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April 29

formalism vs logicism

in the philosophy of mathematics, what is the difference between formalism and logicism? They look like very similar ideas, in that they both seem to reduce mathematics to an abstract game of some sort. Are they actually incompatible, or can one be both a formalist and a logicist? Thanks in advance. 203.221.126.206 (talk) 06:37, 29 April 2008 (UTC)[reply]

I think they're rather different, though perhaps not incompatible. Logicism, as I understand it, wants to reduce all mathematics to logically necessary truths. So for example if P and Q are properties such that P holds of exactly one x, and Q holds of exactly one y, and for every x you can't have both P(x) and Q(x), then there are exactly two x such that P(x)∨Q(x) holds. This could be taken as a logicist interpretation of "1+1=2", and it is logically necessary.

Formalism, on the other hand, at least in its most extreme form, treats "1+1=2" as an uninterpreted string, and attempts to derive it from completely arbitrary formal axioms, which are themselves uninterpreted. Now it should be said that this is the cartoon version of formalism -- it's sort of obviously stupid, and a great many people who call themselves formalists are obviously not stupid, so they have some more nuanced view than that. But when we're looking to explain the distinctions among the major schools (or, in the case of logicism, schools that have been major in reasonably recent memory), the cartoon versions perhaps show up the differences most clearly. --Trovatore (talk) 07:06, 29 April 2008 (UTC)[reply]

It is possible to take a formalist position on logic, and a logicist who espouses that is then also a formalist. A completely formal system for deriving mathematical truths is a logic, and the desire to reduce mathematics to such a system is therefore, essentially, a form of logicism. (See also Hilbert's program.) But one can, conceivably, take a logicist position without being a formalist. In view of Gödel's incompleteness theorems, it can be argued that that is the only reasonable position for someone who is a (logicist OR formalist). --Lambiam 10:08, 2 May 2008 (UTC)[reply]

Hmm, just saw the first part of your response here. No, I don't think a "formal system" is a logic in the relevant sense, mainly because it contains non-logical axioms. Logicism wants to dispense with axioms entirely, and make all mathematical truths into purely logical truths that simply reflect correct inference, and contain no non-discharged assumptions whatsoever. --Trovatore (talk) 22:51, 3 May 2008 (UTC)[reply]

Continuous functions on [0,1]

Suppose $\{f_{n}:[0,1]\to [0,1]\}$ is a sequence of continuous functions such that $\lim f_{n}(x)=0$ for all x. Prove that

\lim \int _{0}^{1}f_{n}=0.

Can this be done simply without resorting to measure theory? — merge 10:52, 29 April 2008 (UTC)[reply]

If memory serves, that's only true if they converge uniformly. If I'm right, then that article is probably the place to start. --Tango (talk) 12:49, 29 April 2008 (UTC)[reply]

I assume they need to take all values in that range, and so I can't simply state that f_n = 1/n. -mattbuck (Talk) 12:50, 29 April 2008 (UTC)[reply]

Uniform convergence is a sufficient condition, however it is not strictly necessary, for example also dominated convergence is sufficient.--Pokipsy76 (talk) 13:16, 29 April 2008 (UTC)[reply]

Yes, of course the result is true by dominated convergence. The OP's question is whether this can be proven (for the Riemann/Darboux integral, presumably) without measure theory and the Lebesgue integral. I can't immediately see how to do this, though you can do Lebesgue's monotone convergence theorem (for continuous functions) without measures, since the convergence is uniform by Dini's theorem. Algebraist 13:30, 29 April 2008 (UTC)[reply]

And the boundedness of the range [0,1] is important, because the functions

f_{n}(x)={\begin{cases}n^{2}x\,:\,0\leq x<{\frac {1}{n}}\\2n-n^{2}x\,:\,{\frac {1}{n}}\leq x<{\frac {2}{n}}\\0\,:\,{\frac {2}{n}}\leq x\leq 1\end{cases}}

with peaks of height n centred at 1/n converge pointwise to 0, but all have an integral of 1. Gandalf61 (talk) 13:41, 29 April 2008 (UTC)[reply]

Could pointwise convergence to 0 on [0,1] + uniform boundedness imply uniform convergence?--Pokipsy76 (talk) 14:28, 29 April 2008 (UTC)[reply]

No it doesn't. Counterexample: the sequence of the characteristic functions of an oscillating and narrowing interval in [0,1]. (It's not difficult to imagine also continuous counterexample with the same behaviour)--Pokipsy76 (talk) 14:32, 29 April 2008 (UTC) I was wrong: it doesn't converge to 0.--Pokipsy76 (talk) 14:33, 29 April 2008 (UTC)[reply]

No. Gandalf61's moving hump, tweaked to have peaks of height 1 instead of n, converges pointwise but not uniformly. You really do need monotonicity, not just uniform boundedness, in Dini's theorem. Algebraist 15:26, 29 April 2008 (UTC)[reply]

I don't see how it converges to 0 on the boundary points.--Pokipsy76 (talk) 16:11, 29 April 2008 (UTC)[reply]

It's always zero at the boundary points! Algebraist 16:24, 29 April 2008 (UTC)[reply]

Anyone have any thoughts? I'm not sure where to go with monotone convergence, since what you really want is Fatou's lemma. The pointwise liminf functions won't necessarily be continuous, so you can't apply Dini. You can assume WLOG that each function is a polynomial using the Weierstrass theorem (approximating the nth function uniformly within 1/n), and I was thinking about finding a uniform bound on the derivatives to apply Arzela-Ascoli. But I couldn't get that to work. 134.173.93.127 (talk) 10:23, 30 April 2008 (UTC)[reply]

You cannot use the monotone convergence theorem except when the sequence is monotone, and you don't have that here. And the proposition is clearly false if you don't have continuity, since there are simple counterexamples, e.g. n times the indicator function of the interval (0, 1/n) converges pointwise to 0 on [0, 1] although the sequence of integrals converges to 1 in that case. More later..... Michael Hardy (talk) 02:36, 1 May 2008 (UTC)[reply]

Remember, f_n(x) is supposed to be in [0,1] for all n and x; otherwise you'd get the same result (pointwise convergence to 0, integral converges to 1) with the sequence of continuous functions f_n(x) = n max(0, 1 - abs(1 - nx)), which I believe is the same as the sequence suggested by Gandalf61 above. —Ilmari Karonen (talk) 03:22, 1 May 2008 (UTC)[reply]

OK, I overlooked the uniform bound. So you can use Lebesgue's dominated convergence theorem.

Here's another way. For any ε > 0, let C_n, ε = {x ∈ [0,1] : f_n(x) ≤ ε}. Notice I said "≤", not "<". That implies the set C_n, ε is closed. The intersection of C_n, ε over all values of n is empty, since any point in the intersection would be a point in the domain at which the sequence of functions fails to converge to 0. So we have a decreasing family of closed subsets of a compact space whose intersection is empty. From basic topology, since this sequence is decreasing, it will be empty for large enough n. That means for large enough n, we have f_n ≤ ε everywhere in the domain. Since ε was arbitrary, the sequence must therefore converge uniformly to 0. Then apply the usual result about uniform convergence and integrals. Michael Hardy (talk) 03:45, 1 May 2008 (UTC)[reply]

Are you sure? f_n(x) = max(0, 1 - abs(1 - nx)) is continuous and bounded, and doesn't converge uniformly even though it also converges pointwise to 0. (Of course, its integral still converges to 0, so it's not a counterexample to the original problem.) I think you're making an additional assumption of the sequence being decreasing. —Ilmari Karonen (talk) 03:55, 1 May 2008 (UTC)[reply]

OK, that argument works for monotone sequences, but won't work here for the same reason the monotone convergence theorem won't work here. The dominated convergence theorem will do it, but I suspect there's a more elementary way. Michael Hardy (talk) 11:31, 1 May 2008 (UTC)[reply]

That's exactly what's happening. This is the proof of Dini's theorem, which requires monotonicity. My earlier comment about applying the MCT wasn't to apply it directly, but to use it to prove a Riemann integral version of Fatou's lemma (from which our problem follows easily). However, that doesn't quite work. The article on Fatou says that there is a direct proof. Can anyone think of one that works for Riemann integrals? 134.173.93.127 (talk) 07:08, 1 May 2008 (UTC)[reply]

Even the statement of Fatou's lemma will have to be modified to work with Riemann integration, since the lim inf of a sequence of R-integrable functions need not be R-integrable (set f_n to be 0 at the first n rationals (in some listing) and 1 everywhere else). Algebraist 13:38, 1 May 2008 (UTC)[reply]

That sounds like "no." ;) Thanks to all for the thoughts! — merge 11:11, 3 May 2008 (UTC)[reply]

Equation of a line given the gradient and a point on the line

Now i get the first part of it which is:

Find the equation of a line with the gradient 6 that passes through point (5,6)
y-y1=m(x-x1) [M being gradient and Y1 and X1 being the points supplied]
Y-6=6(x-5) [Substitute the values in]
Y-6=6x-30 [Operations]
Y-6+6=6x-30+6 [Operations and making it equal y]
Y=6x-24

Can this method be applied to the below problem? , i remember hearing the gradient of two perependicular lines has to be 1 i think ( or negative 1, he wasnt very clear)

Find the equation of the line perependicular to the line y=-3x+6 and passing through point (-2,8)

Cheers, 124.180.254.74 (talk) 12:56, 29 April 2008 (UTC) .[reply]

I don't think you mean gradient, in the mathematical sense; I think you mean slope, although possibly in construction, topology etc the words are interchangeable. Anyway, the product of the slopes of two perpendicular lines is -1, (This site explains why reasonably well). So what you want to do is:

Find the slope, $m_{1}$ of the first (given) line.
Find the slope, $m_{2}=-{\frac {1}{m_{1}}}$ of the second (solution) line.
Construct the equation for the second line using the slope you now know, and the given point on the line, using the method you did for the first part of the problem.

moink (talk) 13:17, 29 April 2008 (UTC)[reply]

"Gradient" is the word commonly used for the slope of a line, especially at lower levels where it's not going to be confused with the vector calculus term - the gradient of a scalar field is just a generalisation of the gradient of a line to multiple dimensions. --Tango (talk) 16:04, 29 April 2008 (UTC)[reply]

Imaginary numbers

Normally I'd be answering questions here, but I've got coursework which I don't understand (and isn't included in the $65 textbook!). It relates to radicals of imaginary numbers (2isquare root of 7) and such, like FOILing imaginary numbers and how they work in equations including real numbers. Can someone help a brother out? Ziggy Saw dust 15:16, 29 April 2008 (UTC)[reply]

I'll help, but I don't yet understand exactly what you're trying to do. Perhaps you could give a specific example, and explain what you don't understand about it? Algebraist 15:20, 29 April 2008 (UTC)[reply]

Like these:

(3-4i)²

(-5+2i square root symbol 7) - (2-7i square root symbol 7) —Preceding unsigned comment added by Ziggy Sawdust (talk • contribs) 15:25, 29 April 2008 (UTC)[reply]

These questions are essentially unchanged by the use of complex numbers. Can you do these questions:

(3-4x)²

(-5+2x{\sqrt {7}})-(2-7x{\sqrt {7}})

?

If so, you can also do them with i in there. Just remember that i²=-1. Algebraist 16:05, 29 April 2008 (UTC)[reply]

i c wut u did thar... Thanks. Ziggy Saw dust —Preceding comment was added at 16:38, 29 April 2008 (UTC)[reply]

Question about graphing equadratic equation

If you are looking at a graph of a quadratic equation, how do you determine where the solutions are? I know sometimes you can just look at it and guess but how can you be sure by looking at it? thank you —Preceding unsigned comment added by Lighteyes22003 (talk • contribs) 15:18, 29 April 2008 (UTC)[reply]

If you have a graph of a function y=f(x) and you want to know the solutions to f(x)=0, then they are simply the (x co-ordinates of the) points where the graph crosses the x-axis. You can thus read the solutions straight off the graph (up to error due to imprecision of the graph image). Algebraist 15:24, 29 April 2008 (UTC)[reply]

Is that really all to it. Theres nothing else, poor thats simple. —Preceding unsigned comment added by 71.225.190.113 (talk) 14:28, 30 April 2008 (UTC)[reply]

3 way intersection flow equation

I'm trying to locate an equation for determining the total bidirectional flow on one leg of a 3-way intersection when the flow both ways is known for the other two legs. Any help appreciated. 71.100.11.39 (talk) 23:49, 29 April 2008 (UTC) [reply]

BTotal = SQT((CSouth-AEast)^2) + SQT((AWest-CNorth)^2) 71.100.11.39 (talk) 04:10, 30 April 2008 (UTC) [reply]

I don't think you're quite "done" here, unfortunately. Evaluating SQT(x^2) just gives you the absolute value of x; you may have wanted that, but since you didn't use an "abs" function I'm suspicious. Moreover, I don't think your problem is well-posed. Consider the case where there is no traffic on the third road at all (and no U-turns): then

a_{1}=d_{2},a_{2}=d_{1},a_{3}=d_{3}=0

(where the names mean arriving and departing and the numbers pick a road). But we could also have

d_{3}=a_{1}=a_{3}=d_{2},a_{2}=d_{1}

which corresponds to (at least the possibility of) everyone arriving on road 1 turning onto road 3, and an equal number of people arriving on road 3 and turning onto road 2. In both cases I could set, say,

a_{1}=d_{2}=10,a_{2}=d_{1}=5

, so you can't tell me even

a_{3}+d_{3}

based only on the values of

a_{1},d_{1},a_{2},d_{2}

. --Tardis (talk) 15:31, 30 April 2008 (UTC)[reply]

71.100.11.39 (talk) 00:16, 1 May 2008 (UTC) [reply]

Yes, that's the answer you gave before. What's your reasoning? Even if that is a correct answer, it isn't unique as Tardis has explained. --Tango (talk) 09:36, 1 May 2008 (UTC)[reply]

Well... The City's engineering department was asked to do a study to determine traffic volume on a street which was one of two inlet/outlets for a cul-de-sac. The purpose of the study was to determine volume to see if it was sufficient enough to open a street that crossed a drainage ditch which is grown up and no one cares about - about 30 feet of conduit pipe, backfilled and paved. Instead of placing the recorder on the actual leg that serves as the inlet/outlet they put recorders on the two legs the inlet/outlet leg runs into. Not only did they do only two legs of the 3 way they placed the recorders out on different weeks and different days of the week.

They do not show this as the method they used to calculate the volume on the inlet/outlet leg and if you eliminate the cross traffic between A and C and add the total for both directions you get 958 which is only 4% short of the 1,000 vehicles per day used as the approval point for opening the cut through to relieve the volume. The tricky part is that 90% of the volume is between 7 AM and 10 PM which is 15 hour not 24. Even without this reason for higher volume per hour there is only a ~3.4 second difference in the vehicle per 86.4 second interval between vehicles for the whole 24 hour period. We think the 1,000 vehicle per 24 hour approval point should be a per hour rate based on a normal distribution and based on a percentile rather than the mean but aside from this we think they are screwing with the process or the numbers under someone's order who does not want to open the street and provide the residents with fast ambulance service using the cut through. The cul-de-sac is filled with old people. We see ambulances three times a week, etc. No formula proves their determination is in error and will give us grounds for appeal. 71.100.11.39 (talk) 12:43, 1 May 2008 (UTC) [reply]

I don't know if this will help you, but "1000 vehicles/day" is a ridiculous criterion for judging congestion. If the cars arrive uniformly, that's less than one per minute and the road could very well look deserted. But if the 1000 cars are 500 in and then out of, say, an opera house, they'll all want to arrive and leave at the same time, so additional paths would be very well recommended. --Tardis (talk) 15:41, 1 May 2008 (UTC)[reply]

That's what you get for not consulting a statistician before starting the study. I don't know why, but people seem to assume that you don't need any kind of expertise to do a statistical analysis. (Just a minor point: If there are 2 ways in and out, it's not a cul-de-sac, surely?) --Tango (talk) 18:01, 1 May 2008 (UTC)[reply]

It depends upon the distance the two inlet/outlet streets are apart. The start and finish of a loop may be so close together or offer so much extraneous travel to a route that for all practical purposes they are the same inlet/outlet, hence only one road. 71.100.11.39 (talk) 20:49, 1 May 2008 (UTC) [reply]

Done?

So everyone then is agreed there is no formula to compute the third leg volume from the other two legs? 71.100.11.39 (talk) 20:47, 1 May 2008 (UTC) [reply]

Right. You can derive some lower and (if U-turns are not allowed) upper bounds from the condition that the number of cars entering the crossing should equal the number leaving it (and that the traffic flow cannot be negative on any lane), but those bounds will generally be so loose as to be practically useless. Sorry. —Ilmari Karonen (talk) 06:57, 2 May 2008 (UTC)[reply]

Deriving the upper and lower bounds is simply a matter of adding and subtracting the totals for the other legs. However, the problem that this presents is that the person conducting the study can then say the volume lies somewhere in between and since these values are 958 and 212 respectively anyone hearing this can quickly dismiss the whole idea of more frequent volume than one vehicle every 86.4 seconds, whereas this volume can be one car every 60 seconds if only 90 percent of the trigger volume occurs within the 15 hour period of greatest use which translates to a "wind chill" factor of 1,440 vehicles per 24 hour rather than just 1,000.

In any case the study needs to be based on a percentile of vehicles per hour rather than on 1,000 per 24 hours so I have sufficient grounds to demand a recorder be placed on the actual inlet/outlet, left there for several week days and a percentile of the 86.4 seconds per vehicle be used. 71.100.11.39 (talk) 15:03, 2 May 2008 (UTC) (UTC) [reply]