Talk:Roots of a quadratic function: Difference between revisions

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→‎obscure algorithm ?: added a reference to show how different computer programming languages deal with infix operators with exponentiation.

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rosettacode>Gerard Schildberger

Revision as of 23:14, 14 October 2009 (view source) Tikkanz (talk \| contribs) m (→‎J example) ← Older edit		Latest revision as of 09:16, 13 November 2020 (view source) rosettacode>Gerard Schildberger (→‎obscure algorithm ?: added a reference to show how different computer programming languages deal with infix operators with exponentiation.)
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Line 1: == obscure algorithm ? == <cite>Forsythe, Michael Malcolm and Cleve Mole suggest to try it on a=1, b=-10<sup>5</sup>, c=1</cite>, but Ada sample code uses -10e5, which is indeed -1e6 (-10<sup>6</sup>), if I was not wrong since I knew the "e" notation... --[[User:ShinTakezou\|ShinTakezou]] 21:24, 22 June 2009 (UTC)▼ I don't like being pushed to use obscure (but more stable) algorithm over algorithm everyone instantly recognises. Said that, I could live with it. But I want to point that: * - link for "What Every Scientist Should Know About Floating-Point Arithmetic" is dead; * - one can google for it, but formatting is different, so how to find what "page 9" actually is? * - quoted in the text, "Suggested by Middlebrook (D-OA), a better numerical method: to define two parameters q=sqr(ac)/b (...)" naturally fails over if a>0 and c<0. "do it better" ? I clearly see <strike>come</strike> some of programs (Ada, and some marked as "translation of Ada") use (b^2-4ac)/2a, there correct way is (b^2-4ac)/(2a). I wonder why noone spotted this earlier! ▲<cite>Forsythe, Michael Malcolm and Cleve ~~Mole~~Moler suggest to try it on a=1, b=-10<sup>5</sup>, c=1</cite>, but Ada sample code uses -10e5, which is indeed -1e6 (-10<sup>6</sup>), if I was not wrong since I knew the "e" notation... --[[User:ShinTakezou\|ShinTakezou]] 21:24, 22 June 2009 (UTC) Basically all the test cases had a = 1, :-) So I added some test cases especially where a ≠ 1, hence 2a ≠ 2/a.... [[User:NevilleDNZ\|NevilleDNZ]] 14:28, 16 September 2010 (UTC) : In the example above (2nd line) (-10<sup>6</sup>) isn't the same as -1e6. The former is -1,000,000 and the latter is +1,000,000. --[[User:Gerard Schildberger\|Gerard Schildberger]] 18:11, 25 June 2011 (UTC) :: Other way round surely: <math>-10^n</math> will always be positive when n is even. --Laurie Alvey 10:45, 19 May 2015 (UTC) ::: Nope. -1e6 = -1000000. And -10^6=-1000000 as well, except in some insane languages. [[User:Eoraptor\|Eoraptor]] ([[User talk:Eoraptor\|talk]]) 17:35, 13 August 2020 (UTC) :::: It all depends if the infix (negative) operator has more or less priority then the exponent operator.     -- [[User:Gerard Schildberger\|Gerard Schildberger]] ([[User talk:Gerard Schildberger\|talk]]) 09:14, 13 November 2020 (UTC) :::: See:   [[Exponentiation with infix operators in (or operating on) the base]]. == J example == Dumontier, hope you don't mind me replacing your example code. I understood that you were trying to illustrate the generality of <code>p.</code> however your example used a quadratic, that had already been shown above. If you were trying to illustrate some other point I apologise! --[[User:Tikkanz\|Tikkanz]] 23:14, 14 October 2009 (UTC) == C examples == I just provided a (more) correct C version, and am now tempted to remove other C examples, because the task specifically mentioned the shortcoming of the naive method, yet they went on that route anyway. Opinions? --[[User:Ledrug\|Ledrug]] 09:08, 25 June 2011 (UTC) <s>Wait, what? The existing C and C++ code gives 10^20 and 10^-20 as roots to equation <code>x^2 - 10^-20 x + 1 == 0</code>? What the? --[[User:Ledrug\|Ledrug]] 09:22, 25 June 2011 (UTC)</s> Eh never mind that, and sorry about making a mess on the incorrect tags--I need sleep... --[[User:Ledrug\|Ledrug]] 09:40, 25 June 2011 (UTC) == Clojure example == Why are there 4 functions for clojure? What I mean is that all of that can be simplified into: <lang clojure>(defn quadratic "Compute the roots of a quadratic in the form ax^2 + bx + c = 0 Returns any of nil, a float, or a vector." [a b c] (let [sq-d (Math/sqrt (- ( b b) (* 4 a c))) f #(/ (% b sq-d) (* 2 a))] (cond (neg? sq-d) nil (zero? sq-d) (f +) (pos? sq-d) [(f +) (f -)] :else nil))) ; maybe our number ended up as NaN</lang> I find it ridiculous to have that much verbiage on 1 example.