User talk:Fly by Night/Archive Feb 11

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Ian R. Porteous‎

Don't know if your aware but Ian R. Porteous‎ passed away reciently and there is a new article about him. You might want to have a look at the article.--Salix (talk): 01:16, 5 March 2011 (UTC)

Thanks Salix. I already knew. I was going to dedicate an upcoming talk to him. Thanks for letting me know. — Fly by Night (talk) 02:26, 5 March 2011 (UTC)

Talkback

Hello, Fly by Night. You have new messages at Bgwhite's talk page.
Message added 02:12, 10 March 2011 (UTC). You can remove this notice at any time by removing the {{Talkback}} or {{Tb}} template.

clopen for business

I'm the IP who has been asking you questions at clopen sets, and I wanted to let you know, I'm curious to hear your next response on that talk page. 64.202.138.67 (talk) 14:38, 14 March 2011 (UTC)

Done. — Fly by Night (talk) 18:06, 14 March 2011 (UTC)

Mathematics Reference Desk: Simultaneous Equations

I agree that the discussion on http://en.wikipedia.org/wiki/Wikipedia:Reference_desk/Mathematics#Simultaneous_equation should not take place. Did you consider it helpful to the OP in the first place to grumble over my answer? Bo Jacoby (talk) 12:51, 27 March 2011 (UTC).

Very much so, because it was incorrect. The values x = −2.55 and y = −2.74 are not solutions to the simultaneous equations 7x − 12y = 15 and y − 5x = 10. Notice that 7×(−2.55) − 12×(−2.74) = 15.03 ≠ 15 and (−2.74) − 5×(−2.55) = 10.01 ≠ 10. The correct answers, as I supplied are x = −135/53 = −2.5471698113207 and y = −145/53 = −2.7358490566037. The fact of the matter is that you used the wrong method to solve what is an elementary problem. For some reason you used your computer to give numerical solutions to some other, more difficult, derived problem. Instead of just admitting you'd made a mistake you went on the attack and started to make farcical claims. Don't just take my word for it; if you know another mathematician then email him/her and ask them which of the above is the correct answer. — Fly by Night (talk) 17:04, 27 March 2011 (UTC)

See for example [[1]]. The solution to x^5+x^4-3=0 is written 1.09404. If you request exact form you get x = [root of x^5+x^4-3 near x = 1.09404]. So you get the equation and the approximation, and that's all. I did the same thing: expressed the equation and the approximate solution. The use of numerical fractions and infinite decimal fractions are less general and outdated. See also numerical analysis and algebraic equation. I wanted to provide to the unknown OP the modern approach. Your know-it-all attitude is in my opinion not improving my answer. Bo Jacoby (talk) 21:21, 27 March 2011 (UTC).

No, the solution is written as x ≈ 1.09404 which means that x is approximately 1.09404. A quick calculation shows that

$(1.09404)^{5}+(1.09404)^{4}-3=-0.24025\neq 0\,.$

A modern approach? Is the modern approach to maths to supply incorrect solutions?! Why do you always insist on moving the goal posts? The OP's question was to solve two linear simultaneous equations which can, and should, be solved exactly. Of course there are degree five equations with no such solution; but the OP didn't ask about degree five equations. I'm guessing that you're not actually a mathematician or you wouldn't be writing such rubbish. Do me a favour: go and ask another mathematician (or a school boy on a bus) for the correct answer before you continue to be so obtuse. — Fly by Night (talk) 21:38, 27 March 2011 (UTC)

Can't you provide a more precise reference to support your point of view than "another mathematician (or a school boy on a bus)"? I answered OP's question in a general way, not using the fact that the equations are linear. Writing the solution as x=-135/53 is no progress from the equation 53x+135=0. The approximation x=-2.55 is progress. I provided both the equation and the approximation, which is the proper thing to do, as shown by the wolframalpha example. You are the one who went on the attack by adressing me rather than adressing the OP. You were (and are) being rude and undiplomatic. Bo Jacoby (talk) 06:58, 28 March 2011 (UTC).

You really need to stop now, the discussion has ceased to be productive. --Salix (talk): 08:02, 28 March 2011 (UTC)

Point taken… will do. — Fly by Night (talk) 08:08, 28 March 2011 (UTC)

Your exchanges may not be entirely wasted because they did raise the interesting subject of Repeating decimals, in this case having a longish 13-digit repeating pattern. The calculators that most of us use are not equipped to detect such repetitions and are limited by their own quantisation (word length). If a long division is done manually by a person who is both tireless and observant then in theory all repeating decimals would be discoverable. However human limitations must intrude at some point because there is no limit to the lengths of repeating pattern that exist, so insisting on quoting the entire recurring pattern of digits in every case, no matter how long, is an impractical stricture. I am equally critical of numerical answers stated without explicit approximation. Example: it is reasonable to let a junior school class use "pi = 3.14" in solving circle problems (and tempting for the examiner to pose questions like "Find the diameter of a circle of circumference 314." whose solution seems easy to mark ") but IMO an examiner must reject all such answers that neglect to state their accuracy as a number of Significant figures. Just showing some digits after the decimal point leaves unknown whether the quoted value is exact, or rounded, or (Heaven Forbid!) truncated. Cuddlyable3 (talk) 19:25, 29 March 2011 (UTC)

The length of the repeating decimal expansion of a rational number is limited by the size of the denominator (provided the numerator and denominator are coprime). When you conduct a long division the remainders must be less than the divisor. Once the remainders repeat, the decimal expansion will repeat. Your scathing message on the maths reference talk page attributes the 12 place decimal expansion of the number to Bo when, in reality, you'll see that that was my post and that that was my decimal expansion. My point was that the rational solutions and the remainder theorem allowed for a rigorous proof of the decimal expansion; where a computer solution does not. Fly by Night on Tour (talk) 22:24, 29 March 2011 (UTC)

The length of the repeating pattern must by definition be finite but I don't know whether it's possible to calculate its maximum length in advance of discovering it. I apologise for not attributing the solution to you and have corrected my post by striking. It would be an interesting programming challenge to write an integer division code that can report recurring decimals within its limited wordlength (that can be made arbitrary). Cuddlyable3 (talk) 09:15, 30 March 2011 (UTC)

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The above content exists solely as an archive.

PLEASE DO NOT MODIFY IT IN ANY WAY.