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November 9[edit]

rationalize more than two[edit]

Is there a procedure, analogous to the method of continued fractions, to find small-integer ratios among more than two given numbers? —Tamfang (talk) 02:15, 9 November 2008 (UTC)[reply]

Yes, several procedures have been developed in the field of experimental mathematics - see integer relation algorithm and its references and external links. I believe the on-line Inverse Symbolic Calculator uses at least one such algorithm. Gandalf61 (talk) 10:12, 9 November 2008 (UTC)[reply]

If you are looking for a software implementation of such an algorithm, the PSLQ algorithm is available in mpmath:

    >>> from mpmath import *
    >>> pslq([pi, acot(5), acot(239)])
    [1, -16, 4]

(Sorry for the plug.) - Fredrik Johansson 10:31, 9 November 2008 (UTC)[reply]

Thank you. —Tamfang (talk) 07:06, 12 November 2008 (UTC)[reply]

(after trying it) I was hoping for something that would yield a sequence of approximations (as the continued fraction algo does) even where there is no true integer relation. Oh well. —Tamfang (talk) 16:26, 12 November 2008 (UTC)[reply]

You can do that too, by asking for a successively lower tolerance (this requires the SVN version):

>>> pslq([pi, e, 1], tol=0.1)
[0, 1, -3]
>>> pslq([pi, e, 1], tol=0.01)
[-3, 2, 4]
>>> pslq([pi, e, 1], tol=0.001)
[-2, -1, 9]
>>> pslq([pi, e, 1], tol=0.0001)
[37, -24, -51]
>>> pslq([pi, e, 1], tol=0.00001)
[124, -211, 184]
>>> pslq([pi, e, 1], tol=0.000001)
[-267, 416, -292]
>>> pslq([pi, e, 1], tol=0.00000001, maxcoeff=10**6)
[-1985, 4341, -5564]
>>> pslq([pi, e, 1], tol=0.000000001, maxcoeff=10**6)
[-523, 9165, -23270]

Fredrik Johansson 23:44, 13 November 2008 (UTC)[reply]

if someone suddenly became as smart as humanly possible at math, what could they do?[edit]

If someone one day suddenly in a flash has the ability and interest to do any math any human could ever do (no matter how much time or reference materials or drugs they're given) what could they do with it? They could solve a few million-dollar-prizes, but anything else? —Preceding unsigned comment added by 82.124.214.224 (talk) 03:05, 9 November 2008 (UTC)[reply]

It's kind of a vague question, but you could probably get a general idea by looking at what some particularly brilliant and hardworking people have already done. All of, for instance, everything, just to pick one example. Almost everything that we rely on required someone to do something clever or difficult with mathematics at some point. This person would be able to do that. We already have irrigation and farming, for instance, but they could (and do) plan automated systems to maintain farms at a trivial fraction of the original cost. Everything in the tech industry, from wireless internet to flatscreen TVs, is in a constant state of flux because clever people are contributing innovations in mathematics and engineering. They could add something to that. Space travel, medicine, the statistics and modeling used in scientific research and economics, everything is open, and much of it is profitable to someone. Not necessarily to the person who came up with it, but that's business for you. The cleverness and hard work of any one person is usually obscured by the cleverness and hard work of countless other people in the same industry, but that doesn't mean they aren't making a difference. It's just that everyone else is too. Black Carrot (talk) 04:36, 9 November 2008 (UTC)[reply]

You could do some Block transfer computation and help keep the universe in existence :) Dmcq (talk) 12:36, 9 November 2008 (UTC)[reply]

It's a difficult question to answer because maths isn't just about being really clever and able to do things other people can do, it's about creativity as well. You need to invent new maths that's never been done before. Solving the various million dollar problems requires a certain amount of basic inspiration in addition to cleverness. --Tango (talk) 14:04, 9 November 2008 (UTC)[reply]

Read the book Új észjárások by Mérő László (orig. title is probably Habits of Mind but I'm not completely sure for they don't give that in the translation for some reason). It gives the answer in chapter 9. – b_jonas 19:20, 14 November 2008 (UTC)[reply]

Ignore the above answer. I haven't read your question properly, so I've answered the wrong question. Sorry. – b_jonas 19:21, 14 November 2008 (UTC)[reply]

Interblags[edit]

Can anyone recommend some good internet math communities? I haven't been able to find much besides this reference desk. Black Carrot (talk) 04:55, 9 November 2008 (UTC)[reply]

I know one but I can't remember its name... If I recall correctly, it is something like 'Ask a topologist'. Perhaps try google searching 'ask a topologist' and you will come up to a forum (something like at.yorku as part of its web adress). From there you can get to many other math subjects such as group theory, analyis, algebraic topology etc...

Hope this helps.

Topology Expert (talk) 10:48, 9 November 2008 (UTC)[reply]

The people on the mathematics board of the xkcd forum[1] are fairly good, and I think there's a tiny overlap with this one (at least, I know *I'm* on there, and I occasionally see questions posted to both). Confusing Manifestation(Say hi!) 23:12, 9 November 2008 (UTC)[reply]

sci.math on Usenet. -- BenRG (talk) 00:38, 11 November 2008 (UTC)[reply]

Physicsforums's mathematics section. Mac Davis (talk) 22:42, 12 November 2008 (UTC)[reply]

Equations of Motion: Why is velocity sometimes given as ^-1 and sometimes with ^-2[edit]

Please see if you can explain this to me - I've googled and checked Equations of Motion:
In some examples I've encountered, velocity is written to the power of negative 1, and sometimes to the power of negative 2.
When should it be ^-1 and when should it be ^-2 ?

Here are two examples using one of the four equations of motion, namely v^-2 = u^-2 + 2as

However, in the first example the answer ends in ^-1 and in another it ends in ^-2.

Note: Examples are included for illustrative purposes only

---

Example 1:
Answer: The car will be travelling at 10 m s^-1 after accelerating at 18 m.

---

Example 2:
Answer: The driver will have to achieve a deceleration of 3.75 m s^-2.

Rfwoolf (talk) 15:00, 9 November 2008 (UTC)[reply]

Do you mean in the units? ms^-1 and ms^-2? The former are the units of velocity, the latter at the units of acceleration, which is change in velocity over time, so has an extra s^-1 in there. --Tango (talk) 15:24, 9 November 2008 (UTC)[reply]

Yes, I do mean those units. I think you've helped me a lot. Let me think it through. Thanks a million! Rfwoolf (talk) 15:30, 9 November 2008 (UTC)[reply]

Er... No I don't quite get it. I'm going to read velocity and acceleration, however, you've successfully explained why there's a difference in power, but I still don't understand why a negative power (or any power) in the first place. Rfwoolf (talk) 15:34, 9 November 2008 (UTC)[reply]

Scratch that. Thanks so much. It all makes sense now. Velocity is m/s. If we eliminate the '/' then we have to put in ^-1. With acceleration it's m/s². If we eliminate the '/' then we have to do ^-2. Why this textbook decided to eliminate the '/' is anybody's guess. Rfwoolf (talk) 15:41, 9 November 2008 (UTC)[reply]

Precisely. x^-1 is just another way of writing 1/x, that applies to units just as it applies in algebra. It's often easier to write it without the slash because it can avoid confusion, for example would m/sN be (m/s)N or m/(sN)? If you write ms^-1N or ms^-1N^-1 it's clear what you mean. --Tango (talk) 15:58, 9 November 2008 (UTC)[reply]

Done}

This function[edit]

How would I represent this function as an equation:

|
|
| 
|/\    /\    /\    /\    /\        
------------------------------
|   \/    \/    \/    \/    \/ 
|
|
|

--Melab ^±¹ 19:11, 9 November 2008 (UTC)[reply]

It can't be done as a single equation. The best you can do is something like:

f(x)={\begin{cases}(x-n)-1,&n\leq x<n+2,\,n\equiv 3\,(\mathrm {mod} \,4)\\1-(x-n),&n\leq x<n+2,\,n\equiv 1\,(\mathrm {mod} \,4)\end{cases}}

--Tango (talk) 19:23, 9 November 2008 (UTC)[reply]

I'm pretty sure it can be using things like mod and the floor function. --Melab ^±¹ 19:30, 9 November 2008 (UTC)[reply]

I don't see how, those functions aren't periodic. If there is some way of doing it, it will be very convoluted, I think my way is better. Is there some reason you absolutely need it as a single equation? --Tango (talk) 19:36, 9 November 2008 (UTC)[reply]

f(x)=({1 \over 2}-\lfloor x\rfloor {\text{ (mod 2)}})*(x-\lfloor x\rfloor -{1 \over 2})

You need to take x -> x + 0.5 if you really care about starting at f(x) = 0. Dragons flight (talk) 19:45, 9 November 2008 (UTC)[reply]

Ah, that mod, I thought the OP was talking about absolute value. Yes, using that kind of mod it might work. I'm not sure that formula works, though, I get f(1)=1/4 when I think it should be 1/2 (assuming the scale on the graph is one character equalling 1). --Tango (talk) 19:54, 9 November 2008 (UTC)[reply]

Also:

f(x)=({1 \over 2}+2\lfloor {x \over 2}\rfloor -\lfloor x\rfloor )*(x-\lfloor x\rfloor -{1 \over 2})

It oscillates between -1/4 and +1/4, and starts at -1/4 for x = 0. Scale and stretch to taste. Dragons flight (talk) 20:04, 9 November 2008 (UTC)[reply]

How about arcsin(sin(x)) ? Dmcq (talk) 21:10, 9 November 2008 (UTC)[reply]

bravo! How did you come up with that? (I tried it here http://www.walterzorn.com/grapher/grapher_e.htm - though firefox crashed the first time) —Preceding unsigned comment added by 82.124.214.224 (talk) 21:29, 9 November 2008 (UTC)[reply]

I just stuck that formula into google and it comes up with loads of entries so there's lots of people just as twisted or more than me.Dmcq (talk) 23:39, 9 November 2008 (UTC)[reply]

I mean, how did you come up with the formula -- how did you think it through from first principles or reduce it from a similar equation above or try a lot of things like that (trial and error) or.... —Preceding unsigned comment added by 82.124.214.224 (talk) 01:37, 10 November 2008 (UTC)[reply]

Also consider the possibility of representing your function by its Fourier series; a series of sines indeed:

\arcsin(\sin(x))={\frac {4}{\pi }}{\Big \{}\sin(x)-{\frac {\sin(3x)}{9}}+{\frac {\sin(5x)}{25}}-{\frac {\sin(7x)}{49}}+{\frac {\sin(9x)}{81}}-{\frac {\sin(11x)}{121}}...{\Big \}}

--PMajer (talk) 13:26, 10 November 2008 (UTC)[reply]

I see if I'd actually looked up wikipedia I'd have got both the formula I gave and the Fourier series in Triangle wave. Dmcq (talk) 14:27, 10 November 2008 (UTC).[reply]

Amazing, wikipedia contains everything...--PMajer (talk) 14:37, 10 November 2008 (UTC)[reply]

At this rate, we might as well just replace the Ref Desks with a link to Special:Search... --Tango (talk) 14:58, 10 November 2008 (UTC)[reply]

I tend towards Inclusionism but I guess there's a point of deleting things otherwise it'll become like Jorge Luis Borges's The Library of Babel Dmcq (talk) 16:13, 10 November 2008 (UTC)[reply]

Deleting things just to keep ref deskers in business would be a questionable policy, though! My comment was in jest. --Tango (talk) 16:38, 10 November 2008 (UTC)[reply]

You've already got answers, but here's a reference. This is given as an exercise in the book Jurij B. Csernyak – Robert M. Rose, A minszki csirke, Akkord Kiadó, 1999, orig. Yuri B. Chernyak – Robert M. Rose, Chicken from Minsk. It gives detailed answers. This problem is nr. 8 in chapter 11, but the book is so good you'd probably read the whole thing anyway if you get it. – b_jonas 19:08, 14 November 2008 (UTC)[reply]