w

+

w

3

3

+

O

⁢

(

w

5

)

)

5

+

O

⁢

(

z

7

)

(7)

=

w

+

1

3

⁢

(

w

3

+

(

3

1

)

⁢

w

3

3

⁢

w

2

+

O

⁢

(

w

7

)

)

-

1

5

⁢

(

w

5

+

O

⁢

(

w

7

)

)

+

O

⁢

(

w

7

)

(8)

=

w

+

1

3

⁢

w

3

+

2

15

⁢

w

5

+

O

⁢

(

w

7

)

.

(9)

And this gives our desired expansion of

z

=

tan

⁡

w

of degree

<

7

.

Summary

To summarize the procedure in general,
we start with the expansion

w

=

f

⁢

(

z

)

=

a

1

⁢

z

+

a

2

⁢

z

2

+

a

3

⁢

z

3

+

⋯

,

a

1

≠

0

,

(10)

and rearrange it to,

z

=

b

1

⁢

(

w

-

a

2

⁢

z

2

-

a

3

⁢

z

3

-

⋯

-

a

n

⁢

z

n

)

+

O

⁢

(

w

n

+

1

)

,

b

1

=

1

a

1

.

(11)

So we know that

z

=

b

1

⁢

w

+

O

⁢

(

w

2

)

,
and we can substitute
this into the term

z

2

of equation (

11

).
At the end we will get an equation of the form

z

=

b

1

⁢

w

+

b

2

⁢

w

2

+

O

⁢

(

w

3

)

, and we can substitute this
into the terms

z

2

and

z

3

of (

11

).
And what ever results we will substitute back into the
terms

z

2

,

z

3

,

z

4

of equation (

11

).
We can repeat this process until
we have all the terms of

z

=

f

-

1

⁢

(

w

)

that we need.

We probably should

normalize

the functions so that

a

1

=

b

1

=

1

to make the computations easier.

References

1

Lars V. Ahlfors.

Complex Analysis

. McGraw-Hill, 1979.

Title

series inversion

Canonical name

SeriesInversion

Date of creation

2013-03-22 15:39:09

Last modified on

2013-03-22 15:39:09

Owner

stevecheng (10074)

Last modified by

stevecheng (10074)

Numerical id

8

Author

stevecheng (10074)

Entry type

Derivation

Classification

msc 30B10

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