On Taylor series expansion of $(1+ z)^{A}$ for $|z|>1$

0.02 MB·English

by  Akhila Raman

On Taylor series expansion of (1 + z)A for |z| > 1
AkhilaRaman
UniversityofCaliforniaatBerkeley,CA-94720.Email:[email protected]. Ph:510-540-5544
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0 Abstract
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ItiswellknownthattheTaylorseriesexpansionof(1+z)Adoesnotconvergefor|z|>1where
n
Aisarealnumberwhichisnotequaltozeroorapositiveinteger. Alimitedseriesexpansionof
a
J thisexpressionisobtainedinthispaperfor|z|>1asaproductofconvergentseries.

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Keywords:
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1. Introduction
H
.
h
ItiswellknownthattheTaylorseriesexpansionof(1+z)Aisgivenby
t
a
m ∞
A
(1+z)A = zn (1)
[ n!
Xn=0

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v for |z| < 1 where A is the binomial choose function. It is well known that the series ex-
n
9 pansiondoesnotconv(cid:16)er(cid:17)gefor|z| > 1whereA isarealnumberwhichisnotequaltozeroora
4 positiveinteger.
2 We could obtain a limited series expansion for |z| > 1 by writing the above expression as
0
follows
.
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0 z z z z z z
0 (1+z)A =(1+ 2 + 2)A =(1+ 2)A(1+ (1+2 z))A =(1+ 2)A(1+ z+2)A (2)
1 2
: The second term in the above equation has a convergent series representation, given that
v
Xi |z+z2|<1. If|2z|>1,wecanwrite
ar (1+ 2z)A =(1+ 4z)A(1+ z+z4)A (3)
Repeatingthisprocedureiteratively,ifm istheminimumvalueforwhich| z | < 1,wecan
0 2m0
write
z m0 z
(1+z)A =(1+ )A (1+ )A (4)
2m0 Yr=1 z+2r
Eachofthetermsintheaboveproductoftermshasaconvergentseriesrepresentation.Given
that we can write the convergentseries expansionfor each of the terms aboveas (1+ z )A =
2m0
PreprintsubmittedtoNumberTheory January11,2010
∞ A ( z )nand(1+ z )A = ∞ A ( z )m,where A representstheChoosefunction[1],
n=0 n 2m0 (z+2r) m=0 m (z+2r) n
wPeha(cid:16)ve(cid:17)theseriesexpansionforP(1+z(cid:16))A(cid:17)expressedasa(cid:16)pr(cid:17)oductofconvergentseries,which
convergesfor |z|>1asfollows:
∞ A z m0 ∞ A z
(1+z)A =[ ( )n][ [ ]m] (5)
Xn=0 n! 2m0 Yr=1Xm=0 m! z+2r
2. Section2
Let us take the case of m = 1 for 1 < |z| < 2. Expanding the term 1 in the above
0 (z+2r)m
equation5asTaylorseriesaroundapointz=0,wehaveform>0
∞
1
= b(j,r,m)zj (6)
(z+2r)m
Xj=0
whereb(0,r,m)= 1 andb(j,r,m)isgivenasfollowsfor j=1,2,3...
(2r)m
m+ j−1 (−1)j
b(j,r,m)= ; (7)
j !(2r)m+j
Form=0, 1 =1.Nowwecanwritethetheseriesexpansionof(1+z)Awhichconverges
(z+2r)m
for1<|z|<2,asaproductoftermsexpandedinTaylorseriesasfollows:
∞ ∞ ∞
A z A
(1+z)A =[ ( )n][ zm b(j,1,m)zj] (8)
n! 2 m!
Xn=0 Xm=0 Xj=0
Forthecaseofm >1for|z|>2,wecanwriteasfollows:
0
∞ A z ∞ A ∞ m0−1 ∞ A 1
(1+z)A =[ ( )n][ zm b(j,m ,m)zj][ zm[ ]m] (9)
Xn=0 n! 2m0 Xm=0 m! Xj=0 0 Yr=1 Xm=0 m! z+2r
Thelasttermintheaboveequation( 1 )mcanbeexpressedasfollows:
z+2r
1 z
( )m =(z+2r)−m =2−r∗m(1+ )−m (10)
z+2r 2r
The term(1 + z)−m can be recursively expanded using Eq.9 by substituting z → z and
2r 2r
A→−mandm →m −rtoobtaintheseriesexpansionof(1+z)A whichconvergesfor|z|>1
0 0
asaproductoftermsexpandedinTaylorseries.
3. Section3
Letusconsiderthefollowingbinomialexpression
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(x+y)A (11)
whereAisarealnumberwhichisnotequaltozeroorapositiveintegerandz= x and|z|>1.
y
Writing(x+y)A =(1+ x)AyA =(1+z)AyA,wecanwritetheseriesexpansionofthisexpression
y
usingresultsobtainedinequations5and9asfollows:
∞ A z m0 ∞ A z
(x+y)A =yA ( )n [ ]m (12)
Xn=0 n! 2m0 Yr=1Xm=0 m! z+2r
∞ A z ∞ A ∞ m0−1 ∞ A 1
(x+y)A =yA[ ( )n][ zm b(j,m ,m)zj][ zm[ ]m] (13)
Xn=0 n! 2m0 Xm=0 m! Xj=0 0 Yr=1 Xm=0 m! z+2r
4. Conclusion
IthasbeenshownthattheTaylorseriesexpansionof(1+z)A canbeexpandedasaproduct
ofconvergentseries,for|z|>1whereAisarealnumberwhichisnotequaltozeroorapositive
integer.
5. Acknowledgements
The author would like to thank Michael Schlosser(Universitat Wien) for his constructive
commentsonthepaper.
6. References
[1] Abramowitz, M. and Stegun, I. A. (Eds.). Handbook of Mathematical Functions with
Formulas,Graphs,andMathematicalTables,9thprinting.NewYork:Dover,1972.
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