2007-07-28 notes: [ If you have come to this entry without having read the one above it (sw00547math), please read it, and then return to this one. ] [ Tomorrow will be a better day to enter my--if I am viable tomorrow--1983 "Konen Hart's Chain Method for Determining the Coefficients of a Binomial". ] [ Konen Hart is a pen name I used for the original version of most of what is below. ] # Determining Binomial Coefficients (the complex way) The Method Case in point: 1//12//(12/2 = 6, - 1/2 = 5&1/2, x 12 =) 66// - It is presently Sunday, July 29, 2007. (66 - 12 = 54, 54/3 = 18, + 1/3 = 18&1/3, x 12 =) 220// (220 - 54 = 166, 166/4 = 41&1/2, - 1/4 = 41&1/4, x 12 =) 495// (495 - 166 = 329, 329/5 = 65&4/5, + 1/5 = 66, x 12 =) 792// (792 - 325 = 463, 463/6 = 77&1/6, - 1/6 = 77, x 12 =) 924. # Comments I was led to the above discovery (or creation, if you wish) by a desire to find a method for determining the coefficients of a binomial that was different from (and "better" than) either Pascal's Triangle or the Binomial Theorem. It was a difficult struggle. It continues to be a difficult struggle. But on 9-15-83 the long hours spent investigating the Binomial Theorem and--especially--Pascal's Triangle (and trying this & trying that) resulted in the breakthrough here exemplified. I hope that this is only a beginning. It is the beginning of my understanding of interrelated arithmetic sequences, and of the 2nd and higher orders of arithmetic sequences. Securely locked within each other, they do not allow one easy access. If, for instance, I should want to know only the coefficient for the 7th term of (x + y)¹², my searches have so far revealed to me that merely a minor shortening of the Binomial Theorem will suffice. It happens that—as a large enough Pascal's Triangle will show See this site (title below) inserted in my 1983 paper on 09.29.07— 7 is at the apex of the triangle with the Triangle that it and 12 (column two) and 12's 7th term define. So it's five to the right and five up column two: that is, (11 x 10 x 9 x8 x 7) divided by (2 x 3 x 4 x 5 x 6), and the quotient then multiplied by 12. If you noted the - 1/2, + 1/3, - 1/4, + 1/5, - 1/6 in the chain, you know where the (2 x 3 x 4 x 5 x 6) is coming from.An easy way to remember why comes from knowing that in the going across/ each of these numbers represents the order of the arithmetic sequence the coefficient at that point is a member of. So the coefficient of the 3rd term of a binomial is a member of an arithmetic sequence of the 2nd order, and the others follow in like manner. Another view of the problem proposed here is: ( 1/720 (n-1) (n-2) (n-3) (n-4) (n-5) ) n. The Binomial Theorem definitely is a beautiful gift. In regards, though, to my chain method, I have nothing to say about the beauty of it; but there are a couple to facts about it I found intriguing. The first was—once I found I needed that sequence—the need to alternately minus and plus the fractions required. The second was the need to subtract the power of the binomial/ from the determined coefficient of the 3rd term at the start of determining the coefficient for the 4th term, and then the following corresponding need to subtract the remainder of this first subtraction/ from the determined coefficient of the 4th term at the start of determining the coefficient for the 5th term, etcetera. There is much yet to learn; but whatever the future of this method I've devised, what most now pleases me is that I was able to devise a new viable method at all. Now to finish: //(924 - 465 = 461, 461/7 = 65&6/7, + 1/7 = 66, x 12 =) 792// (792 - 461 = 331, 331/8 = 41&3/8, - 1/8 = 41&1/4, x 12 =) 495// (495 - 331 = 164, 164/9 = 18&2/9, + 1/9 = 18&1/3, x 12 =) 220// (220 - 164 = 56, 56/10 = 5&6/10, - 1/10 = 5&1/2, x 12 =) 66// (66 - 56 = 10, 10/11 = 10/11, + 1/11 = 1, x 12 =) 12// (12 - 10 = 2, 2/12 = 2/12 or 1/6, - 1/12 = 1/12, x 12 =) 1. September 17, 1983 Konen Hart This last: By way of correction, the highest order of arithmetic sequence in a given binomial is that order to which its middle term (or pair of terms) belongs. # Have decided to include an odd-power example. Have chosen (x + y) to the 9th power - 1//9//(9/2= 4&1/2, -1/2 = 4, x 9 =) 36// (36 - 9 = 27, 27/3 = 9, + 1/3= 9&1/3, x 9 =) 84// (84 - 27 = 57, 57/4 = 14&1/4, - 1/4 = 14, x 9 =) 126// (126 - 57 = 69, 69/5 = 13&4/5, + 1/5 = 14, x 9 =) 126// (126 - 69 = 57, 57/6 = 9&1/2, -1/6 = 9&1/3, x 9 =) 84// (84 - 57 = 27, 27/7 = 3&6/7, + 1/7 = 4, x 9 =) 36// (36 - 27 = 9, 9/8 = 1&1/8, - 1/8 = 1, x 9 =) 9// (9 - 9 = 0, 0/9 = 0, + 1/9 = 1/9, x 9 =) 1. - Naturally, I like to think that the algorithm exemplified in both this working out and/ in the prior one/ will prove to be of more than just play value. Obviously, it is a method that does do what it is meant to do. I know, Newton's raised dots did too. Still, I am pleased. * © 2006 Brian Salchert Thinking Lizard - Afternote: This evening (some evening in 2006) and late this morning of July 29, 2007, I visited a hidden-author Pascal's Triangle site at Tripod wherein what are there called Triangular Numbers and Square Numbers (even-odd pairs beginning with 0 and 1) are in fact consecutive terms in the zero-included nnss (natural number summation sequence). About the rows in the Triangle/ this hidden author makes some observations I was quite taken by, the one regarding primality especially; but this author does not point out that these numbers are Bernoulli numbers. Even so, it is worth perusing. Read only. - See also Rasko Jovanovic's World of mathematics and mathforum.org's pascal page # A Collection of Identities and Formulas Involving Binomial Coefficients (title of site linked to in paragraph 2 of Comments above) # Brian A. J. Salchert
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Saturday, July 28, 2007
sw00546math-binom.coeffs.at65
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