Talk:Fibonacci sequence: Difference between revisions

→‎Alternative: Thanks - you caught the what I hope is one of the last of *many* formulae with Gerard accidentally made invisible :(
(→‎Alternative: Thanks - you caught the what I hope is one of the last of *many* formulae with Gerard accidentally made invisible :()
 
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<!-- Sorry for the "changing" of someone's comment, but I couldn't read the original formula, so I though an enlarged version would be beneficial to everyone. --- Gerard Schildberger. -->
Fibonacci sequence can also be calculated using this formula.
<big><big><big><math> Fib[n] = \frac{(1+\sqrt{5})^n - (1-\sqrt{5})^n}{2^n \sqrt{5}} </math></big></big></big>
Size of the floating-point type (float, double, long double etc..) will limit how high n can be calulated.
 
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:OK; I'll take a look into it [[User:Telephon|Telephon]].
:I can reproduce it, yes. I had just tried and it worked on Firefox. Is there an alternative that would work on other browsers?--[[User:Telephon|Telephon]] ([[User talk:Telephon|talk]]) 14:12, 15 October 2016 (UTC)
 
:: Good question - for the set membership 'drawn from', we can use the HTML entity (ampersand '''isin''' semicolon), though it comes out smallish on its own <pre><big>&isin;</big></pre> In isolation, neither <pre><math>n</math></pre> nor <pre><math>\mathbb{N}</math></pre> choke the preprocessor, but it does seem to have difficulty with both <pre><math>\mathbb{N_0}</math></pre> and <pre><math>\mathbb{Z}</math></pre>
 
:: Would you be happy to fall back to English descriptions of the sets ? It's possible that these limitations will eventually be overcome in MediaWiki or the Math extension. [[User:Hout|Hout]] ([[User talk:Hout|talk]]) 14:59, 15 October 2016 (UTC)
 
 
Yes, a description in English is really easy here. Thank you for looking into it in detail! --[[User:Telephon|Telephon]] ([[User talk:Telephon|talk]]) 15:34, 15 October 2016 (UTC)
 
==Possibly misleading use of 'iterative' in Haskell subsection ?==
 
Hi [[User:WillNess|WillNess]] , I notice that the ('''fibonacci by folding''') Haskell example has just been moved under the heading 'Iterative'. I wonder if that doesn't risk confusing a little, or even possibly misleading ?
 
Folds are implemented recursively (either directly or indirectly) in the Prelude, and are generally understood as 'recursion schemes' in the sense of Meijer et al. (See, for example, http://blog.sumtypeofway.com/an-introduction-to-recursion-schemes/ and the much-read paper which it references http://maartenfokkinga.github.io/utwente/mmf91m.pdf).
 
I also notice that other examples which have ended up in the 'Iteration' section might risk compounding a reader's confusion – they are either implemented by direct and immediate recursion on helper functions like '''go''' and '''next''', or are expressed in terms of '''zipWith''', '''scanl''' etc, which are also implemented as recursive functions.
 
Perhaps 'iteration' is not quite the clearest or best-fitting term to use here ?
[[User:Hout|Hout]] ([[User talk:Hout|talk]]) 19:31, 2 February 2017 (UTC)
 
FWIW, I would argue that if we need to subdivide, then the main distinction here is '''memoising vs not memoising'''. If we feel a need to subdivide further, and perhaps capture something like the category now labelled "Iterative", then perhaps what we really mean here is closer to ''direct and indirect recursion'' ? [[User:Hout|Hout]] ([[User talk:Hout|talk]]) 19:44, 2 February 2017 (UTC)
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