Wikipedia:Reference desk/Archives/Mathematics/2006 September 28

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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)[reply]

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

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

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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

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)[reply]

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

dont know what it means?