International Journal of Advanced Research in Mathematics
ISSN: 2297-6213, Vol. 7, pp 1-9
doi:10.18052/www.scipress.com/IJARM.7.1
© 2016 SciPress Ltd., Switzerland
Submitted: 2016-08-30
Revised: 2016-09-13
Accepted: 2016-10-04
Online: 2016-12-15
Expansion of Function z ln z in the Quasi-Reciprocal Continued Fraction
Mykhaylo Pahirya
Department of natural sciences and information technology,
Mukachevo state university,
Mukachevo, Ukraine
[email protected]
Keywords: Quasi-reciprocal continued fractions, approximation functions, functions of complex variable, reciprocal derivatives of 2-nd type, Thiele type formula, quasi-reciprocal continued C-fraction.
Absract. Expansion of function z ln z in the quasi-reciprocal continued fraction has been obtained.
Convergence region of expansion has been established.
Introduction
It is well known that function of complex or real variable can be approximated by polynomials [1],
by generalized polynomials, by rational functions [2], by Padé approximation [3] or by continued
fractions. Approximation of functions is used in the calculation of the function [4, 5]. Basic methods of
expansion of functions in continued fractions are considered in monographs [6, 7]. Danish mathematician T.N.Thiele in his book [8] presented for the first time reciprocal derivatives and proposed a
formula which is analogous to the Taylor formula in the theory of continued fractions.Further research
were performed Nörlund in [9].
The reciprocal derivatives of 2-nd type are introduced into consideration in the paper [10] and
analogue of Thiele formula for quasi-reciprocal continued fractions is received. Problem of expansion
of function z ln z in the Thiele continued fraction and regular continued C-fraction have been studied
in the [11]. Convergence regions of expansions have been obtained. This paper is a continuation of
previous studies. Problem of expansion function z ln z in the quasi-reciprocal continued fraction is
discussed and convergence region of expansion is determined.
Quasi-reciprocal continued fractions
In [10] has been shown that if f (z) is analytic function on the compact Z ⊂ C then function can be
expanded in the Thiele-type quasi-reciprocal continued fraction (T-QCF) of the form in a neighborhood of the point z∗ ∈ Z
(
)−1
z − z∗ z − z∗
z − z∗
f (z) = d0 +
,
(1)
d1 + d2 + · · · dn + · · ·
where
d0 =
1
,
f (z∗ )
dn = (
d1 = {1}f (z∗ ),
n
)′ = {n}f (z∗ ) − {n−2}f (z∗ ).
f (z∗ )
{n−1}
(2)
Here {n}f (z) is a reciprocal derivative of 2-nd type. A reciprocal derivative of 2-nd type is defined by
the recurrence relation
1
f (z) =
,
f (z)
{0}
−f 2 (z)
f (z) = ′
,
f (z)
{1}
f (z) = (
{n}
n
)′ + {n−2}f (z),
f (z)
{n−1}
n = 2, 3, . . . . (3)
It is well known [12] that (1) can be written in the form of equivalent continued fraction
(
)−1
en (z − z∗ )
e1 (z − z∗ ) e2 (z − z∗ )
f (z) = e0 +
,
1
+
1
+ ···
1
+ ···
SciPress applies the CC-BY 4.0 license to works we publish: https://creativecommons.org/licenses/by/4.0/
(4)
2
Volume 7
where
1
1
, dn =
,
n = 2, 3, . . . .
d1
dn−1 dn
Continued fraction (4) is called quasi-reciprocal continued C-fraction (C-QCF).
If repeated consideration of [12] we can prove the following theorem.
e0 = d0 ,
e1 =
(5)
Theorem 1. Let
Ra = {z : | arg(a(z − z∗ ) + 1/4)| < π}
and elements of C-QCF
e1 (z − z∗ ) e2 (z − z∗ ) e3 (z − z∗ )
1
e0 +
1
+
1
+
1
+ ···
(6)
such that
lim en = a ̸= 0,
n→∞
a ∈ C.
(7)
Then
(A) continued fraction convergence to a function f (z), which is meromorphic in Ra ;
(B) if compact Z ⊂ Ra doesn’t contains poles of function f (z) then convergence is uniform;
(C) function f (z) is holomorphic in point z = z∗ .
A function ln(c + z), c = const, has expansion in the T-QCF of following form
ln(c + z) =
∗
S0∗ (z − z∗ )S−1
(z − z∗ )S1∗ (z − z∗ )S0∗ (z − z∗ )S2∗
1 −
zS0∗
+
2
+
3zS1∗ +
1
+
∗
(z − z∗ )Sk∗ (z − z∗ )Sk−1
(z − z∗ )S1∗
,
+
5zS2∗ + · · · +
2/k
+ (2k + 1)zSk∗ + · · ·
(8)
where
∗
z = c + z∗ , S−1
= 1, S0∗ = ln z, Sn∗ = 2Hn + ln z, H0 = 0, Hn =
n
∑
1/k,
n = 1, 2, . . . . (9)
k=1
In accordance with (5), coefficients of expansion function ln(c + z) in the C-QCF in a neighborhood
of the point z∗ will be equal
e0 =
∗
nSn−1
1
−1
nSn∗
,
e
=
,
e
=
,
e
=
,
1
2n
2n+1
∗
S0∗
z(S0∗ )2
2(2n − 1)zSn−1
2(2n + 1)zSn∗
n = 1, 2, . . . .
That is we have expansion function in the C-QCF
∗
/zS0∗ (z − z∗ )S1∗ /2zS0∗ (z − z∗ )S0∗ /6zS1∗
S0∗ (z − z∗ )S−1
ln(c + z) =
1 −
1
+
1
+
1
+
(z − z∗ )2S2∗ /6zS1∗ (z − z∗ )2S1∗ /10zS2∗
1
+
1
+ ···+
+
∗
∗
∗
(z − z∗ )nSn /2(2n − 1)zSn−1 (z − z∗ )nSn−1 /2(2n + 1)zSn∗
.
+
1
+
1
+ ···
(10)
Theorem 2. (A) C-QCF (10) and equivalent T-QCF (8) are converges to the function ln(c + z) on
the set
R(c, z∗ ) = {z ∈ C\{−c} : | arg(z + c) − arg(c + z∗ )| < π}
(B) C-QCF (8) and T-QCF (10) are converges uniformly on the arbitrary compact Z ⊂ R(c, z∗ ).
International Journal of Advanced Research in Mathematics Vol. 7
3
Expansion function z ln(z) in the quasi-reciprocal continued fraction
The main result of this paper will be proved.
Theorem 3. (A) Function z ln z, where z ̸= 0, has reciprocal derivatives of 2-nd type of arbitrary
order which are defined including the formulae
z 2 ln2 z
,
ln z + 1
{0}
(z ln z) =
1
,
z ln z
{2n}
(z ln z) =
−1
,
z(n(n + 1) ln z + αn ln z + βn )
(11b)
(z ln z) =
z 2 (An ln3 z + Bn ln2 z + Cn ln z + Dn )
,
ln z + γn
(11c)
{2n+1}
{1}
(z ln z) = −
2
(11a)
where
αn = 4n(n + 1)Hn − (2n2 + 2n + 1), γn = 2Hn + 1/(n + 1),
βn = 4n(n + 1)Hn2 − 2(2n2 + 2n + 1)Hn + 2n(n + 1),
(12)
n(n+1)2 (n+2)
(n+1)(5n3 +12n2 +6n+2)
2
An =
, Bn = 3n(n+1) (n+2)Hn −
,
2
2
Cn = 6n(n + 1)2 (n + 2)Hn2 − 2(n + 1)(5n3 + 12n2 + 6n + 2)Hn +
Dn = 4n(n + 1)2 (n + 2)Hn3 −
+n(21n3 + 64n2 + 57n + 18)/4,
−2(n + 1)(5n3 + 12n2 + 6n + 2)Hn2 + n(21n3 + 64n2 + 57n + 18)Hn /2−
−n(16n3 + 43n2 + 27n + 2)/4,
n = 1, 2, . . . , Hn defined in (8).
(B) Coefficients of expansion function z ln z in the neighborhood of point z = z∗ are equals
d0 =
1
− z∗2 z2
−2(z+1)2
3z∗2 (2z2 + 3z + 2)2
, d1 =
, d2 =
,
d
=
,
3
z∗ z
z+1
z∗ z(2z2 +3z+2)
2(z+1)(z+ 52 )
2n(z + γn−1 )2
d2n =
,
z∗ ((n − 1)nz2 + αn−1 z + βn−1 )(n(n + 1)z2 + αn z + βn )
(
)2
(2n + 1)z∗2 n(n + 1)z2 + αn z + βn
(
)(
)
d2n+1 =
,
n(n + 1) z + γn−1 z + γn
(13a)
(13b)
(13c)
where z = ln z∗ , n = 2, 3, . . . .
Proof. (A) Relationship (11a) follows from the definition of reciprocal derivative of 2-nd type. We
use inductive method. It is easy to show that
{2}
{3}
(z ln z) =
(z ln z) =
−1
,
z(2 ln z + 3 ln z + 2)
2
z 2 (6 ln3 z + 11 ln2 z + 12 ln z + 6)
.
ln z + 5/2
Thus (11b) and (11c) holds at n = 1. Let (11b) and (11c) holds at n = k − 1. Thus from (3) follows
that
)′
/( z 2 (A
3
2
k−1 ln z + Bk−1 ln z + Ck−1 ln z + Dk−1 )
{2k}
(z ln z) = 2k
−
ln z + γk−1
4
Volume 7
)
(
/ (
−1/z (k − 1)k ln2 +αk−1 ln z + βk−1 = 2k(ln z + γk−1 )2 z 2Ak−1 ln4 z + (2Bk−1 +
+2Ak−1 (1 + γk−1 )) ln3 z + (2Ck−1 + Bk−1 (1 + 2γk−1 ) + 3Ak−1 γk−1 ) ln2 z+
)
+(2Dk−1 + (2Ck−1 + 2Bk−1 )γk−1 ) ln z + (2Dk−1 + Ck−1 )γk−1 − Dk−1 −
−1/z((k − 1)k ln2 +αk−1 ln z + βk−1 ).
Substituting the values of αk−1 , βk−1 , γk−1 , Ak−1 , Bk−1 , Ck−1 we obtain
/ (
(
{2k}
(z ln z) = 2k(ln z + 2Hk−1 + k1 )2 z (k 2 − 1)k 2 ln4 z + 8(k 2 − 1)k 2 Hk−1 −
)
(
2
−2k(2k 3 − 2k 2 − k + 2) ln3 z + 24(k 2 − 1)k 2 Hk−1
− 12(2k 3 − 2k 2 − k + 2)Hk−1 +
)
(
2
3
+
− 24k(2k 3 − 2k 2 − k + 2)Hk−1
+8k 4 − 12k 3 + 6k − 5 ln2 z + 32(k 2 − 1)k 2 Hk−1
)
+4(8k 4 − 12k 3 + 6k − 5)Hk−1 − k2 (4k 5 − 8k 4 + 2k 3 − 2k + 1) ln z+
4
3
2
+16(k 2 − 1)k 2 Hk−1
− 16k(2k 3 − 2k 2 − k + 2)Hk−1
+ 4(8k 4 − 12k 3 + 6k − 5)Hk−1
−
)
− k4 (4k 5 − 8k 4 + 2k 3 + 4k 2 − 2k + 1)Hk−1 + 4(k − 1)(k 3 − k 2 + 1) −
(
(
)
2
−1/z (k − 1)k ln2 z + 4(k − 1)kHk−1 − 2k 2 + 2k − 1 ln z + 4(k − 1)kHk−1
−
)
−2(2k 2 − 2k + 1)Hk−1 + 2(k − 1)k .
The denominate of the fist fraction can be presented as the following product
(
(
)
2
z ((k − 1)k ln2 z + 4(k − 1)kHk−1 − 2k 2 + 2k − 1 ln z + 4(k − 1)kHk−1
−
)(
(
−2(2k 2 − 2k + 1)Hk−1 + 2(k − 1)k) k(k + 1) ln2 z + 4k(k + 1)Hk−1 −
)
−2k + 2k + 3 ln z + 4k(k +
2
2
1)Hk−1
− 2(2k − 2k − 3)Hk−1 +
2
2
(k 3
k
)
− k + 1) .
2
We reduce to a common denominator and take replacement Hk−1 = Hn − k1 . We finally get
(
(
)
{2k}
(z ln z) = −1/z k(k + 1) ln2 z + 4k(k + 1)Hk − 2k 2 − 2k − 1 ln z+
)
+4k(k + 1)Hk2 − 2(2k 2 + 2k + 1)Hk + 2k(k + 1) =
−1
.
k(k + 1) ln z + αk ln z + βk
2
Analogously
/(
)′
−1
(z ln z) = (2k + 1)
+
z(k(k + 1) ln2 z + αk ln z + βk )
/
+z 2 (Ak−1 ln3 z + Bk−1 ln2 z + Ck−1 ln z + Dk−1 ) (ln z + γk−1 ) =
{2k+1}
=
(2k + 1)z 2 (k(k + 1) ln2 z + αk ln z + βk )2
+
k(k + 1) ln2 z + (αk + 2k(k + 1)) ln z + αk + βk
+
z 2 (Ak−1 ln3 z + Bk−1 ln2 z + Ck−1 ln z + Dk−1 )
.
ln z + γk−1
International Journal of Advanced Research in Mathematics Vol. 7
5
If in denominates of fractions substitutes values of αk , βk and γk−1 then we have
k(k + 1) ln2 z + (αk + 2k(k + 1)) ln z + αk + βk = k(k + 1) ln2 z + (4k(k + 1)Hk − 1) ln z+
)(
)
(
+4k(k + 1)Hk2 − 2Hk − 1 = ln z + 2Hk − k1 k(k + 1) ln z + 2k(k + 1)Hk + k ,
ln z + γk−1 = ln z + 2Hk − k1 .
Now substituted in numerators values αk , βk , Ak−1 , Bk−1 , Ck−1 , Dk−1 and reducing to a common
denominator. We get
( 2
(
3
2
{2k+1}
(z ln z) = z 2 k (k+1)2 (k+2) ln4 z + 4k 2 (k + 1)3 (k + 2)Hk − k(k+1)
(5k 3 + 13k 2 +
2
)
(
+9k + 4) ln3 z + 12k 2 (k + 1)3 (k + 2)Hk2 − 3k(k + 1)2 (5k 3 + 13k 2 + 9k + 4)Hk +
)
(
+ k+1
(21k 5 + 74k 4 + 91k 3 + 54k 2 + 16k + 4) ln2 z + 16k 2 (k + 1)3 (k + 2)Hk3 −
4
−6k(k + 1)2 (5k 3 + 13k 2 + 9k + 4)Hk2 + (k + 1)(21k 5 + 74k 4 + 91k 3 + 54k 2 + 16k + 4)Hk −
)
− k(k+1)
(16k 4 + 64k 3 + 91k 2 + 59k + 18) ln z + 8k 2 (k + 1)3 (k + 2)Hk4 − 4k(k + 1)2 ×
4
×(5k 3 + 13k 2 + 9k + 4)Hk3 + (k + 1)(21k 5 + 74k 4 + 91k 3 + 54k 2 + 16k + 4)Hk2 − k(k+1)
×
2
)
k × (16k 4 + 64k 3 + 91k 2 + 59k + 18)Hk + k(k+1)
(16k 3 + 43k 2 + 27k + 2) ×
4
((
)(
))−1
1
× ln z + 2Hk − k k(k + 1) ln z + 2k(k + 1)Hk + k
.
We divide numerate and denominate to k(k + 1) ln z + 2k(k + 1)Hk − k − 1. We finally obtain
(
(
2
{2k+1}
(z ln z) = k(k+1)2 (k+2) ln3 z + 3k(k + 1)2 (k + 2)Hk − k+1
(5k 2 + 12k 2 + 6k+
2
)
(
+2) ln2 z + 6k(k + 1)2 (k + 2)Hk2 − (k + 1)(10k 3 + 24k 2 + 12k + 4)Hk + k4 (21k 3 + 64k 2 +
)
+75k + 18) ln z + 4k(k + 1)2 (k + 2)Hk3 − 2(k + 1)(5k 3 + 12k 2 + 6k + 2)Hk2 +
)/(
)
3
2
3
2
k
k
1
+ 2 (21k + 64k + 57k + 18)Hk − 4 (16k + 43k + 27k + 2)
.
ln z + 2Hk + k+1
In other words
{2k+1}
(z ln z) =
z 2 (Ak ln3 z + Bk ln2 z + Ck ln z + Dk )
.
ln z + γk
Formulae (11) have been proven.
(B) Formulae (13a) follows directly from (11a) and (11b)-(11c) at n = 1.
According to (11b) we get that
(z ln z) − {2n−2}(z ln z) =
{2n}
+
1
−1
+
z(n(n + 1) ln z + αn ln z + βn )
2
(
)
= 2n ln2 z + (αn − αn−1 ) ln z + βn − βn−1 /
z((n − 1)n ln2 z + αn−1 ln z + βn−1 )
(
)
z((n − 1)n ln2 z + αn−1 ln z + βn−1 )(n(n + 1) ln2 z + αn ln z + βn ) .
Substituting values of αn−1 , βn−1 , αn , βn from (2) we obtain that in the neighborhood of point z = z∗
d2n
2n(z + γn−1 )2
=
.
z∗ ((n − 1)nz2 + αn−1 z + βn−1 )(n(n + 1)z2 + αn z + βn )
6
Volume 7
Thus (13b) holds.
Similarly
(z ln z) −
{2n+1}
−
z 2 (An ln3 z + Bn ln2 z + Cn ln z + Dn )
(z ln z) =
−
ln z + γn
{2n−1}
z 2 (An−1 ln3 z + Bn−1 ln2 z + Cn−1 ln z + Dn−1 )
.
ln z + γn−1
Reducing to common denominate and substituting the values of An , Bn , Cn , Dn , γn , An−1 , Bn−1 ,
Cn−1 , Dn−1 , γn−1 , will be
(
(
{2n+1}
{2n−1}
2
(z ln z) −
(z ln z) = (2n + 1)z n(n + 1) ln4 z + 8n(n + 1)Hn −
)
(
2 +2n+1)+1 )
−2(2n2 + 2n + 1) ln3 z + 24n(n + 1)Hn2 − 12(2n2 + 2n + 1)Hn + 4n(n+1)(2n
×
n(n+1)
(
2 +2n+1)+1)
× ln2 z + 32n(n + 1)Hn3 − 24(2n2 + 2n + 1)Hn2 + 4(4n(n+1)(2n
Hn −
n(n+1)
)
2 +2n+1)+1)
×
−4(2n2 + 2n + 1) ln z + 16n(n + 1)Hn4 − 16(2n2 + 2n + 1)Hn3 + 4(4n(n+1)(2n
n(n+1)
)/((
)
)(
)
×Hn2 − 8(2n2 + 2n + 1)Hn + 4n(n + 1)
ln z + γn−1 ln z + γn .
If multiple numerate and denominate to n(n + 1) and substitutes z = z∗ than we have
d2n+1
(
)2
(2n + 1)z∗2 n(n + 1)z2 + αn z + βn
(
)(
)
=
.
n(n + 1) z + γn−1 z + γn
Relationship (13c) has been proven too. Also, formulae (13) are true.
From theorem 3 follows that function z ln z in the neighborhood of point z = z∗ is expanded into
T-QCF of the form
(
1
z − z∗
z − z∗
+
z ln z =
2
2
2
z∗ z −z∗ z /(z + 1)+ −2(z + 1) /z∗ z(2z2 + 3z + 2)+
+
z − z∗
2
2
3z∗ (2z +3z+2)2 /2(z+1)(z+ 25 )+
+ ···+
z − z∗
( 5 )2
4 z+ 2 /z∗ (2z2 +3z+2)(6z2 +23z+27) +
z − z∗
2
2
2n(z+γn−1 ) /z∗ ((n−1)nz +αn−1 z+βn−1 )(n(n+1)z2 +αn z+βn )+
)−1
z − z∗
.
+ (2n+1)z∗2 (n(n+1)z2 +αn z+βn )2 /n(n+1)(z+γn−1 )(z+γn )+ · · ·
(14)
In the particular case at z∗ = e and z = 1 we have
(
1
z−e
z−e
z−e
z−e
z−e
z ln z =
+
e −e2 /2+ −8/7e + 21e2 /2+ 1/8e + 2240e2 /13+ · · · +
z−e
+
2n(2Hn +(n−1)/n)2
2
e(4(n−1)nHn−1
−2Hn−1 +n2 −n−1)
z−e
+
2 −2H +n2 +n−1)2
(2n+1)e2 (4n(n+1)Hn
n
n(n+1)(2Hn +(n−1)/n)(2Hn+1 +n/(n+1))
)−1
+ ···
.
International Journal of Advanced Research in Mathematics Vol. 7
7
According to (5) and (13) coefficients of expansion function z ln z into C-QCF in the neighborhood of
point z = z∗ are equal
1
−(z+1)
2z2 +3z+2
−z(z+5/2)
, e1 =
,
e
=
,
e
=
,
2
3
z∗ z
z∗2 z2
2z∗ z(z+1)
3z∗ (z+1)(2z2 +3z+2)
(n − 1)(z + γn−2 )(n(n + 1)z2 + αn z + βn ))
e2n =
,
2(2n − 1)z∗ (z + γn−1 )((n − 1)nz2 + αn−1 z + βn−1 )
(n + 1)(z + γn )((n − 1)nz2 + αn−1 z + βn−1 ))
e2n+1 =
, n = 2, 3, . . . ,
2(2n + 1)z∗ (z + γn−1 )(n(n + 1)z2 + αn z + βn )
e0 =
(15)
where z = ln z∗ and values αn , βn , γn determined in (12).
We introduce the notation
pn =z + γn = ln z∗ + 2Hn + 1/(n + 1),
n = 0, 1, . . . ,
(
qn =n(n + 1)z2 + αn z + βn = n(n + 1) ln2 z∗ + 4n(n + 1)Hn − 2n2 −
)
−2n−1 ln z∗ +4n(n+1)Hn2 −2(2n2 +2n+1)Hn +2n(n+1).
(16)
If used the notation (16) in the coefficient (15) than we obtain expansion of function z ln z in the
neighborhood of point z = z∗ into C-QCF
z ln z =
z∗ ln z∗ (z − z∗ )p0 /z∗ q0 (z − z∗ )q1 /2z∗ ln z∗ p0
1 +
1
+
1
+
(z − z∗ )p1 q0 /3z∗ p0 q1 (z − z∗ )p0 q2 /6z∗ p1 q1 (z − z∗ )p2 q1 /10z∗ p1 q2
1
+
1
+
1
+
+
(z − z∗ )(n − 1)pn−2 qn /2(2n − 1)z∗ pn−1 qn−1
+ ···+
1
+
(z − z∗ )(n + 1)pn qn−1 /2(2n + 1)z∗ pn−1 qn
.
+
1
+ ···
If z = e than the expansion (17) takes the form
z ln z =
(17)
e (z − e)pe0 /eqe0 (z − e)qe1 /2epe0 (z − e)pe1 qe0 /3epe0 qe1
1+
1
+
1
+
1
+
(z − e)pe0 qe2 /6epe1 qe1 (z − e)pe2 qe1 /10epe1 qe2
+
1
+
1
+ ···+
e
e
e
e
(z − e)(n − 1)pn−2 qn /2(2n − 1)epn−1 qn−1
+
1
+
(z − e)(n + 1)pen qen−1 /2(2n + 1)epen−1 qen
,
1
+ ···
+
where
pen = 2Hn+1 + n/(n + 1),
qen = 4n(n + 1)Hn2 − 2Hn + n2 + n − 1,
n = 0, 1, . . . .
Theorem 4. (A) C-QCF (17) and T-QCF (14) are convergent to the function z ln z on set
R(0, z∗ ) = {z ∈ C\{0} : | arg(z) − arg(z∗ )| < π}
(B) C-QCF (17) and T-QCF (14) are convergent uniformly on the arbitrary compact Z ⊂ R(0, z∗ ).
8
Volume 7
Proof. Take limit
1
z + 2Hn−2 + n−1
(n − 1)pn−2 qn
n−1
lim e2n = lim
= lim
lim
×
n→∞
n→∞ 2(2n − 1)z∗ pn−1 qn−1
n→∞ 2(2n − 1)z∗ n→∞ z + 2Hn−1 + 1
n
(
(
)
× lim n(n+1)z2 + 4n(n+1)Hn −2n2 −2n−1 z+4n(n+1)Hn2 −2(2n2 +2n+1)Hn +
n→∞
)/(
(
)
2
+2n(n + 1)
n(n − 1)z2 + 4n(n − 1)Hn−1 − 2n2 + 2n − 1 z + 4n(n − 1)Hn−1
−
)
−2(2n2 − 2n + 1)Hn−1 + 2n(n − 1) .
Since limn→∞ Hn /Hn−1 = 1 than
1
n−1
2Hn−1 + n1
z/2Hn−2 + 1 + 1/2(n − 1)Hn−2
= 1.
n→∞ z +
n→∞ z/2Hn−2 + Hn−1 /Hn−2 + 1/2nHn−2
(
(
)
lim n(n+1)z2 + 4n(n+1)Hn −2n2 −2n−1 z+4n(n+1)Hn2 −2(2n2 +2n+1)Hn +
n→∞
)/(
(
)
2
+2n(n + 1)
n(n − 1)z2 + 4n(n − 1)Hn−1 − 2n2 + 2n − 1 z + 4n(n − 1)Hn−1
−
(
)
n(n + 1)z2 4n(n + 1)Hn z
+
−
−2(2n2 − 2n + 1)Hn−1 + 2n(n − 1) = lim
2
2
n→∞
4n2 Hn−1
4n2 Hn−1
)
(2n2 + 2n + 1)z
4n(n + 1)Hn2 2(2n2 + 2n + 1)Hn 2n(n + 1)
−
z+
−
+
:
2
2
2
2
4n2 Hn−1
4n2 Hn−1
4n2 Hn−1
4n2 Hn−1
(
2
n(n − 1)z2 4n(n − 1)Hn−1 z (2n2 − 2n + 1)z 4n(n − 1)Hn−1
+
−
+
−
:
2
2
2
2
4n2 Hn−1
4n2 Hn−1
4n2 Hn−1
4n2 Hn−1
)
2(2n2 − 2n + 1)Hn−1 2n(n − 1)
−
+
= 1.
2
2
4n2 Hn−1
4n2 Hn−1
In that case limn→∞ e2n = 1/4z∗ . Similarly is proved that limn→∞ e2n+1 = 1/4z∗ .
Thus
1
.
lim en =
n→∞
4z∗
Further
(z − z
1)
∗
arg
+
= arg(z) − arg(z∗ ).
4z∗
4
The conditions of theorem 1 are held than this theorem is valid.
lim
z + 2Hn−2 +
= lim
In the expansion function ln(c + z) in C-QCF (10) in the neighborhood of point z = z∗ let’s choose
z = z∗ and multiply to z. Then we have
z ln z∗ (z − z∗ )/z∗ ln z∗ (z − z∗ )(2 + ln z∗ )/2z∗ ln z∗
1 +
1
+
1
+
(z − z∗ ) ln z∗ /6z∗ (2 + ln z∗ ) (z − z∗ )(3 + ln z∗ )/3z∗ (2 + ln z∗ ) ln z∗
1
+
1
+ · · ·+
+
(z −z∗ )nS̄n /2(2n−1)z∗ S̄n−1 (z −z∗ )nS̄n−1 /2(2n+1)z∗ S̄n
,
(18)
+
1
+
1
+ ···
where S̄n = 2Hn + ln z∗ , n = 1, 2, . . . .
Approximants of continued fraction (18) create sequence
[ n+1P̄
] n (z)/Q̄n (z), where
[ n+2 ]of rational functions
polynomials degree satisfying inequality deg P̄n (z) ≤ 2 , deg Q̄n (z) ≤ 2 .
Conversely function z ln z in the neighborhood of point z = z∗ expanded into C-QCF of form (17).
Approximants of this continued fraction create sequence[ of] rational functions
[ P̂n](z)/Q̂n (z), where
. These continued
polynomials degree satisfying conditions deg P̂n (z) ≤ n2 , deg Q̂n (z) ≤ n+1
2
fractions are not equivalent, but have a common area of convergence.
z ln z =
International Journal of Advanced Research in Mathematics Vol. 7
9
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