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August 19

Mathematicians do not use sliderule??

Is it true that Mathematicians do not use sliderule? I'm talking about back in the days where there is no such thing as pocket (electronic) calculators, and the only device for calculations are sliderule. Heard that Mathematicians do not use them and that most purchasers of sliderule are engineers. 211.28.51.233 (talk) 09:02, 19 August 2008 (UTC)[reply]

That is right. See Mathematics, it tends to be about general rules rather than particular numbers and you don't need a slide-rule for that. Engineers are more concerned about the concrete ;-) Dmcq (talk) 09:31, 19 August 2008 (UTC)[reply]

By the way you might like Human computer, they'd do the actual calculation very often when someone wanted to work some formulas or statistics out. Even engineers didn't use slide-rules all that much, many used mechanical calculators or could do just as good or better an approximation in their heads. They looked good though. Dmcq (talk) 09:49, 19 August 2008 (UTC)[reply]

It's not just a matter of sliderules, mathematicians don't use electronic calculators much either. As an example, my Uni Maths department offers 59 exams, only 23 of which allow the use of a calculator (and those are mostly statistics and probability), and it's not that they're trying to make the exams harder, you don't don't have any need for a calculator. --Tango (talk) 17:09, 19 August 2008 (UTC)[reply]

Right. In our department calculators are normally allowed for all exams, but (at least within pure mathematics) there's usually little point in bringing one; for some reason they don't tend to do topology at all well, and axiomatic set theory only slightly better. :-) —Ilmari Karonen (talk) 22:41, 19 August 2008 (UTC)[reply]

Try one of the new Möbius calculators. Wanderer57 (talk) 02:23, 20 August 2008 (UTC)[reply]

You need a better calculator. Just get one that's Turing-complete, program it to generate all the theorems of ZF, and wait for the result you want to appear on the screen. Algebraist 08:53, 20 August 2008 (UTC)[reply]

You may joke but the annual CADE ATP Systems competition which was on the 13th of this month says we'll soon be licking up to our robotic overlords and masters. Automated theorem proving says more about this competition. Dmcq (talk) 11:11, 20 August 2008 (UTC)[reply]

Pronumerals that satisfy an equation

I need to find all the combinations that satisfy this equation x² + y² + z² = 390. Is there a fast way to do it? 220.244.75.174 (talk) 10:14, 19 August 2008 (UTC)[reply]

You will probably notice that all such combinations are points that lie on the surface of a sphere with radius √390. I am not sure what kind of answer you are after but one nice way of characterizing the solutions to this is by using spherical co-ordinates:

x = √390*(cos(θ))*(sin(Φ))

y = √390*(sin(θ))*(sin(Φ))

z = √390*(cos(Φ))

for Φ belonging to [0,π] and θ belonging to [0,2π). Therefore, just choose any Φ and θ in the given range, and you can generate a triple that satisfy your solution. Of course, there may be other ways of characterizing the solutions to your equation but this is one way. For more information, you might like to have a look at Spherical co-ordinates.

I hope this helps.

Topology Expert (talk) 11:20, 19 August 2008 (UTC)[reply]

Firstly decide if the order of the numbers and signs matter. Then first simple method is to write a small program to find all the solutions and count them - or just list them all irrespective of order and then decide if the order matters. Second less simple method is to go through by hand taking the squares of 1 to 19 from 390 and check by hand if any of these is the sum of two squares. Third and most complex - learn up some maths Fermat's theorem on sums of two squares. The article doesn't say it but for two squares one can work out from the factorization how many ways of making up the number there are. There's also a theorem by Lagrange which says anything not of the form

4^{a}(8b+7)

is okay. So with this one can be sure there is some such x,y,z and you just take the squares of 1 to 19 off and use Fermat's theorem. You don't have to work out the numbers but the problem is this last method only works if the order of the numbers matters. Dmcq (talk) 11:07, 19 August 2008 (UTC)[reply]

It is not specified whether the solutions to the equation have to be integers or not. However, now that I look at this, it seems more likely that only integer solutions are required. Otherwise, there would be uncountably many solutions.

My method still works though; just work when 390 multiplied with the square of cos (Φ) is a perfect square; this is an easy thing to do since the square of cos (Φ) has an absolute value less than 1. Then for each such angle Φ (there are only 19 such values), work out whether there exists an angle θ in the given range such that (sin (θ))*(sin(Φ)) squared times 390 is a perfect square; i.e work out all such numbers with absolute value less than 1 with the property of sin (θ); from there you can work out θ by using the inverse sine function. From here you reduce the possibilites of θ to ensure that x is also an integer.

I hope this helps.

Topology Expert (talk) 11:38, 19 August 2008 (UTC)[reply]

Hyperbolic dynamics problem

Suppose x₀ is an hyperbolic fixed point of the map T.

Let T_ε be a map with the same hyperbolic fixed point such that for any x

|T_{\epsilon }(x)-T(x)|<\epsilon

.

Question: can we say that for ε small the unstable manifold W^u(T,x₀) is close to the unstable manifold W^u(T_ε,x₀)? How close should they be? --Pokipsy76 (talk) 14:17, 19 August 2008 (UTC)[reply]

Pronumerals?

What are pronumerals? I don't think they were invented when I went to school. Wanderer57 (talk) 23:20, 19 August 2008 (UTC)[reply]

I've never heard the term before (perhaps it's commoner in some other language?) but Wikipedia and Google agree that it's another word for variable (presumably specifically a variable that represents a number). Algebraist 23:25, 19 August 2008 (UTC)[reply]

Thank you. I think mathematicians should be careful with this word invention business. Otherwise it could get as bad as with the physicists. Wanderer57 (talk) 23:57, 19 August 2008 (UTC)[reply]

Pronumeral is a fairly restrained example; it has a clear etymology which makes its meaning clear. Have you ever glanced at Morphism#See also? Algebraist 00:17, 20 August 2008 (UTC)[reply]

Thank you! That is jargoneering at its best. Idempotent endomorphism! Wanderer57 (talk) 02:09, 20 August 2008 (UTC)[reply]

Not quite the same thing. All the foomorphisms are (presumably) distinct concepts that someone (presumably) needed a name for. Pronumeral on the other hand looks like a silly name for something that already has a name. Just possibly, I suppose, it's intended to make a philosophical point (say, something that stands for an undetermined numeral, rather than an undetermined number, a numeral being a symbol for a number rather than the underlying noumenon). --Trovatore (talk) 07:35, 20 August 2008 (UTC)[reply]

I doubt any mathematician was involved in inventing the term pronumeral. It sounds to me like some maths teacher thing. <RANT>It seems that many of them are happier defining new words, saying how important their new words are and that parent's aren't capable of helping because they don't know them, and complaining about how badly the children spell the new words. I'm not a words person myself so I find it doubly annoying, I feel like saying why are you teaching maths if you're only interested in messing around with words?</RANT> Dmcq (talk) 08:49, 20 August 2008 (UTC)[reply]

*cough* improper fraction. Also check out "Lockhart's Lament" on K-12 math education in the US: [[1]]. Eric. 90.184.71.189 (talk) 10:46, 20 August 2008 (UTC)[reply]

Thanks for that. Can't fully agree - I think some rote learning is good, even I have to do it, but yeah I see here he's coming from. Dmcq (talk) 13:52, 20 August 2008 (UTC)[reply]

I haven't read any of the referenced articles, so will just ask a rhetorical question...

Pronumerial is to numeral/number as pronoun is to noun -- true or false? At least, that's what it looked like to me. --Danh, 70.59.119.73 (talk) 23:53, 20 August 2008 (UTC)[reply]

Sounds like a plausible explanation to me. --Tango (talk) 00:38, 21 August 2008 (UTC)[reply]

There's a whole lot of other ones like that under Pro-form, I'd protest if I were tested on them though! :) Dmcq (talk) 09:53, 21 August 2008 (UTC)[reply]

Just struck me, if they had pronumerals then presumably they'll also have proangles, procircles, prosets.... Perhaps we should call any answer to a question a coquestion instead to get more like modern jargon? Dmcq (talk) 12:53, 21 August 2008 (UTC)[reply]

What do you call the device that you use to measure a proangle? -- Coneslayer (talk) 12:55, 21 August 2008 (UTC)[reply]

Well there are pro-objects in category theory (and by extension, within whatever domain of mathematics you care to formalise in a category), but they are something else again. -- Leland McInnes (talk) 13:31, 21 August 2008 (UTC)[reply]