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Taylor series via division∗
pahio†
2013-03-22 0:10:22
Let the real (or complex) functions f and g have the Taylor series
f (x) = a0 + a1 (x − a) + a2 (x − a)2 + . . .
g(x) = b0 + b1 (x − a) + b2 (x − a)2 + . . .
on an interval I (or a circle in C) centered at x = a. If b0 6= 0, then also the
f (x)
quotient
apparently has the derivatives of all orders on I. It is not hard
g(x)
to justify that if one divides the series of f by the series of g, the obtained series
f (x)
= c0 + c1 (x − a) + c2 (x − a)2 + . . .
g(x)
(1)
is identically same as the Taylor series of f (x)/g(x) on I.
We consider the coefficients cn of (1) as undetermined constants. 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
of x−a with the corresponding coefficients of the series of f . Accordingly, we
have the conditions
a0 = b0 c0 ,
a1 = b0 c1 + b1 c0 ,
a2 = b0 c2 + b1 c1 + b2 c0 ,
...
(2)
(n = 0, 1, 2, . . .).
(3)
Since for every index n, the equation
an = b0 cn + b1 cn−1 + b2 cn−2 + . . . + bn c0
holds and b0 6= 0, we get the recurrence relation
cn = −
b1
b2
bn
an
cn−1 − cn−2 − . . . − c0 +
b0
b0
b0
b0
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Example. We will calculate the Bernoulli numbers, which are the numbers
x
Bn appearing in the Taylor series of x
expanded with the powers of x:
e −1
∞
X
x
Bn n
=
x
ex − 1
n!
n=1
(4)
This function has really all derivatives in the point x = 0, since in this point
x
2
x
the inverse e x−1 = 1 + 2!
+ x3! + . . . naturally has the derivatives and the value
1 distinct from zero. Let us think the division of x by the Taylor series of ex−1.
Corresponding to (1), we denote the right side of (4) as c0 + c1 x + c2 x2 + . . ..
n
2
When we now think this series and the series x + x2! + . . . + xn! + . . . of the
x
denominator of x
to be multiplied, the result must be x, i.e. the coefficients
e −1
of all powers of x except the first power are 0. So the two first conditions
corresponding to (2) are c0 = 1, c1 + 21 c0 = 0; thus
c0 = B0 = 1,
1
c1 = B1 = − .
2
Setting the coefficient of xn equal to zero gives the formula
c0
c1
cn−2
+
+ ... +
+ cn−1 = 0
n! (n − 1)!
2!
for n ≥ 2. Putting here ci =
Bi
i!
(5)
to (5) we obtain
B1
B2
Bn−2
cn−1
B0
+
+
+ ... +
+
= 0,
0!n! 1!(n − 1)! 2!(n − 2)!
(n − 2)!2! (n − 1)!
and multiplying this by n!,
n
n
n
n
n
B0 +
B1 +
B2 + . . . +
Bn−2 +
cn−1 = 0.
n
n−1
n−2
2
1
This yields, by substituting the values of B0 and B1 and recalling that the odd
Bernoulli numbers are zero (n > 2), the recursion formula
1−2k
2k+1
2k+1
2k+1
2k+1
+
B2 +
B4 + . . . +
B2k−2 +
B2k = 0
2
2
4
2k−2
2k
for the even Bernoulli numbers B2k (k = 1, 2, . . .). It gives successively
1
− + 3B2 = 0,
2
3
− + 10B2 + 5B4 = 0,
2
5
− + 21B2 + 35B4 + 7B6 = 0,
2
1
From here we obtain B2 = 61 , B4 = − 30
, B6 =
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
2
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
1
2
2
expand 1+x
2 with the powers of x−1, we write 1 + x = 2+2(x−1)+(x−1) . The
two first conditions corresponding to (2) are 2c0 = 1 and 2c1 +2c0 = 0, whence
c0 = 12 and c1 = − 12 . The coefficient of (x−1)n gives the condition 2cn +2cn−1 +
cn−2 = 0, whence the simple recursion formula cn = −cn−1 − 12 cn−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).
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