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October 29

About the Word 'Rigorize'

Hello! I saw the word 'rigorize' widely used in wiki, but when I refer to dictionaries, I can't find it anywhere. So I wonder whether this is an official word in English. — Preceding unsigned comment added by 113.68.24.244 (talk) 09:11, 29 October 2013 (UTC)

Wiktionary contains an entry on "rigorize" and defines it as "To make rigorous; for example, to add further detail or elaborate on a proof or demonstration". Gandalf61 (talk) 10:26, 29 October 2013 (UTC)

I asked several people, but they did not reach an agreement. I wonder if this word is often used by mathematicians. Does it appear in some publications? — Preceding unsigned comment added by 113.67.114.215 (talk) 12:23, 29 October 2013 (UTC)

Google only finds 4,000-and-something instances of "rigorize", so it is certainly not a common word. Choice of words is very subjective, but I would probably use a synonym such as "formalise" (or "formalize" in American English). Gandalf61 (talk) 12:52, 29 October 2013 (UTC)

There is no official body that recognizes words in English so it's a matter of what is commonly accepted as proper. Given that it's not found in most dictionaries I would hesitate to use it in a formal setting, but the meaning is self-evident so there's no reason not to use it otherwise. I'm not sure what you (the OP) mean by 'widely used in wiki', I got only 7 matches on WP including talk pages. --RDBury (talk) 17:07, 29 October 2013 (UTC)

I don't think rigorous means the same thing as formal. I would say "make more rigorous". StuRat (talk) 21:28, 30 October 2013 (UTC)

convergence order

if G(x)= x -[x-g(x)]/[1-g'(x)/r] has convergence order r+1 ,prove that g(x) has convergence order r . how we can prove this please give some idea.True path finder (talk) 17:51, 29 October 2013 (UTC)

October 30

Logistic Map

File:Logistic map examples.gif This image does not seem to depict the Logistic map as it claims. Various features appear wrong, most apparent is that the asymptote appears at x=1 and decreases before the onset of periodic behaviour, whereas we know from File:LogisticMap BifurcationDiagram.png which is correct that the asymptote move upwards from x=0 before the fixed points begin bifurcating. — Preceding unsigned comment added by 109.157.219.5 (talk) 20:43, 30 October 2013 (UTC)

Sorry,I don't understand what you're saying. What asymptote are you referring to? I can't see anything wrong with it, though admittedly it is hard to follow because it moves from frame to frame so fast. Duoduoduo (talk) 21:19, 30 October 2013 (UTC)

For small r (prior to the first bifurcation) the long term behaviour of a system governed by the logistic map is to tend towards a specific value x_inf (the asymptotic behaviour of the system). For r=0 x_inf=0, and as r is increased x_inf increases too, until it reaches about x_inf=0.65, and the system begins a series of repeated bifurcations before the onset of chaos. The .gif linked to shows that x_inf remains at x_inf=0 before suddenly jumping to x_inf=1 and then decreasing before the onset of chaos. The two files I linked to clearly do not describe the same behaviour. — Preceding unsigned comment added by 109.157.219.5 (talk) 23:53, 30 October 2013 (UTC)

I see what you mean. I'll have to think about it tomorrow. Have you tried leaving a message on the talk page of the person who created the .gif? Duoduoduo (talk) 01:11, 31 October 2013 (UTC)

It seems to me that each graph, displaying an iteration of a logistic function and used as a single frame in the animation, is independently scaled to the frame's height. As long as iterations are monotonic, the asymptote lies on the frame's horizontal edge, but as soon as osciallations appear the highest oscillation peak touches the upper line of a frame. That's why the asymptote seems to decrease: it simply gets lower relative to the maximum value reached.
Watch also the starting point. It is probably constant, i.e. the same for all iteration runs. However in frames 1 through 19 the graph starts at the frames' top, then suddenly jumps to the bottom (frame 20 in yellow, where iteration reaches its limit point in the first step, meaning f(x₀) = x₀) and remains there till the frame 32, and then seems to slowly shift up as iterations span increasing interval (compare animated File:LogisticCobwebChaos.gif). --CiaPan (talk) 19:59, 1 November 2013 (UTC)

"solving" a circuit for voltages and currents

In any university physics text, you can find statements to the effect that, given a circuit diagram with resistances labeled, you can determine the voltage or the current at any point as a rational function of the input voltage and the resistances. But there is never any proof of existence or uniqueness. I can see how to write down a system of linear equations that describe the situation, but I do not see how to justify the fact that this system will always have a unique solution.

Any suggestions? Thanks. 50.103.239.50 (talk) 22:57, 30 October 2013 (UTC)

I suspect that they rely on empirical observations to determine if their answer is correct, rather than mathematical theory. That is, they measure the voltage and current in an actual circuit. Indeed, math often can provide extraneous answers. For example, the simple problem of asking what size square room will have an area of 100 gives answers of 10×10 and also (-10)×(-10). Since wall lengths of -10 make no sense, this answer is ignored. StuRat (talk) 23:12, 30 October 2013 (UTC)

While not a direct proof, Thévenin's theorem shows there is a unique equivalent circuit. --Mark viking (talk) 23:24, 30 October 2013 (UTC)

StuRat, good point about the extraneous solution. I hadn't thought of that. The reason I'm interested in this is because certain rectangle tiling problems can be analyzed via "circuit" graphs, as in Stein, Mathematics: The Man-Made Universe. (Of course, Stein does not include a proof...) I was hoping to be more rigorous. 50.103.239.50 (talk) 00:04, 31 October 2013 (UTC)

A single resistor with neither end connected to anything can have its ends at any voltage at all with respect to a reference ground node (though the two voltages are necessarily the same). Any linear network with entirely disjoint parts will have an indeterminate solution. If your network can include reactive components -specifically capacitors- then even a connected network could be indeterminate at d.c. catslash (talk) 00:45, 31 October 2013 (UTC)

For a general linear network, if the nodal admittance matrix is singular then by Cramer's rule, there won't be a unique solution for the voltages. For the floating resistor described above, the nodal admittance matrix is

${\begin{pmatrix}-{\frac {1}{R}}&{\frac {1}{R}}\\{\frac {1}{R}}&-{\frac {1}{R}}\end{pmatrix}}$

(or according to the article, minus that). Even if you have a unique solution for the voltages, you may not have a unique solution for the current, as any zero-resistance loop in the circuit (such as two super-conducting wires running in parallel between the same two points) can support an arbitrary circulating current. --catslash (talk) 02:02, 31 October 2013 (UTC)

For a classical linear electrical network (ie with idealized resistors, capacitors and inductors) the problem of determining the currents through the elements given all the applied emfs always has a solutions and the solution is unique. This has been known/proven for decades (starting I think from Gustav Kirchhoff's work) and is not a empirical result. You should be able to find the proof in any univ. level electrical network analysis book; for example look up Mac Van Valkenburg's text. Since I don't have access to it right now, here is a more mathematically sophisticated proof of the result. Abecedare (talk) 02:27, 31 October 2013 (UTC)

The above-linked text Mathematical Aspects of Electrical Network Analysis does not claim that every classical linear electrical network has a unique solution for the currents, only those networks which are what it calls ohmic. Also, the definition it gives for ohmicity seems to be little different from being soluble for the currents - at any rate, any definition in terms of the circuit topology or component values is well hidden. --catslash (talk) 18:56, 31 October 2013 (UTC)

You are right,I should have been clearer.

• The result I stated for an idealized RLC network and (independent) voltage sources is true but needs one topology condition to eliminate the degenerate case of connecting say two identical voltage sources to each other and asking what current is flowing through that loop (for idealized sources the current is indeterminate). So the result holds as long the circuit doesn't have any loops with only voltage sources.
• The result also holds when we introduce independent current sources into our circuit, as long as there is no cut set containing only current sources.
• The result can be further extended for circuits with mutual inductances (transformers), with the additional conditional that no set of these inductances be "perfectly coupled".
• Roth considers an even more general case with dependent sources and any general linear elements and formulated the condition of "ohmicity" (which can be shown to reduce the more intuitive forms for the special cases listed above).

All of this is covered (with proofs) in modern electrical network analysis textbook, and the simplest case asked by the OP will be found even in older "classical" texts like the ones Valkenburg, Seshu & Reed, Chen, Balabanian & Bickart etc. What I want to emphasize, is that none of this is taken on faith or considered magical like magnets. :) Abecedare (talk) 00:01, 1 November 2013 (UTC)

Your conditions dispose of the overdetermined cases of incompatible sources, and so guarantee the existence of a solution. What about the underdetermined cases? --catslash (talk) 17:25, 2 November 2013 (UTC)

The listed conditions in fact guarantee a unique solution.

Further: for the "Roth" case the ohmicity condition is just sufficient for the existence and uniqueness of the result (it becomes necessary too with some additional assumptions). For the rest, the given conditions are both necessary and sufficient. I found an online source (see pages 125-132) which discusses/derives results of the sort listed above; note that for pedagogical reasons, it starts with the general ohmicity condition and then reduces to simpler conditions for the special cases. IIRC some classical texts take the opposite approach: state and prove the RLC result and then relax conditions for more general cases. The latter may be easier for the OP to follow, but unfortunately I didn't find any of them accessible online. Abecedare (talk) 20:18, 2 November 2013 (UTC)

Sorry, I misread your third condition (imperfect coupling) as being necessary only when coupled inductors are present.

Wai-Kai Chen's Applied Graph Theory which you link to seems to define perfectly coupled as: having a zero eigen value. If I look for and find a zero eigen value, then I have already found a set of non-unique solutions, namely any multiple of the corresponding eigen vector (possibly plus some other stuff, if there are sources). So the uniqueness part of the theorem reduces to: the solution is unique unless you can find a non-unique set of solutions.

Whether or not there is a simple test for ohmicity/perfect-coupling/uniqueness that does not amount to solving the system (unlikely), it is nevertheless easy to construct circuits that have non-unique solutions (at any given set of discrete frequencies (which could include d.c.)), so in general a circuit will not have a unique solution.

It should however be possible to put a bound on the number of frequencies at which non-unique solutions might exist, based on the number of capacitors and number of inductors in the network. --catslash (talk) 02:02, 3 November 2013 (UTC)

"the solution is unique unless you can find a non-unique set of solutions": That is of course tautologically true but not a useful or suggested way to analyze the problem; and the above tests do not do so. And developing a "simple test for ohmicity/perfect-coupling/uniqueness that does not amount to solving the system" far from being unlikely, is the whole raison d'etre of the above discussion and the book section! For example, for an RLCM case, note that:

1. The matrix L whose non-singularity needs to be tested is only a sub-matrix of the bigger matrix Z_pp, which itself is only a sub-matrix of the big "system matrix" in equation 2.271 that incorporates all the information about the specific circuit along with the KCL and KVL constraints.
2. Checking for non-singularity (or bounding a positive-semidefinite matrix's smallest eigenvalue value away from zero), is much easier computationally than computing its eigen-decomposition explicitly.
3. As the book suggests in the last paragraph on page 131, even if the naive impedance matrix turns out to be singular, there are workarounds for obtaining unique solutions for the current through branches of practical interest.
4. Even in instances when the solution is non-unique (eg, is say a transmission line or microwave circuit with multiple frequency modes), the results listed above are useful to characterize the solution space, and engineering techniques have been developed to damp the undesirable modes of operations.

Obviously the textbook doesn't go into details of the last two points, but such methods are old-hat by now and used routinely in electronic circuit and electrical network simulation, design, and operation. Frankly, the basic existence/uniqueness problem is so well-solved that this is somewhat of a boring area for fresh research, because one needs to invent really exotic scenarios to break existing methods. That's the reason this 2012 book is not citing any work later than 1967 in the existence and uniqueness section! Abecedare (talk) 06:29, 3 November 2013 (UTC)

You are entirely right and I was wrong about the ease of calculation of perfect-coupledness. However the proof seems to require that the circuit be initially relaxed (p127), and is actually merely demonstrating that the solution is unique if you ignore voltages and currents that (for a real frequency) are not delivering power from the sources into the resistors - a result which is entirely expected. A circuit with no sources might support any number of degenerate modes, but under this criterion of admissibility only the trivial (zero) solution would be recognized. Unfortunately, the viewable section of the book is not self-contained, so it's possible that I'm still misunderstanding something. --catslash (talk) 13:09, 5 November 2013 (UTC)

October 31

Calculus homework - related rates

Hi, we are learning how to solve "related rates" word problems in calculus. So far I know how to do problems involving the changing area of a circle, Pythagorean distance problems involving right triangles and the classic problem involving the changing water level in a conical tank.

Generally, the solutions involve taking a geometrical equation, then taking the derivative of that equation using implicit differentiation then plugging the given values into the results in order to solve the problem (usually there is one missing derivative, which we solve for algebraically).

But we were given a different kind of related rates problem that I don't know how to solve.

An object is moving along a circle of radius 1 . When the object is at the point (4.5,−3.5) , the x value of its position is changing at the rate of 9 units per second. How fast is the y value of its position changing at that moment?

I looked up the answer and it is 12, but I don't know the method of solving this problem. I'm guessing it has something to do either with the equation of the circumference of the circle or the (x-a)^2 + (y-b)^2 = r^2 equation. But I really don't know where to go from here (i.e., how to set up the implicit differentiation, so I can substitute the radius 1 and the rate of change of x, which is 9).

How do you go about solving this type of problem?--Jerk of Thrones (talk) 09:47, 31 October 2013 (UTC)

There is insufficient information when we don't know anything about the center of the circle. For example, if the center is (4.5,−4.5) then the object is moving horizontally right above the center and the answer would have been 0. PrimeHunter (talk) 10:38, 31 October 2013 (UTC)

Indeed. Implicit differentiation gives us:

$2{\dot {x}}(x-a)+2{\dot {y}}(y-b)=0$

which we can re-arrange to get

${\dot {y}}=-{\dot {x}}\left({\frac {x-a}{y-b}}\right)$

We know ${\dot {x}}$, x and y, but unless we are also given one of a or b (from which we could infer the other, as we know r) then we are stuck. Gandalf61 (talk) 10:57, 31 October 2013 (UTC)

............Seems reasonable to assume the coordinates were meant to be (4/5,–3/5) instead of (4.5,–3.5) and the circle's center is (0,0) by default. --CiaPan (talk) 14:19, 31 October 2013 (UTC)

Ah, yes ! Good catch. Gandalf61 (talk) 14:36, 31 October 2013 (UTC)

Then the velocity vector is orthogonal to the position vector, so ${\frac {\dot {y}}{\dot {x}}}={\frac {v_{y}}{v_{x}}}=-{\frac {r_{x}}{r_{y}}},$ which implies $v_{y}=-v_{x}\cdot {\frac {r_{x}}{r_{y}}}=-9\;\mathrm {unit/s} \cdot {\frac {\frac {4}{5}}{-{\frac {3}{5}}}}=9\;\mathrm {unit/s} \cdot {\frac {4}{3}}=(3\cdot 4)\;\mathrm {unit/s} =12\;\mathrm {unit/s}$ --CiaPan (talk) 17:58, 31 October 2013 (UTC)

Dot-notation corrected. --CiaPan (talk) 08:15, 4 November 2013 (UTC)

Yes, sorry, It was (4/5, -3/5). I think when I copied and pasted the problem, the fractions didn't copy correctly, so I accidentally changed it to 4.5 and -3.5. Thanks for your help!--Jerk of Thrones (talk) 06:10, 6 November 2013 (UTC)

Distribution functions

Suppose I have two functions f() and g(), where f() returns either 0 or 1, with probability 0.5 for each, and g() always returns 0.5. The question is, how would I construct a function h(x) where x can be at least any real between 0 and 1, and returns h(0) -> f() and h(1) -> g()? Obviously you can now say "well, apply coefficients to f and g, or return f or g with a certain probability", but that is not what I mean, I mean is there some sort of abstract mathematical concept I should read more about, if I want to know how to go smoothly from a fair coin flip to a 100% chance of the coin landing on its side? Ginsuloft (talk) 20:24, 31 October 2013 (UTC)

Hm, I'm a bit unclear what you mean. I think you are looking for random variables, rather than functions (although a random variable are functions from the sample space into the reals. If f, g are random variables with the properties as above, then you can simply set $h=(1-x)f+xg$ where $0\leq x\leq 1$. Such a random variable would return f for x=0 and g for x=1. For any other x it would be the RV that returns 0.5+x/2 and 0.5-x/2 with probability 0.5 each.

A different way to achieve a smooth transition would be to consider 3 events, namely 1 (heads), 0 (tails) and 0.5(coin on side) and define h as $P(h=1)=(1-x)/2,P(h=0)=(1-x)/2,P(h=0.5)=x$. Don't know if any of this is close to what you are after. 86.179.30.226 (talk) 22:26, 31 October 2013 (UTC)

No, I'm afraid that's not it. To clarify, the points h(0) and h(1) should be extreme cases; usually the function should return a pretty uniform distribution, meaning that normally you shouldn't be able to guess what the returned function returns. This is for the purposes of experimenting with connectionism. Anyway, it probably isn't possible to get an answer here, as I can't really put my wants into words without giving implementation details (and therefore solving the problem myself). Thanks anyway to you and anyone else who tries to help. Ginsuloft (talk) 22:39, 31 October 2013 (UTC)

I think I basically understand what you are asking for. The most common approach is to feed the output of h into a sigmoid function, with a parameter setting the slope of the sigmoid. A commonly used choice of sigmoid is

${\frac {1}{1+e^{\alpha (1/2-x)}}}$

If α is zero, the output is always 1/2. If α is infinite, the result is either 0 or 1 with equal probability. Intermediate values of α give you a graded distribution. Looie496 (talk) 23:04, 1 November 2013 (UTC)

Yes, thank you for making sense of my nonsense. That was exactly what I was looking for. I never realized the answer was just a sigmoid function. I knew that sigmoids are widely used in neural nets but never would have thought that even some improvised probabilistic model I came up by myself would have made use of them. Ginsuloft (talk) 17:11, 2 November 2013 (UTC)

November 1

Descriptive Complexity and FMT- a really simple question

Hi:-) This question is, probably, borderline stupid, but I can't reason it out now that I'm thinking about it. For a logic L (F0, LFP, whatever), define the set of L-expressible problems as {Finite_Models(u): there is a vocabulary t and u is in the L-language of t}. I've frequently seen it mentioned that you can get away with just considering the vocabulary that amounts to binary strings, so my question is: if I restricted the definition of L-expressible to the case where t has a 1-place relation and a linear order, then if I showed (in the this restriction) that L-expressible = L'-expressible, does that mean that the general case would follow? More generally, can we just fix t to be any vocabulary?Phoenixia1177 (talk) 06:46, 1 November 2013 (UTC)

Im guessing that in the case of strings it's because you encode models to strings, but whqt about the general case? Could you just consoder graphs with a single binary relation? Essentially, can you represent all strings in any other vocabulary or are they special?Phoenixia1177 (talk) 11:07, 1 November 2013 (UTC)

There's a standard folklore result in model theory that every structure can be represented by a graph (directed or undirected). Obviously every graph can be coded by a binary string, but I don't see that they have the same logical content. For example, it's easy to say in the language of graphs "there is a 5-clique". How would you say "the graph encoded by this string has a 5-clique" in first-order logic? — Preceding unsigned comment added by 80.109.80.78 (talk) 11:34, 1 November 2013 (UTC)

As long as you could state something equivalent to E (the edge relation) in FO, you can do the rest. (In DC it is assumed we have the natural order and the ability to check bits) To encode our string, starting at (0, 0) and working lexicographically, if E(x, y) then write 00BINN(x)11BINN(y) where BINN(z) is z encoded in binary using 10 and 01 instead of 1 and 0. There are FO relations L and R so that L(x) iff x is a position immediately after two 0's, and R(x) iff x is a position immediately after two 1's. Then, using those you can get E(x, y) that holds iff L(x) and R(y) and there is exactly one instance of 11 between x and y. While tedious, any statement about the graph translates into a statement in the string. You can probably do this more simply; especially if you step beyond FO to anything including FO(TC).Phoenixia1177 (talk) 06:04, 2 November 2013 (UTC)

I'm a little off in the above, there's goofiness checking parity in FO and you can get 00 in the BINN strings; so you might need to do something like 001BIN(x)0001BIN(y) and use 011 and 01 in place of 10 and 01 (or some such), then you can be sure that you're grabbing the right bits.Phoenixia1177 (talk) 06:15, 2 November 2013 (UTC)

The issue I see here is defining equality. Sure, I can make a first order predicate that recognizes when I'm at the beginning of a code (L and R in the above). But how can I make a predicate to recognize when x and y are the beginning of the same code (in different locations)? It seems you would need this to translate a statement about graphs. For example, how would you translate the statement "there is a 3-cycle", i.e. $\exists x\exists y\exists zE(x,y)\wedge E(y,z)\wedge E(z,x)$.--80.109.80.78 (talk) 10:46, 2 November 2013 (UTC)

You can do this using the ordering: even if you don't know the length of strings encoding vertices, you can write a predicate that holds iff both are equal at each position between 001 and 0001. Again, it's a horrible pain- and I don't know the math tags to write it out- but it can be done.Phoenixia1177 (talk) 10:57, 2 November 2013 (UTC)

I'm going to remain skeptical without a little more detail. The issue seems to be that a first order sentence should only be able to examine boundedly many bits at once. Whatever the bound, I can make two distinct strings that look the same up to that bound: any interval in one of length that bound occurs as an interval of the other.--80.109.80.78 (talk) 12:34, 2 November 2013 (UTC)

I'll need to think on this- I've seen things that suggest otherwise, but I have yet to see the logic spelled out (I just checked a few references). Would it suffice to consider the case of a finite bit string with a constant indicating where the second substring beings; that is, would it work to show there is a relation that is true iff the part before the constant and the part after, and including, the constant are equal? You can easily constrain variables in the above case of encoding a graph to whittle it down to such.Phoenixia1177 (talk) 15:52, 2 November 2013 (UTC)

I don't understand what you're asking here. It's easy to come up with formulae that describe the beginning and end of a code for a point. The task is given BIN(x) and BIN(y), you need to find a formulae that holds if and only if they are equal. Where are you suggesting placing the constant?--80.109.80.78 (talk) 16:09, 2 November 2013 (UTC)

I was looking at them as part of one string Bin(x)Bin(y) with the constant marking where Bin(y) starts, since we were working with one big string above. I wasn't separating them, essentially. I'm sleepy, so I probably garbled what I was saying.Phoenixia1177 (talk) 16:47, 2 November 2013 (UTC)

In DC terms, you can do plus(x, y, z) is true iff x + y = z in FO. So, going with the one string above and Bin(x) starting at 0, Bin(y) at a: you could check that for all positions u and v that if plus(u, a, v), then they stored the same bit; and also that for the maximal position max that plus(a, a, max) so that you know both were the same length.Phoenixia1177 (talk) 16:57, 2 November 2013 (UTC)

I'm not sure what DC is, but you were asking about First Order logic earlier, and addition is definitely not FO-definable in finite linear orders. So this won't do it for FO-logic.--80.109.80.78 (talk) 18:03, 2 November 2013 (UTC)

Descriptive Complexity, that's the main motivation for the question.Phoenixia1177 (talk) 18:19, 2 November 2013 (UTC)

Number lines

Why are number lines virtually always horizontal? I never thought of this until I looked at our article and saw "The number line is usually represented as being horizontal", but thinking about it, I can't remember ever seeing a simple number line that was vertical. Nyttend (talk) 22:34, 1 November 2013 (UTC)

Might have to do with it being a teaching tool and blackboards being longer in the horizontal direction, and kids being able to reach the entire number line to show an answer on the board. StuRat (talk) 22:45, 1 November 2013 (UTC)

That makes sense. Another possibility is that when drawing the X-Yplane, X is normally horizontal. Bubba73 ^{You talkin' to me?} 01:43, 2 November 2013 (UTC)

I think that's backwards. X is horizontal because we make the plane adding the Y-axis to the standard number line.--80.109.80.78 (talk) 03:33, 2 November 2013 (UTC)

If I had to guess (and I absolutely am), I'd say it's because we read right to left.Phoenixia1177 (talk) 06:18, 2 November 2013 (UTC)

.elbisualp sdnuos tahT PrimeHunter (talk) 12:26, 2 November 2013 (UTC)

Now that's interesting. In the old days of Arabic contributions to mathematics, given that Arabic is written right to left, was there ever a convention of having the number line with numbers rising as we go right to left? Duoduoduo (talk) 14:21, 2 November 2013 (UTC)

Everything I could find, which is little, seems to indicate that the number line came about in relation to understanding negative numbers and is a fairly modern western invention (modern meaning the last 3-4 centuries, in this case). That could, just as well, be utterly false, the few mentions were not high quality and were focused in other directions (ha!). I did come across a suggestion that it had some relation to lines used in construction by Egyptians- there was also an implication in some random forum that depending on how it is written, Hebrew number lines can run right to left (I don't know how to check that, though.)Phoenixia1177 (talk) 16:02, 2 November 2013 (UTC)

November 2

How confident are we that basic axiomatic systems are consistent?

How confident are we that basic axiomatic systems mathematicians work with today are consistent? How hard is it to imagine that 4000 years from now, people will say that the poor people of third millennium spent time exploring primitive axiomatic systems each of which was totally inconsistent and could - though these people did not know it - be used to prove either assertions or their opposites.

Is that a reality that's within the realm of something we can imagine? I mean, I can imagine people saying that, 4000 years from now. But what about more rationally? Is it in the realm of the imaginable that we are living in deeply inconsistent axiomatic systems through some benightedness that we are totally unaware of? 212.96.61.236 (talk) 01:46, 2 November 2013 (UTC)

Well, this is a question on which reasonable people can disagree. Edward Nelson thinks that even Peano arithmetic is inconsistent, and he's a serious, respected mathematician. In my view you might as well ask whether you're a brain in a vat. However, this in itself is a major slippage from the old view that the consistency of mathematics was an apodeictic certainty, even more certain than that you're not a brain in a vat. --Trovatore (talk) 02:02, 2 November 2013 (UTC)

I doubt Nelson has a proof of that. Bubba73 ^{You talkin' to me?} 02:06, 2 November 2013 (UTC)

Not yet :-) --Trovatore (talk) 02:55, 2 November 2013 (UTC)

I'm not holding my breath. Bubba73 ^{You talkin' to me?} 03:10, 2 November 2013 (UTC)

Aw, isn't that cute! Brain-in-a-vat thinks it has breath to hold! --Trovatore (talk) 03:28, 2 November 2013 (UTC)

Consistency of even the axiomatic approach and logic itself is essentially a matter of faith: confidence it it is empirically built, and does not correspond to any identifiable part of physics. Logic is a construct that we have evolved/learned to understand and manage the world, and only has applicability to concepts and categories of the models that we have constructed (not the "real" world itself). Just like Newtonian physics was found to be inaccurate in an unexpected but fundamental way with the discovery of special relativity, logic may have limited applicability in its own domain. Add to this that axiomatic systems are varied, that consistency is difficult to prove even in the simplest systems of axioms, and that we need more than the simplest axiomatic systems to deal with most useful mathematics, and that this has brought with it logical incompleteness, I'd suggest that it would be foolish to claim a high level of confidence at the 4000-year level. — Quondum 16:38, 2 November 2013 (UTC)

I get this, but it is interesting as an 'act of faith' because it is very falsifiable. Someone could very strongly prove the inconsistency of an axiomatic system, and that would kind of be the 'last word' on that system, no? So at the moment it seems like such a proof would have to be inordinately nuanced or extensive (since the systems we work with seem consistent enough) but once made or discovered they would no longer allow any mathematician who could not point out a flaw in them to practice rigorous mathematics as normal. In this sense it is quite different from matters of faith. I found the rest of your comment highly interesting. 212.96.61.236 (talk) 19:40, 2 November 2013 (UTC)

Right - if it is inconsistent, that can be proven. The statement that it is consistent is falsifiable. Bubba73 ^{You talkin' to me?} 19:42, 2 November 2013 (UTC)

See also here:

"So I deny even the existence of the Peano axiom that every integer has a successor. Eventually we would get an overflow error in the big computer in the sky, and the sum and product of any two integers is well-defned only if the result is less than p, or if one wishes, one can compute them modulo p. Since p is so large, this is not a practical problem, since the overflow in our earthly computers comes so much sooner than the overflow errors in the big computer in the sky.

However, one can still have `general' theorems, provided that they are interpreted correctly. The phrase `for all positive integers' is meaningless. One should replace it by: `for finite or symbolic integers'. For example, the statement: (n + 1)^2 = n^2 + 2n + 1 holds for all integers" should be replaced by: (n + 1)^2 = n^2 + 2n + 1 holds for finite or symbolic integers n" . Similarly, Euclid's statement: `There are infinitely many primes' is meaningless. What is true is: if p1 < p2 < ... < pr < p are the first r finite primes, and if p1p2...pr + 1 < p, then there exists a prime number q such that pr + 1 <= q <= p1p2 ... pr + 1. Also true is: if pr is the `symbolic rth prime', then there is a symbolic prime q in the discrete symbolic interval [pr + 1; p1p2 ... pr + 1]. By hindsight, it is not surprising that there exist undecidable propositions, as meta-proved by Kurt Godel. Why should they be decidable, being meaningless to begin with! The tiny fraction of first order statements that are decidable are exactly those for which either the statement itself, or its negation, happen to be true for symbolic integers. A priori, every statement that starts "for every integer n" is completely meaningless." Count Iblis (talk) 20:31, 2 November 2013 (UTC)

The problem (or, at least, one problem) with this ultrafinitistic philosophy is that it fails to give an explanation for the observed consistency (that is, failure to find an inconsistency) in mathematical theories.

Now, you could say that that doesn't need an explanation — there may be an inconsistency but we just haven't found it yet. In other words, you reduce it to a "brute fact". But any physical observations could equally be brute facts, so that's not terribly satisfying.

On the other hand, mathematical realism (sometimes called Platonism) explains the observations perfectly well. --Trovatore (talk) 20:37, 2 November 2013 (UTC)

Right, that's why I said that an inconsistency proof of basic axiomatic systems would have to be extraordinarily nuanced or complicated: it couldn't be simple at all, since on a simple scale the systems we work with seem very consistent.

On the other hand I have a question for you guys: one respondent above (Bubba73) said that if it is inconsistent, this can be proven. Is this necessarily the case? Maybe an axiomatic system could be inconsistent without either it, or the system with which it is being examined, being strong enough to prove this inconsistency? After all, we no longer expect to be able to 'prove' everything that is true (incompleteness theorem). So I think the person above (Bubba73) might be wrong in saying that if it is inconsistent, this can be proven. 212.96.61.236 (talk) 00:12, 3 November 2013 (UTC)

If an axiomatic system is inconsistent, then by definition, there is a proof in that system of both a statement and its negation. In principle, those proofs can be found, and therefore, yes, the inconsistency can be proved. In practice, it is possible that the proofs may not be humanly discoverable (they might even be too long to be expressed in the observable universe). --Trovatore (talk) 00:51, 3 November 2013 (UTC)

If it is inconsistent then you can derive a contradiction from the axioms (in a finite number of steps). That is what it means. Bubba73 ^{You talkin' to me?} 01:29, 3 November 2013 (UTC)

One thing that seems to be missing from this discussion is Russel's paradox. Before it was discovered, set theory seemed to be a fairly natural extension of logic and therefore on sound footing. Basing the rest of mathematics on set theory was seen as a way of making it more rigorous. People probably reasoned, as above, that a contraction extremely unlikely to be found even if one existed. There were high hopes that a proof of consistency would be found. So when a contraction was found, in particular one that could be expressed in a single paragraph, it was very unexpected. Russel found (in programming lingo) a rather kludgy work-around to remove the paradox, improved upon since but still a bit of a kludge. Which is why a consistency proof seemed more important and why Gödel's theorem was more of a problem than it seems now. Right now the status is "so far so good" operating on the assumption that if there was a contradiction then it would have been found by now. But by that reasoning there is a higher probability that set theory will be proven inconsistent than that the Riemann hypothesis will be proven. If there is a contradiction what will probably happen is what happened with Russel's paradox, someone will weaken the axioms enough so that the contradiction disappears while keeping enough so that set theory still works as a basis for the rest of mathematics. So accountants and engineers can sleep peacefully in the knowledge that they won't wake up to find 2+2=5 or square wheels work better than round ones. --RDBury (talk) 09:29, 3 November 2013 (UTC)

So it is true that the RP is a historical instance of falsification in mathematics (which is good, it proves it's falsifiable!). But there are some problems with your narrative, just the same.

Mainly, it is not clear that Cantor ever tried to unify set theory and logic, and if he did, he realized his mistake long before Russell. Remember how Cantor got started: He was not coming from abstract logic at all; he started with real analysis, specifically, the points of discontinuity of Fourier sums. His first "sets" were specifically sets of real numbers. If you look at Contributions to the Founding of the Theory of Transfinite Numbers today (BTW you can get it at Amazon for twelve bucks — just go do it; you'll remember it better than anything else you're likely to have spent it on), there is nothing at all that is jarring compared to a modern approach based on the iterative hierarchy.

So summing up, RP relies on a conflation of the extensional and intensional notion of set, and it's not clear Cantor ever did that. (The historiography on the question is mixed.)

As for Russell's "kludgy" fix, that would be the system of Principia Mathematica, which is a historical dead-end. No one uses it today. Probably the biggest barrier to entry is the awful notation; personally, I have never seriously tried to get over that barrier, so I don't know how much of a kludge it actually is.

But it really doesn't matter, because modern set theory does not descend from Russell, but from Ernst Zermelo, who got Cantor right. I do not see Zermelo as having worked around a flaw in Cantor as exposed by Russell — what he did was go back carefully to the iterative idea, arguably already implicit in Cantor, and formalize that.

So summing up a second time, yes, there is an a posteriori component to all this, but the idea that modern set theory is a shaky workaround to an original flawed design is not accurate. --Trovatore (talk) 18:43, 3 November 2013 (UTC)

Having looked up the dates, I should say that the Contributions translation I'm talking about is from 1915, fourteen years after the Russell paradox and eight after Zermelo's first work on the subject, and I don't know how much Philip Jourdain, who translated it, might have tweaked the translation in response. I should try to work through the original Beiträge (1895,1897) sometime and see. --Trovatore (talk) 19:29, 3 November 2013 (UTC)

It's Frege, not Cantor, that you want to look to for a unification of set theory and logic (and it was in response to a preprint of Frege's manuscript that Russell first conceived of the paradox). Also Russell's system was that of type theory (in practice ramified type theory to get around other related paradoxes) which has been influential in its own way, and lives on in New Foundations which still kicks around of the outskirts. -- Leland McInnes (talk) 02:55, 4 November 2013 (UTC)

An exponential question

Dear Wikipedians:

For the following exponential function:

I was able to successfully deduce the original equation is $-{\frac {3}{2}}(2)^{x+5}+7$. However, I have arrived at this answer via trial and error. I am wondering if there is a more systematic way of deducing this answer.

Ed: The +7 part I was able to derive from the fact that the function's asymptote is y=7. The coefficient and exponent parts I had to do via trial and error.

Thanks,

L33th4x0r (talk) 21:37, 2 November 2013 (UTC)

You could use a Nonlinear regression and if a general equation of that form is tested, those five points will determine the coefficients. Bubba73 ^{You talkin' to me?} 23:03, 2 November 2013 (UTC)

Step by step simplification. You subtracted the y values from the guessed asymptote to get (x,7-y)=(-6,3/4), (-5,3/2), (-4,3), (-3,6), (-2,12). Note that the y-7 values are doubled each time x is increased by one. So (x,(7-y)*2^(-x))=(-6,48), (-5,48), (-4,48), (-3,48), (-2,48). Note that the dependent variable has the constant value (7-y)*2^(-x)=48. Solve the equation to get the final result y=7-48*2^x. This is slightly more systematic than mere trial and error. Bo Jacoby (talk) 14:04, 3 November 2013 (UTC).

Right. Your correct equation $y=-{\frac {3}{2}}(2)^{x+5}+7$ can be written more simply as $y=7+(-48)(2)^{x}.$ In parametric form you would postulate $y=a+bc^{x}.$ If you have the hypothesis that a=7, then you have $(y-7)=bc^{x}$, so $(7-y)=(-b)c^{x}$ hence $\ln(7-y)=\ln(-b)+x\ln c$ where ln is the natural logarithm. So now you can run the linear regression Y=B+Cx where $Y=\ln(7-y),$ and see if you get a perfect R² of 1.0. Then since $B=\ln(-b)$ and $C=\ln c,$ you can compute $-b=e^{B}$ and $c=e^{C}$. [corrected version]Duoduoduo (talk) 15:42, 3 November 2013 (UTC)

Thank you all for your help! L33th4x0r (talk) 00:06, 4 November 2013 (UTC)

Resolved

November 3

Entropy of uniform distribution

Melbourne, Australia

Looking at the entries for uniform distribution the entropy is given as ln(n) for the discrete case, and ln(b-a) for the continuous case. For a normalised distribution in the interval [0,1] the discrete case agrees with Shannon's definition of entropy, but the continuous case produces entropy of 0. Is this correct? Rjeges (talk) 10:06, 3 November 2013 (UTC)

I think it would be more accurate to say that the entropy in the continuous case is undefined. Looie496 (talk) 14:30, 3 November 2013 (UTC)

From Differential entropy#Definition:

Let X be a random variable with a probability density function f whose support is a set $\mathbb {X}$. The differential entropy h(X) or h(f) is defined as

$h(X)=-\int _{\mathbb {X} }f(x)\log f(x)\,dx$.

and

One must take care in trying to apply properties of discrete entropy to differential entropy, since probability density functions can be greater than 1. For example, Uniform(0,1/2) has negative differential entropy

$\int _{0}^{\frac {1}{2}}-2\log(2)\,dx=-\log(2)\,$.

Thus, differential entropy does not share all properties of discrete entropy.

Duoduoduo (talk) 15:58, 3 November 2013 (UTC)