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October 21

Formula for converting solid volume to liquid volume

This may sound like a strange question but this has bugged me for a long time

is there a mathematical formula for predicting the change change in volume of substance after it changes states.

for example if i was to take a 5'x5' block of ice and melt it how would i estimate the volume of liquid water i would have?

thanks in advance —Preceding unsigned comment added by 209.167.165.2 (talk) 03:00, 21 October 2010 (UTC)[reply]

This is really more of a chemistry/physics question, but the ideal gas law is a good approximation to allow you to predict gas volume given the number of particles (equivalently, initial/solid mass or weight), atmospheric pressure, and gas temperature. There are of course more refined approximations if you're really interested, but it doesn't seem like that's what you're after. There are probably similar formulas for other states; I purposefully forgot the chem I learned :). 67.158.43.41 (talk) 05:11, 21 October 2010 (UTC)[reply]

Water is different from most other substances in that it expands when it freezes; a block of ice is approximately 10% larger, by volume, than the water it froze from (see Properties of water#Density of water and ice, in which this is explained in terms of density). So your 5′ × 5′ × 5′ block of ice (I assume you meant to have three dimensions there), which has a volume of 125 cubic feet, would melt into about 114 cubic feet of water. (Keep in mind that the volume of liquid water expands a bit as it heats up, too.) These calculations are specific to water, though—other substances behave differently. Most substances shrink when they freeze. —Bkell (talk) 05:31, 21 October 2010 (UTC)[reply]

Is Coefficient of thermal expansion relevant, or does that only apply when there's no change of state? Rojomoke (talk) 11:18, 21 October 2010 (UTC)[reply]

The thermal expansion coefficient assumes that the state is the same. It would not make sense to have one for state changes, since the phase change takes place without any change in temperature. –Henning Makholm (talk) 14:32, 21 October 2010 (UTC)[reply]

Bachmuth representation

I'm currently reading a paper of Chein ([1]), and he states that if F is the free group on three generators (a,b,c), and $\Phi =F/F''$ is the free metabelian group on three generators, then we can regard a,b and c as generators of $\Phi$ , and we have a mapping sending

$a\mapsto {\begin{pmatrix}s_{1}&t_{1}\\0&1\end{pmatrix}},b\mapsto {\begin{pmatrix}s_{2}&t_{2}\\0&1\end{pmatrix}},c\mapsto {\begin{pmatrix}s_{3}&t_{3}\\0&1\end{pmatrix}},$

which induces a representation of $\Phi$ (where the $s_{i},t_{i}$ are commuting indeterminates in some ring R). Perhaps I'm being dense, but I don't see how this gives a homomorphism. I'm assuming there's something I'm not understanding about multiplication in $\Phi$ . The problem I'm having is with the (1,2) entry of the image of, say, ab, which is $t_{1}t_{2}$ , but in the product of the images of a and b, the (1,2) entry is $s_{1}t_{2}+t_{1}$ .

Chein says that Magnus has proved this, but Magnus' paper is in German, and I've not managed to find an English translation. I've checked other sources, and they all agree that this map is a representation, but I just can't see it. Any help would be great, thanks, Icthyos (talk) 11:54, 21 October 2010 (UTC)[reply]

The multiplication reminds me of a semi-direct product. Not sure if it's just a coincidence, or not. — Fly by Night (talk) 12:12, 21 October 2010 (UTC)[reply]

No coincidence. If R is a ring, then the matrix group

{\begin{pmatrix}R^{\times }&R\\0&1\end{pmatrix}}

is isomorphic to the semidirect product

(R,+)\rtimes (R^{\times },\cdot )

, where

R^{\times }

acts on R by multiplication.—Emil J. 15:04, 21 October 2010 (UTC)[reply]

What makes you believe that the (1,2) entry in the image of ab is

t_{1}t_{2}

?—Emil J. 13:18, 21 October 2010 (UTC)[reply]

(e/c) Anyway: any function from {a,b,c} to any group G extends to a unique homomorphism f from F to G (that's the definition of a free group). In order to get a homomorphism from Φ to G, you need f to factor through the quotient map F → Φ, which happens if and only if Ker(f) includes F'', so one would need to check that. Alternatively, since Φ is the free metabelian group, it would suffice to check that G (or just the range of f as a subgroup of G) is metabelian. I can't tell how difficult is this to establish in your case, in fact, I don't even understand the description of the mapping. (What is an "indeterminate in a ring"? And when you write that "

s_{i},t_{i}

are commuting", do you only mean that s_i commutes with t_i for every i=1,2,3, or that any pair of elements of {s₁,t₁,s₂,t₂,s₃,t₃} commutes?)—Emil J. 13:37, 21 October 2010 (UTC)[reply]

Oh, I see what you mean about the (1,2) entry, I don't know what I was thinking. Your statement about the unique extension of the homomorphism was one of the first things I learned about free groups, it should've been obvious. I agree that the map is not defined in a clear way. Another source I have (Lyndon & Schupp's "Combinatorial Group Theory", Chapter 1.4) defines the

s_{i}

and

t_{i}

as being 'independent invertible elements' in a commutative ring, R, which isn't really much better. Thanks for pointing out my mistake; I shall attempt to soldier on! Icthyos (talk) 14:12, 21 October 2010 (UTC)[reply]

I had a look on Chein's paper, and the description there is easier to understand: the elements are apparently taken in the rational function field

\mathbb {Q} (s_{1},s_{2},s_{3},t_{1},t_{2},t_{3})

(or rather, its subring

\mathbb {Z} [s_{1},s_{2},s_{3},t_{1},t_{2},t_{3},s_{1}^{-1},s_{2}^{-1},s_{3}^{-1}]

). The only important point is that the ring is commutative (and the s_i are invertible). The mapping goes into the group G of matrices of the form

\left({\begin{smallmatrix}x&y\\0&1\end{smallmatrix}}\right)

, where x,y are elements of the ring, and x is invertible. Then the existence of the homomorphism is clear, since G is metabelian: when the matrices are multiplied, the (1,1) entry of the result is just the product of the individual (1,1) entries, which are taken in a commutative ring, hence G' only contains matrices of the form

\left({\begin{smallmatrix}1&y\\0&1\end{smallmatrix}}\right)

. The group of such matrices is commutative (they are multiplied by adding the y's), hence G'' is trivial.—Emil J. 14:32, 21 October 2010 (UTC)[reply]

Could you please provide the reference to Magnus'es paper? JoergenB (talk) 13:25, 21 October 2010 (UTC)[reply]

The reference to Magnus is [2], and Chein says Magnus proves that the above is a representation in Theorem 6.

He actually says that he proved it to be a faithful representation. I gather that the faithfulness is the hard part.—Emil J. 14:38, 21 October 2010 (UTC)[reply]

Actually, I found the Mathematische Annalen 111... and there Magnus is discussing a (superficially) quite different representation (working with finite upper left hand corners of certain upper triangular infinite matrices, seemingly). This could be slightly confusing.

I do not understand your discussions about in which rings to find the representations. In the definition on page 606, Chein just talks about commuting indeterminates (i. e., elements like the variables x and y in a polynomial expression such as

5x^{2}+3xy^{2}-y^{5}

), without complicating matters by embedding them in a ring of high enough transcendence degree; right? JoergenB (talk) 15:07, 21 October 2010 (UTC)[reply]

When one just refers to something as commuting indeterminates in a context asking for a ring, it is understood that one takes the ring of polynomials in said indeterminates (or simply the free commutative ring in these indeterminates, i.e., polynomials over Z). I'm not embedding anything anywhere, I'm using directly that polynomial ring (more or less: in this particular case, one obviously needs the s_i to have inverses, so plain

\mathbb {Z} [s_{1},s_{2},s_{3},t_{1},t_{2},t_{3}]

is not enough). He actually explicitly refers to "

\mathbf {Z} (s_{1}^{\pm 1},s_{2}^{\pm 1},s_{3}^{\pm 1})

" on the next page.—Emil J. 15:26, 21 October 2010 (UTC)[reply]

Good; I agree with that. Ichtyos, did you essentially mean the same thing with

"which induces a representation of $\Phi$ (where the $s_{i},t_{i}$ are commuting indeterminates in some ring R)"? JoergenB (talk) 19:36, 21 October 2010 (UTC)[reply]

Yes; I was essentially just quoting Chein, but that's what I took him to mean.Icthyos (talk) 11:19, 22 October 2010 (UTC)[reply]

Thanks for the discussion, both of you! Icthyos (talk) 11:19, 22 October 2010 (UTC)[reply]

What is Mathematica doing?

So, I am putting an improper integral on a quiz, $\int _{-1}^{8}x^{-1/3}\,dx$ , and I plugged it in to Mathematica to make sure I didn't make any mistakes. The answer Mathematica gave me was $6-{\frac {3}{2}}(-1)^{2/3}$ , which is what I got. But, I then simplified $(-1)^{2/3}=1$ , whereas finding the numerical form of this in Mathematica apparently picks a complex cube root of unity here. So, I tried plotting the function from -8 to -2 and Mathematica shows nothing at all, Plot[x^(-1/3), {x, -8, -2}]. Wolfram Alpha shows two graphs, one labeled real part and one labeled imaginary part. What's the deal? StatisticsMan (talk) 16:52, 21 October 2010 (UTC)[reply]

Just a guess, but it's probably doing the principal value. Basically, you can define a cube root function that behaves well except for when z is on the negative real axis. The value of the function on the axis depends on how you define it, but it would have to be one of the complex values. You might try breaking up the interval and using something like

-(-x)^{-1/3}

where x is negative.--RDBury (talk) 17:41, 21 October 2010 (UTC)[reply]

(ec) Well, if you want

x\mapsto x^{-1/3}

to make sense for nonzero complex numbers (you may not care, but how would Methematica know that?), and get the real branches for both positive and negative

x

, you need cuts in both the upper and the lower half-plane. It seems reasonable for Mathematica to always always use the same single cut (say, infinitesimally below the negative real axis) for all non-integral powers. That's the best you can do for almost all exponents anyway. If you want

(-1)^{y}=-1

whenever

-1

is a possible value, then

y\mapsto (-1)^{y}

would need to be discontinuous on the entire real line. –Henning Makholm (talk) 18:08, 21 October 2010 (UTC)[reply]

The problem is that x^–1/3 is multivalued. Mathematica is using one of the complex roots as a solution. You should be able to tell it to work over the reals, or to assume that x is a real variable. — Fly by Night (talk) 17:49, 21 October 2010 (UTC)[reply]

The IEEE floating point standard treats the problem by having a special function rootn(x,n) where n has to be an integer. If you put in pow(-1,1/3) it would give an error. Dmcq (talk) 22:32, 21 October 2010 (UTC)[reply]

I put that equation into my TI-89 Titanium calculator and, after freezing up my calculator for about a minute or two, and saying "questionable accuracy," (to 4 floating points) it gave me 4.5. I am not a "math expert," so I couldn't begin to tell you if that's correct or not.

---WikiHelper46

what do we know about large primes?

we know they are odd. What other properties do they have? (other than the obvious one!) —Preceding unsigned comment added by 85.181.49.255 (talk) 17:31, 21 October 2010 (UTC)[reply]

All of the cool properties of division/multiplication. If you reverse the digits, it will not be divisible by 9. If you add the digits up (recursively until there is only 1 number left), it will not be 9. See divisibility rule for more. -- kainaw ™ 17:55, 21 October 2010 (UTC)[reply]

(What kind of "property" are you looking for? Something like Fermat's little theorem? Or deterministic primality tests? Number theory is full of things that have been proved to be true specifically for primes. They are too numerous to list. –Henning Makholm (talk) 17:58, 21 October 2010 (UTC)[reply]

For a large prime, it is true that no positive integer other than 1 and itself divides it. That's a cool property shared by all large primes. Also, no large, odd prime is divisible by 3, 5, 7, 11, or 13 (saying they are odd is just saying they are not divisible by 2). StatisticsMan (talk) 18:08, 21 October 2010 (UTC)[reply]

For that matter, there are no large even primes that are divisible by 3, 5, 7, 11, or 13 either (I have a wonderful proof of this, which unfortunately is WP:OR). –Henning Makholm (talk) 18:12, 21 October 2010 (UTC)[reply]

No large prime is divisible by any positive integer other than 1 and itself; not just 3, 5, 7, 11, or 13. If it were then it wouldn't be prime! — Fly by Night (talk) 13:42, 22 October 2010 (UTC)[reply]

Well, the fact that only 1 and itself divides is what I meant above with "other than the obvious one". What I'm really interested is what Makholm has said "Number theory is full of things that have been proved to be true specifically for primes. They are too numerous to list." Surely he could have listed like 10 of them!! I am really interested in things that are going to be true for large primes, but not for every large number, other than the fact that they have two factors... 85.181.49.255 (talk) 18:49, 21 October 2010 (UTC)[reply]

So things like Fermat's little theorem, Wilson's theorem, Goldbach's conjecture, the prime number theorem, or the fact that the order of every finite field is a power of a prime? You can find more in Category:Prime numbers. —Bkell (talk) 19:08, 21 October 2010 (UTC)[reply]

OK, let us list 10 random prime properites:

Fermat's little theorem.
Wilson's theorem.
Euclid's lemma
Primes are the only nonzero numbers that can be the characteristic of a field.
Primes are ~~the only numbers~~ such that two finite groups with this many elements are necessarily isomorphic.
The Sylow theorems
Eisenstein's criterion
Only if p is a prime will the p-adic numbers be a field (rather than a ring).
Quadratic reciprocity
Touchard's congruence

–Henning Makholm (talk) 02:35, 22 October 2010 (UTC)[reply]

The Finite group article gives infinitely many counterexamples to #5: Every group of order pq is cyclic when q < p are primes with p-1 not divisible by q. Matthew Auger (talk) 11:07, 22 October 2010 (UTC)[reply]

Ooops! –Henning Makholm (talk) 14:58, 22 October 2010 (UTC)[reply]

They are hard to factor, which is the basis of RSA encryption. They get rarer and rarer as a consequence of the prime number theorem. 67.158.43.41 (talk) 22:23, 21 October 2010 (UTC)[reply]

Actually primes are very easy to factor. If you give me a large prime I'll factor it for you immediately, even without a computer. :-) —Bkell (talk) 22:39, 21 October 2010 (UTC)[reply]

I bow down to you. I'm not worthy. :) --Kinu ^t/_c 22:42, 21 October 2010 (UTC)[reply]

Haha, whoops. I of course meant numbers "near" them are hard to factor, ex. pq for large primes p and q., Interestingly, though, primality testing is "easy" inasmuch as it can be done in polynomial time in the number of digits in the input using the AKS primality test. That is, you can test whether or not a given large number is prime "quickly". 67.158.43.41 (talk) 00:19, 22 October 2010 (UTC)[reply]

Bertrand's postulate is an interesting property, there's another prime below 2p. 22:38, 21 October 2010 (UTC)

The primes contain arbitrarily long arithmetic progressions, which usually occur for very large primes; see the Green–Tao theorem. —Anonymous Dissident^Talk 22:48, 21 October 2010 (UTC)[reply]

According to http://primes.utm.edu/largest.html, some large primes are Mersenne primes.

—Wavelength (talk) 01:04, 22 October 2010 (UTC)[reply]

Depends on your definition of 'large'. There may be no 'large' Mersenne primes if they are finite in number. Dmcq (talk) 10:01, 22 October 2010 (UTC)[reply]

All primes are small, because most primes are bigger. Bo Jacoby (talk) 12:25, 22 October 2010 (UTC).[reply]

If you take all the prime numbers, square them, take the reciprocal of the numbers, take one minus these numbers, and then multiply all the numbers, you get exactly 6/pi^2:

$\prod _{p}\left(1-{\frac {1}{p^{2}}}\right)={\frac {6}{\pi ^{2}}}$

Count Iblis (talk) 15:01, 22 October 2010 (UTC)[reply]

The OP's question may have been better than I though at first. I understand it to ask for properties of each prime individually, that is, something you can do with a singe large number that happens to be prime. Most of what people suggest here are interesting properties of the set of all prime numbers in general. I really had to reach to fill out the 10 properties I quote above without including properties that are trivial rephrasings of each other (and one of them even turned out erroneous, stupid me). It seems to be hard to get very far when you cannot even use the fundamental theorem of arithmetic (which is a collective property of all the primes, rather than a property of each prime individually). –Henning Makholm (talk) 15:18, 22 October 2010 (UTC)[reply]

If the OP's question was really, what is the practical use of knowing that some medium-large number (with, say, hundreds of digits) is prime, then the best answer may be to point to the use of prime numbers in various public-key cryptography schemes. The most significant mathematical underpinnings of these seem to be variations on Fermat's little theorem, the theory of finite fields, and in particular the assumed difficulty of computing discrete logarithms. –Henning Makholm (talk) 15:33, 22 October 2010 (UTC)[reply]