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September 30

Bin packing problem

Hi all,

for my webshop I need some kind of algorithm, which given the size and weight limits of postal packages and the sizes/weights determines the best (cheapest) way to distribute the goods into postal packages. Does anyone know a solution for that problem? The stuff on google is filled with university-level math and allows only for one standard "bin" size... 85.181.101.47 (talk) 12:55, 30 September 2012 (UTC)[reply]

You're going to need to define the problem precisely. I assume more than one item is allowed per package, but is there a maximum number ? Do we have a table of package dimensions and shipping rates, and another of items dimensions ? Does weight matter ? Are there any rules about what must be in what packages (like refrigerated items in insulated packages) ? If the total number of items and packages is low, I suggest a brute-force method, where you try every combo to determine which is best. StuRat (talk) 16:23, 30 September 2012 (UTC)[reply]

Yes, there is more than 1 item per package, but no maximum limit (although there will be a practical one ;)); there is a table of package dimensions and the associated rates, and the list of items to be sent contains the size/weight data of each element. There's no rules about combinations of items and packages, and the orientation of goods doesn't matter in nearly all cases, so it can be left out. 85.181.101.47 (talk) 18:06, 30 September 2012 (UTC)[reply]

If they give weights for each item, that implies that there's a weight limit on each package to consider. Is there ? It would help if you could give us the exact wording of the problem, as slight variations in the setup can change what method is optimal. But, to outline a general method, you might want to sort items from largest to smallest, and place the largest item first, in the smallest package which will accommodate it, then place the second largest item in the smallest package in which it will fit, and continue through the list, trying to pack items along with other items as you go. This isn't guaranteed to give you the optimal solution, but should be a "good" one. If there's no maximum number, it may not be possible to always get the optimal solution, as calculating it would take an infinite amount of time. You might also want to create a hybrid program, which, when the number of items is low enough, does the brute-force calculations to find the optimal solution, but, when the number is too high for that, uses the method I just outlined. StuRat (talk) 04:00, 1 October 2012 (UTC)[reply]

Hard to give advice without an example of the table and typical items; I looked at http://pe.usps.com/text/dmm300/Notice123.htm but there are way too many options. If all your items come in boxes that should stay in pristine condition, and you want a way to stack them so the amount of empty space is minimised, I'm not sure if any algorithm can help you much with that. A computer controlled robot hand can stack things in order but filling a box by hand with say 40 rectangular items, based on a print-out of a 3-D arrangement seems quite challenging, and computing time might be excessive.

If you're sending items the way most Chinese webshops do it, everything wrapped in plastic foil, where the sum of the volumes is a good enough measure to know if they would fit a box, then it becomes more a knapsack problem (for volume or weight).

Usually price per unit of volume or weight is lower for larger parcels, so you'd want to use as few parcels as possible.

One possible approach would be: Start with the total volume plus (or minus) a certain percentage based on experience, find cheapest combination of parcels for the volume needed, check if largest items will fit in largest parcel (for long thin items, putting them diagonally might be an option), then use a knapsack algorithm based on the volume to fill the parcels, starting with smallest parcel (only using items whose length would fit). Verify if weight limits are respected.

In it's most general, it's a bounded multiple constraints (with m=2) multiple knapsack problem (and that doesn't include the item length), but the details would determine the best practical way to go about it. Sites like paypal offer solutions for calculating shipping costs but you'd have to integrate their system in your website I suppose. Ssscienccce (talk) 09:16, 1 October 2012 (UTC)[reply]

October 1

http://en.wikipedia.org/wiki/Ellipse#General_polar_form

In the section of this article "General polar form", the equaitons are given for expressing the polar form of the equation for an ellipse if we know a few parameters like the semi major and semi minor axes etc. To this end, the equation is given defining Q(theta). This equation contains the square root of the difference of two terms. Those terms are R(theta) and another term that varies independently of R(theta).

R(theta) is itself the sum of two terms, one of which includes (b**2 - a**2) and the other is (a**2 + b**2).

Since a and b can be arbitrarily close to one another, the first term involving (b**2 - a**2) can be arbitrarily small. also, a and b can be arbitrarily small, so that (a**2 + b**2) can be arbitrarily small. Consequently, R(theta) can be arbitrarily small.

Returning our attention to the square root term above, it seems that the argument under the square root can become negative since R(theta) can be arbitrarily small and the term subtracted from it can be positive and larger than R(theta). But then the square root is not defined, at least in terms of real numbers.

This leads me to wonder if there is some mistake in the given equations, since in physical reality the ellipse exists in the real plane, nd we should not expect to encounter negative square roots.

I am not expert in this area, so I might misunderstand and perhaps there is no problem, but I cannot seem to resolve this. I cannot seem to find other articles the cite the same equations (although I will keep looking). I would appreciate any clarification of any confusion on my part, or correction of a typo etc in the article. — Preceding unsigned comment added by 68.98.184.146 (talk) 00:17, 1 October 2012 (UTC)[reply]

I haven't verified the equations in detail, but, in all cases where the origin does not lie inside the ellipse, it is inevitable that r will have no real values for certain values of theta. 81.159.110.168 (talk) 01:42, 1 October 2012 (UTC)[reply]

That's right. You'll notice that if we put a plus minus sign before the radical, the number of real solutions for r(theta) equals the number of real solutions for the square root in the expression for Q(theta) -- namely either zero or two, or one in a razor's edge case. If the origin is outside the ellipse and you draw a ray from the origin, it either does not intersect the ellipse (no real solution; expression under the radical is negative), or it intersects the ellipse twice (expression under the radical is positive, giving two real solutions), or it has a single tangency point with the ellipse (expression under the radical is zero, so the number of distinct real roots is one). On the other hand, if the origin is inside the ellipse, and you allow the ray to have positive or negative length, then there are two intersections of the ray (actually line in both directions) with the ellipse; presumably the formula always gives real values for the square root when the origin is inside, and presumably the formula for r(theta) then gives one positive and one negative solution.

That section could use a drawing for clarity. The best I could find was in angular diameter

The angles the two tangent lines make with the horizontal are the minimum and maximum value for θ, outside of that, the value under the square root is negative. Ssscienccce (talk) 21:51, 1 October 2012 (UTC)[reply]

Like the previous respondent, I haven't verified the equations, but they appear to be consistent with the above analysis. I'll tag that sub-section as citation-needed. Duoduoduo (talk) 15:06, 1 October 2012 (UTC)[reply]

Complex functions

Can someone give an explicit 1:1 conformal map from the square to the half disk? Money is tight (talk) 08:06, 1 October 2012 (UTC)[reply]

I'd guess "no". A conformal map is supposed to preserve angles between crossing lines in the plane. The corners of the square can be thought of as the intersection point of two orthogonal straight lines (restricted to the square) crossing each other at an angle of 90 degrees. The images of these lines under the presumed conformal map (wich must be parts of the circle with a common point on the circle) can't possibly cross at 90 degrees. (It would be 180 degrees because its a circle.)

But, there is a homeomorphism (continuous and continuously invertible map) from the filled square in the real 2-dimensional plane to the filled circle. Divide each vector in the filled square with its length. YohanN7 (talk) 09:25, 1 October 2012 (UTC)[reply]

There is a map, it's guaranteed by the Riemann mapping theorem. But the proof is non constructive and I need an explicit function. Money is tight (talk) 11:20, 1 October 2012 (UTC)[reply]

I guess I was wrong then. But you must probably replace "square" by "interior of the square", and likewise for the circle. Better: use "open disk". Edit: "Open" is in the hypothesis of the Riemann mapping theorem.

I now recall something vaguely similar from a complex analysis text. [Churchill & Brown?]. I'll have a look later this evening. YohanN7 (talk) 11:46, 1 October 2012 (UTC)[reply]

The maps you need are discussed at http://www.encyclopediaofmath.org/index.php/Conformal_mapping (examples 5 and 9). Both are of the Schwarz–Christoffel mapping type (a digon and a tetragon), and the upper half-plane to square is given in our article here "explicitly". A source is Zeev Nehari's book "Conformal mapping". —Kusma (t·c) 12:07, 1 October 2012 (UTC)[reply]

You can get an idea of why something simple doesn't work from a little puzzle you might like at The Problem of the Right Wiggly Triangles. Dmcq (talk) 12:28, 1 October 2012 (UTC)[reply]

I don't agree with the solution for nr 2, that's not a wiggly triangle, that's a wiggly pentagon! Ssscienccce (talk) 22:00, 1 October 2012 (UTC)[reply]

I believe Paul Daniels catchphrase at the Bunko Booth about covers that "You'll like this...not a lot, but you'll like it." ;-) Dmcq (talk) 23:40, 1 October 2012 (UTC)[reply]

YohanN7, yes I do mean open square and open half disk, sorry for not pointing that out. Kusma, in Schwarz–Christoffel mapping which specific square does the half plane get mapped to (i.e. what are the coordinates of the vertices)? The integral is very difficult to evaluate. Money is tight (talk) 00:46, 2 October 2012 (UTC)[reply]

You might be able figure this out using the Jacobi elliptic functions, which map a rectangle into the upper half plane. But I have no experience doing this, I can just point you to the literature :( —Kusma (t·c) 10:12, 2 October 2012 (UTC)[reply]

October 2

Time derivative of the anomalies

Are there three types of angular velocity in motion on a ellipse, each corresponding to the time derivative of mean anomaly, true anomaly and eccentric anomaly?--188.26.22.131 (talk) 12:41, 2 October 2012 (UTC)[reply]

Yes. See Kepler's laws of planetary motion. Bo Jacoby (talk) 13:13, 2 October 2012 (UTC).[reply]

That shows how you use them to calculate the motion, but I'm not sure that's what the OP is looking for. Or maybe I just don't see it... Ssscienccce (talk) 17:35, 2 October 2012 (UTC)[reply]

The mean anomaly is

M=nt={\frac {2\pi }{P}}t.

So the mean motion is

{\frac {dM}{dt}}=n={\frac {2\pi }{P}}

where P is the period.

The eccentric anomaly E satisfies

M=E-\varepsilon \cdot \sin E.

So

{\frac {dE}{dt}}={\frac {n}{1-\varepsilon \cdot \cos E}}.

The true anomaly θ satisfies the equation

\tan({\frac {\theta }{2}})={\sqrt {\frac {1+\varepsilon }{1-\varepsilon }}}\cdot \tan({\frac {E}{2}})

Differentiate to obtain an equation for ${\frac {d\theta }{dt}}.$ Bo Jacoby (talk) 06:35, 4 October 2012 (UTC).[reply]

How much mathematics is essential?

My question comes down to what math should cover- is math the fundamental operations of addition, subtraction, division, and multiplication, coupled in applications such as equations etc.

And is calculus, differential equations etc. mostly experimentally derived workings that are useful in other subjects such as physics and finances?

So if i were to pick up a calculus book, most of the operations will be in the form of subtraction, division, miltiplication and division and just mostly operations of what we know? I know in linear algebra, there are directional shifts in matrices in regards to division, subtraction, addition, multiplication; but what is the real thing I should I know? — Preceding unsigned comment added by 108.50.249.125 (talk) 16:58, 2 October 2012 (UTC)[reply]

Know for what purposes ? If you get a job which doesn't require additional math skills, then the core set that everyone needs is addition, subtraction, multiplication, division, fractions, conversions, percentages, and interest rate calculations (using a formula). Some basic geometry, like calculating areas and volumes, would also be useful. I've never found the need to use algebra, trigonometry or calculus in "real life" (outside of work). StuRat (talk) 18:23, 2 October 2012 (UTC)[reply]

And I use all three routinely in my hobbies. —Tamfang (talk) 02:27, 3 October 2012 (UTC)[reply]

http://xkcd.com/1050/ . Just sayin'. --Trovatore (talk) 02:36, 3 October 2012 (UTC)[reply]

Calculus generally involves integration and differentiation which are more advanced mathematical operations than the arithmetic ones you have listed. Understanding those more advanced operations is essential to much of physics and most forms of engineering, though many people lead perfectly happy and successful lives while having no idea what calculus is about. Any decent introductory calculus text can teach you integration and differentiation, but these are not concepts that any student would be expected to already know before reaching calculus. So, in that sense, calculus is certainly not something that people already know simply because they have learned arithmetic. As others have said, what you "should" know can't really be defined without some idea of what you want to be able to do. Dragons flight (talk) 11:07, 3 October 2012 (UTC)[reply]

I agree with StuRat that his list is a good list of what everyone needs to know about math. And I agree with Tamfang that for many people, life can be immeasurably enriched by hobbies that use algebra, trig, and calculus. It's not clear to me what the context of your (the OP's) question is -- if you are a high school or college student wondering what areas of math to expose yourself to, one thing you should know is that the more math you learn, the higher-paying job you'll probably be able to get, the more your employment is likely to be shielded from economic recessions, and the more comfortable your life is likely to be. Duoduoduo (talk) 14:58, 3 October 2012 (UTC)[reply]

Another possibility for the list: I think that knowing some basic statistics is also extremely important for "real life", at least up to the point where you can comprehend that there are better methods of judging hypotheses than recounting a few anecdotes ("I took <insert quack cure> and my <insert malady> got better!"), but that these methods are not infallible ("Well, those 'scientists' were wrong about X, so why should I listen to what they say about Y?"). Though, of course, many would argue that statistics isn't a subset of maths. I think you could make the case that a bit of exposure to how "real" maths works (looking for patterns, proof, abstraction, etc.) is also important.

As to the question about calculus being "experimentally derived workings that are useful in other subjects", it is true that much of calculus was developed for the purpose of solving problems in other fields, and initially in a very informal way, but the advent of mathematical analysis put calculus on a firm theoretical footing and led to lots of beautiful mathematics. 130.88.99.231 (talk) 13:08, 4 October 2012 (UTC)[reply]

I agree that some basic knowledge of statistics is valuable. However, it's not the math that everyone needs to know (calculating standard deviations and such). What people need to know about statistics, they don't seem to teach in math class. There's how to ask an unbiased question (not "In the US Presidential election, do you intend to vote for the Muslim candidate or the Christian ?"), the importance of having a high level of participation among the people you poll (so no mailing out surveys, getting 1% back, and drawing conclusions from that 1%), the importance of polling a sample representative of the population (no surveys at country clubs), etc. StuRat (talk) 04:31, 6 October 2012 (UTC)[reply]

Calculus and differential equations

How do both relate? Are differential equations a part of Calculus? Are differential equations based on Calculus? Are differential equations related to, but independent of Calculus? — Preceding unsigned comment added by 80.58.205.34 (talk) 16:59, 2 October 2012 (UTC)[reply]

Differential equations are part of calculus, but they are ordinarily taught in a separate course, which students take after Calculus I and Calculus II. In other words, they are an advanced topic in calculus. Looie496 (talk) 19:53, 2 October 2012 (UTC)[reply]

Integral problems are a subset of differential equations, in particular those of the type

y'(x)=g(x)

, so that the solution is

y(x)=\int g(x)dx,

, and the main challenge is in solving the integral. In a generic differential equation, the left hand side is more complex. Sjakkalle (Check!) 20:13, 2 October 2012 (UTC)[reply]

"There is no such subject as calculus." — Paul Halmos Sławomir Biały (talk) 20:28, 2 October 2012 (UTC)[reply]

It's all very well to deny the existence of calculus, but there are limits! Marnanel (talk) 20:38, 2 October 2012 (UTC)[reply]

I find your humor to be quite derivative. Sławomir Biały (talk) 21:36, 2 October 2012 (UTC)[reply]

I found it partially derivative, but I am not so anti-derivative as you seem to be. 86.176.211.202 (talk) 02:41, 3 October 2012 (UTC)[reply]

Fast test for non-factorable polynomials of a special form

I have an arbitrary degree polynomial where every coefficient is one. Like x^117+x^87+x^30+x^23+x^13+x^4+x^2+x+1. Is there a fast, straightforward method for determining the factorability of such polynomials? Sebastian Garth (talk) 19:44, 2 October 2012 (UTC)[reply]

Just as an example, x^7+x^5+x^4+x^3+x^2+1 = (x^2-x+1) (x^5+x^4+x^3+x^2+x+1) and in general x^2-x+1 times (x^k+x^(k-1)+...+x+1) is (x^(k+2)+x^k+x^(k-1)...x^3+x^2+1) so there is at least one factorable polynomial with only unit coefficients for any positive integer degree greater than two.Naraht (talk) 19:57, 2 October 2012 (UTC)[reply]

Similarly (x^4-x^3+x^2-x+1) times (x^k+x^(k-1)+...+x+1) gives a similar result.Naraht (talk) 20:22, 2 October 2012 (UTC)[reply]

You may want to start at http://www.math.sc.edu/~filaseta/irreduc.html and some of the papers referenced.Naraht (talk) 20:40, 2 October 2012 (UTC)[reply]

Thanks, yes that looks somewhat promising. Or how about this: is there an efficient method for detecting whether or not there is a real root? Or is that just as difficult as factoring? Sebastian Garth (talk) 01:01, 3 October 2012 (UTC)[reply]

I can see some cases where the existence of a real root is clear:

If the constant term is 0 there is obviously a real root at x = 0. So from now on let's assume the constant term is 1, so f(0) = 1.
If the leading term (highest degree) has odd degree then there is a real root < 0.
If there are more terms with odd degree than with even degree then f(-1) < 0 so there is a real root between -1 and 0.
If there are the same number of terms with odd degree as with even degree then there is a real root at x = -1.
At the other extreme, if all terms have even degree, or if there is only one term with odd degree and it is not the leading term, then there are no real roots.

So the ambiguous case is when the leading term has even degree and there are more terms with even degree than odd degeer, but at least two terms with odd degree. I can't think of a simple method of handling that case (at least, not without using a root finding algorithm). Gandalf61 (talk) 09:36, 3 October 2012 (UTC)[reply]

If you sort the terms and find that every odd power is bracketed both above and below by an even power (i.e. the 5th power in x¹² + x⁵ + x⁴ is bracketed), then one can group terms and conclude there are no real roots. That reduces Gandalf's unresolved case to polynomials having two or more consecutive odd powers. Actually, one can use the same argument to say that if every set of N consecutive odd powers is bracketed both above and below by at least N consecutive even powers, then there are no real roots. Dragons flight (talk) 10:47, 3 October 2012 (UTC)[reply]

The number of real roots of the polynomial can be computed easily using Sturm's theorem.—Emil J. 12:44, 3 October 2012 (UTC)[reply]

That article looks like it could *really* use an example to be more clear. Say counting the number of real roots on a quintic equation..Naraht (talk) 13:09, 3 October 2012 (UTC)[reply]

I've added this to the references of Sturm's Theorem: Baumol, William. Economic Dynamics, section "Sturm's Theorem". Not a web resource, but I remember it has a very good example--maybe you can find it in your library. Duoduoduo (talk) 15:08, 3 October 2012 (UTC)[reply]

I've added an example of using Sturm's Theorem here. Thanks for the suggestion, Naraht. Duoduoduo (talk) 20:01, 4 October 2012 (UTC)[reply]

See also Descartes' rule of signs, which can sometimes prove that there exists at least one real root. Duoduoduo (talk) 15:12, 3 October 2012 (UTC)[reply]

Back to the original question, on factorization: see Factorization of polynomials and the various wikilinks therein. Duoduoduo (talk) 15:36, 4 October 2012 (UTC)[reply]

October 3

Log- Linear Plot given the coordinates?

When logy is graphed as a function of x, a straight line results. Graph straight lines, each given by two points, on a log linear plot, and determine the functional relationship (The original x-y coordinates are given)

(x1,y1) = (0,5) and (x2,y2) = (3,1)

The correct answer is y= 5 * (0.58)^x Could someone tell me how to arrive at this answer? thank you! — Preceding unsigned comment added by 24.86.170.175 (talk) 00:41, 3 October 2012 (UTC)[reply]

Let's see. You start with:

y = a(b)^x

You plug in x_₁ for x and y_₁ for y:

5 = a(b)⁰ = a(1) = a

So, a = 5:

y = 5(b)^x

Now plug in x_₂ for x and y_₂ for y:

1 = 5(b)³

Divide both sides by 5:

0.2 = b³

Take the cube root of both sides:

0.5848 = b

Also, since you are dealing with logarithms, I suppose I should put that last "finding the cube root" step in terms of logs. Here you find the antilog of log(0.2)/3, or the antilog of -.233, to get 0.5848. StuRat (talk) 01:16, 3 October 2012 (UTC)[reply]

This process worked with this question but when i tried applying it to other questions it does not work any longer. For example, the same question, except with coordinates (-2,3) and (1,1). I first substituted the first set of coordinates into the equation to get 3b^2= a Then i substituted the second set of coordinates to get a = b. After a bit of algebra my answer was y = 1/3 * (1/3)^x. However, this is incorrect. I have spent some time trying to figure this out but I still do not understand. — Preceding unsigned comment added by 24.86.170.175 (talk) 01:40, 3 October 2012 (UTC)[reply]

The second given point tells you ab=1, not a=b. —Tamfang (talk) 02:26, 3 October 2012 (UTC)[reply]

Oh! that was a very silly mistake. Thank you to the both of you for helping — Preceding unsigned comment added by 24.86.170.175 (talk) 02:40, 3 October 2012 (UTC)[reply]

You're quite welcome. StuRat (talk) 02:57, 3 October 2012 (UTC)[reply]

Resolved

Generating function of a particular sequence

I have a sequence in n which is $\sum _{g=0}^{n}{\binom {g+m-1}{m-1}}z^{n-g}$ where m is a natural number and z is some indeterminate, and I'm trying to find a rational generating function for this sequence. In other words, a rational function f(x), such that when expressed as a power series,

$f(x)=\sum _{n=0}^{\infty }\sum _{g=0}^{n}{\binom {g+m-1}{m-1}}z^{n-g}x^{n}.$

With z = 1, this is just $f(x)=1/(1-x)^{m+1}$ , but the z term complicates things.

If it's helpful, one way to think about the sequence is that the nth element is equal to the sum of all monomials of degree n in m+1 variables with one variable evaluating to z and the rest to 1. Another way is that it counts the number of monomials of degree ≤ n in m variables but weighting the sum based on their degree. I'm not totally sure that the sequence does have a rational generating function, but I have some reasons to believe it should.

Anyone have any ideas? Rckrone (talk) 04:51, 3 October 2012 (UTC)[reply]

Ok now I'm feeling a bit silly. I spent a while on this earlier and got nowhere, and then 30 min after I post it here the answer occurs to me. It should just be

f(x)={\frac {1}{(1-x)^{m}(1-zx)}}

, right? Rckrone (talk) 05:24, 3 October 2012 (UTC)[reply]

Model theory

Does every (first order) proposition having a model, constitute a part - of a finite class of (first/second order) propositions that has a unique model (up to isomorphism) for the individuals? HOOTmag (talk) 16:23, 3 October 2012 (UTC)[reply]

Why trig doesn't work?

let say I have this triangle. Why I didn't get the correct answer when I used inversed of cosine or inversed of tangent or inversed of sine? I saw the answer key used cosine law. Why only cosine law gives the correct answer?Pendragon5 (talk) 22:42, 3 October 2012 (UTC)[reply]

This is not a right triangle, so none of the side-length ratios will be equal to the sine, cosine or tangent of the angle. However the law of cosines can be applied to any triangle (not just right triangles) so you can use it to compute cos θ. Rckrone (talk) 23:34, 3 October 2012 (UTC)[reply]

October 4

Finitely generated abelian groups

I'm stumped on what should be an easy problem. The book defines a free finitely generated abelian group as one with a basis, i.e., a linearly independent generating set. Then an exercise defines a torsion-free group in the usual way, and asks me to prove that a finitely generated abelian group that is torsion-free must be free. I know this follows from the fundamental theorem of finitely generated abelian groups, but I get the impression I'm not supposed to import a big gun to solve this problem.

Can I just start from the fact that I've got a finite generating set, add the fact that there's no torsion, and show that there must be a basis? I know I can't count on the basis being a subset of my generating set, because I've got a simple counter-example: the set {2,3} generates the integers, but neither {2} nor {3} does it alone. Help! Thanks in advance. -GTBacchus^(talk) 20:07, 4 October 2012 (UTC)[reply]

October 5

Mathematics degrees

Is there any stand alone degree of Algebra, Analysis, Combinatoric, Geometry? In the same way that you can study computational science independently. — Preceding unsigned comment added by 83.41.157.164 (talk) 09:29, 5 October 2012 (UTC)[reply]

Well, there are certainly lots of postgraduate degrees covering specific branches of mathematics. Googling gave me this course for a "Bachelor of Arts in Mathematical Sciences - Combinatorics and Optimization" at the University of Montana (Googling for "degree geometry" turns out not to be very helpful). You might need to be a bit more specific about what kind of degrees you are interested in, and which countries. 130.88.99.231 (talk) 17:54, 5 October 2012 (UTC)[reply]

Combinatorics and Optimization is an emphasis at Montana, not a stand-alone degree. From their website: The department offers a Bachelor's Degree with emphases in Pure Math, Applied Mathematics, Mathematics Education, Combinatorics and Optimization, Combined Mathematical Sciences-Computer Science and Statistics.' [my bolding]. I strongly doubt that any American undergraduate department offers such a specialized degree as Combinatorics and Optimization, Algebra, etc.--to fill up the required in-major credit hours would require courses that get so deep that they would not be undergraduate level. But some do offer separate degrees in Pure Mathematics and in Applied Mathematics and in Statistics. Duoduoduo (talk) 18:14, 5 October 2012 (UTC)[reply]

The University of Waterloo offers several specialized undergraduate majors in mathematics, including Combinatorics and Optimization. See [1]. —Bkell (talk) 21:02, 5 October 2012 (UTC)[reply]

Incompleteness theorem

Once when I was feeling brave and tried to understand Godel's incompleteness theorem, I thought the essential result was that it is possible to formulate syntactically valid statements (in some applicable system) that can be neither proved nor disproved. However, in an article I'm currently reading, it says that the theorem "revealed that mathematics must contain true but unprovable propositions", and again in the intro to the Wikipedia article "there will always be statements about the natural numbers that are true, but that are unprovable within the system".

How does the notion of "true" get involved in this? How do we know that the statements that cannot be proved or disproved are "true"? Why is the theorem described in those terms rather than in my original terms? 86.176.210.107 (talk) 20:28, 5 October 2012 (UTC)[reply]

The crucial point is "... within the system". Thus, we may be able to prove the truth of a statement that is expressible in a given formal system (under some semantic model), but is not a theorem of the system. Meaning that using the axioms and rules of the system, the statement is not provable, but reasoning outside the rules of the system is not constrained in the same way. Any formal system of symbols and rules has no inherent meaning; it requires a semantic model to match an interpretation (and hence what a string expresses as we would understand it). — Quondum 21:08, 5 October 2012 (UTC)[reply]

I see, thanks. Is there any reason why it is important to emphasise the existence of true statements that are not provably true (within some system), rather than any other combination, such as false statements not provably false, or statements that we can never know the truth of (in any way) not provable one way or the other (in some particular system)? 86.176.210.107 (talk) 21:17, 5 October 2012 (UTC)[reply]

I would phrase things somewhat differently from the way Quondam did. The point, which is often obscured in popular treatments, is that the Goedel sentence of a formal theory is an assertion about a very clear collection of objects, the natural numbers. If you agree that (i) there are such things as natural numbers and (ii) it is well-specified what they are, then you must agree that the Goedel sentence of (say) Peano arithmetic is either really true or really false.

The Goedel sentence of a consistent theory T is necessarily true. This is because what the sentence asserts is of the form "for every natural number n, n is not the Goedel number of a proof of (something complicated that we don't need to worry about the details of here) within T ", and if that were false, then there would be such a natural number n, and it would follow, by the arguments of the proof of the incompleteness theorem, that T was inconsistent.

The reason it is important to emphasize the truth of the Goedel sentence of a consistent theory is that, otherwise, it is easy to make the mistake that many popularizers make, and suppose that the "choice" between adding the Goedel sentence as an axiom, and adding its negation as an axiom, is a neutral one. It is not neutral; it is very clear — the Goedel sentence of a consistent theory is true. --Trovatore (talk) 21:27, 5 October 2012 (UTC)[reply]

(i) When you say "the Goedel sentence", you mean a constructed statement that is not provable or disprovable within T, right? (ii) How do we know that things are really "true" about the natural numbers without invoking some other system that itself suffers from the flaw (or incompleteness) that Godel exposed? 81.159.104.36 (talk) 22:55, 5 October 2012 (UTC)[reply]

For your (i), yes, but something a bit more specific than that. I mean specifically a sentence that encodes "I am not provable in T".

For your (ii), it is easy to show, assuming very little, that if T is consistent, then the Goedel sentence of T (call it G_T) is true. Now, T itself does not prove that G_T is true, but that is only because T also does not prove that T is consistent (this is the second incompleteness theorem, and in fact that's basically how it's proved).

So as to how we know that G_T is true — well, we know that only to the extent that we know that T is consistent, which is a whole 'nother discussion. Depending on what T is, perhaps we don't know that T is consistent. But that doesn't change the fact that the Goedel sentence of any consistent theory is true. --Trovatore (talk) 01:22, 6 October 2012 (UTC)[reply]

October 6

Root problem

Can someone explain for me how to do problem 3? I have the answer key but unable to understand what is it saying. I don't get how (p^2 - 1)/4 is the product of the roots. And the part where it say "which checks vs. the coefficient of the x term" --> What is this even mean? 65.128.190.136 (talk) 03:05, 6 October 2012 (UTC)[reply]

They are talking about simplifying this expression:

x² - px + (p²-1)/4

into these roots:

[x + (p+1)/2] [x + (p-1)/2]  (See correction below.)

The first step is to resolve:

(p²-1)/4

which can be rewritten as:

p²/4-1/4

into it's roots:

[(p/2)+(1/2)] [(p/2)-(1/2)]

or:

[(p+1)/2] [(p-1)/2]

You might recognize this as a difference of squares problem, where (a+b)(a-b) = a² - b². StuRat (talk) 03:35, 6 October 2012 (UTC)[reply]

Why when I multiplied this out. I got everything the same except x^2 + px instead of - px. How did they get -px? Can you show me? Thanks!65.128.190.136 (talk) 03:57, 6 October 2012 (UTC)[reply]

Oops, I had the signs wrong on the roots. Here's the corrected version:

[x - (p+1)/2] [x - (p-1)/2]

StuRat (talk) 04:11, 6 October 2012 (UTC)[reply]

Okie! I think everything is clear! Thanks!65.128.190.136 (talk) 04:20, 6 October 2012 (UTC)[reply]