Wikipedia:Reference desk/Archives/Mathematics/2014 July 25

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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: $d^{2}y/dx^{2}+y=0$. 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 Cℓ_d,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 Cℓ_0,1(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 A_n 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 A_n 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.

Some numerical results

n = 1 0 / 10000000 0.000% n = 2 499453 / 10000000 4.995% n = 3 247150 / 10000000 2.472% n = 4 125081 / 10000000 1.251% n = 5 67303 / 10000000 0.673% n = 6 36118 / 10000000 0.361% n = 7 19967 / 10000000 0.200% n = 8 11317 / 10000000 0.113% n = 9 6402 / 10000000 0.064% n = 10 3660 / 10000000 0.037% n = 11 2094 / 10000000 0.021% n = 12 1204 / 10000000 0.012% n = 13 744 / 10000000 0.007% n = 14 451 / 10000000 0.005% n = 15 264 / 10000000 0.003% n = 16 152 / 10000000 0.002% n = 17 83 / 10000000 0.001% n = 18 49 / 10000000 0.000% n = 19 31 / 10000000 0.000% n = 20 19 / 10000000 0.000%

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:

A non-monotonic minimal path. "X" on path; "x" filling out remaining black squares.
XXooxx oXooxx oXooox oXoooo oXoXXX oXXXoX

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)

Sadly, the paper I have appears to be a method I don't see as easily applying here...it is an interesting problem, though, I'm going to keep fiddling with it - I doubt I'll arrive at an answer, as mentioned, combinatorics isn't my strong point. At any rate, if anyone else works up a proof, even if it is after this is archived, please inform me of it on my talk page - reading up on related problems, I'm rather interested and would like to see a worked out version.:-) Phoenixia1177 (talk) 04:26, 30 July 2014 (UTC)

One obvious, simplistic bound is that at most 3/4 of the grids will have at least one black square adjacent to the upper left corner, and at most 9/16 of the grids will have this satisfied for both the upper left and lower right corners, providing an upper bound on the number of paths, but while this excludes the 1-1/e guess, it does little else, particularly as the fraction of successful paths seems to approach 0 exponentially. Real Life^TP has keep me from doing any more with this, but I will eventually post the exact number of paths for 2n=6, and some minimum path numbers as well. Add my name to those who'd like to be pinged if anything gets added to this in archive. -- ToE 14:55, 31 July 2014 (UTC)

Probably totally useless information, but I calculate the number of corner-to-corner minimal paths as follows: 2x2 - 2; 3x3 - 6; 4x4 - 20; 5x5 - 92; 6x6 - 832; 7x7 - 20164; 8x8 - 1008708. 86.179.112.162 (talk) 03:20, 30 July 2014 (UTC)

I assume that you used a backtracking algorithm to feel out all possible paths, backtracking from those with excessive adjacency. Very nice! I hadn't though of building these up from blow, but was instead planning on testing all paths for minimality, which would not be practical beyond a 4x4 6x6 grid. For this particular problem, paths on a 2n x 2n grid must be at most (2n)²/2 long. It might be useful to see the number of minimal paths vs path length (perhaps further broken down into monotonic, non-monotonic in only one axis, and non-monotonic in both axes), as those figures could give some insight into developing a formula. There is, then, the question of how to go from minimal paths to a count of all successful grids, as the same successful grid could be formed by adding remaining squares to different minimal paths. -- ToE 14:55, 31 July 2014 (UTC)

I'd be quite surprised if anything that could be termed a "formula" exists. One could hope for a counting procedure that is a lot more efficient than brute force. 86.179.113.14 (talk) 19:28, 31 July 2014 (UTC)