Applied Mathematics
Vol.07 No.03(2016), Article ID:64069,7 pages
10.4236/am.2016.73030
Series Representation of Power Function
Petro Kolosov
Marine Engineering Department, Odessa State Maritime Academy, Odessa, Ukraine

Copyright © 2016 by author and Scientific Research Publishing Inc.
This work is licensed under the Creative Commons Attribution International License (CC BY).
http://creativecommons.org/licenses/by/4.0/



Received 22 December 2015; accepted 26 February 2016; published 29 February 2016
ABSTRACT
This paper presents the way to make expansion for the next form function:
to the numerical series. The most widely used methods to solve this problem are Newtons Binomial Theorem and Fundamental Theorem of Calculus (that is, derivative and integral are inverse operators). The paper provides the other kind of solution, except above described theorems.
Keywords:
Series Expansion, Series Representation, Binomial Theorem, Power Function, Cube Number, Number to Power

1. Introduction
Let basically describe Newtons Binomial Theorem and Fundamental Theorem of Calculus and some their properties. In elementary algebra, the binomial theorem (or binomial expansion) describes the algebraic expansion of powers of a binomial. According to the theorem, it is possible to expand the power
into a sum involving terms of the form
where the exponents b and c are nonnegative integers with
, and the coefficient a of each term is a specific positive integer depending on n and b. The coefficient a in the term of
is known as the binomial coefficient. The main properties of the binominal theorem are next:
1) The powers of x go down until it reaches
starting value is n (the n in
).
2) The powers of y go up from 0 (
) until it reaches n (also n in
).
3) The n-th row of the Pascal’s Triangle will be the coefficients of the expanded binomial.
4) For each line, the number of products (i.e. the sum of the coefficients) is equal to 
5) For each line, the number of product groups is equal to
.
By using binomial theorem for our case, we obtain next type function [1] :

We can reach the same result by using Fundamental Theorem of Calculus, according it we have [2] :

by means of the addition of integrals

For presented in this paper method, the properties of binomial theorem are not corresponded and prime function (i.e function, which we use with sum operator) has the recursion structure for x basic view is the next:
. Below is represented theoretical algorithm deducing such a function, which, when substituted to the sum operator, with some k number of iterations, returns the correct value of a number 




where deltas for 
going from it we can get next property of the powers function:
According Table 1 (for case
Note that upper sign shows the rank of the difference and doesn’t mean power sign. As we can see, according Table 1, the values of third rank difference are equal to 
Table 1. Numbers according third power.
make conclusion of the next power functions property:

where
Let use sum operator for expression [3] , we obtain [4] :
Or
Now, we have successful formula, which disperses any natural number 



2. Change over to Higher Powers Expression
In this section are reviewed the ways to change obtained in previous annex expression [4] to higher powers i.e
Expression [4] has the follow view:
As we can see, iteration limits for [2] and [4] are:
is not possible. Let change the formula [2] by the next way [5] :
Next, give the formula [2] follow changes:
Going from expression [5] for 
By means of main property of the powers function

According to the above property from the expression [4] for
For 
Expression [7] has the next property (as well right for [4] [6] -[8] ):
3. Binomial Theorem Representation
By means of Binomial theorem for
According expression [2] , we have the next corresponding:
Let, going from expression [4] , change the binomial expansion for
So, for
oing from it, by means of power function properties, we can only to multiply by x every product of the series, by this way, for 
4. ex Representation
According above method we have right to present function 
By means of general 
5. Difference from Binomial Theorem
To show changes from binomial theorem, let use other algorithm for
We have right to integrate the 


For third derivative we have next equation:
Let derive the
Let be
First derivative is next:
Let calculate the 
And for 
In case of 
So,
and corresponds to binomial expansion. Main difference is adjustable limits of the function [4] , see page 3 and 4.
6. Conclusion
The paper presented a method of expansion of the function of the form 
Cite this paper
PetroKolosov, (2016) Series Representation of Power Function. Applied Mathematics,07,327-333. doi: 10.4236/am.2016.73030
References
- 1. Weisstein, E.W. (2002) Binomial Theorem. Wolfram MathWorld.
http://mathworld.wolfram.com/BinomialTheorem.html - 2. Coolidge, J.L. (1949) The Story of the Binomial Theorem. The American Mathematical Monthly, 56, 147-157.
http://dx.doi.org/10.2307/2305028 - 3. Martzloff, J.-C., Wilson S.S., Gernet, J. and Dhombres, J. A History of Chinese Mathematics. Springer, Berlin.
- 4. Biggs, N.L. (1979) The Roots of Combinatorics. Historia Mathematica, 6, 109-136.
http://dx.doi.org/10.1016/0315-0860(79)90074-0 - 5. O’Connor, J.J. and Robertson, E.F. (1953) Abu Bekr Ibn Muhammad Ibn Al-Husayn Al-Karaji. MacTutor History of Mathematics Archive, University of St Andrews, University of St Andrews.
http://www-history.mcs.st-andrews.ac.uk/Biographies/Al-Karaji.html - 6. Landau, J.A. (2007) Historia Matematica Mailing List Archive: Re: [HM] Pascal’s Triangle. Archives of Historia Matematica. Retrieved 2007-04-13.
- 7. Kline, M. (1972) History of Mathematical Thought. Oxford University Press, Oxford, 273.
- 8. Bourbaki, N. Elements of the History of Mathematics Paperback. J. Meldrum (Translator).
- 9. Barth, N.R. (2004) Computing Cavalieri’s Quadrature Formula by a Symmetry of the n-Cube. The American Mathematical Monthly (Mathematical Association of America), 111, 811-813.
http://dx.doi.org/10.2307/4145193
Application 1. Visual Basic 6.0 Program Codes
Expression [4] :
j = 6 x = Val(Text1.Text) n = Val(Text2.Text) r = 0 For k = 1 To x 
Expression [6] :
j = 6 x = Val(Text1.Text) n = Val(Text2.Text) r = 0 For k = 1 To x Step 1 For m = 0 To k − 1 Step 1 

Expression [7] :
j = 6
x = Val(Text1.Text)
n = Val(Text2.Text)
r = 0
For k = 1 To x Step 1
For m = 0 To k − 1 Step 1
Next m
Next k
ex Representation:
j = 6
e = 0
x = Val(Text1.Text)
r = Val(Text2.Text)
f = 0
For m = 0 To r Step 1
If m = 0 Then
f = 1
Else
f = f × m
End If
For k = 1 To x
Next k
Next m












































