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Task: Develop a fraction equivalent to $$ 1\over{\sum\limits_{i=0}^{n-1}c_in^{i/n}} $$ in which the denominator is rational.

Did
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Senex Ægypti Parvi
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  • If you're trying to do what I think yoiu're trying to do, the way to do it is to post the question, and to post the answer as an answer. – Gerry Myerson Jun 18 '14 at 09:51
  • Mr. @Gerry Myerson: I have to high-tail it back to the sandbox: Somehow, even though I posted my answer, it sems to have gotten kiboshed. Back in a minute. – Senex Ægypti Parvi Jun 18 '14 at 10:01
  • Mr. @Gerry Myerson: Thank you for your advice. By the way, I admire your constant attitude of helpfulness; it's greatly appreciated. All respect, Donn S. Miller – Senex Ægypti Parvi Jun 18 '14 at 10:15
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    Can the people voting to close perhaps say why? This question seems okay to me - the OP is answering the question themself, so there is no issue with it being homework or similar! – user1729 Jun 18 '14 at 10:16
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    I proposed a task, then I posited a method for accomplishing the task. Please tell me that this has not been criticized simply because, as in the Jeopardy quiz show, I did not frame it in the form of a question. – Senex Ægypti Parvi Jun 18 '14 at 11:16
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    @user1729PhD This just came up in the review queue (just the question). My immediate reaction was to close because I can't understand question due to the terminology equivalent fraction, then I noticed the comments and opened the question to see the answer. I saw equivalent is supposed to mean equal and I chose to keep it closed. The question is not clear to me. Whether I was ignorant in my decision is of no concern here, I'm merely pointing out why I kept it closed and why I could close it if I saw it before it was closed. – Git Gud Jun 18 '14 at 11:35
  • So it boils down to my choice of the word "equivalent," as opposed to the practically synonymous "equal?" Oh, please ... – Senex Ægypti Parvi Jun 18 '14 at 11:43
  • @GitGud I disagree with your pedantry, because I think it might actually be incorrect pedantry. If my memory serves me correctly, $\frac12$ and $\frac24$ are equivalent rather than equal, in the same way that $1$ and $13$ are equivalent modulo $12$ rather than equal. This interpretation is backed up by a superficial google search. – user1729 Jun 18 '14 at 11:46
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    @SenexÆgyptiParvi Synonymous? I don't think so. Statements are equivalent, mathematical objects are equal (or different, if they are not equal), not equivalent. I honestly couldn't understand your question. Only by coming here and seeing the answer did I understand it. Once I gained this information, I understood the question and you could have opted to open it. But I chose not to because I really think it's not clear. – Git Gud Jun 18 '14 at 11:46
  • (But words are not my thing so I may be entirely incorrect. If I am then I still disagree with your pedantry because it is precisely that - pedantry!) – user1729 Jun 18 '14 at 11:47
  • @user1729PhD "Whether I was ignorant in my decision is of no concern here". And I meant to imply that maybe I'm being ignorant here. Maybe it's widely used. But I must say, it's very bad terminology. And again: I really couldn't understand the question. I've often been described as pedantic. I take it as a compliment. I don't see pedantry as a bad thing, quite the contrary. Why wouldn't one want to be as clear as possible, I do not understand. – Git Gud Jun 18 '14 at 11:50
  • Hey, man, look! I wasn't asking anything; I was sharing. Close it, open it, blow it up -- I really don't care. Goodbye! – Senex Ægypti Parvi Jun 18 '14 at 11:51
  • @SenexÆgyptiParvi Don't take it personally. At least on my part this isn't a fight at all. We're just communicating. For what it's worth, I think your question will eventually be reopened and I'm glad it will. I just couldn't bring myself to do it. – Git Gud Jun 18 '14 at 11:53
  • @GitGud I don't see how you could not understand the question. I mean, it has a whiff of poor English about it but it is unfair to hold that against the OP. They are just doing something like $\frac1{\sqrt2}=\frac{\sqrt2}{2}$ in an odd setting...Anyway, I think we have picked this apart enough and I need my lunch! (and I apologise for jumping on you like this when all you did was take the time to explain why you didn't re-open the question. So thanks for explaining.) – user1729 Jun 18 '14 at 11:54
  • @user1729 PhD: I couldn't let this go without thanking you. I followed the "superficial Google search" to which you alluded; and to my mind you achieved Game, Set, Match and Championship! I hope I die before I develop such a chromosomal-level inability to admit to being wrong as I have seen in other parts of this kerfluffle. Best regards. – Senex Ægypti Parvi Jun 18 '14 at 14:44

1 Answers1

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Step 1: Develop the denominator. The objective will be a fraction in which the numerator and denominator are two nearly-similar determinants which differ only in their top rows. The initial form of the denominator is

$\begin{vmatrix} c_0&c_{n-1}n&c_{n-2}n&\cdots&c_3n&c_2n&c_1n\\ c_1&c_0&c_{n-1}n&\cdots&c_4n&c_3n&c_2n\\ c_2&c_1&c_0&\cdots&c_5n&c_4n&c_3n\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\vdots\\ c_{n-3}&c_{n-4}&c_{n-5}&\cdots&c_0&c_{n-1}n&c_{n-2}n\\ c_{n-2}&c_{n-3}&c_{n-4}&\cdots&c_1&c_0&c_{n-1}n\\ c_{n-1}&c_{n-2}&c_{n-3}&\cdots&c_2&c_1&c_0\\ \end{vmatrix}$.

Step 2: The numerator is merely the same determinant as the denominator, except that its top row is replaced by

$\begin{matrix} 1&\sqrt[n]n&\sqrt[n]{n^2}&\cdots&\sqrt[n]{n^{n-3}}&\sqrt[n]{n^{n-2}}&\sqrt[n]{n^{n-1}}\\ \end{matrix}$.

This gives rise, for example, to the following:

${1\over{A+B\sqrt n}}= {{\begin{vmatrix} 1&\sqrt n\\ B&A\\ \end{vmatrix}}\over {\begin{vmatrix} A&Bn\\ B&A\\ \end{vmatrix}}}={{A-B\sqrt n}\over{A^2-B^2 n}}$ and
${1\over{A+B\sqrt[3] n+C\sqrt[3]{n^2}}}= {{\begin{vmatrix} 1&\sqrt[3]n&\sqrt[3]{n^2}\\ B&A&Cn\\ C&B&A\\ \end{vmatrix}}\over{ \begin{vmatrix} A&Cn&Bn\\ B&A&Cn\\ C&B&A\\ \end{vmatrix}}}={{A^2-BCn+(C^2n-AB)\sqrt[3]n+(B^2-AC)\sqrt[3]{n^2}}\over{A^3+B^3 n+C^3 n^2-3ABCn}}$.

An example of how a typical denominator (between the bars) is set up:

$\begin{array}{ccccc|cccccc|ccccc} F&E&D&C&B&A&Fn&En&Dn&Cn&Bn&&&&&\\ &F&E&D&C&B&A&Fn&En&Dn&Cn&Bn&&&&\\ &&F&E&D&C&B&A&Fn&En&Dn&Cn&Bn&&&\\ &&&F&E&D&C&B&A&Fn&En&Dn&Cn&Bn&&\\ &&&&F&E&D&C&B&A&Fn&En&Dn&Cn&Bn&\\ &&&&&F&E&D&C&B&A&Fn&En&Dn&Cn&Bn\\ \end{array}$

Senex Ægypti Parvi
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