Explicit Expressions for Free Components of. Sums of the Same Powers

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1 Applied Mathematical Sciences, Vol., 27, no. 53, HIKARI Ltd, Explicit Expressions for Free Components of Sums of the Same Powers Alexander I. Nikonov Samara State Technical University Faculty of Automation and Information Technology, Samara, 443, Russia Copyright 27 Alexander I. Nikonov. This article is distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Abstract Present article gives another proof of explicit expressions for free components related to sums of the same powers. These powers are represented by natural numbers. We begin the proof from consideration of equality between specified explicit expression and the term in sequence of elementwise and matrix products. The advantage of this proof over the known consists in its simplification. Keywords: sums of the same powers, free components, explicit expressions, elementwise and matrix products Introduction Let the initial expression for the sum of weight powers with the same natural values be given: p v Φ ( p,v) = b l ; p,v,l, b. l N l R () l= The combinatorial form of this sum was presented by us earlier [4] as the sum of products of its free α = α ( v) and weighted Φ =Φ ( ) components: p p v max α p N, = Φ ( p,v) = Φ ; p: p v, max = v : v p. (2)

2 264 Alexander I. Nikonov For both of these components, we obtained a recurrent representation. Further investigations allowed to express them explicitly. In particular, free components were expressed according to the following statement [5, pp ]. Theorem.. If v, N and ν = =, then + q q v α( v) = ( ) C q. (3) Let us briefly consider the well-known proof. The sum initially specified for transformation [4], looks like expression (). The result of this transformation is expression (2) with different values of max. The values of α ( v) and Φ p ( v ) are revealed in a recurrent manner. Our task is to investigate α ( v), and we give here the formulas that serve as the basis for its recurrent finding: α i j( ) j( ) = α j( i) C j( i) ;i = (4) j( i)= j( )+ Therefore j( i ) {,..., }, = v i; j ( ) {,..., max j( )}, max j( ) = v ; : j() < v, α j () = : j() = v. α = j( i) = For the proof we use the method of mathematical induction [6]. The induction base provides calculation of the component α () for the value of ν = through expression (4): α α i j ( i ) α () = C () =. This result corresponds to the special case (3) for = = :. ν 2 α () = ( ) C =.

3 Explicit expressions for free components of sums of the same powers 264 Within the induction step we consider the sequence with terms of the form v v, each of which can be represented as Φ( v,v)= Φ ( v,v, β ( v)); β ( v)= ( b = δ, τ =,, ν ); δν τ ν Kronecker Delta. ν ν τ ν At transition of the value ν from natural n- to n transformations Φ( n,n ) + Φ n, n n s + q q q n q q n n+ q q n n ( ) ( n n q n n ) ( ) n Φ = C C q C C q + C q, q, q, {,...,n } s translates ( n,n ) ( n ) n n Φ = to level Φ ( n,n) = n. This corresponds to a change in the free component from inductive hypothesis level to the level α ( n ) = ( ) + q q n C q + q q n α ( n) = ( ) C q. It is easy to see that these actions are valid for all consecutive changes in the natural numbers n >. Thus, Theorem.. is proved. Details are presented in [5]. However the substantiation of the explicit expression for free values forced us to apply weighted components Φ ( v,v) which considerably expanded volume of produced operations. The purpose of this article is to consider another proof of explicit expressions for free components. It has an advantage over the known proof, which consists in its simplification. 2 The main result of researches Let's deduce expression (3) in another way.

4 2642 Alexander I. Nikonov Alternative proof of Theorem.. We represent the right part (3) as an arbitrary term in the sequence of elementwise and matrix products: ν ( ) aν ( ) ( a ( ) g ) α ν = = ; (5) ( ) ( ) + ( ) + =[ ] ; a C C g q q,, ν v = ; = ; T = [ ] ; times the symbol ο represents the known elementwise multiplication; the result of given multiplication is the Hadamard product [2,3]; [] is sum - vector (column) [] with elements; the number α ( ν ) represents the result of ordinary matrix multiplication; here natural argument is ν. It is easy to make sure that right parts (3) and (5) are equal among themselves: Besides + q q v ν ( ) C q = ( a ( ) g ). a ν + ( ) = a ν ( ) g ; a ν ( ) = a ( ) gν ; (6) g = q is an analogue of the common ratio concerning to an ordinary geometric progression. Next we'll use the method of mathematical induction [6]. The induction base is formed for v = and, respectively, for =, i =. Equality (5) for the specified conditions appears as follows: α () = ( a () g ) =. Its validity can be confirmed by using expression (4) as an independent source of the information α j( ) = i j( ) α j ( i ) C j ( i ) ; = = j( i)= j( )+ i.

5 Explicit expressions for free components of sums of the same powers 2643 Hence we have i j ( i ) α = α () = j ; So, when v = the equalities (3), (5) are true. α () =. The induction step is based on the assumption that (3), (5) are satisfied for v = k: k ( ) ( ( ) ) α k = a g. (7) Let us now prove that (5) is valid for v = k + : k ( + ) = ( ( ) ) α k a g. According to assumption (7) and formula (6), we have k k ( + ) = ( ( ) ) = ( ( ) ) α k a g g a g. This implies the validity of equality (3) revealed by Theorem.. Remark 2.. An arbitrary term in the sequence of elementwise and matrix products from right part (5) can have an alternative form Besides ν ( ) = aсν ( ) ( a c( ) gc ) α ν = ; T + + T c( ) ( ) = ( ) ( ) a = a [ C C ] ; g g q q,, ν ν T v c = ( ) = ; = ; = [ ] is sum - vector (row) with elements. times сν сν с сν a + ( ) = a ( ) g ; a ( ) = c gc ν ; ( a ( ) ) T gс = q. However in our proof it practically will change nothing. 3 Numerical results Example 3.. Find explicit values of α ( v) for v = 5 and =,, 5. Verifying values should be taken from the results of recurrent calculation of free components.

6 2644 Alexander I. Nikonov We find recurrence results according to expressions (4), see Table. Table. Recurrent results. i α i α 2 i α 3 i α 4 i α Then, on the basis of expression (3) we calculate explicit values of free components, see Table 2. Table 2. Еxplicit values. + q q v ( ) C q α + C C2 C C3 C3 2 C C4 C4 2 C4 3 C C5 C5 2 C5 3 C5 4 C

7 Explicit expressions for free components of sums of the same powers 2645 As we see, explicitly calculated free components coincide with verification values respectively in Tables 2 and (columns α ). 4 Conclusion The well-known justification for explicit expressions for free components of sums of the same powers has a more complex structure than the justification considered in present article. The reason for this is the need for an extended investigation of the sum Φ (ν, ν) and the increment of the sum Φ (n -, n - ). Therefore the present proof is much more simple. This proof was carried out by the method of mathematical induction, with use of the apparatus of elementwise and matrix multiplication. Within considered numerical example we compare results of recurrent and explicit calculation of free components. Given comparison illustrates correctness of performed proof. References [] G. Hadley, Linear Algebra, Addison-Wesley Publishing Company, Inc. Reading, Massachusetts, US, 977. [2] R. Horn, Ch.R. Johnson, Matrix Analysis, Cambridge University Press, Cambridge, England, [3] H. Lutkepohl, Handbook of Matrices, John Wiley & Sons Ltd, Chichtster, England, 996. [4] A.I. Nikonov, Converting the Sum of Weight Degrees of Natural Numbers with the Same Parameters, Vestn. Samar. Gos. Tekhn. Univ. Ser. Fiz. Mat. Nauki, (2), no. 2, (In Russian, the summary in English). [5] A.I. Nikonov, Reduction of the sum of the weight equal powers to explicit combinatorial representation, Vestn. Samar. Gos. Tekhn. Univ. Ser. Fiz.-Mat. Nauki, 3 (22), no. 28, (In Russian, the summary in English). [6] A. Soifer, Mathematics as Problem Solving, 2nd ed. Springer Science, Business Media, NY, US, Received: September 8, 27; Published: October 5, 27