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# July 23

## Is every infinite field with cardinality aleph-0 isomorphic to the rationals?

Just wondering. --2404:2000:2000:5:0:0:0:C2 (talk) 00:31, 23 July 2014 (UTC)

No. Consider the field Q(sqrt(2)) (I'm no longer sure of the notation; what I mean is the smallest field containing the rationals and sqrt(2)). Any isomorphism between them would have to fix the rationals, so there's nowhere for sqrt(2) to go on the Q side. --Trovatore (talk) 00:47, 23 July 2014 (UTC)
Your notation is pretty standard, sometimes rendered with the "blackboard bold" or square brackets, e.g. as $\mathbb{Q}[\sqrt{2}]$. Our relevant article is Algebraic_number_field. SemanticMantis (talk) 15:52, 23 July 2014 (UTC)
I think I learned square brackets for the extension as a ring, parentheses (round brackets) for the field. --Trovatore (talk) 17:59, 23 July 2014 (UTC)
That is also what I learned. However, $\mathbb{Q}[\sqrt{2}]=\mathbb{Q}(\sqrt{2})$, so it is not very important in this case. —Kusma (t·c) 18:02, 23 July 2014 (UTC)
Even worse: the algebraic closure of a finite field is countable but doesn't even contain an isomorphic image of the rationals. —Kusma (t·c) 07:29, 23 July 2014 (UTC)
But any field of characteristic 0 contains the rational as a subfield --77.126.41.10 (talk) 14:53, 24 July 2014 (UTC)

## equation

I'm trying to figure out a way to equitably pay off debt with my spouse. She makes 71% of what I make.

Let's say debt d=1750 and d=m+h (mine and her contribution)
Does h=0.71m??
And then does d=0.71m+m??
Does m=1232.39?

I'm doubting because then h=517.61 and I think h/m=0.71, but it doesn't...

I want to ultimately make an excel spreadsheet. Thanks

You're right up to d=0.71m+m, but you must have made a mistake after that, because it gives m = d/1.71 = 1023.39, and so h = 726.61. AndrewWTaylor (talk) 16:34, 23 July 2014 (UTC)

Wow, thanks for the quick answer. That makes sense. Maybe you could help me spot my mistake, too?

1. d=0.71m+m
2. Divide d by 0.71 and cancel from other side
3. so 2464.79=2m
4. m=1232.39

Step 2, somewhere I think. Thank you again! — Preceding unsigned comment added by 166.147.123.165 (talk) 16:53, 23 July 2014 (UTC)

If you want a simple formula for a spreadsheet for proportionally dividing a quantity, think in terms of each income as a fraction of the total income. So, say you each earn a = $1 and b =$0.71 respectively. Then an apportionment multipliers would be a′ = a/(a+b) = 0.585 and b′ = b/(a+b) = 0.415, so m = ad and h = bd, and these factors a′ and b′ can be reused for splitting other amounts. —Quondum 17:54, 23 July 2014 (UTC)
And as to the specific algebraic error, yes, step 2. You can divide both sides of an equation by 0.71, but the right hand side would become (0.71m + m)/0.71 = m + m/0.71, not m + m. What you should have done is to see that 0.71m + m = 1.71m, and then divide both sides by 1.71. -- ToE 21:21, 23 July 2014 (UTC)

This is how you should solve the problem. Very simple ratio.

You should think like this

1. For every $100 I earn, she earns$71
2. Thus we earn a total of $171 dollars ($100 + $71) for every$100 that I earn
3. Thus my ratio of the debt is 100/(100+71)
4. Thus her ratio of the debt is 71/(100+71)
• d = 1750
• m = 100/(100+71) * d
• h = 71/(100+71) * d

Very easy. You don't need an excel spreadsheet. You just need to think clearly. 202.177.218.59 (talk) 04:12, 24 July 2014 (UTC)

I'm not at all certain why the paying off of the debt is to be divided up according to how much each earns. There are lots of ideas on how it should be done, e.g. see fair division and airport problem and see how complex and confused these sorts of things can be made at Entitlement (fair division)#Entitlement in the Talmud. Basically you sre agreeing with proportional tax rather than a progressive tax for income. Dmcq (talk) 11:24, 25 July 2014 (UTC)

# July 25

## What's the use of complex numbers?

After reading Complex number I still don't get what makes the invention of imaginary numbers so special. Is it like Syntactic sugar for mathematicians so they can use less text to get to a proof? Are there proofs that wouldn't be possible without inventing i at the spot? I remember reading a book on fractals with surprisingly little mathematics in it, yet it had some formulas using the magical i as well. The BASIC code that was also in the book to actually draw the fractals didn't need such magic and was completely understandable (besides the astonishing pictures generated by such simple code, of course). Are there things that wouldn't have been discovered/invented/proven by now if no one ever had been thinking out of the box to by writing down the square root of -1? Joepnl (talk) 00:23, 25 July 2014 (UTC)

No, you can represent the complex numbers using pairs of real numbers. Dealing with such pairs doesn't, however, have a natural touch an feel like the set of complex numbers have once the -1 = i is accepted. At any rate, many discoveries and applications would have been delayed many years without the complex numbers. YohanN7 (talk) 00:40, 25 July 2014 (UTC)
Edit: As a matter of fact, you can dispose of -1 = i as well. Just regard complex numbers as a particularly efficient notation for a certain field consisting of pairs of real numbers with extraordinary properties. It is the field with these properties we can't do without, whether it's represented by complex numbers or pairs of real numbers. YohanN7 (talk) 00:56, 25 July 2014 (UTC)
You need complex numbers to have a "Closed Field" or algebraically closed field. What this means is that any mathematical operations on a "complex number" will always result in another "complex number". As a comparison, an mathematical operation on a Real number can result in a number that is NOT REAL. For example the square root of negative one. 202.177.218.59 (talk) 00:50, 25 July 2014 (UTC)
It's useful in appreciating the bigger picture, which is always a good thing to focus on. Because we're typically taught complex numbers later in our mathematical education, it might seem that they're curious exceptions to the real numbers. The truth is sort of the reverse. ALL numbers can be expressed as a complex number, but not all complex numbers can be expressed as a real number. That means that the real numbers, the ones we know and love and are familiar with, are the real curiosities, being merely an infinitesimally small subset of the complex numbers. Henceforth, when you order 3 hotdogs, you'll be asking for "3 + 0i" hotdogs. Right? -- Jack of Oz [pleasantries] 01:38, 25 July 2014 (UTC)
See Complex numbers#Applications.—Wavelength (talk) 01:47, 25 July 2014 (UTC)
Hello, I'm not an expert in mathematics, but only an enthusiast. I think that I have enough knowledge to provide an answer. There are 2 separate issues in this question, the first one is that complex numbers aren't an elementary concept as taken by most mathematic treatises, but rather they're usually defined as pairs of real numbers, one of which is the real part and the other is the imaginary part, then arithmetic operations and other properties are defined as well. You can replace all references for complex numbers in theorems and their proofs for the corresponding definition and they will still be valid, in this case complex numbers only make the role of syntactic sugar; likewise it would be possible to deal away with intermediate theorems on proofs by replacing them with their proofs and so on until only axioms and inference steps remain, but it would make intractably huge proofs, and consisting mostly of redundant information.
However, as far as human reasoning is concerned, the concept of complex number embodied in its definition (Or axioms, if you're treating them as a an elementary concept) is absolutely necessary because when we think, we do so on the properties of complex numbers as a structure standing on its own and the definition is abstracted away. There are results for which complex numbers are necessary which aren't about complex number themselves. For instance, the prime number theorem; using an area of mathematics to prove results in another is commonplace. Note that the PNT uses not only complex numbers, but complex analysis. (I'm not knowledgeable enough to understand those proofs themselves or results of complex analysis, however).
Thinking (Apart from writing proofs) in terms of complex numbers makes some things easier, even when they're not indispensable. For instance, a Fourier transform convert a signal from time domain (Values represent immediately intensity as a function of time) to frequency domain (Values represent the intensity of pure sinusoidal frequencies, which sum to the original signal and hence are another representation of it). Each frequency component has 2 components which are in quadrature (90° out of phase), even if the input is real; at this point it's trivial to see each of them as a pair of real numbers, rather than a single complex number, and there's little if any practical difference. However, the FT has very nice properties that are only intuitive when expressed with complex numbers. For instance, the convolution theorem says that convolution in time domain equals multiplication in frequency domain. It makes sense to see one operand as the signal and the other as the filter in a convolution (Specifically, its impulse response). Intuitive, when you pass a signal through a filter, it may attenuate some frequencies in different degrees (Multiply them), but it also may rotate the phase of the frequencies, and complex numbers define this multiplication for both components of each frequency (Either expressed as sine waves in quadrature or the magnitude and phase of a single one, AKA rectangular and polar form) while real numbers by themselves only explain intuitively the magnitudes.
It's exactly the same with real numbers, which can be defined as Dedekind cuts or Cauchy sequences, or rational numbers, which can be defined as pairs of integer numbers, which in turn are usually defined in set theory treatises in terms of sets. Of course, the reason you're asking about complex numbers and not any of the just listed number sets is because you're likely not used to them since your childhood, as you're with the R, N and Q sets (I wasn't, either, but maybe parents and elementary schools should begin teaching them). There's no fundamental distinction, your question can also be justifiably made for them and also for any other mathematical definition.
Also, note that complex numbers are defined and the question of the square root of negative numbers has no sense whatsoever in real numbers. To manipulate expressions containing $\sqrt{-1}$ pretending that it's a real numbers has no more validity than the pseudo-proofs of any other Mathematical fallacy and may give contradictory results as well.
I hope that it helps, regards.
QrTTf7fH (talk) 02:47, 25 July 2014 (UTC)
By the way, QrTTf7fH, parentheses within a sentence do not call for a capital. —Tamfang (talk) 01:06, 26 July 2014 (UTC)

A very important use of complex numbers is for solving the differential equation of a linear harmonic oscillator: $d^2y/dx^2+y=0$. An exponential function $y=e^{ax}$ is a solution if $a^2+1=0$. Without complex numbers you are stuck, but using complex numbers you find $a^2+1=(a-i)(a+i)$, and so the solutions are $y=Ae^{ix}+Be^{-ix}$. The differential equation can be written $(d/dx-i)(d/dx+i)y=0$, and so the second order differential equation breaks up into two first order differential equations, $(d/dx+i)y=0$ and $(d/dx-i)y=0$. This trick is used in quantum mechanics: the Schrödinger equation is of order one in time. Bo Jacoby (talk) 23:43, 26 July 2014 (UTC).

Well, you aren't stuck without the complex numbers, since you can just guess $A\sin x + B\cos x$ instead of $e^{ax}$, but yes, this is an important application of complex numbers.
An interesting point here, though, is that the complex-number approach fails (as far as I know) to generalize to the harmonic field equation in d+1 spacetime dimensions. What does work for all values of d is the degree-0-and-1 part of the Clifford algebra Cℓd,1(R), which is isomorphic to the complex numbers when d=0 (the harmonic oscillator case) but not when d>0. The usefulness of complex numbers in the harmonic oscillator seems to be an "accident" inasmuch as none of their interesting properties (like algebraic completeness, or even being a field) matter, just their isomorphism to Cℓ0,1(R). This makes me wonder if the complex numbers in quantum mechanics would likewise disappear in an approach that didn't break the spacetime symmetry by treating the time coordinate specially. But I've never found a paper supporting that idea. -- BenRG (talk) 03:53, 27 July 2014 (UTC)

## elementary mathematics

what are the main basics in mathematics.what are the elementary based questions that appear in the competitive exams117.204.70.51 (talk) — Preceding undated comment added 05:47, 25 July 2014 (UTC)

The meaning of "elementary" varies widely depending on context. I assume that you want to know about some examinations set in India, but you will have to tell us the level of the examination before anyone can help. Dbfirs 20:09, 25 July 2014 (UTC)

## Percentage of Grids with connected pathways?

Let An be the Universe of grids of 2n by 2n black and white squares where half of the squares are white and half are black. *and* both the upper left and lower right square are black. (so A1 only has one grid, A2 has 14C6 grids (the other 6 black squares among the other 14 spots. Let a grid be successful if there is a path of black squares joined on edges from the black square on the upper left to the black square on the lower right. As n goes to infinity, does the percentage of successful grids in An go to 0%, go to 100% or something else?Naraht (talk) 14:39, 25 July 2014 (UTC)

Purely intuitively, I'd guess that the proportion is either 1/e or 1 - 1/e.86.146.61.61 (talk) 22:42, 25 July 2014 (UTC)
I ran a numerical simulation where I generated random grids meeting your criteria and tested them for successful paths. Here are the numbers for ten million iterations for each value of n from 1 through 20.
For n=2 it is easy to hand-enumerate the 150 distinct successful grids (there are twenty distinct paths (each of total length seven), with nine remaining spots for each path to place the eighth black square, though some remaining square choices need to be excluded to avoid duplicating earlier paths), and the actual ratio of 150 / 3003 = 4.995%, matches my simulation closely. This gives me hope that my code may be correct.
If so, this suggests that the percentage goes to 0% as n goes to infinity, and for these first few numbers, at least, it appears to do so roughly exponentially, with the ratio of successive values not so far from the inverse of the golden ratio, though I don't know if there is anything to that. -- ToE 03:37, 26 July 2014 (UTC)
Okay, I worked the following out on the back of a piece of paper during a business meeting, stuffed it in my pocket, and forgot about it till now - it is a slightly dirty use of estimation, but should bear out. (in what follows, Bin(a,b) is the binomial coeff. a over b) Define a minimal path to be a path of black squares (from the fixed corners) so that removing any square breaks connectivity, let k(n) be the number of minimal paths in an n x n grid. Then, k(2n) = Bin(2(2n - 1), 2n - 1). Given a fixed minimal path, the number of ways to colour the rest of the grid to the desired parameters is Bin(4n2 - 4n + 1, 2n2). Since every successful grid contains a minimal path, there are at most Bin(2(2n - 1), 2n - 1)Bin(4n2 - 4n + 1, 2n2) such grids (in fact, should be quite a bit less). The total number of grids, in general, is Bin(4n2 - 2, 2n2), taking the quotient gives, [(4n-2)!(4n2 - 4n + 1)!(2n2 - 2)!] / [(2n - 1)!2(2n2 - 4n + 1)!(4n2 - 2)!]. Using that ln(n!) is approx. n*ln(n) - n + 1; applying ln to the previous quotient and taking n -> infinity, if we may assume (for purposes of the limit) that ln(an + b) is small enough to disregard and that ln(an2 + bn + c) is ln(an2), then we arrive at (4n - 3) ln(1 / 2), which goes to -infinity, and, hence, the quotient limits to 0. Since the quotient approx. bounds the percentage above, this should also go to 0. --since we will have several minimal paths in any successful grid, we are overestimating quite a bit, so even if there is a little fudging involved, I'd feel confident in asserting that it does, indeed, go to 0; obviously, though, this is nothing like a proof, and "back of the envelope" no less, so take it with a grain of salt. -- the approx. bounds on the % is smaller than the above numerical, however, a few random square root factors are being left out and ln of linear growth discarded, thus, for small n, it isn't surprising that it is off.Phoenixia1177 (talk) 07:35, 28 July 2014 (UTC)
"Then, k(2n) = Bin(2(2n - 1), 2n - 1)" -- that's definitely correct, is it? I'm surprised that this formula is so simple. 86.179.112.162 (talk) 11:53, 28 July 2014 (UTC)
k(2n) = Bin(2(2n - 1), 2n - 1) is only counting monotonic paths. For n=2, k(2*2) = 20 gives the correct total number of minimal paths because there is not enough room in a 4x4 grid to turn back and forth, but consider this path on a 6x6 grid:
As n grows, these non-monotonic paths should become more common, and I suspect that they may dominate for large n. -- ToE 13:43, 28 July 2014 (UTC)
You're absolutely right, I can't believe I missed that...I feel kind of like a jackass. Thank you for catching my mistake, sorry for the wrong answer. I still feel that there will be a fairly tractable formula that deals with these, I don't think they are too unwieldy - yes, now that you point them out, I agree that they should end up dominating as n goes up. I'll think more about this tonight, unless someone else posts an answer first. Thank you again:-)Phoenixia1177 (talk) 15:45, 28 July 2014 (UTC)
Don't beat yourself up over it; questions would be archived before they were ever answered if we all waited until we were absolutely certain of our responses. I'd be very interested to see a formula which just even set some bounds on the number of paths. Numerically, the n=3 6x6 grid has only 34C16 ≈ 2.2 billion valid combinations, and thus only about 54 million paths (2.472%, based on my Monte Carlo run), so I should be able to write some code to precisely determine the actual number of paths and also the number of minimal paths vs path length for lengths from 11 through 18. I'll try to do the coding tonight. That's as far as I'll be able to go numerically, as the n=4 8x8 grid has 62C30 ≈ 4.5x1017 valid combinations with about 5.6x1015 paths (1.25%). -- ToE 14:01, 29 July 2014 (UTC)
I believe that a path should only be able to have an *odd* number of squares in its path, so the valid paths for the 6x6 grid should only be able to be 11, 13, 15 & 17, hope that helps.Naraht (talk) 15:16, 29 July 2014 (UTC)
I do not seem to have any luck finding a sensible bounds - I've come up with some, but they all end up with "the probability <= 1", which is useless, or are intractable to work with. However, I think if one were very clever with some logic, you could get the above grids as a class of finite models, then express connectivity in transitive closure logic or finite variable infinitary logic (, or some such, maybe MSO (?)), and show that a 0-1 law is satisfied by that class for that logic - then you only need show that it does not converge to 1 to get that it converges to 0. It's not hard to come up with a theory of such grids, it does seem to difficult to get the right logic and axioms to do it and easily show the 0-1 law part, at least, off the top of my head. At any rate - it seems fun - I'm looking into some stuff from Compton, and some others, to see if this might be a more viable avenue of attack; combinatorics isn't really my area, that part seems to be hampering me a bit.Phoenixia1177 (talk) 20:36, 29 July 2014 (UTC)

# July 26

## Type conversion

Is there any standard mathematical notation for specifying the type (e.g. scalar, vector, matrix) and dimensions of an otherwise ambiguous expression? For example, can the zero matrix of some unknown dimensions x×y and the scalar zero be represented by separate symbols that are standard and unambiguous, given that 0 can mean either? (Intuitively I'd think $\left( 0 : 0 \in \mathbb{R} \right)$ and $\left( 0 : 0 \in \mathbb{R}^{x \times y} \right)$ would be comprehensible, but possibly more awkward than necessary.) Also, is it possible to distinguish the empty set whose sum is scalar zero from the empty set whose sum is a zero matrix? NeonMerlin 13:00, 26 July 2014 (UTC)

Computer programmers distinguish, but mathematicians do not. The zero scalar is identified with the zero matrix, and no distinction is called for. Bo Jacoby (talk) 23:02, 26 July 2014 (UTC).
That's completely wrong. --Trovatore (talk) 02:09, 27 July 2014 (UTC)
The Mathematics_of_general_relativity#Energy_conservation has
$T^{ab}{}_{;b} \, = 0$
not
$T^{ab}{}_{;b} \, = 0^a$
Bo Jacoby (talk) 07:26, 27 July 2014 (UTC).
But surely by 0, the RHS means the zero vector, rather than the zero scalar. No identification has occurred. Sławomir Biały (talk) 14:51, 27 July 2014 (UTC)
Exactly. The zero vector $0^a$ is written 0. No distinction is called for. Bo Jacoby (talk) 14:58, 27 July 2014 (UTC).
It is clear from the context here that the notation 0 means the zero vector rather than the zero scalar. No one seriously believes there is no difference between these (which is what "is identified with") would mean, and it would be perfectly reasonable to write 0^a if there were any risk of confusion. You would not see the vacuum equation written as $R_a^b=1/2R$. Sławomir Biały (talk) 15:05, 27 July 2014 (UTC)
Also, if you were to believe that scalars and vectors were naturally identified, then you would write the Einstein vacuum equation as $R_a^b=1/2R$. As far as I know, no one writes it this way, even though scalars do naturally embed into the endomorphism algebra. Since no one writes it this way, it would be rather mystifying if they regarded the zero on the right hand side of the equation $G_a^b=0$ as a scalar, rather than the zero endomorphism. And for your example, there isn't a natural embedding of the scalars into the space of contravariant vectors to begin with, so claiming that the zero on the RHS of the equation is the same as the zero scalar is obvious nonsense. Sławomir Biały (talk) 21:47, 27 July 2014 (UTC)
I respectfully disagree. I, for one, seriously believe that "there is no difference between these". $0^a-0 = 0$. Where do I find the vacuum equation? Bo Jacoby (talk) 18:35, 27 July 2014 (UTC).
You may believe that there is no difference between the zero vector and the zero scalar. But you are wrong. It only makes sense to "identify" scalars with vector quantities when there is a natural embedding of the scalars into the vector space (e.g., working over an algebra). The zero vector and zero scalar, in this setting, live in completely different spaces. Do not be confused by the fact that the same symbol is used for these different mathematical objects. And this is not the place to push your crank theory that everything denoted by the symbol 0 is the same thing. That has already been conclusively refuted by others. Sławomir Biały (talk) 19:52, 27 July 2014 (UTC)
Bo, to answer your question: at Einstein field equations#Vacuum field equations. Actually, in the context of tensor component equations, there is room for interpretation: interpreted as a whole bunch of equations on numeric components, the problem does not arise (each component is just a real number); however this interpretation of the notation rapidly loses value, for example when the covariant derivative is used (as in your example), for which the component-wise equations make no real sense. Sławomir's argument is entirely unaffected by this notational thing. If you want to be able to equate a zero scalar with a zero vector, you need to embed them both in the same algebra (which is possible: see Tensor algebra), but in this case you are formally treating them as part of the same algebra, and permitting addition of tensors of differing order. —Quondum 22:28, 27 July 2014 (UTC)
Thanks. The vacuum field equation is written
$R_{\mu \nu} = 0 \,.$
The zero on the right hand side is to be understood as a zero tensor field. The OP asked: "can the zero matrix of some unknown dimensions x×y and the scalar zero be represented by separate symbols", and my answer is that zero is written 0, no matter if it is the zero scalar or zero vector or zero matrix or zero function or whatever. My examples show that this is correct. I did not mean, (and I hope I did not write), that scalars are generally embedded in vectors. Bo Jacoby (talk) 05:52, 28 July 2014 (UTC).
Ok. But the zero scalar, zero vector, and zero matrix are not identified (your phrasing, that you vigorously defended). They are different mathematical objects that are often denoted by the same symbol.
Also, the Einstein field equations can be written in a number of equivalent ways. I brought it up only because you seemed comfortable enough with relativity to use it as an example. Obviously that is not the case. The vacuum equation can be written as $G_{ab}=0$ or, raising an index with the metric, as $G_a^b=0$, or. using the definition of the Einstein tensor, as $R_a^b-1/2R\delta_a^b=0$, or as $R_a^b=1/2R\delta_a^b$. By taking traces, one would normally see that the Ricci scalar is zero at this point, but it is still a valid tensor equation. It would not be written as $R_a^b=1/2R$ however. Sławomir Biały (talk) 13:48, 28 July 2014 (UTC)
Thanks. Summarizing: The scalars are not embedded in vectors and so the zero scalar is not the same object as the zero vector even if the two objects share the same symbol 0. The scalars are embedded in (say) 4×4 matrices and the zero matrix is written 0 and the unit matrix $\delta_a^b$ may be written 1. But an equation like $R_a^b=1/2R\delta_a^b$ is not written $R_a^b=1/2R$ because $\delta_a^a=4$ is more obvious than trace(1)=4. Do you agree? Bo Jacoby (talk) 06:39, 29 July 2014 (UTC).
I've seen 1n and Idn for the n×n identity matrix. I may have seen 0x×y, or I may be imagining it. You don't see zero matrices in any notation very often.
Abstract index notation is widely used in relativistic physics. It distinguishes scalars, vectors and matrices (properly rank-2 tensors) by the number of indices, and often their size is implicitly encoded in the letters used for the indices (μν for spacetime indices, ij for spatial indices, etc.). The identity matrix/tensor is sometimes written δμν or δij and called the Kronecker delta.
I've never seen a notation for a typed empty set, except in programming languages. -- BenRG (talk) 23:29, 26 July 2014 (UTC)
Mathematicians do actually distinguish. There are times when "up to isomorphism" is meant, and no ambiguity results. At other times, two structures can be isomorphic but are intended to be regarded as distinct sets, such as the set of complex numbers being isomorphic to distinct subrings of the quaternions (there is no canonical embedding of C in H, unlike with Z in R). However, it seems pretty normal to "expect" the reader to understand what is "meant", with occasional verbal disambiguation. This is somewhat frustrating to the more literal-minded and to newcomers. The closest seems to be set membership, usually appended as a qualifying statement (e.g. " where xQ). —Quondum 00:46, 27 July 2014 (UTC)

# July 27

## Where internationally is USD valued highest?

In which major metro cities in the world is USD valued highest? AKA where will your dollar work harder for you outside the US? — Preceding unsigned comment added by 2602:306:8051:4D60:918B:EBE3:685C:C84C (talk) 17:15, 27 July 2014 (UTC)

This is going to vary over time and with the basket of goods you want to purchase. A general measure to help evaluate this is the purchasing power parity; a light-hearted version of this is the Big Mac index. --Mark viking (talk) 17:39, 27 July 2014 (UTC)
This seems like the wrong reference desk section for this question, but I agree with the former Wikipedian. QrTTf7fH (talk) 17:42, 27 July 2014 (UTC).

I was wondering where the dollar goes further and the refreshingly simple Big Mac Index answers my question completely. Thanks. — Preceding unsigned comment added by 2602:306:8051:4D60:918B:EBE3:685C:C84C (talk) 18:53, 27 July 2014 (UTC)

The answer might actually be in a place where the US dollar is illegal, but continues to be used for black market transactions. StuRat (talk) 19:48, 27 July 2014 (UTC)
I'm curious what exchange rate they use. In Argentina, the government sets an official exchange rate for dollars to argentine pesos, but the black market will give you substantially better. I forget exactly how much better, but it might be enough to push Argentina into the lead.--80.109.80.78 (talk) 22:26, 27 July 2014 (UTC)
I doubt if there's a single black market exchange rate, unless there's one powerful black market leader who can set a universal rate. More likely, the rate is set at each individual transaction, via haggling. StuRat (talk) 12:55, 28 July 2014 (UTC)
No, they don't haggle. In fact, they openly post their rates. There is some small variation between merchants, but they're all pretty much the same.--80.109.80.78 (talk) 19:20, 28 July 2014 (UTC)
I would think those engaged in black market sales would use the same "one hand giveth and the other taketh away" strategy as car dealers, where they offer you a great deal on the purchase price of the new car, but you get soaked on the trade-in, finance rate, extended warranty, etc. In this case, they could offer you a great price for the black market item, then you decide to buy, only to find out the exchange rate is bad. Or, conversely, they might offer a great exchange rate to sucker you in, only to find out the purchase price is high. StuRat (talk) 20:15, 28 July 2014 (UTC)
That would require them to be selling actual items. They're simply selling currency.--80.109.80.78 (talk) 21:42, 28 July 2014 (UTC)

# July 28

## Rotational invariants?

I am not a mathematician but a long standing project drags me in operating with objects I do not completely understand. I need help here.

I have a 2-Sphere (radius r = 1) and a function F(θ,φ) defined on it. The function F is well behaved, it is continuous and differentiable everywhere. The function F is therefore a function in the Hilbert space on 2-Sphere with the basis consisting of Spherical Harmonics with two indices $l$ & $m$: $Y_l^m$(θ,φ). I also want to introduce another basis $Y_l^m$(α,β) which is the result of rotation of the first basis via Euler angles ε, γ & ω. The latter are fixed.

Then I expand the function F(θ,φ) into the first basis: F(θ,φ) = $\sum_{l=0}^N$$\sum_{m=-l}^l$ $f_l^m$$Y_l^m$(θ,φ) where N is in fact a very large number.

My next step is to expand F(θ,φ) into the second basis: F(θ,φ) = $\sum_{l=0}^N$$\sum_{m=-l}^l$$g_l^m$$Y_l^m$(α,β)

I want to know if the expression: $\boldsymbol{\Iota}$ = $\sum_{m=-l}^{l}$$f_l^m\bar g_l^m$, where the bar denotes a complex conjugate, will be invariant under rotations? In other words if I arbitrarily rotate the function F(θ,φ) via three Euler's angles and keep calculating the expression $\boldsymbol{\Iota}$, the latter will remain invariant?

I also want to makes sure that the expression $\boldsymbol{\Iota}$ is the inner product in the Hilbert space?

My next question is this. Is the expression: $\sum_{m=-l}^l$$f_l^m\bar f_l^m$ not equal zero? If so, is it going to be invariant under rotations?

It may be easiest to answer the last question first. If you can write the expression as a function of Y alone rather than Y and the F's then it must be invariant. In this case I believe the sum is equal to $\int_S F\bar F$, which follows from multiplying out the expansion of F in terms of the Y's and using orthonormality. So the answer to the last question is yes. More generally, if G is any function which can be written $\sum_{l=0}^N$$\sum_{m=-l}^l$$g_l^m$$Y_l^m (\theta,\phi)$ then the $\boldsymbol{\Iota}$ you defined will indeed be the inner product $\int_S F\bar G$ and so invariant. So to answer the first question plug in G(θ,φ)=F(α,β) so show that it's invarient, though you have to be careful in your definition to apply the rotations (θ,φ)->(α,β) and rotation on the sphere in the correct order because it makes a difference here. (In the 1-sphere case, basically Fourier analysis, the rotations commute so it's not an issue.) Note there's nothing special about the sphere here, the same applies so any orthonormal bases for the functions defined on some base set and the group of inner product preserving symmetries of the base set. If it makes it more intuitive, let your base set consist of three points, then your function space is just Euclidean 3-space. --RDBury (talk) 03:19, 28 July 2014 (UTC)

Thank you very much, Sir. I will need some time to think about what you said. --AboutFace 22 (talk) 11:30, 28 July 2014 (UTC)