Nested Derivatives: A simple method for computing

arXiv:math/0501052v2 [math.CA] 4 Jan 2005

series expansions of inverse functions.

Diego Dominici∗
Department of Mathematics
State University of New York at New Paltz
75 S. Manheim Blvd. Suite 9
New Paltz, NY 12561-2443, USA
February 1, 2008

Abstract
We give an algorithm to compute the series expansion for the inverse of a given
function. The algorithm is extremely easy to implement and gives the first N terms of
the series. We show several examples of its application in calculating the inverses of
some special functions.

MSC-class: 30B10 (Primary) 30D10, 33E20 (Secondary)

1 Introduction
“One must always invert.”
Carl G. J. Jacobi

The existence of series expansions for inverses of analytic functions is a well-known result
of complex analysis [17]. The standard inverse function theorem, a proof of which can be
found, for example, in [12], states that

Theorem 1.1 Let h(x) be analytic for |x − x0 | < R where h′ (x0 ) 6= 0. Then z = h(x) has
an analytic inverse x = H(z) in some ε-neighborhood of z0 = h(x0 ).
∗
dominicd@newpaltz.edu

1
 In the case when x0 = z0 = 0, |h(x)| ≤ M for |x| < R, and h′ (0) = a, R. M. Redheffer
2
[25] has shown that it is enough to take ε = 41 (aR)
M
.
However, the procedure to obtain the actual series is usually very difficult to implement
in practice. Under the conditions of Theorem 1.1, the two standard methods to compute the
coefficients bn of X
h−1 (z) ≡ H(z) = bn (z − z0 )n
n≥0

are reversion of series [16], [26], [33], and Lagrange’s theorem. The first one requires to
expand h(x) around x0 X
h(x) = an (x − x0 )n
n≥0

and then solve for bn in the equation

" #n
X X
z= an bn (z − z0 )n − x0
n≥0 n≥0

by equating powers of z and taking into account that a0 = z0 and b0 = x0 . This method
is especially useful if all that is known about h(x) are the first few an . When x0 = z0 = 0
and a1 = a, it was shown by E. T. Whittaker [34] that
 
1 a2 1  3a2 a 
b1 = , b2 = − 3 , b3 = , ...
a a 3!a5  6a3 4a2 
 
 na2
 a 0 0 ··· 

 2na3 (n + 1)a2 2a 0 ··· 
(−1)n−1  3na (2n + 1)a (n + 2)a 3a ···

bn = 4 3 2

n!a2n−1  4na5 (3n + 1)a4 2(n + 1)a3 (n + 3)a2
 
··· 
 .. .. .. ..
 
..
.

 . . . . 

where |·| ≡ det (·) . In Example 10, we show how to get the bn in term of the an using our
method.
A computer system like Maple can reverse the power series of h(x), provided h(x) is not
too complicated, by using the command

> Order := N + 1;
> solve ( series ( h(x) , x = x0 , N + 1) = z , x);

where N is the number of terms wanted. Fast algorithms of order (n log n)3/2 for reversion
of series have been analyzed by Brent and Kung [6], [5]. The multivariate case has been
studied by several authors [4], [8], [14], [21] and Wright [35] has studied the connection
between reversion of power series and “rooted trees”.

2
 The second and more direct method is Lagrange’s inversion formula [1],
n 
1 dn−1

x − x0 
bn = n−1
 . (1.1)
n! dx h(x) − z0 
x=x0

Unfortunately, more direct doesn’t necessarily mean easier and, except for some simple cases,
Lagrange’s formula (1.1) is extremely complicated for practical applications. The q-analog
(a mathematical expression parametrized by q which generalizes an expression and reduces
to it in the limit q → 1+ ) of (1.1) has been studied by various authors [2], [18], [19], [20]
and a unified approach to both the regular and q-analog formulas have been obtained by
Krattenthaler [23]. There has also been a great deal of attention to the asymptotic expansion
of inverses [27], [28], [31], [32].
In this note, we present a simple, easy to implement method for computing the series
expansion for the inverse of any function satisfying the conditions of Theorem 1.1, although
the method is especially powerful when h(x) has the form
Zx
h(x) = g(x)dx
a

and g(x) is some function simpler than h(x). Since this is the case for many special functions,
we will present several such examples. This note is organized as follows:
In section 2 we define a sequence of functions Dn [f ] (x), obtained from a given one f (x),
that we call “nested derivatives”, for reasons which will be clear from the definition. We give
a computer code for generating the nested derivatives and examples of how Dn [f ] (x) look
for some elementary functions. Section 3 shows how to compute the nested derivatives by
using generating functions. We present some examples and compare the results with those
obtained in Section 1.
Section 4 contains our main result of the use of nested derivatives to compute power
series of inverses. We test our result with some known results and we apply the method for
obtaining expansions for the inverse of the error function, the incomplete Gamma function,
the sine integral, and other special functions.

2 Definitions
Definition 2.1 We define Dn [f ] (x), the nth nested derivative of the function f (x), by the
following recursion:

D0 [f ] (x) = 1
d 
Dn [f ] (x) = f (x) × Dn−1 [f ] (x) ,

n ≥ 1. (2.1)
dx
Proposition 2.2 The nested derivative Dn [f ] (x) satisfies the following basic properties.

3
 (1) For n ≥ 1, Dn [κ] ≡ 0, κ constant.

(2) For n ≥ 0, Dn [κf ] (x) = κn Dn [f ] (x), κ constant.

(3) For n ≥ 1, Dn [f ] (x) has the following integral representation:

n−1
1 f (zn ) Y f (zk )
I I I
n
D [f ] (x) = ··· dzn . . . dz1 ,
(2πi)n (zn − x) k=1 (zk − zk+1 )2
2
C1 C2 Cn

where Ck is a small loop around x in the complex plane.

Proof. Properties (1) and (2) follow immediately from the definition of Dn [f ] (x).
To prove (3) we use induction on n. For n = 1 the result follows from Cauchy’s formula

1 f (z1 )
I
1 df
D [f ] (x) ≡ = dz1 .
dx 2πi (z1 − x)2
C1

Assuming that the result is true for n and using (2.1), it follows that

1 f (zn+1 )Dn [f ] (zn+1 )

I
n+1 d n
D [f ] (x) ≡ [f (x) × D [f ] (x)] = dzn+1
dx 2πi (zn+1 − x)2
Cn+1

1 f (zn+1 ) f (zn )
I I
= n+1
···
(2πi) (zn+1 − x) (zn − zn+1 )2
2
C1 Cn+1
n−1
Y f (zk )
× dzn+1 . . . dz1
(zk − zk+1 )2
k=1
n
1 f (zn+1 ) Y f (zk )
I I
= n+1
··· dzn+1 . . . dz1 .
(2πi) (zn+1 − x) k=1 (zk − zk+1 )2
2
C1 Cn+1

Algorithm 2.3 The D algorithm. The following Maple procedure implements the recurrence
relation (2.1). We define d(k) = Dk [f ] (x), where N is the number of terms desired.

> d(0) := 1;
> for k from 0 to N do:
> d(k + 1): = simplify ( diff ( f (x) ∗ d(k), x )): (2.2)
> print ( k + 1, d(k + 1) ):
> od:

4
Example 2.4 The function f (x) = x.

D1 [f ] (x) = 1
D2 [f ] (x) = 1
..
.
n
D [f ] (x) = 1.

Example 2.5 The power function f (x) = xr , r 6= 1.

D1 [f ] (x) = rxr−1
D2 [f ] (x) = r(2r − 1)x2(r−1)
D3 [f ] (x) = r(2r − 1)(3r − 2)x3(r−1)
..
.
Yn
n
D [f ] (x) = [jr − (j − 1)] xn(r−1)
j=1
1


Γ n + 1 + r−1
= (r − 1)n 1

xn(r−1) .
Γ 1 + r−1
k
Notice that when r = k+1
, k = 1, 2, . . . , the sequence of nested derivatives has only
k + 1 non-zero terms
n
k!
x− k+1 ,

n (k−n)! (k+1)n
0≤n≤k
D [f ] (x) = .
0, n≥k+1

Example 2.6 The exponenetial function f (x) = erx .

D1 [f ] (x) = rerx
D2 [f ] (x) = 2r 2 e2rx
D3 [f ] (x) = 6r 3 e3rx
..
.
D [f ] (x) = n!r n enrx .
n

3 Generating functions
Generating functions provide a valuable method for computing sequences of functions defined
by an iterative process; we will use them to calculate Dn [f ] (x). In the sequel, we shall
implicitly assume that the generating function series converges in some small disc around
z = 0.

5
 R 1
Theorem 3.1 Given h(x) = f (x) dx, its inverse H(x) = h−1 (x) and the exponential gen-
n
D [f ] (x) zn! , it follows that
P n
erating function G(x, z) =
n≥0

1
G(x, z) = (f ◦ H) [z + h(x)] . (3.1)
f (x)
Proof. Taking (2.1) into account gives
∂ X d zn
[f (x) × G(x, z)] = [f (x) × Dn [f ] (x)]
∂x n≥0
dx n!
X zn X n z n−1
= Dn+1 [f ] (x) = D [f ] (x)
n≥0
n! n≥1
(n − 1)!
∂ X n zn ∂
= D [f ] (x) = G(x, z).
∂z n≥0 n! ∂z

Hence, the generating function G(x, z) satisfies the PDE

∂(f × G) ∂G
=
∂x ∂z
with general solution
1
G(x, z) = g [z + h(x)] (3.2)
f (x)
where g(z) is an arbitrary analytic function. Invoking the boundary condition G(x, 0) =
D0 [f ] (x) = 1, (3.2) gives
1
1= g [h(x)]
f (x)
and therefore
f (x) = (g ◦ h) (x).
If x = H(w), then
(f ◦ H) (w) = (g ◦ h ◦ H) (w) = g(w)
and the theorem follows.

Example 3.2 The

R function f (x) = x.
Here h(x) = x1 dx = ln(x), H(x) = ex , and from (3.1) it follows that
1
G(x, z) = exp [z + ln(x)] = ez .
x
We could obtain the same result from Example 2.4 by summing the series
X zn
G(x, z) = 1 = ez .
n≥0
n!

6
Example 3.3 The power function f (x) = xr , r 6= 1.
1−r 1
Now h(x) = x−r dx = x1−r , H(x) = [(1 − r)x] 1−r , and we get
R

( 1 )r  r
1−r 1−r 1−r
  1−r 
x (1 − r)z + x
G(x, z) = x−r (1 − r) z + =
1−r x1−r
 r
= 1 + (1 − r)xr−1 z 1−r .


Expanding in series around z = 0, we recover the result from Example 2.5.

r k
If 1−r = k, i.e. r = k+1 , k = 0, 1, . . ., then G(x, z) is a polynomial of degree k in z
and hence
Dn [f ] (x) = 0, n ≥ k + 1
as we have already observed in Example 2.5.
Given the particular form of the function h(x) in Theorem 2, we can get alternative
expressions for (3.1) which sometimes are easier to employ.
R 1
Corollary 3.4 Let h(x) = f (x) dx, its inverse H(x) = h−1 (x) and the exponential gener-
n
D [f ] (x) zn! . Then,
P n
ating function G(x, z) =
n≥0

(i)
1
G(x, z) = H ′ [z + h(x)] (3.3)
f (x)
and
(ii)
d
G(x, z) = H [z + h(x)] .
dx
Proof.
(i) By definition (h ◦ H)(x) = x, so
h′ [H(x)] H ′ (x) = 1.
1
R
Since h(x) = f (t)
dt,
1
H ′ (x) = 1
f [H(x)]
or
(f ◦ H)(x) = H ′ (x)
and therefore
1
G(x, z) = (f ◦ H) [z + h(x)]
f (x)
1
= H ′ [z + h(x)] .
f (x)

7
 (ii) Using the chain rule

d
H [z + h(x)] = H ′ [z + h(x)] h′ (x)
dx
1
= H ′ [z + h(x)]
f (x)

and the conclusion follows from part (i).

4 Applications
We now state our main result.
Rx 1
Theorem 4.1 Let h(x) = f (t)
dt, with f (a) 6= 0, ±∞, and its inverse H(x) = h−1 (x). Then,
a
X zn
H(z) = a + f (a) Dn−1 [f ] (a) (4.1)
n≥1
n!

where |z| < ε, for some ε > 0.

Proof. Let’s first observe that since h(a) = 0, it follows that H(0) = a and from (3.3)
1 1
G(a, z) = H ′ [z + h(a)] = H ′ (z)
f (a) f (a)

where X zn
G(a, z) = Dn [f ] (a) .
n≥0
n!
Hence,
Zz X tn
H(z) = H(0) + f (a) Dn [f ] (a) dt
n≥0
n!
0
X z n+1
= a + f (a) Dn [f ] (a)
n≥0
(n + 1)!
X zn
= a + f (a) Dn−1 [f ] (a) .
n≥1
n!

8
Example 4.2 The natural logarithm function. Let f (x) = e−x , with a = 0. We have f (0) =
1,
Zx
h(x) = et dt = ex − 1, H(x) = ln(x + 1)
0

and from Example 2.6

Dn [f ] (0) = (−1)n n!.
Hence, from (4.1) we get the familiar formula
X zn
ln(z + 1) = (−1)n−1 .
n≥1
n

Example 4.3 The tangent function. Let f (x) = x2 + 1, with a = 0.Now f (0) = 1,
Zx
1
h(x) = dt = arctan(x), H(x) = tan(x)
t2 +1
0

and (4.1) implies that

X zn
tan(z) = Dn−1 [x2 + 1] (0) .
n≥1
n!

Therefore,

D2k+1 [x2 + 1] (0) = 0, k≥0

2 k k+1
D2k [x2 + 1] (0) =

 
4 4 − 1 B2(k+1)  , k≥1 (4.2)
k+1
where Bk are the Bernoulli numbers [1].

Remark 4.4 From Example 2.5, we recall that

Dn [x2 ] (x) = (n + 1)!xn

and consequently
Dn [x2 ] (0) = 0, n ≥ 1. (4.3)
Comparing (4.2) and (4.3) we can see the highly nonlinear behavior of the nested deriva-
tives, since even the addition of 1 to f (x) creates a completely different sequence of values,
far more complex than the original.

We now start testing our result on some classical functions.

9
 p
Example 4.5 Elliptic functions. Let f (x) = 1 − p2 sin2 (x), 0 ≤ p ≤ 1, a = 0. We
have, f (0) = 1 and

Zφ
dθ
h(φ) = p = F (p; φ), H(p; x) = am(p; x)
1 − p2 sin2 (θ)
0

where F (p; φ) is the incomplete elliptic integral of the first kind, and am(p; x) is the elliptic
amplitude [29]
"p #
1 − dn2 (p; x)
am(p; x) = arcsin [sn(p; x)] = arccos [cn(p; x)] = arcsin
p

with sn(p; x), cn(p; x), and dn(p; x) denoting the Jacobian elliptic functions.
Computing Dn [f ] (0) with (2.2) gives

D2k+1 [f ] (0) = 0, k≥0

D2k [f ] (0) = (−1)k p2 Qk (p), k≥1

where Qk (p) is a polynomial of degree 2(k − 1) of the form

Qk (p) = p2(k−1) + · · · + 22(k−1) .

The first few Qk (p) are

Q1 (p) = 1
Q2 (p) = p2 + 4
Q3 (p) = p4 + 44p2 + 16
Q4 (p) = p6 + 408p4 + 912p2 + 64
Q5 (p) = p8 + 3688p6 + 307682p4 + 15808p2 + 256

and (4.1) implies that

z3 z5 z7
am(p; x) = z − p2 + p2 (p2 + 4) − p2 (p4 + 44p2 + 16) + · · · (4.4)
3! 5! 7!
in agreement with the known expansions for am(p; x) [11].

Example 4.6 The Lambert-W function. Let f (x) = e−x (x + 1)−1 , a = 0, f (a) = 1.
Here
h(x) = xex , H(x) = LW (x)

10
where by LW (x) we denote the Lambert-W function [9], [10], [22]. In this case, (2.2) gives

D1 [f ] (0) = −2
D2 [f ] (0) = 9
D3 [f ] (0) = −64
..
.
D [f ] (0) = [−(n + 1)]n .
n

From (4.1) we conclude that

X zn
LW (z) = (−1)n−1 nn−1 .
n≥1
n!

Example 4.7 We now derive a well known result [1] about reversion of series. If we take

h(x) = a1 x + a2 x2 + a3 x3 + a4 x4 + a5 x5 + a6 x6 + a7 x7 + · · ·

where a1 6= 0, then
1 1
f (x) = =
h′ (x) a1 + 2a2 x + 3a3 x2 + 4a4 x3 + 5a5 x4 + 6a6 x5 + 7a7 x6 + · · ·
1
a = 0, f (0) = a1
and from (2.2) we get
a2
D1 [f ] (0) = −2
(a1 )2
2 2 (a2 )2 − a1 a3
D [f ] (0) = 6
(a1 )4
3 5a1 a2 a3 − (a1 )2 a4 − 5(a2 )3
D [f ] (0) = 24
(a1 )6
4 6(a1)2 a2 a3 + 3 (a1 a3 )2 + 14(a2 )4 − (a1 )3 a5 − 21a1 (a2 )2 a3
D [f ] (0) = 120 .
(a1 )8

Hence,

1 a2 2 2 (a2 )2 − a1 a3 3 5a1 a2 a3 − (a1 )2 a4 − 5(a2 )3 4

H(z) = z − z + z + z
a1 (a1 )3 (a1 )5 (a1 )7
6(a1 )2 a2 a3 + 3 (a1 a3 )2 + 14(a2 )4 − (a1 )3 a5 − 21a1 (a2 )2 a3 5
+ z +··· .
(a1 )9

11
Remark 4.8 An explicit formula for the nth term is given in Morse and Feshbach [24, Part
1 pp. 411–413],
 s  t
1 X s+t+u+··· n(n + 1) · · · (n − 1 + s + t + u + · · · ) a2 a3
bn = n (−1) ···
n (a1 ) s,t,u,... s!t!u! · · · a1 a1
s + 2t + 3u + · · · = n − 1.

Example 4.9 The Error Function, erf(x). We now have

Zx
2 2
h(x) = erf(x) = √ e−t dt
π
0
√ √
π x2 π
f (x) = e , a = 0, f (a) =
2 2
and (2.2) gives (
Dn [f ] (0) =  √0, n n = 2k + 1, k ≥ 0
π
2
Ak , n = 2k, k ≥ 0
where

A0 = 1, A1 = 2, A2 = 28, A3 = 1016, A4 = 69904

A5 = 7796768, A6 = 1282366912, A7 = 291885678464, . . . .

From (4.1) we get

√ 2n+1
X π z 2n+1
H(z) = An .
n≥0
2 (2n + 1)!

which agrees with other authors calculations previously published [3], [7], [13], [15], [30].

We will now extend (4.1) to a more general result.

Rx 1
Corollary 4.10 Let h(x) = f (t)
dt, z0 = h(b), with f (b) 6= 0, ±∞ and its inverse H(x) =
a
h−1 (x). Then,
X (z − z0 )n
H(z) = b + f (b) Dn−1 [f ] (b) (4.5)
n≥1
n!

where |z − z0 | < ε, for some ε > 0.

Proof. We consider the function

u(x) = h(x) − z0

12
which satisfies u(b) = 0, and its inverse U(x) = u−1 (x). Since f (b) 6= 0, ±∞, we can apply
(4.1) to u(x) and conclude that
X zn
U(z) = b + f (b) Dn−1 [f ] (b) .
n≥1
n!

All that is left is to see the relation between U(z) and H(z).
Suppose that u(x) = y. Then

y = u(x) = h(x) − z0
h(x) = y + z0
x = H(y + z0 )

and therefore
U(y) = H(y + z0 )
or
H(z) = U(z − z0 )
and (4.5) follows.
We will now use our results to get some power series expansions that have not been
studied before.

Example 4.11 The incomplete Gamma function, γ(ν; x). We have

Zx
h(ν; x) = γ(ν; x) ≡ e−t tν−1 dt, ν > 0, x≥0
0
x 1−ν
f (ν; x) = e x , a = 0.

Since 
0, 0 < ν < 1
f (ν; 0) =
∞, ν > 1
we can’t apply (4.1). Choosing b = 1, z0 (ν) = γ(ν; 1), f (ν; b) = e, we conclude from
(4.5) that
X [z − z0 (ν)]n
H(ν; z) = 1 + e Dn−1 [f ] (1) .
n≥1
n!

We use (2.2) to compute the first few Dn [f ] (1) and obtain

Dn [f ] (1) = en Qn (ν)

13
where Qn (ν) is a polynomial of degree n

Q1 (ν) = 2 − ν
Q2 (ν) = 7 − 7ν + 2ν 2
Q3 (ν) = 36 − 53ν + 29ν 2 − 6ν 3
Q4 (ν) = 245 − 474ν + 375ν 2 − 146ν 3 + 24ν 4
Q5 (ν) = 2076 − 4967ν + 5104ν 2 − 2847ν 3 + 874ν 4 − 120ν 5

and we can write

X
n [z − z0 (ν)]n
H(ν; z) = 1 + e Qn−1 (ν) .
n≥1
n!

Example 4.12 The sine integral function, Si(x). In this case

Zx
sin(t) x
h(x) = Si(x) ≡ dt, f (x) = , a = 0.
t sin(x)
0

For this example f (a) is well defined, but to simplify the calculations we choose b =
π
2
, z0 = Si( π2 ) ≃ 1.370762. Then,
π π 
f (b) = , Dn [f ] = Qn (π)
2 2
where Qn (x) is once again a polynomial

Q1 (x) = 1
1
Q2 (x) = 1 + x2
4
7
Q3 (x) = 1 + x2
4
9 4
Q4 (x) = 1 + 8x2 + x
16
61 159 4
Q5 (x) = 1 + x2 + x
2 16
423 2 1671 4 225 6
Q6 (x) = 1 + x + x + x.
4 16 64
and from (4.5) we obtain

π πX (z − z0 )n
H(z) = + Qn (π) .
2 2 n≥1 n!

14
Example 4.13 The logarithm integral function, li(x). From the definition
Zx
1
h(x) = li(x) ≡ dt, f (x) = ln(x), a = 0.
ln(t)
0

In this case f (a) = −∞, so we must choose b. A natural candidate is b = e, which gives
f (b) = 1, z0 = li(e) ≃ 1.895117816
Dn [f ] (e) = e−n An
with
A1 = 1, A2 = 0, A3 = −1, A4 = 2, A5 = 1
A6 = −26, A7 = 99, A8 = 90, A9 = −3627
and we have
X (z − z0 )n
H(z) = e + An .
n≥1
n!

Example 4.14 The incomplete Beta function, B(ν, µ; x). By definition

Zx
h(ν, µ; x) = B(ν, µ; x) ≡ tν−1 (1 − t)µ−1 dt, 0≤x<1
0

and hence
f (ν, µ; x) = x1−ν (1 − x)1−µ , a = 0.
To avoid the possible singularities at x = 0 and x = 1 we consider b = 21 , and therefore
1 1
f (ν, µ; b) = 2ν+µ , z0 (ν, µ) = B(ν, µ; ).
4 2
The D algorithm now gives
 
n 1
D [f ] = 2n(ν+µ−1) Qn (ν, µ)
2
with Qn (ν, µ) a multivariate polynomial of degree n
Q1 (ν, µ) = µ − ν
Q2 (ν, µ) = −2 + ν − 4νµ + µ + 2µ2 + 2ν 2
Q3 (ν, µ) = (µ − ν)(6µ2 − 12νµ + 7µ − 12 + 6ν 2 + 7ν)
Q4 (ν, µ) = 16 − 46νµ2 − 46ν 2 µ − 63µ2 − 22µ + 154νµ − 96νµ3 − 96ν 3 µ+
144ν 2 µ2 − 22ν − 63ν 2 + 24ν 4 + 46ν 3 + 24µ4 + 46µ3
Q5 (ν, µ) = (µ − ν)(120µ4 + 326µ3 − 480νµ3 + 720ν 2 µ2 − 323µ2 − 326νµ2
− 362µ − 480ν 3 µ + 1154νµ − 326ν 2 µ − 323ν 2 + 240 + 120ν 4
+ 326ν 3 − 362ν).

15
and we have
1 1 ν+µ X n(ν+µ−1) (z − z0 )n
H(z) = + 2 2 Qn (ν, µ) .
2 4 n≥1
n!

Conclusion 4.15 We have presented a simple method for computing the series expansion
for the inverses of functions and given a Maple procedure to generate the coefficients in these
expansions. We showed several examples of the method applied to elementary and special
functions, and stated the first few terms of the series in each case.
Acknowledgement 4.16 I wish to express my gratitude to the referees for their extremely
valuable suggestions and comments on previous versions of this work.

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