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Series Representation of Power Function Series Representation of Power Function Kolosov Petro May 7, 2017 Abstract This paper presents the way to make expansion for the next form function: y = n x ; 8(x; n) 2 N to the numerical series. The most widely used methods to solve this problem are Newton's Binomial Theorem and Fundamental Theorem of Calculus (that is, derivative and integral are inverse operators). The paper provides the other kind of solution, based on induction from particular to general case, except above described theorems. Keywords. power, power function, monomial, polynomial, power series, third power, series, finite difference, divided difference, high order finite difference, derivative, bi- nomial coefficient, binomial theorem, Newton's binomial theorem, binomial expansion, n-th difference of n-th power, number theory, cubic number, cube, Euler number, ex- ponential function, Pascal triangle, Pascal's triangle, mathematics, number theory 2010 Math. Classification Subject. 40C15, 32A05 e-mail: kolosov [email protected] ORCID: http://orcid.org/0000-0002-6544-8880 Social media links Twitter - Kolosov Petro Youtube - Kolosov Petro Mendeley - Petro Kolosov Academia.edu - Petro Kolosov LinkedIn - Kolosov Petro Google Plus - Kolosov Petro Facebook - Kolosov Petro Vimeo.com - Kolosov Petro VK.com - Math Facts Publications on other resources HAL.fr articles - Most recent updated ArXiV.org articles Archive.org articles 1 ViXrA.org articles Figshare.com articles Datahub.io articles 1 Introduction Let basically describe Newton's Binomial Theorem and Fundamental Theorem of Cal- culus and some their properties. In elementary algebra, the binomial theorem (or binomial expansion) describes the algebraic expansion of powers of a binomial. The theorem describes expanding of the power of (x + y)n into a sum involving terms of the form axbyc where the exponents b and c are nonnegative integers with b + c = n, and the coefficient a of each term is a specific positive integer depending on n and b. The coefficient a in the term of axbyc is known as the binomial coefficient. The main properties of the binominal theorem are next: n I. the powers of x go down until it reaches x0 = 1 starting value is n (the n in (x + y) ) II. the powers of y go up from 0 (y0 = 1) until it reaches n (also n in (x + y)n) III. the n-th row of the Pascal's Triangle (see [1]) will be the coefficients of the ex- panded binomial. IV. for each line, the number of products (i.e. the sum of the coefficients) is equal to x + 1 V. for each line, the number of product groups is equal to 2n According to the theorem, it is possible to expand any power of x + y into a sum of the form (see [2]): n X n (x + y)n = xn−kyk (1) k k=0 By using binomial theorem for our case we obtain next form function: x−1 X n n xn = nkn−1 + kn−2 + ::: + k + 1 (2) 2 n − 1 k=0 We can reach the same result by using Fundamental Theorem of Calculus, we have, respectively: x Z xn = ntn−1dt 0 by means of addition of integrals x−1 x−1 X Z k+1 X = ntn−1dt = (k + 1)n − kn (3) k=0 k k=0 For presented in this paper method the properties of binomial theorem are not corre- sponded and the function, which we use summation operator on, has the recursive n P structure depending on x, basic view is the next: x = k θ(x; k). Below is repre- sented theoretical algorithm deducing such a function, which, when be substituted to 2 the summation operator, with some k number of iterations, returns the correct value of the number xi 2 R to power n = 3. The main idea that the law of basic elements distribution of the value xi to third power seen in finding the 3-rd order finite dif- n k ference (note that n-order difference is written as ∆ (xi )) between nearest two items 3 3 3 ∆(xi ) = xi+1 − xi ; xi = i∆x; i 2 N; ∆x = xi+1 − xi; ∆x 2 R. For example, let be a set of numbers, which xi has constant difference between numbers, constant difference is a key of this method (example for ∆x = 1; i.e xi = i): 3 3 2 3 3 3 i xi ∆(xi ) ∆ (xi ) ∆ (xi ) 0 0 1 6 6 1 1 7 12 6 2 8 19 18 6 3 27 37 24 6 (4) 4 64 61 30 6 5 125 91 36 6 216 127 7 343 Figure 1: Difference table of third power k 3 Where differences ∆ (xi ) of orders k 2 [1; 3], for xi = i · ∆x distribution are defined as: 3 def 3 3 3 3 3 3 ∆(xi ) = xi+1 − xi = (i + 1) · ∆x − i · ∆x 2 3 3 3 ∆ (xi ) = ∆(xi+1) − ∆(xi ) 3 3 2 3 2 3 ∆ (xi ) = ∆ (xi+1) − ∆ (xi ) 1 More generally, the n-th difference of sequence ff(xi)gi=1 could be written as (see [10]): n X n ∆nf(x ) = ∆n−1f(x ) − ∆n−1f(x ) = (−1)k · f(x ) i i+1 i k i+n−k k=0 m Basically, the n-order difference of the function f(xi) = xi has the follow view: 8 m! · ∆xm; if n = m > > > > > n n m < X n k m m ∆ (xi ) = (−1) · i + n − k · ∆x ; if n < m (5) > k > k=0 > > > : (i + 1)m · ∆xm − im · ∆xm; if n = 1 3 According to Figure 1, in case of f(xi) = xi ; ∆x = 1, the first order finite difference has next regularity (sequence A008458 in OEIS): 3 3 3 ∆(x0) = x1 − x0 = 1 + 3! · 0 3 3 3 ∆(x1) = x2 − x1 = 1 + 3! · 0 + 3! · 1 3 3 3 3 ∆(x2) = x3 − x2 = 1 + 3! · 0 + 3! · 1 + 3! · 2 . 3 3 3 ∆(xm) = xm+1 − xm = 1 + 3! · 0 + 3! · 1 + 3! · 2 + ··· + 3! · m n Theorem 1. For each function f(xi) = xi ; xi = i∆x; ∆x = const; ∆x 2 R; 8(i; n) 2 n n n n d (x ) ∆ (xi ) N; xi 6= x, holds the following equality: dxn = (∆x)n = n!, i.e for any power function with natural number as exponent holds the equality between finite difference and derivative with order respectively to exponent and equals to exponent under factorial sign. k Proof. Since, the (xn)0 = nxn−1 (see [3]), we can say: d · f(x) = n · (n − 1) × x dxk f(x)=xn n−k · · · × (n − k + 1) · x , n 2 N. Using limit notation, we have: 0 1 dm(#(x)) dn−0(#(x)) lim @ A = = n! m!n− dxm dxn−0 #(x)=xn | {z } φ(m) By means of expression (5), in case of m < n; n 2 N, we have the limit: m n m ! ∆ (x ) X m m lim i = lim (−1)k · i + m − k m!n− (∆x)m m!n− k k=0 | {z } !(m) n X n n−0 = (−1)k · i + n − k = n! k k=0 Obviously, if φ(m) = n!;!(m) = n! −! φ(m) = !(m), so we have right to equal dn(xn) ∆n(xn) = i = n! dxn (∆x)n This completes the proof. 3 As we can see, according to Figure 1, the values of third rank differences of xi , when ∆x = 1 are equal to 3! (see [6], [7]) and constant for each index i. According to the values from Figure 1, the difference functions are corresponding to next recursively defined expressions: i−1 3 3 3 2 3 X 3 3 ∆ (xi ) = j · (∆x) ; ∆ (xi ) = ∆ (xi ) · m m=0 i−1 ! 3 3 X 2 3 ∆(xi ) = (∆x) + ∆ (xk) k=0 4 Where j, according to theorem 1 and ([6], [7]), equals to 3!. Hence, we have follow expansion for natural cube numbers (also the sum of sequences A008458 over k from 0 to x − 1): x3 = (1 + 3! · 0) + (1 + 3! · 0 + 3! · 1) + (1 + 3! · 0 + 3! · 1 + 3! · 2) + ··· ··· + (1 + 3! · 0 + 3! · 1 + 3! · 2 + ··· + 3! · (x − 1)) (6) x3 = x + (x − 0) · 3! · 0 + (x − 1) · 3! · 1 + (x − 2) · 3! · 2 + ··· ··· + (x − (x − 1)) · 3! · (x − 1) Note that step ∆x from equation (6) is set as ∆x = 1 and not displayed. Using summation notation over k from 0 to x − 1 on (6), we obtain: x−1 X x3 = x + j mx − m2 (7) m=0 Or x−1 m ! x−1 X X X x x3 = 1 + j u ≡ j mx − m2 + ; x 2 (8) j(x − 1) N m=0 u=0 m=1 x>1 P Property 1. Let be expression (7) written as (Ω(x)) = x + m2Ω(x) χ(m); Ω(x) ⊂ Ξ(x) z }| { N0. Let be set Ω(x) defined as: Ω(x) := ff0; f1; 2; 3; 4; : : : ; x − 1g; xgg Since the | {z } Λ(x) χ(0) = χ(x) = 0, we can say that 8(7) : (Ω(x)) ≡ (Λ(x)) ≡ (Ξ(x)); (Λ(x); Ξ(x)) ⊂ Ω(x); Λ(x) 6⊂ Ξ(x). It shows that changing iteration sets of expression (7) the function value is not changed. Now we have successful expression, which disperses any natural number x3 to the numerical series (this example shows only expansion for any number x 2 N to power n = 3, but this method works for floats numbers also, it depends of start set of numbers, function's form also depends of the chosen set, step ∆x between numbers should be constant all the time). Going from it, in the next section we introduce changing over the function (7) to power n > 3; n 2 N.
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