Series Representation of Power Function

16 Pages·2017·0.38 MB·English

by  Kolosov Petro

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SERIES REPRESENTATION OF POWER FUNCTION
KOLOSOVPETRO
Abstract. In this paper we discuss a problem of generalization of binomial
distributed triangle, that is sequence A287326 in OEIS. The main property
of A287326 that it returns a perfect cube n as sum of n-th row terms over
k, 0≤k≤n−1or1≤k≤n,bymeansofitssymmetry. Inthispaperwehave
derived a similar triangles in order to receive powers m = 5,7 as row items
sum and generalized obtained results in order to receive every odd-powered
monomialn2m+1, m≥0assumofrowtermsofcorrespondingtriangle.
2010 Math. Subject Class. 30BXX
ORCID: 0000-0002-6544-8880
e-mail: [email protected]
Contents
1. Structure of the manuscript 1
2. Introduction 2
3. Generalization of sequence A287326 7
3.1. Properties of L (n,k) and A 13
m m,j
3.2. Example of use 14
4. Acknowledgements 14
5. Conclusion 15
References 15
1. Structure of the manuscript
The problem of finding expansions of monomials, binomials, trinomials, etc. is
classical and a lot of theorems have been found, the most prominent examples
are Binomial Theorem [2], Multinomial theorem, Wozpitsky Identity [30], Stirling
numbers of second kind identity, etc. In this paper we try to solve the classical
problem of finding expansions of monomials. We start from binomial distributed
triangle A287326 [11] in OEIS. The main property of A287326 that it returns a
perfect cube n as n-th row sum, starting from 0,...,n−1 or from 1,...,n by means
of its symmetry. Therefore, the following question stated:
• Can we find similar to A287326 triangles in order to receive monomial
nt, t>3assumofrowterms? Inotherwords,canA287326begeneralized
in order to receive monomial nt, t>3 as sum of row terms?
Date:May25,2018.
Key words and phrases. Power function, Monomial, Binomial coefficient, Binomial Theorem,
Finitedifference,Perfectcube,Pascal’striangle,Seriesrepresentation,Faulhaber’sformula.
1
© 2018 by the author(s). Distributed under a Creative Commons CC BY license.
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2 KOLOSOVPETRO
Findingananalogsfort=5,7insection3,weanswertoabovequestionspositively.
Could this process be continued for each t = 1,3,5,7... similarly? Positive answer
to this question is given by theorem (3.29).
2. Introduction
LetdescribethederivationofthesequenceA287326inOEIS.SequenceA287326
returns the perfect cube n as row sum over k, 0 ≤ k ≤ n−1, as well as sum over
1 ≤ k ≤ n, by means of its symmetry. First, consider a difference table of perfect
cubes ([4], eq. 7)
n ∆0(n3) ∆1(n3) ∆2(n3) ∆3(n3)
0 0 1 6 6
1 1 7 12 6
2 8 19 18 6
3 27 37 24 6
4 64 61 30 6
(2.1)
5 125 91 36 6
6 216 127 42 6
7 343 169 48 6
8 512 217 54
9 729 271
10 1000
Table 1: Difference table of perfect cubes n, 0≤n≤10 up to 3rd order.
Reviewing above table, we have noticed that
(2.2) ∆(03) = 1+6·0=6(cid:0)1(cid:1)+(cid:0)1(cid:1)
2 0
∆(13) = 1+6·0+6·1=6(cid:0)2(cid:1)+(cid:0)2(cid:1)
2 0
∆(23) = 1+6·0+6·1+6·2=6(cid:0)3(cid:1)+(cid:0)3(cid:1)
2 0
∆(33) = 1+6·0+6·1+6·2+6·3=6(cid:0)4(cid:1)+(cid:0)4(cid:1)
2 0
.
.
.
∆(n3) = 1+6·0+6·1+6·2+···+6·n=6(cid:0)n+1(cid:1)+(cid:0)n+1(cid:1)
2 0
Above difference identity is closely related to Faulhaber’s sum of cubes, where
n3 = 6(cid:0)n+1(cid:1)+(cid:0)n+1(cid:1), see ([21], p. 9). Note that ∆2(n3) could be found similarly
3 1
using above identity ∆2(n3)=6(cid:0)n+1(cid:1)+(cid:0)n+1(cid:1).
3−2 1−2
Property 2.3. (Generalized finite difference of power using Faulhaber’s formula).
Consider the identities, ([21], p. 9).
n1 = (cid:0)n(cid:1)
 1
n3 = 6(cid:0)n+1(cid:1)+(cid:0)n(cid:1)
3 1
n5 = 120(cid:0)n+2(cid:1)+30(cid:0)n+1(cid:1)+(cid:0)n(cid:1)
5 3 1
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SERIES REPRESENTATION OF POWER FUNCTION 3
We can find the first order finite difference of odd power as decreasing the variable
of corresponding binomial coefficients by 1, for example
∆n1 = (cid:0)n(cid:1)
 0
∆n3 = 6(cid:0)n+1(cid:1)+(cid:0)n(cid:1)
2 0
∆n5 = 120(cid:0)n+2(cid:1)+30(cid:0)n+1(cid:1)+(cid:0)n(cid:1)
4 2 0
Continue similarly, we can express each difference of order t≥1. The coefficients
{1,6,1,120,30,1} in above identities are generated by
r (cid:18) (cid:19)
1(cid:88) 2r
(2.4) V = (−1)j (r−j)2n,
n,k r j
j=0
where r =n−k+1, this formula was provided by Peter Luschny in [27]. Therefore,
for every odd t>0 and m≥0, we have
(cid:18) (cid:19)
(cid:88) n+m−k
∆tn2m+1 = V , if t >0 and odd
m,k l
0≤k≤m
l≤2(m−k)+1−t
l is even
Let be m≥0, t>1 and even, then
(cid:18) (cid:19)
(cid:88) n+m−k
∆tn2m+1 = V , if t >1 and even
m,k l
0≤k≤m
l≤2(m−k)+1−t
l is odd
Let show finite differences, set m ≥ 1, t > 1, then we have finite difference
identity
(cid:18) (cid:19)
(cid:88) 1 n+m−k
∆tn2m = V , if t >0 and odd
n m,k l
0≤k≤m
l≤2(m−k)+1−t
l is even
And
(cid:18) (cid:19)
(cid:88) 1 n+m−k
∆tn2m = V , if t >1 and even
n m,k l
0≤k≤m
l≤2(m−k)+1−t
l is odd
By the identity (cid:80)n−1∆nm =nm, we have right to represent perfect cube n as
k=0
(2.5) n3 =6(cid:0)1(cid:1)+(cid:0)1(cid:1)+6(cid:0)2(cid:1)+(cid:0)2(cid:1)+6(cid:0)3(cid:1)+(cid:0)3(cid:1)+···+6(cid:0)n+1(cid:1)+(cid:0)n+1(cid:1)
2 0 2 0 2 0 2 0
Let rewrite it again and display every binomial coefficient as summation (cid:0)n+1(cid:1) =
2
1+2+···+n, then
n3 =(1+6·0)+(1+6·0+6·1)+···+(1+6·0+···+6·(n−1))
Particularizing above expression, we get
(2.6) n3 =n+(n−0)·6·0+(n−1)·6·1+···+(n−(n−1))·6·(n−1)
Provided that n is natural. Now we apply a compact sigma notation on (2.6), thus
(cid:88)
(2.7) n3 =n+ 6k(n−k)
1≤k≤n
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4 KOLOSOVPETRO
(cid:80)
As sum 6k(n−k) consists of n terms, we have right to move n in (2.7)
1≤k≤n
under sigma notation, we get
(cid:88)
(2.8) n3 = 6k(n−k)+1
1≤k≤n
Property2.9. (Proofofsymmetry). LetbeasetsA(n):={1, 2, ..., n}, B(n):=
{0, 1, ..., n}, C(n):={0, 1, ..., n−1}, let be expression (2.8) defined as
def (cid:88)
M(n, C(n)) = 6k(n−k)+1
k∈C(n)
where x is natural-valued variable and C(n) is iteration set of (2.8), then we have
equality
(2.10) M(n, A(n))=M(n, C(n))
Let review and define expression (2.6) as
def (cid:88)
U(n, C(n)) = n+6· k(n−k)
k∈C(n)
then
(2.11) U(n, A(n))=U(n, B(n))=U(n, C(n))
Other words, changing of iteration sets of(2.6) and (2.8) by A(n), B(n), C(n) and
A(n), C(n), respectively, doesn’t change resulting value for each natural x.
Proof. Let be a plot y(n,k)=6k(n−k)+1, k ∈R, 0≤k ≤10, given n=10
150
1
+
)
k 100
−
n
(
k
6
=
50
)
k
n,
(
y
0
0 2 4 6 8 10
k ∈R, 0≤k ≤10
Figure 2. Plot of 6k(n−k)+1, k ∈R, 0≤k ≤n, where n=10.
Obviously, being a parabolic function, it’s symmetrical over n, hence equivalent
2
M(n, A(n)) = M(n, C(n)) follows. Reviewing (2.6) and denote u(n,k) = kn−
k2, we can conclude, that u(n,0) = u(n,n) = 0, then equality of U(n, A(n)) =
U(n, B(n))=U(n, C(n)) immediately follows. This completes the proof. (cid:3)
Review above property (2.9). Let be an example of triangle built using
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SERIES REPRESENTATION OF POWER FUNCTION 5
Definition 2.12. For every n≥0
def
(2.13) L (n,k) = 6k(n−k)+1, 0≤k ≤n
1
over n from 0 to n = 4, where n denotes corresponding row and k shows the
item of row n.
Row0: 1
Row1: 1 1
Row2: 1 7 1
Row3: 1 13 13 1
(2.14) Row4: 1 19 25 19 1
Figure 3. Triangle generated by L (n,k) from 0 to n=4, sequence A287326 in
1
OEIS, [11].
Note that n-th row sum of Triangle (2.14) over 0 ≤ k ≤ n−1 returns perfect
cube n. We can see that each row with respect to variable n = 0, 1, 2, 3, 4,...,
has Binomial distribution of row terms. One could compare Triangle (2.14) with
Pascal’s triangle [1], [12]
Row0: 1
Row1: 1 1
Row2: 1 2 1
Row3: 1 3 3 1
Row4: 1 4 6 4 1
Figure 4. Pascal’s triangle read by rows, sequence A007318 in OEIS, [1].
Let us approach to show a few properties of triangle (2.14) and L (n,k).
1
Properties 2.15. Properties of triangle (2.14).
(1) Summation of items L (n,k) of n-th row of triangle (2.14) over k from 0
1
to n−1 returns perfect cube n≥0 as follows
(cid:88)
(2.16) L (n,k)=n3
1
1≤k≤n
(2) Relation between α and α
0,n 1,n
α =α , n≥1
0,n+1 1,n
(3) Firstitemofeachrow’snumbercorrespondingtocentralpolygonalnumbers
sequence a(n) = n2+n+2 (sequence A000124 in OEIS, [13]) returns finite
2
difference of consequent perfect cubes. For example, let be a k-th row of
triangle (2.14), such that k = n2+n+2, n=0,1,2,..., then item
2
(cid:18)n2+n+2 (cid:19)
(2.17) L ,1 =(n+1)3−n3
1 2
(4) Items of (2.14) have Binomial distribution over rows.
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6 KOLOSOVPETRO
(5) Linear recurrence, for every k and n>0
(2.18) 2L (n,k)=L (n+1,k)+L (n−1,k)
1 1 1
This linear recurrence is direct result of second order binomial transform of
L (n,k) over n.
1
(6) Linear recurrence, for each n>k
(2.19) 2L (n,k)=L (2n−k,k)+L (2n−k,0)
1 1 1
(7) From (1.24) for every n≥0 follows
(cid:88) (cid:88) (cid:18)n2+n+2 (cid:19)
(2.20) L (n,k)= L ,1 =n3
1 1 2
1≤k≤n 1≤k≤n
(8) Triangle (2.14) is symmetric, i.e
(2.21) L (n,k)=L (n,n−k)
1 1
Property 2.22. (Generalized binomial series by means of identity (2.16). Let
review identity (2.16) in sense of
(cid:88)
L (n,k)=α n−β
1 0,t 0,t
1≤k≤t
By property (2.9) we rewrite above expression as
(cid:88)
L (n,k)=α n−β
1 1,t 1,t
0≤k≤t
wheresubscripts0,tand1,tdenotetherangesofsummation, respectively. Running
over t>0 above identities produce sets of coefficients {α } , {β } , {α } and
0,t t 0,t t 1,t t
{β } . Below table shows initial terms of these sequences
1,t t
t α β α β
0,t 0,t 1,t 1,t
1 1 0 6 5
2 6 4 18 28
3 18 27 36 81
4 36 80 60 176
5 60 175 90 325
6 90 324 126 540
7 126 539 168 833
8 168 832 216 1216
9 216 1215 270 1701
10 270 1700 330 2300
Table 5. Array of coefficients α , β given n=1,...,10.
0,1,n 0,1,n
Therefore, perfect cube n could be rewritten as binomials of the form
(cid:40)
α n−β , if t=n−1;
n3 = 0,n−1 0,n−1
α n−β , if t=n
1,n 1,n
By the main power property, for every m∈N
(cid:40)
α nm−2−β nm−3
nm = 0,n−1 0,n−1
α nm−2−β nm−3
1,n 1,n
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SERIES REPRESENTATION OF POWER FUNCTION 7
We denote above equation as
nm =α nm−2−β nm−3
0,1,n−1,n 0,1,n−1,n
Let rewrite the right part of above expression regarding to itself as recursion
nm = α (α nm−4−β nm−5)
0,1,n−1,n 0,1,n−1,n 0,1,n−1,n
− β (α nm−5−β nm−6)
0,1,n−1,n 0,1,n−1,n 0,1,n−1,n
= α2 nm−4−2α β nm−5+β2 nm−6
0,1,n−1,n 0,1,n−1,n 0,1,n−1,n 0,1,n−1,n
We can observe corresponding binomial coefficient present before each α
0,1,n−1,n
times β . Continuous j-times recursion gives
0,1,n−1,n
∞ (cid:18) (cid:19)
(cid:88) j
nm = (−1)k αj−k β nm−2j−k, j ≥0
k 0,1,n−1,n 0,1,n−1,n
k≥0
Sequences α , α are generated by 3n2+3n, sequence A028896 in OEIS, [23].
1,t 0,t>1
Sequence β is generated by 2n3+3n2, sequence A275709 in OEIS, [20].
1,t
In this section we have reached binomial distributed triangle (2.14), such that
perfect cube n could be found as sum of n-th row terms of (2.14). Therefore, the
follow question is stated
Question2.23. CanwefindsimilartoA287326trianglesinordertoreceivemono-
mial nt, t>3 as sum of row terms? Is it exist L (n,k), v (cid:54)=1, such that
v
(cid:88)
nt ≡ L (n,k), v (cid:54)=t ?
v
1≤k≤n
3. Generalization of sequence A287326
InordertogetanalogsofTriangle(2.14)oneshouldsolveasystemofequations,
whereunknownsarecoefficientsofpolynomialandvariableofpolynomialisk(n−k).
LetshowatrianglegeneratedbyL (n,k),suchthatsumofn-throwtermsreturns
2
n5.
Example 3.1. We suspect that n-th row of triangle is generated by
(3.2) L (n,k)=A (n−k)2k2+A (n−k)k+A
2 2,2 2,1 2,0
whereA ,A ,A areunknowncoefficientsandn≥0, 0≤k ≤n. Assumethat
2,2 2,1 2,0
for every n≥0, m≥0 holds
(cid:88)
(3.3) L (n,k) ≡ n5
2
1≤k≤n
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8 KOLOSOVPETRO
In more explicit view
(cid:88) (cid:88)
(3.4) A k2(n−k)2+A k(n−k)+A n
2,2 2,1 2,0
1≤k≤n 1≤k≤n
(cid:88) (cid:88)
= A k2(n2−2nk+k2)+A kn−k2+A n
2,2 2,1 2,0
1≤k≤n 1≤k≤n
(cid:88) (cid:88)
= A k2n2−2nk3+k4+A kn−k2+A n
2,2 2,1 2,0
1≤k≤n 1≤k≤n
(cid:88) (cid:88) (cid:88) (cid:88)
= A n2 k2−2A n k3+A k4+A n k
2,2 2,2 2,2 2,1
1≤k≤n 1≤k≤n 1≤k≤n 1≤k≤n
(cid:88)
− A k2+A n
2,1 2,0
1≤k≤n
Thus, we have received expression containing sums of powers of successive natural
numbers,wherepowersare{1,2,3,4}. BytheFaulhaber’sformula[7],thefollowing
identities hold
(cid:88) n2+n
(3.5) k = ,
2
1≤k≤n
(cid:88) 2n3+3n2+n
(3.6) k2 = ,
6
1≤k≤n
(cid:88) n4+2n3+n2
(3.7) k3 = ,
4
1≤k≤n
(cid:88) 6n5+15n4+10n3−n
(3.8) k4 = .
30
1≤k≤n
Now we substitute above identities to (3.4), respectively, we get
2n3+3n2+n n4+2n3+n2 6n5+15n4+10n3−n
A n2 −2A n +A
2,2 6 2,2 4 2,2 30
n2+n 2n3+3n2+n
+ A n −A +A n
2,1 2 2,1 6 2,0
Particularizing the elements of above expression and moving them under the com-
mon divisor, we get
A n5−A n+30A (cid:18)n3−n(cid:19)
(3.9) 2,2 2,2 2,0 +A
30 2,1 6
We have to remember that expression (3.9) is the left side of the input equation
(2.2). Therefore,
A n5−A n+30A (cid:18)n3−n(cid:19)
(3.10) 2,2 2,2 2,0 +A =n5, n≥0
30 2,1 6
In order to satisfy (3.10) for each natural n, coefficients A ,A ,A should be
2,0 2,1 2,2
a solutions of following system of equations

1 A =1
30 2,2
A =1
2,1
30A −A =0
2,0 2,2
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SERIES REPRESENTATION OF POWER FUNCTION 9
The only solution of above system is A = 30, A = 0, A = 1. Hereby,
2,2 2,1 2,0
L (n,k) takes the form
2
(3.11) L (n,k)=30k2(n−k)2+1
2
And for each natural n holds
(cid:88)
(3.12) 30k2(n−k)2+1 = n5
1≤k≤n
Let show initial rows of triangle built by L (n,k)
2
1
1 1
1 31 1
(3.13) 1 121 121 1
1 271 481 271 1
1 480 1081 1081 481 1
...
Figure 6. Triangle generated by L (n,k), 0≤k ≤n, sequence A300656 in OEIS,
2
[15].
Similarly, finding the coefficients A ,A ,A ,A in
3,0 3,1 3,2 3,3
(3.14) L (n,k)=A k3(n−k)3+A k2(n−k)2+A k(n−k)+A
3 3,3 3,2 3,1 3,0
wegetA =140, A =−14, A =0, A =1, therefore, foreachn≥0holds
3,3 3,2 3,1 3,0
(cid:88)
(3.15) 140k3(n−k)3−14k2(n−k)2+1 = n7
1≤k≤n
Below we show a few initial rows of triangle built by L (n,k)
3
1
1 1
1 127 1
(3.16) 1 1093 1093 1
1 3793 8905 3793 1
1 8905 30157 30157 8905 1
...
Figure 7. Triangle generated by L (n,k), 0≤k ≤n, sequence A300785 in OEIS,
3
[16].
We assume now that generalization of A287326 holds for odd powers only. To gen-
eralizeoursequencesA287326,A300656,A300785foreveryoddpower2m+1, m=
0,1,2...wehavetoreviewthegeneratingfunctionsofcorrespondingsequences,that
is
(cid:88) (cid:88)
(3.17) A kj(n−k)j =n2m+1, m=1,2,3
m,j
1≤k≤n0≤j≤m
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10 KOLOSOVPETRO
Where A are unknown coefficients of polynomials (2.1) and (2.13).
m,j
Definition 3.18. Let define the part of (2.1) as
(cid:88) A kj(n−k)j d=ef L (n,k) d=ef (cid:88) A Tj(n,k)
m,j m m,j
0≤j≤m 0≤j≤m
where
def
T(n,k) = k(n−k).
Note that L (n,k) is generalization of definitions (2.12) for m = 1 and (3.11) for
m
m=2, respectively.
For example, generating functions of sequences A287326, A300656, A300785 are

L (n,k)=1+6k(n−k), for A287326
 1
L (n,k)=1−0k(n−k)+30k2(n−k)2, for A300656
2
L (n,k)=1−14k(n−k)−0k2(n−k)2+140k3(n−k)3, for A300785
3
Where coefficients A , for m = 1,2,3 are {A }1 = {1,6}, {A }2 =
m,j 1,j j=0 2,j j=0
{1,0,30}, {A }3 = {1,−14,0,140} in definitions of generating functions of
3,j j=0
A287326, A300656, A300785, respectively. To generalize above result in order to
receive monomial n2m+1 as (cid:80) L (n,k) = n2m+1, m = 0,1,2,... one has to
1≤k≤n m
solvethesystemofequations. Completesetofcoefficients{A , ..., A }such
m,0 m,m
that (cid:80) L (n,k) = n2m+1, m ≥ 0 holds can be found solving follow system
1≤k≤n m
of equations
L (1,0)=12m+1
Lmm(2,0)+Lm(2,1)=22m+1
(3.19) Lm(3,0)+Lm(3,1)+Lm(3,2)=32m+1
...L (r,0)+L (r,1)+···+L (r,r−1)=r2m+1, r ≥m
m m m
List of solutions1 of system (2.4) is split and assigned to OEIS under the numbers
A302971 (numerators of A ) and A304042 (denominators of A ). To reach
m,j m,j
recurrentformulaofA , firstletfixtheunusedvaluesA =0, forj <0orj >
m,j m,j
m, so we don’t need to care about the summation range for j, then by expanding
(n−k)j and using Faulhaber’s formula [7], we get
n−1 n−1 ∞ (cid:18) (cid:19)
(cid:88) (cid:88)(cid:88) j
(3.20) (n−k)jkj = nj−i(−1)iki+j
i
k=0 k=0 i
(cid:88)∞ (cid:18)j(cid:19) (−1)i (cid:34)(cid:88)∞ (cid:18)i+j+1(cid:19) (cid:35)
= nj−i B ni+j+1−t−B
i i+j+1 t t i+j+1
i t
(cid:88)∞ (cid:18)j(cid:19) (−1)i (cid:18)i+j+1(cid:19) (cid:88)∞ (cid:18)j(cid:19) (−1)i
= B n2j+1−t− B nj−i
i i+j+1 t t i i+j+1 i+j+1
i,t i
(cid:124) (cid:123)(cid:122) (cid:125) (cid:124) (cid:123)(cid:122) (cid:125)
((cid:63)) ((cid:5))
1One can produce a list of solutions of system (2.4) up to t = 11 using Mathematica code
solutions system 2 4.txt,[24].