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September 27

Math Question

Jody paid $44 bill at the hardware store using a combination of $10, $5, and $1 bills. If she paid with 13 bills in all, how many of each bill did she use?

Let i, j and k stand for the number of $10, $5, and $1 bills, respectively. Then we have 10i + 5j + k = 44 and i + j + k = 13. Subtracting the two (corresponding to the substitution k := 13 − i − j) gives 9i + 4j = 31. Modulo 4 this is i ≡ 3, so substitute i := 4m + 3, giving 36m + 27 + 4j = 31, or 36m + 4j = 4, or 9m + j = 1. Since m and j can't be negative, this is only solved by (m, j) = (0, 1). Backsubstitution gives i = 4·0 + 3 = 3 and k = 13 − 3 − 1 = 9, so (i, j, k) = (3, 1, 9). --Lambiam^Talk 01:23, 27 September 2006 (UTC)

Yo dude, I know that Lambian basically spoonfed you the answer, but make sure you can do this. An A on the homework doesn't make up for an F on a quiz. Representin', --AstoVidatu 02:46, 27 September 2006 (UTC)

Note that there aren't very many ways to get $44 from those 3 denominations, so you could just try all the possibilities, too. Obviously, you can exclude those possibilities with 14 or more $1 bills:

10+10+10+10+ 1+ 1+ 1+ 1
10+10+10+ 5+ 5+ 1+ 1+ 1+ 1
10+10+ 5+ 5+ 5+ 5+ 1+ 1+ 1+ 1
10+ 5+ 5+ 5+ 5+ 5+ 5+ 1+ 1+ 1+ 1
 5+ 5+ 5+ 5+ 5+ 5+ 5+ 5+ 1+ 1+ 1+ 1

10+10+10+ 5+ 1+ 1+ 1+ 1+ 1+ 1+ 1+ 1+ 1
10+10+ 5+ 5+ 5+ 1+ 1+ 1+ 1+ 1+ 1+ 1+ 1+ 1
10+ 5+ 5+ 5+ 5+ 5+ 1+ 1+ 1+ 1+ 1+ 1+ 1+ 1+ 1
 5+ 5+ 5+ 5+ 5+ 5+ 5+ 1+ 1+ 1+ 1+ 1+ 1+ 1+ 1+ 1

StuRat 09:44, 27 September 2006 (UTC)

Use a reasoning like: you need 4 $1 bills to get from $40 to $44 (why?), so the problem reduces to getting $40 from 9 bills. You can have 0 or 5 $1 bills (why?), 0 is impossible (why?), so you are left with $35 and 4 $5/$10 bills. Now it should be obvious.--gwaihir 10:00, 27 September 2006 (UTC)

Plural of 'series'

On the urging of another Wikipedia member, after reversions of my ideas regarding this topic, I have decided to post here and ask your opinions.

In his Principia Mathematica, Isaac Newton, when discussion series, used the latin plural 'seriei' instead of our ambiguous english habit. While the habit of singular/plural duality is tolerated for words of english origin such as 'deer', I do not believe this ambiguity belongs in Wikipedia mathematics articles.

Quite simply, the singular/plural duality of 'series', as we somewhat haphazardly have defined it, is confusing. It would not be as confusing if the word did not end with an 's', however it does. Anyone reading these articles has to do a double take to check the verb when reading an article on a series. I have asked several collegues about this already, and they voiced support to me for a clearer, less ambiguous, more grammatically correct spelling.

I thus propose the changing of 'series' in the plural to 'seriei'. One series, several seriei; again, a distinction made by Isaac Newton and lacking the ofttimes confusing singular/plural duality we have assigned to 'series' at present. We use 'criterion/criteria', and 'nucleus/nuclei', and 'formula/formulae', and 'curriculum/curricula', many of which are not exactly common distinctions made by your average english speaker, yet which go a long way towards the goal of precision and accuracy that we strive for here on Wikipedia. I propose simply extending this customary observance to another latin borrow-word which has its own well-established plural. Dbsanfte 05:19, 27 September 2006 (UTC)

I think there must be some mistake here. In Latin, the plural of series is series, and we preserve this in English. Seriei is the genitive singular. Latin nouns ending in -es do not form their plurals in -ei. Maid Marion 07:46, 27 September 2006 (UTC)

Now I feel absolutely silly. You're correct, I misread my dictionary. Please disregard this. Dbsanfte 13:03, 27 September 2006 (UTC)

Consider it thoroughly ignored. Black Carrot 14:37, 27 September 2006 (UTC)

But just to indulge in a little bit of hypothesis, if "seriei" were the Latin plural, do you seriesly (error intended) think that people of today would catch on to either the spelling or whatever the pronunciation of "seriei" is? I rather doubt it (but than, I'm a serial doubter). JackofOz 20:50, 27 September 2006 (UTC)

In general it is not incorrect to pluralize a word with S in English. There are words with latin roots which have alternative pluralization. You should find both "nebulae" and "nebulas" in your dictionary, for instance. - Rainwarrior 04:21, 29 September 2006 (UTC)

What is this question called?

I heard this question a long time ago, but I have forgotten what it is called. Suppose you have 2 types of stamps, one with a denomination of p dollars, and another of q dollars. What is the largest amount that cannot be made with a linear combination of them? The solution is (p-1)(q-1)-1 if I recall, but I'd like to find more background about this. Thoughts? --HappyCamper 17:12, 27 September 2006 (UTC)

Coin problem. Chuck 17:28, 27 September 2006 (UTC)

Largest Number Contest.

What is the largest number you can create using 38 characters? Use any notation.- R_Lee_E (talk, contribs) 17:56, 27 September 2006 (UTC)

9!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

(an iterated factorial, exclamation points are small) –Joke 18:00, 27 September 2006 (UTC)

I interpret this as saying that the expression must have 38 chars, so you may be cheating. I'm not sure what is the best way to use Conway chained arrow notation, but I'll place my bets on:

9→9→9→9→9→9→9→9→9→9→9→9→9→9→9→9→9→99

-- Meni Rosenfeld (talk) 18:10, 27 September 2006 (UTC)

A word of caution is in order here. This is a really, really large number. For example, the factorials above are, if I'm not mistaken, much less than 2→4→3. And this grows unimaginably large for longer chains. -- Meni Rosenfeld (talk) 18:24, 27 September 2006 (UTC)

That number gives me headaches :-) --HappyCamper 18:26, 27 September 2006 (UTC)

Better B(B(B(B(B(B(B(B(B(B(B(B(B(B(B(B(B(B(9)))))))))))))))))) where B(n) is the Busy beaver function. Gives a much larger number than that with the chained arror notation. (Even only 4 or 5 or busy beavers will almost certainly surpass the chained arrow number given above. This number should give even worse headaches. Note that it is arguably not in the same category as the above numbers since it is uncomputable where one could at least in principal compute the other numbers. JoshuaZ 18:33, 27 September 2006 (UTC)

To the contrary, B(...(B(9)...) is just one natural number and is thus trivially computable. The function B(...(B(n)...) would not be computable, but this is just one value. CMummert 02:33, 2 October 2006 (UTC)

This is a pretty big cardinal number:

$2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{\aleph _{\omega }}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}$

But probably someone can write a bigger one. –Joke 18:53, 27 September 2006 (UTC)

Interesting. The rules didn't specify the number has to be finite. Here is a large ordinal number (though if I'm not mistaken, it's less than the ordinal matching the cardinal above):

$\epsilon _{\epsilon _{\epsilon _{\epsilon _{\epsilon _{\epsilon _{\epsilon _{\epsilon _{\epsilon _{\epsilon _{\epsilon _{\epsilon _{\epsilon _{\epsilon _{\epsilon _{\epsilon _{\epsilon _{\epsilon _{\epsilon _{\epsilon _{\epsilon _{\epsilon _{\epsilon _{\epsilon _{\epsilon _{\epsilon _{\epsilon _{\epsilon _{\epsilon _{\epsilon _{\epsilon _{\epsilon _{\epsilon _{\epsilon _{\epsilon _{\epsilon _{\epsilon _{\epsilon _{\epsilon _{\omega }}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}$

Now for a question of my own. Joke's cardinal can be rewritten with Beth number notation:

$\beth _{\beth _{\beth _{\beth _{\beth _{\beth _{\beth _{\beth _{\beth _{\beth _{\beth _{\beth _{\beth _{\beth _{\beth _{\beth _{\beth _{\beth _{\beth _{\beth _{\beth _{\beth _{\beth _{\beth _{\beth _{\beth _{\beth _{\beth _{\beth _{\beth _{\beth _{\beth _{\beth _{\beth _{\beth _{\beth _{\beth _{\beth _{\aleph _{\omega }}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}$

I've heard once of a Gimel number notation, which potentially could be used to write even larger numbers, but am unable to find any information about it. Has anyone ever heard of such a notation? -- Meni Rosenfeld (talk) 19:20, 27 September 2006 (UTC)

(Edit conflict.) According to Beth number, we can rewrite the above as

$\beth _{38}(\aleph _{\omega })$.

A much larger number would be

$\beth _{9\rightarrow 9\rightarrow 9\rightarrow 9\rightarrow 9\rightarrow 9\rightarrow 9\rightarrow 9\rightarrow 9\rightarrow 9\rightarrow 9\rightarrow 9\rightarrow 9\rightarrow 9\rightarrow 9\rightarrow 9\rightarrow 9\rightarrow 9}(\aleph _{\omega })$

But we're starting to get silly. –Joke 19:40, 27 September 2006 (UTC)

Yes. That number is no joke. - R_Lee_E (talk, contribs) 21:36, 27 September 2006 (UTC)

gimel(κ) is sometimes used to mean κ raised to the power cof(κ). I don't think it necessarily blows up hugely, but it is always larger than κ. As I recall the singular cardinal hypothesis can be stated as gimel(&kappa) = κ⁺ for every singular κ. --Trovatore 19:37, 27 September 2006 (UTC)

Unfortunately, this question gets a bit silly when we realise that there is probably nothing stopping me defining a function T(n) to be B(B(B(n))) (B = Busy beaver function) and so on ... - so why do we not make it a bit more of a challenge: what is the biggest number you can create within the 38 character limit if we restrict ourselves to the symbols: 0-9, +, -, x, /, (, ) i.e. no exponentials and no factorials. Any takers? Madmath789 20:14, 27 September 2006 (UTC)

Yes, but the idea is to only use standard notation, or at least one which is defined somewhere other than this discussion. Besides, under the rules you present the answer is trivially 99999999999999999999999999999999999999. -- Meni Rosenfeld (talk) 20:32, 27 September 2006 (UTC)

Wondered how long it would take to spot that :-) Madmath789 20:40, 27 September 2006 (UTC)

Allow me to cheat: $1/0$ —Bromskloss 09:32, 28 September 2006 (UTC)

Nope, 1/0 is, at best, an unsigned infinity, which isn't larger than any of the other numbers suggested here. -- Meni Rosenfeld (talk) 09:43, 28 September 2006 (UTC)

Now the only problem is determining a winner. 207.233.9.240 20:58, 27 September 2006 (UTC)

Am I allowed to use non-notable notation? User:Salix alba/Jonathan Bowers defines some truly hugh finite numbers. --Salix alba (talk) 22:01, 27 September 2006 (UTC)

Please! No!! we had enough of a problem a few weeks ago with a large number of articles he created ... Madmath789 22:05, 27 September 2006 (UTC)

Well, if you don't require the axiom of choice, we could always go with Reinhardt cardinal, which is 18 characters and vastly larger than anything else mentioned here (or derived from the processes mentioned here). For a boggling array of very big things, see list of large cardinal properties. -- Fuzzyeric 03:07, 28 September 2006 (UTC)

If we want to stay within finite numbers, I would use Steinhaus–Moser notation in conjunction with Conway chained arrow notation:

9→9→9→9→9→9→9→9→9→9→9→9→9→9→9→9→9→9 all inside a polygon with the same amount of sides (since that polygon could be drawn as a single symbol, technically it would only count as one ^_^)

--Ķĩřβȳ♥ŤįɱéØ 06:38, 28 September 2006 (UTC)

Bah... My best guess would've been B(B(B(B(B(B(B(B(B(B(B(B(G^G)))))))))))) ☢ Ҡi∊ff⌇↯ 23:06, 28 September 2006 (UTC)

If we aren't limited need to use mathematical notation strictly, len(π) or len(${\sqrt {2}}$) or even len(1/3) in an infintely powerful computer would return the number of digits in a recurring/irrational/transcendental number (ie infinity). smurrayinchester^{(User), (Talk)} 23:20, 29 September 2006 (UTC)

That's only $\aleph _{0}$. We've already done way better than that. -- Meni Rosenfeld (talk) 08:39, 30 September 2006 (UTC)

For any number n there is a notation in which the symbol Q stands for n and thus n is expressible in one symbol using this notation. Similar problems appear in Kolmogorov complexity - for any string, there is some compression algorithm (like PKZIP, but different) that reduces the string to a single character. CMummert 02:28, 2 October 2006 (UTC)

What about (The largest number defined here)+1? (R_Lee_E wrote "Use any notation". And yes, I've heard the name Gödel.) JoergenB 16:02, 2 October 2006 (UTC)

Actually, the largest number defined here is $\beth _{9\rightarrow 9\rightarrow 9\rightarrow 9\rightarrow 9\rightarrow 9\rightarrow 9\rightarrow 9\rightarrow 9\rightarrow 9\rightarrow 9\rightarrow 9\rightarrow 9\rightarrow 9\rightarrow 9\rightarrow 9\rightarrow 9\rightarrow 9}(\aleph _{\omega })$, an infinite cardinal, which of course absorbs addition with natural numbers. So your "new" number is, in fact, equal to it. -- Meni Rosenfeld (talk) 17:20, 2 October 2006 (UTC)

Math Composer ThIng

Does anyone here know of a Windows program that can be used to write out math problems and such. I'm speaking of an application like MS Word only for math and numbers. Hope I'm being specific enough. Deltacom1515 20:53, 27 September 2006 (UTC)

I'm not sure, but LyX is quite good. --HappyCamper 21:02, 27 September 2006 (UTC)

MS Word will actually work for many purposes. In MS Word (I'm assuming MS Word 2003 here; I don't know whether the feature existed in earlier versions, or if so, how good it was), go to Insert - Object... - Microsoft Equation 3.0. The interface takes a bit of getting used to, but you can do quite a bit as far as writing out mathematical formulae. Chuck 21:49, 27 September 2006 (UTC)

I would definitely go with a version of LaTeX, such as the freely available MikTeX. The Word equation editor is OK for fairly trivial uses, but for anything of any complexity or of any length, you really need power and flexibility of TeX. Madmath789 21:53, 27 September 2006 (UTC)

Thanks guys. Question answered. Deltacom1515 00:20, 28 September 2006 (UTC)

Late to the party... I use OpenOffice. Much like Word... Insert|Object|Formula. -- Fuzzyeric 03:10, 28 September 2006 (UTC)

Solids

Is there a name for this solid? It is a square prism with rounded corners, but the top and botton are rounded also

What about this one, an oval prism also with rounded top and bottom. Oblate? Prolate? Maby there is no name for these.

I actually need this info for an article. Herostratus 21:24, 27 September 2006 (UTC)

These are not shapes for which there are standard geometrical names. I've seen "pillow" used for shapes somewhat similar to the first.  --Lambiam^Talk 22:25, 27 September 2006 (UTC)

Rounded rectangular prism? Rounded cylinder? (Or whatever an oval cylinder is called.) - Rainwarrior 04:35, 28 September 2006 (UTC)

Rounded eliptical cylinder perhaps? - Rainwarrior 04:39, 28 September 2006 (UTC)

My best guess is the union between a superellipsoid and a cylinder ☢ Ҡi∊ff⌇↯ 04:51, 29 September 2006 (UTC)

Clouds can form in shapes like these, they are known as lenticular clouds. If you were defining a name for these, you could call these lenticular solids. Dysprosia 08:49, 29 September 2006 (UTC)

Thanks to all for your replies. Herostratus 07:22, 30 September 2006 (UTC)

Soap? – b_jonas 13:18, 1 October 2006 (UTC)

September 28

4D "volume" ?? (length^4)

1D:length

2D:area

3D:volume

4D:??

If anyone knows this term it would be greatly appreciated. Tuvwxyz (T) (C) 02:29, 28 September 2006 (UTC)

4-volume. (Exercise for the reader: generalize to d dimensions.) –Joke 02:38, 28 September 2006 (UTC)

Actually, if I were seriously writing it I would put "four-volume" per the manual of style. –Joke 02:39, 28 September 2006 (UTC)

Volume suggests "content", which I'd never heard before, and doesn't mention "4-volume". Generally "4-volume" will do, as will "volume" if it's clear you're in four dimensions.

(If anything else finds the "content" note odd, feel free to fix it).

RandomP 02:44, 28 September 2006 (UTC)

There's also the term "hypervolume". MathWorld also has "content", and that may be where Wikipedia got it from. It may be a shortened version of Jordan content.  --Lambiam^Talk 05:06, 28 September 2006 (UTC)

4 dimensions doesnt physically exist in our world, and even if it did you'll still see it as a 3D object--RedStaR 08:48, 28 September 2006 (UTC)

• The OP did not mention the word "physical" anywhere.
• It is very common for time to be considered the 4th dimension.
• There are (prominent?) physical theories according to which the universe has as many as 26 dimensions.

-- Meni Rosenfeld (talk) 09:00, 28 September 2006 (UTC)

If we take time as the fourth dimension, it is difficult to delineate dimensions. What time do you call the "start" of your object, or the "end" of it? How does one interpret a measure of how much spacetime an object occupies meaningfully? Maelin 08:11, 29 September 2006 (UTC)

The only problem here is that objects in spacetime tend to be unbounded. If you have a cylinder in 3D with infinite length, its volume will be infinite. But if you clip it to a finite length, you'll get a "normal" cylinder with finite volume. Likewise, if you have a 3D sphere which does not move, it will have an infinite "length" in the time dimension when viewed in 4D, and thus have infinite 4-volume. However, if you take such an object and "clip" it, that is, look only at its portion in a specific time interval, you'll have a finite 4-volume. For example, a sphere of radius 2m which exists for 3s will have a 4-volume of 100.53 m³s. Another example: A ballon which starts out at 0 volume, expands and then shrinks back to 0 volume. It will have a finite 4-volume, calculated as the integral of its spatial volume with respect to time. -- Meni Rosenfeld (talk) 08:35, 29 September 2006 (UTC)

Well it is just the measure whatever the dimension d.Billlion 17:17, 1 October 2006 (UTC)

Why 1+1=2?

Can someone tell me why 1+1=2? Please...I really want to know.Thks.

It's the definition of 2. Conscious 11:26, 28 September 2006 (UTC)

In the article on natural numbers, I suggest you see the section Formal definitions for a definition of the numbers themselves and then the section Properties for a definition of addition. Return here when there is something you don't understand. (Also, please sign your comments.) —Bromskloss 12:15, 28 September 2006 (UTC)

I seem to recall that it was proven in the Principia Mathematica. Why not look there? (Expect to have to read about 300 pages to get to it though.) Myself, I just accept it as an axiom instead. - Rainwarrior 15:50, 28 September 2006 (UTC)

Metamath merely states 1+1=2 as the definition of 2. However, it has a proof that 2+2=4. If expanded fully, the proof consists of 22,607 steps. Fredrik Johansson 16:04, 28 September 2006 (UTC)

Like so much else, it depends on your foundations. Some authors define 2 to be 1+1. Others define 2 to be the successor of 1, so to prove 1+1=2 you'd need to compare that with whatever definition of "+" you're using. Given that counting underlies our intuition about naming numbers, the definition of 2 as "the next number after 1" is pretty natural. A more direct definition of 2 is this many: **. A picture proof would be good enough on that level. Melchoir 19:06, 28 September 2006 (UTC)

Or, if you prefer a proof based on number conservation as a foundation for the cardinality concept: http://www.youtube.com/watch?v=KCIHn5adOnM&NR Melchoir 19:38, 28 September 2006 (UTC)

If we use the Peano axioms for the natural numbers, we can define 1 = S(0), 2 = S(S(0)), 3 = S(S(S(0))) and 4 = S(S(S(S(0)))). Addition is defined by m + 0 = m, m + S(n) = S(m) + n. Then 2 + 2 = S(S(0)) + S(S(0)) = S(S(S(0))) + S(0) = S(S(S(S(0)))) + 0 = S(S(S(S(0)))) = 4. Only four steps needed. I wonder what Metamath is doing in the other 22,603 steps.  --Lambiam^Talk 19:10, 28 September 2006 (UTC)

I'm getting confused. How did you go from here:

S(S(0)) + S(S(0)) aka 2 + 2

to here

S(S(S(0))) + S(0) aka 3 + 1

The step above is an instance of m + S(n) = S(m) + n with m = S(S(0)) and n = S(0).  --Lambiam^Talk 01:52, 29 September 2006 (UTC)

No one has answered the actual question, which is "Why?", because it is not a proper question in mathematics (or in physics). We can show a particular set of axioms and a deduction within a system of logic and proofs, but this is mostly an exercise in definitions. Once we have defined "1", "2", "+", and "=", there is little left to say in a formal demonstration. But as for why we choose these definitions, one might as well ask, "Why pound a nail with a hammer, not pliers?" --KSmrq^T 23:17, 28 September 2006 (UTC)

Ok, so the reason why 1+1=2 is that it follows from a set of axioms that we chose to adopt because they very successfully model our experiences with collections of objects (also our experiences with magnitudes) in the world? -GTBacchus^(talk) 23:21, 28 September 2006 (UTC)

Personally I prefer to say that it follows from a set of definitions that we adopted. Yes, they are axioms, but in the sense of the axioms in the definition of a group. The validity of the conclusion 1+1=2 is independent of the motivation for choosing these definitions. As Einstein said (according to Wikiquote): As far as the laws of mathematics refer to reality, they are not certain; and as far as they are certain, they do not refer to reality. I might add that this does not say much about maths; in fact, as far as anything refers to reality it is not certain. For example, the model does not work very well for frogs in a wheelbarrow, nor for socks in the wash. You might say that these particular "axioms" proved to offer a useful mathematical model for counting objects. But, somewhat conversely, you might say that the model tells us something about us, namely what we mean by "counting" and what we consider "objects", by defining properties "objects" should have before it makes sense to "count" them.  --Lambiam^Talk 02:14, 29 September 2006 (UTC)

Good points. I'm reminded of the book Where Mathematics Comes From, by Lakoff and Nuñez, in which they discuss evidence that very young infants are capable of subitizing, which suggests that our tendency to discretize the manifold of our experience into countable objects is, to some degree, hard wired. This in turn suggests that we exist in some kind of environment that is amenable to being successfully navigated by means of discretization into "objects". Not that I'm saying we really obtain any access to "the thing in itself", but it is pretty easy to believe that there's something out there corresponding to consensual reality. I'm going to continue to assume it, anyway. ;) -GTBacchus^(talk) 18:29, 29 September 2006 (UTC)

Ha ha ha! I've actually tried pounding a nail with pliers when no hammer was available. (No, it doesn't work too well.) - Rainwarrior 04:12, 29 September 2006 (UTC)

valid range

Suppose I have a function such as f(x)=[x*(x-1)]/2 where f(0)=0 and f(1)=0 and f(2)=1 where a value of zero for the function is mathematically accurate and suggests that all values of x below two are invalid. Do I simply add a note of text to state this or is there a mathematical symbol or notation I can or should use? Adaptron 15:50, 28 September 2006 (UTC)

You should be clearer about what this function signifies. If it's just a function then there's no problem with a value of 0 for the function or with values less than 2 for x. If it represents some physical situation where x must be at least 2, you can say, "for every x ≥ 2, $f(x)={\frac {x(x-1)}{2}}$", or you can say that the domain of f is [2, ∞). -- Meni Rosenfeld (talk) 16:12, 28 September 2006 (UTC)

Would I use the same method of qualification to indicate that only integer values of x would be valid? Adaptron 17:00, 28 September 2006 (UTC)

That depends on how technical you want to be. Here are several suggestions:

• For every integer x ≥2, $f(x)={\frac {x(x-1)}{2}}$. Probably the clearest.
• For every n ≥2, $f(n)={\frac {n(n-1)}{2}}$. The letter n is usually only used for integers, and most people who read this will interpret it this way.
• $f\colon \mathbb {Z} \cap [2,\infty )\to \mathbb {Z} \quad \quad n\mapsto {\frac {n(n-1)}{2}}$. Very precise, but will probably not be legible to non-mathematicians. You can also say that $\mathbb {Z} \cap [2,\infty )$ is the domain of the function.

-- Meni Rosenfeld (talk) 17:47, 28 September 2006 (UTC)

To some extent these are matters of taste, but I might write

$f(x)={\frac {x(x-1)}{2}}\qquad ;x\geq 2\,\!$

for a simple boundary restriction, and either

$f(n)=n(n-1)/2\qquad ;x\in \mathbb {Z} \,\!$

or

$f\colon \mathbb {Z} \to \mathbb {Z} \,\!$

$n\mapsto n(n-1)/2\,\!$

to restrict to arbitrary integers. To impose both restrictions, some authors would write

$f(n)=n(n-1)/2\qquad ;2\leq x\in \mathbb {Z} .\,\!$

Personally, I find this uncomfortably compressed. Likewise, I find the intersection form needlessly baroque. I'd have to see the context to decide what works best, but generally I'd keep it simple:

For n an integer greater than 1, let f(n) = n(n−1)/2.

After all, we write mathematics for other humans to read, and the notation is supposed to help, not hinder, our efforts. --KSmrq^T 23:45, 28 September 2006 (UTC)

Venturing into physics for context and acknowledging half-life represents a transformation between two atoms where the daughter increases in direct and absolute proportion to the decrease in parent atoms where the sum total of atoms remains the same, would the above usage be adequate in your opinion to express and clarify the idea that a situation where the number of parent isotopes is less than one but greater than zero can not exist since half-life is not an infinite process (as it might be if atoms were divisible by the half-life process) but terminates when the parent isotope is less than one? 71.100.167.194 23:49, 29 September 2006 (UTC)

I assume that by "the number of parent isotopes" you mean: "the number of atoms of the parent isotope". If by "number of" you mean a (finite) cardinal number, as used for counting: 0, 1, 2, ..., and N represents that number, you can say: N is a non-negative integer. Depending on your belief you might perhaps say: N is a natural number. This applies not only to the number of atoms, but also to the number of times you've been married, or the number of pennies you have in your wallet. Saying that adequately represents the idea that N is then not equal to 0.99 or −1.  --Lambiam^Talk 00:37, 30 September 2006 (UTC)

Since I believe the likelihood of the number of atoms of the parent isotope (thanks for the correction) remaining no less than one for any reasonable multiple of the half-life after one atom is achieved is zero the statement that N is a non-negative integer as opposed to a positive integer or a natural number seems to be the right way to go. I have heard the claim that not achieving zero number of atoms of the parent isotope is possible even in a closed system subject only to decay because half-life measurement is a relative measure rather than being absolute. 71.100.167.194 01:36, 30 September 2006 (UTC)

Graphing an indifference Curve

Suppose we have a Utility function (in economics) where a certain string of baskets of goods equals the same utility, creating an indifference (preference) curve. Normally, I would find this easy to graph, creating the indifference curves eqyal to a fixed C.

However, for some reason, I can't graph U = sqrt(X) + sqrt (Y) correctly. The way I graph it in my head is like so: First, lets do the curve for U = 1. The intercepts would be 1,0 and 0,1. This would mean that the curves intersect the axis. Then, assuming equal proportions of X and Y, so both their roots equal 0.5, both X and Y are about 0.7~. Adding (0.7,0.7) gives me a graph similar to an X squared plus Y squared equals 1 graph...how is this so?

I know the answer should be a normal asymptotic curve along the axis, but I cant logically get it? Help! 152.163.100.11 17:27, 28 September 2006 (UTC)

If √X = 0.5, then X=0.25, not 0.7. -- Meni Rosenfeld (talk) 17:49, 28 September 2006 (UTC)

And it's not really an asymptotic curve. As X → 0, Y → U². In fact, it is a segment of a parabola whose axis is formed by the line x = y, touching the two coordinate axes x = 0 and y = 0.  --Lambiam^Talk 18:17, 28 September 2006 (UTC)

Oh snap thanks, I completely overlooked the possibility of a trivial mistake in my rooting. Thanks 64.12.116.11 20:35, 28 September 2006 (UTC)

Sine of a complex number?

I have looked at complex numbers but the article did not tell me wherether sin(i) is a legal mathematical operations. Or is it illegal like division by zero? 202.168.50.40 21:50, 28 September 2006 (UTC)

It's equivalent to $i\sinh 1\approx 1.1752\,i$. In general, $\cos ix=\cosh x\!$, and $\sin ix=i\,\sinh x$. You can also use Euler's formula and its corollaries. To do complex rather than just imaginary numbers, use Euler or else the sum and difference formulas. --Tardis 22:45, 28 September 2006 (UTC)

There are also more details at Sine#Relationship to exponential function and complex numbers. Melchoir 22:55, 28 September 2006 (UTC)

Once we extend the sine function to complex numbers, there is nothing illegal about feeding it a complex number. The question is, how do we so extend? In this case, we can take the series expansion,

$\sin(x)=x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}-\cdots ,$

and use it unchanged, thus converting a real analytic function into a complex one. Furthermore, as a complex function it is analytic over any open subset of complex numbers. Specifically, it converges at i. To find the value, it is convenient to use Euler's formula,

$e^{ix}=\cos(x)+i\sin(x).\,\!$

This equality is written in terms of complex numbers, and holds for complex x as well as real. Noting that sine is an odd function, so that sin(−x) = −sin(x), we derive

$\sin(x)={\frac {e^{ix}-e^{-ix}}{2i}}.$

Letting x equal iy for real y, this produces

$\sin(iy)\,\!$	${}=(e^{i^{2}y}-e^{-i^{2}y})/2i\,\!$
	${}=(e^{-y}-e^{y})/2i\,\!$
	${}=i(e^{y}-1/e^{y})/2.\,\!$

Thus a pure imaginary argument produces a pure imaginary result, averaging an exponential rise with an exponential decay. For large values of y the decaying term makes a negligible contribution; and the periodicity of sine for real arguments has been completely lost for imaginary arguments. --KSmrq^T 01:46, 29 September 2006 (UTC)

Please visit your new friend. Twma 04:43, 29 September 2006 (UTC)

September 29

Mixed trig / linear terms

Is there an algebraic method for finding solutions to equations of the form ax + bsin(x) + c = 0 ? Maelin 10:50, 29 September 2006 (UTC)

I don't think so. One special case would be $\cos x=x$ (almost, anyway), and there is no clean way to solve that, as far as i know. —Bromskloss 11:38, 29 September 2006 (UTC)

There's probably a way using the series definitions of the trig functions. ☢ Ҡi∊ff⌇↯ 11:59, 29 September 2006 (UTC)

No, that won't help. Replacing the trigonometric function with an infinite series just gives you an equation where one side is an infinite series. This doesn't simplify the problem. It would help if you could do some manipulation on the series and find that the result is a series for some other known function that provides an algebraic solution, but in this case the series is just the sine series with a few terms altered, and that doesn't help. Fredrik Johansson 15:35, 29 September 2006 (UTC)

I am almost certain that (except for special cases) it will be impossible to find a closed expression for the solutions in terms of a finite combination of elementary functions like exponentials, sin, cos, tan, (and inverse trig functions) log, exp, or hyperbolic functions. I suspect that the best you will get (other than a numerical answer to a certain degree of accuracy) will be an infinite series expansion by using the Lagrange inversion theorem. Madmath789 12:28, 29 September 2006 (UTC)

I think that there isn't. If there was a way, then you could construct a cos(1) angle with compass and straightedge; but http://en.wikipedia.org/wiki/Angle_trisection. Sorry, maybe in the morning my brain will work, so I can give you the exact proof.

Functor nOOb200.65.178.127 07:45, 30 September 2006 (UTC)

data

data and graphs from 1990 to 2005 of the following South African macroeconomic variables: Real GDP, Inflation, Unemployment and Balance of payments

The above is not a question, and is not about mathematics, the topic of this reference desk.  --Lambiam^Talk 12:57, 29 September 2006 (UTC)

Hello! You should be able to post your question in another ref. desk page, as the misc. one. And please remember that no mechanical nor computerized parts are tortured to answer you question, we're not googols ;-)

So the title of the question might give : South A. macroeconomics, and the question could start with "hello, where may I find ...". Now, try this link first : South Africa. Thank you. -- DLL ^{.. T} 19:16, 29 September 2006 (UTC)

Economic Bundles

I've solved an equation for a bundle that includes 4 of unit X and 2.5 of unit Y as the utility-maximizing one. How come it is okay to have non-whole numbers as part of an answer...its not like I can go out an actually buy 2 and a half bananas!

Because mathematicians care more about number than whether it possible to have 2.4 children. More seriously, there are many applications where it is useful to have fractional results. So its good practice to start using them now as it will help you a lot in the future. --Salix alba (talk) 21:36, 29 September 2006 (UTC)

Just multiply by 2, then you have 8 of unit X and 5 of unit Y. StuRat 21:45, 29 September 2006 (UTC)

There are also applications where the solution must need be in whole numbers; anything else is totally useless. That is why there is the whole area of integer programming in the field of operations research. However, this is much and much harder.  --Lambiam^Talk 22:55, 29 September 2006 (UTC)

September 30

Inequality Question

Hi, I know that this is homework, but I would be indebted to whoever can point out and explain where I have gone wrong and how I should have gone right:

The area of a rectangle is 12 cm². Find the range of possible values of the width of the rectangle if the diagonal is more than 5cm.

I got (apologies for my inability to use the Latex notation).

height=h, width=w

h^2 + w^2 > 5^2

h^2 + w^2 > 25

hw = 12

h = 12/w

(12/w)^2 + w^2 > 25

144/w^2 + w^2 > 25

144 + w^4 > 25w^2

Then the whole thing falls apart for me.

Thank you for your help, —Daniel (‽) 09:46, 30 September 2006 (UTC)

Define z = w^2. Then your last equation can be rewritten as 144 + z^2 > 25z, or equivalently z^2 − 25z + 144 > 0. This is a standard quadratic inequation. Given the range of z satisfying it, and considering that w must be non-negative, the possible range for w consists of the square roots of the non-negative part of the solution range for z. It may be easier to look at the values that violate the inequation.

A different route to the solution is found from the consideration that you can easily deduce inequations for h^2 + 2hw + w^2 = (h + w)^2 as well as for h^2 − 2hw + w^2 = (h − w)^2.  --Lambiam^Talk 10:21, 30 September 2006 (UTC)

Thanks a lot. I used your first method to come to 0 < w < 3 (or) w > 3. I don't quite understand what you mean by the second method. Thank you again! —Daniel (‽) 16:06, 30 September 2006 (UTC)

Unfortunately, I think you made an error somewhere. Take w = 2√3 = 3.4541... This satisfies w > 3. Then w^2 = 12, so h = 12/w = w. Then h^2 + w^2 = 2w^2 = 24, which is not greater than 25. Unless you made a simple copying error, to get into this situation, you must have concluded before that z > 9 satisfies the inequation in z. But clearly that is incorrect: 144 + 100 < 250.  --Lambiam^Talk 16:25, 30 September 2006 (UTC)

I suspect the problem is meant to suggest a 3-4-5 right triangle, which is half a rectangle with diagonal 5 and area 12. Suppose we let the width be 4, the larger side. If the width increases by a factor of s, the height must be divided by s to maintain the area. Consider the effect on the diagonal. Is an width increase allowed? A decrease? (Hint: if the rectangle is a perfect square, the sides have length √12, making the diagonal too short; while if the width is at least 5 the diagonal is at least 5.) --KSmrq^T 18:34, 30 September 2006 (UTC)

Lambiam: Sorry, the keyboard gremlin got me there; I meant to type 0 < w < 3 (or) w > 4. Hopefully that sorts out your problem. KSmrq: I can vaguelly see where you're going, but to be honest lack the time and energy to follow it up, as I have answered the question and have a huge amount of other stuff to do. :( —Daniel (‽) 20:03, 1 October 2006 (UTC)

Venn diagrams

I was wondering, if all wigs are wags, and some wags are wogs, does that mean that all wigs are wogs? Please hurry, my exam finishes in 10 minutes!

If all students are humans, and some humans are females, does that mean that all students are females?  --Lambiam^Talk 15:51, 30 September 2006 (UTC)

If your exam finishes in 10 minutes, what are you doing writing on Wikipedia? —Daniel (‽) 16:02, 30 September 2006 (UTC)

No. Himanyo 17:27, 30 September 2006 (UTC)

They were doing the Tickle IQ test, and just failed miserably. Don't worry, if you got over 100, you are average. But seriously dude, who tries to cheat on an IQ test? --AstoVidatu 00:04, 1 October 2006 (UTC)

Isn't that IQ test which is so ridiculously unbalanced that the average score is about 120? Laïka 09:42, 1 October 2006 (UTC)

If all students are slackers, and some slackers don't do their own homework, does that mean no students do their own homework ? StuRat 11:34, 1 October 2006 (UTC)

Gold!! Thanks for your help guys, that was a past question from a math test I did. PS. I got it right. —The preceding unsigned comment was added by 83.142.184.86 (talk • contribs) 14:48, October 1, 2006 UTC).

Function question

File:Functions x2 sqrtx 0-10.JPG

I was wondering if I could find an equation to describe a curve in a Cartesian System of Axes, given that the curve I need is the graphs of 2 (or maybe more) functions combined.

i.e. for:

f(x) = x^2
g(x) = sqrt(x)

I would need an equation to describe a curve like the one at the right

Thanks, --Danielsavoiu 15:50, 30 September 2006 (UTC)

You could use the equation (x − y^2)(y − x^2) = 0, which however does not give you a curve. If you want to stay within the first quadrant, gluing these two graphs together, you could use a parametric equation, something like

     x = t^2 if t < 0, x = t  otherwise

     y = − t if t < 0, y = t^2 otherwise

 --Lambiam^Talk 15:59, 30 September 2006 (UTC)

Thanks, Lambiam. I thought about parametric equations, but I would rather have only one solid equation. Your first suggestion ( (x − y^2)(y − x^2) = 0 ) is ok, but what if I wanted to glue the graphs of:

f(x) = x^3
g(x) = 3x

--Danielsavoiu 16:17, 30 September 2006 (UTC)

If I understood your question correctly, then no you can't do that. There's no single function of one variable that can have the values of x^3 and 3x at the same time. You can, however, glue then in different sides of the Y axis: $x^{3}\left({\frac {\operatorname {sgn}(x)+1}{2}}\right)+3x\left({\frac {\operatorname {sgn}(-x)+1}{2}}\right)$.

Where sgn() is the sign function. This works with any function. The functions are "glued", but their domains aren't overlapping like you seem to want. ☢ Ҡi∊ff⌇↯ 17:08, 30 September 2006 (UTC)

If you want to combine the graphs of y = x^3 and y = 3x, use (y − x^3)(y − 3x) = 0.  --Lambiam^Talk 23:28, 30 September 2006 (UTC)

October 1

Missing posts

Has anyone else noticed posts going missing on these pages? Like you click on an item in your watch list by UserXXXX, and it aint there? Im posting this msg on all ref desks.--Light current 11:27, 1 October 2006 (UTC)

Now restored. See Wikipedia_talk:Reference_desk#Archive_dump. --hydnjo talk 14:11, 1 October 2006 (UTC)

Simplifying rational expressions

If i have the expression: x(x-2)/3x(x-1) Can I cancel the 2 x's out and turn the expression into: x-2/2x(x-1)? Why or why not? Jamesino 17:54, 1 October 2006 (UTC)

I assume that multiplication in your formula takes precedence over division, so 3x(x-1) is the denominator of this fraction. Presumably, you mean "(x-2)" in the last formula; what you wrote would normally be interpreted as x-(2/2x(x-1)). What you appear to be doing is divide by x in the numerator (replacing x by x/x = 1, which can be omitted), while you subtract x from a fragment of the denominator (replacing 3x by 3x−x = 2x). There's no way you can do that. One of the simplest sanity checks is to substitute some concrete value and see if things pan out. If we substitute 10 for x in the original fraction, we get (10×8)/(3×10×9) = 80/270 = 8/27. For the second fraction we get 8/(2×10×9) = 8/180. These are clearly different numbers.

What you can do is cancel equal factors against each other by dividing by them. The basis of this rule is that (p/q)×(r/s) = (p×r)/(q×s). (See the rules for multiplying fractions.) Switching the two sides, and replacing q by p, we get: (p×r)/(p×s) = (p/p)×(r/s). Now if p is not equal to 0, p/p = 1, and we get (p×r)/(p×s) = r/s. But if p = 0, the expression (p×r)/(p×s) stands for 0/0, a so-called indeterminate form that has no meaning. You are not allowed to state that that is equal to something that has a meaning, like r/s.

Applied to your original problem: You can simplify x(x-2)/3x(x-1) to (x-2)/3(x-1) under the condition that x ≠ 0.  --Lambiam^Talk 18:38, 1 October 2006 (UTC)

Ah ok thanks. Using your explanation i figured that X also cannot equal 1, -2, or 2. Jamesino 18:48, 1 October 2006 (UTC)

Both x = 2 and x = −2 are fine. For x = 2, both x(x-2)/3x(x-1) and (x-2)/3(x-1) can be simplified to 0. For x = −2, nothing special is going on, and both fractions can be simplified to 4/9. The case x = 1 is indeed problematic: it implies division by zero. In this case both fractions are undefined (meaningless) in the domain of real numbers. Before, for the case x = 0, one was undefined and the other well defined.  --Lambiam^Talk 21:02, 1 October 2006 (UTC)

To avoid potential confusion, write the expression as

${\frac {x(x-2)}{3x(x-1)}}.$

The value of a fraction is unchanged if both numerator and denominator are multiplied by a nonzero quantity. Likewise, a nonzero quantity can be cancelled top and bottom. These are our only possibilities. For example:

${\frac {3}{4}}={\frac {3\times 5}{4\times 5}}={\frac {15}{20}}.$

${\frac {6}{8}}={\frac {3\times 2}{4\times 2}}={\frac {3}{4}}.$

When dealing with expressions involving unknowns, we usually do not know in advance what is nonzero, so any manipulations depending on that assumption must be explicitly so qualified.

In the following example, we see a factor of z in both numerator and denominator. If we stipulate that z is not zero, we are allowed to cancel it:

${\frac {3z}{4z}}={\frac {3}{4}}.$

Polynomials are much like integers; we can write both in terms of "primitive factors". For both, the factors are essentially unique. We say "essentially" because we are not concerned about the order of the factors, nor pairs of sign changes. For example,

$15=3\times 5=(-5)\times (-3).\,\!$

$z^{2}-1=(z+1)(z-1)=(1-z)(-1-z).\,\!$

With a little practice we can cope with fractions of polynomials nearly as easily as common fractions. --KSmrq^T 21:11, 1 October 2006 (UTC)

Is my derivative correct?

I have been using the Calculus Wikibook to teach myself derivative calculus. I get the derivative of $f(x)=2x+3$ to be equal to $2x^{2}+4+3x$. Is this correct? If not, what is the correct answer? Thanks. --80.229.152.246 21:00, 1 October 2006 (UTC)

Nevermind, I have found where I went horribly wrong... --80.229.152.246 21:05, 1 October 2006 (UTC)

BTW, the term commonly used by mathematicians is differential calculus.  --Lambiam^Talk 21:14, 1 October 2006 (UTC)

:::Thanks, I forgot about that. That's what you get for typing a question in a hurry. By the way, can anyone help me with finding the correct derivative of $f(x)=2x+3$? Can you tell me what is it and how you get it? I can't even get the first demonstration to equal to what I work out now! That's what you get for reading stuff like that late at night wen you should really be in bed. I've just realised now I've spent more words talking off the track than on it. I think I better stop now. Thanks in advance. --80.229.152.246 21:17, 1 October 2006 (UTC)

Nevermind, after looking at some articles on Wikipedia, I think I have got the hang of it. Would I be right in saying that the answer is 2? Thanks for your help. --80.229.152.246 21:30, 1 October 2006 (UTC)

Yes, that's more like it.  --Lambiam^Talk 21:31, 1 October 2006 (UTC)

N-Dimensional Geometry question

If I have some n-dimensional plane (a1*x1 + a2*x2 .... = 0) (a's are constants, x's are variables) that passes through the origin, how can I determine if the plane intersects with a corner of the n-dimensional cube defined by ( 0 <= x1,x2 ... <= 1). Thanks for any help with this problem, the only thing I've found is trying the co-orindate of each corner into the plane equation. I'm looking for something more effecient than that. Thanks again. AmitDeshwar 23:29, 1 October 2006 (UTC)

I think I can rephrase the question in a way that may be useful... you want to know if there is a solution to the equation $a_{1}x_{1}+\cdots +a_{n}x_{n}=0$ for which all x's are 0s and 1s. This is the same as asking whether some subset of the coefficients (other than the empty subset) add up to zero. Does that help at all? -GTBacchus^(talk) 23:37, 1 October 2006 (UTC)

This is 0-1 Integer Programming, which is on the original list of Karp's 21 NP-complete problems. So if you find an efficient method, let us know.  --Lambiam^Talk 00:26, 2 October 2006 (UTC)

I see now that this is also the Subset Sum Problem. I'll let you know if I find an effecient solution ;). AmitDeshwar 03:43, 2 October 2006 (UTC)

October 2

I'm taking image requests

This seemed like the ideal place to ask because I know the right people are going to read this.

I've been very bored lately and I feel like contributing with some nice images to Wikipedia. Animations and 3D images are more fun to do, so those are preferred. Anything is cool as long as they're maths\physics related. Oh, and here are some sample images I created to replace the ugly old ones or to add new content:

• 
  Better-looking sphere instead of Image:Sphere.jpg
• 
  Better-looking torus instead of Image:Torus.jpg
• 
  Villarceau circles
• 
  Additive synthesis of a square wave

Also, check my gallery to see what sort of thing I can do.

So, do you guys have any suggestions for images, graphics or diagrams that need to be improved or replaced? :) ☢ Ҡi∊ff⌇↯ 00:19, 2 October 2006 (UTC)

Just the other day there was a complementary posting at Wikipedia talk:WikiProject Mathematics#Suggestion to improve most math articles. I also note that John Reid, member of WikiProject Mathematics, states on the participants' list: "I do have considerable ability in graphic design and an interest in clear illustration. Yes, I take requests.", while Carl Peterson professes: "Will take requests for geometry-based vector images."

Now wouldn't it be cool if there was something like a subproject Wikipedia:WikiProject Mathematics/Images of a number of people who collaborate on improving the imagery of the mathematical articles in a coordinated fashion? A good idea might be to start with the more basic ones that are likely to be read by people without advanced mathemarical background, such as those listed on Wikipedia:Vital_articles#Mathematics.  --Lambiam^Talk 00:47, 2 October 2006 (UTC)

Follow-up: I see that in fact such a subproject exists: Wikipedia:WikiProject Mathematics/Graphics. It's not dead, but it looks like it could use more life.  --Lambiam^Talk 01:03, 2 October 2006 (UTC)

Thanks, I joined the project. I see they want an animation of a torus morphing into a coffee mug. I'll try to give it a shot. :) ☢ Ҡi∊ff⌇↯ 02:59, 2 October 2006 (UTC)