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# July 4

## Prove that a polynomial in n variables and of arbitrary integer degree is continuous

How do I prove that a function P:RnR defined by a polynomial in n variables is everywhere continuous regardless of the highest degree (as long as they are all of integer degree)? I'm thinking about mathematical induction for any monomial (continuity of every single term suffices since the sum of continuous functions is also continuous). For n=1, it's simple because it's trivial to show that any power function (of integer degree) is differentiable (by the power rule), and thus continuous. But what about n>1? Since the partial derivatives' existence alone is not sufficient to guarantee differentiability, and thus continuity, it remains to be shown that all such partial derivatives must also be continuous. Those partial derivatives are also monomials in n variables, hence why mathematical induction comes to mind.--Jasper Deng (talk) 06:27, 4 July 2014 (UTC)

Haven't done this sort of stuff for ages (proving the bleeding obvious) and do not intend to do it now either, so don't take my word for anything. Await reliable replies.
That said, continuity of sums and products should take you far. I'd attack the "any degree" supposing that there is a maximum "continuous degree" and try to derive a contradiction before resorting to induction. Induction should still work. YohanN7 (talk) 07:14, 4 July 2014 (UTC)
Polynomials are built up from projections using sums and products. Those are all continuous. The result follows.--Antendren (talk) 10:39, 4 July 2014 (UTC)
I'd prove the following:
• Constant functions are continuous; that's all polynomials of degree zero.
• If a polynomial P(x1, ... , xn) is continuous, then x1 P(x1, ... , xn) + c, x2 P(x1, ... , xn) + c, ... and xn P(x1, ... , xn) + c are continuous, for all polynomials P and real numbers c.
• Induction. - ¡Ouch! (hurt me / more pain) 07:48, 8 July 2014 (UTC)
Credit where it's due: my method is basically Antendren's, in more words, and I didn't catch all polynomial either. The second step should read
• If a polynomial P(x1, ... , xn) is continuous, then x1 P(x1, ... , xn) + Q(x2, ... , xn) is continuous, for all polynomials Q over Rn-1.
• Two inductions are needed, one by degree and one by n. - ¡Ouch! (hurt me / more pain) 08:10, 8 July 2014 (UTC)

## Test for which is greater, average or product

Given a finite set of positive real values {n1, n2, n3, ...}, most of which are less than 2, is there an easy test to determine whether their arithmetic mean or product is greater, without actually fully calculating the arithmetic mean or product? —SeekingAnswers (reply) 07:08, 4 July 2014 (UTC)

If they are all less than one the arithmetic mean is greater and if they are all greater than one the product is greater. I'm not sure how much more wou;d qualify as not calculating the mean or product. Dmcq (talk) 20:22, 4 July 2014 (UTC)
Yes, but the difficulty is that some values are greater than 1 and some are less than 1. I am looking for some test, preferably one with a simple algorithm that can be done easily and quickly by a human without a calculator or computer, that is computationally less intensive than actually calculating the mean and product. —SeekingAnswers (reply) 20:53, 4 July 2014 (UTC)
Any rigorous test would probably involve as much calculation as is required for the actual mean and product, but a quick manual method is to pair off reciprocal values (e.g. pair 2.0 with 0.5 and pair 1.5 with 0.667), crossing off these pairs. If the remaining values are mainly more than unity then the product is probably greater (this isn't guaranteed because the average of the pairs will be around 1.1 for a rectangular distribution of values). If the remaining values are mainly less than unity then the arithmetic mean is almost certainly greater (assuming no exceptionally large outlier). The method is not infallible (someone will, no doubt, construct an unusual set of values for which it fails), but will be fairly reliable for values randomly distributed in the interval ]0, 2]. Dbfirs 07:26, 5 July 2014 (UTC)

## Set theory: Axiom of extensionality

In ZF, the axiom of extensionality - claims: "Two sets are equal iff they contain the same sets".

My question is about whether it's also provable (from ZF) that: "Two sets are equal iff they are contained in the same sets".

77.127.104.238 (talk) 09:14, 4 July 2014 (UTC)

Yes. The axiom of pairing gets us that for any set $X$, the set $\{X\}$ exists. If $A \neq B$, then $A$ is contained in $\{A\}$ but $B$ is not. For the converse you don't need any of ZFC; substitution using equality is a basic property of first-order logic.--Antendren (talk) 10:43, 4 July 2014 (UTC)
Thankxs. 77.127.104.238 (talk) 11:12, 4 July 2014 (UTC)

## New problem

What's the limit of $(1-\cos(x))/(x^2)$

As $x$ approaches 0?? Georgia guy (talk) 16:57, 4 July 2014 (UTC)

This is most easily done using the Taylor series of the cosine. —Kusma (t·c) 19:08, 4 July 2014 (UTC)
It's 1/2. Just plot the function in any graphing calculator or computer algebra system. —SeekingAnswers (reply) 19:11, 4 July 2014 (UTC)
I highly discourage this approach for finding the limit because appearances can be deceiving. It could be useful for getting an idea of what the limit could be, but plotting it does not rigorously find the limit.--Jasper Deng (talk) 20:49, 5 July 2014 (UTC)
It depends on the application. For some situations, an approximate limit would be fine. For example, if you were dealing with an open water tank, with water flowing in at a the top at a constant rate, and out at the bottom at a rate that varies with depth, you could develop a formula to tell you what the depth should be at each point in time, and the graphic limit of that function would be a perfectly acceptable estimate for how full the tank should get (since both the inflow and outflow rates are approximate, depending on whether you have laminar or turbulent flow, and other factors like evaporation would also play into it). So, the exact mathematical solution would only yield a false precision anyway, and there's more chance of a math error that way. StuRat (talk) 21:31, 5 July 2014 (UTC)
Graphing to find limits is not a good idea for more complicated, perverse functions, but for a very simple elementary trigonometic function like this, graphing is the fastest/easiest way that works in general (power series substitution, L'Hôpital's rule, etc., will only work for specific forms and can involve complicated manipulations in which it's easy to make a mistake). Just look at the overall shape of the function and use common sense. For simple functions consisting of basic trigonometic functions, it's obvious when there's no limit; examples include the vertical asymptotes of 1 / tan x and the rapid oscillatory behavior of sin (1 / x). The curve of (1 - cos x) / x^2 is very well-behaved, without any vertical asymptotes or rapid oscillatory behavior, and there's no doubt that the limit at 0 is 1/2. —SeekingAnswers (reply) 00:08, 6 July 2014 (UTC)
With limits, vagueness is not acceptable, hence why I discourage this. All too often, I've seen situations where folks get into trouble with the limit definition and finding the limit due to things as trivial as a graphing error. Common sense must be justified. While it may not seem such a big deal with "well-behaved" functions (a term that's anyways not well-defined in this context), I've found that it encourages the false notion that graphing and "seeing what the function approaches" is the way to find the limit.
In applications, approximating limits numerically, such as with a graph, may be appropriate. But not in an instructional context like this one.--Jasper Deng (talk) 02:11, 6 July 2014 (UTC)
The original poster did not ask how one finds the limit; he asked what is the limit. So this did strike me as a question originally of application rather than a question originally of instruction. —SeekingAnswers (reply) 03:18, 7 July 2014 (UTC)
I just interpret "what is the limit" as "what is the limit, and how?". That's all.--Jasper Deng (talk) 06:35, 7 July 2014 (UTC)
Apply L'Hôpital's rule twice. The answer is 1/2.Phoenixia1177 (talk) 20:56, 4 July 2014 (UTC)
To do this (more or less) from first principles, without invoking the heavy machinery of the Taylor series or L'Hôpital's rule, rewrite cos(x) in terms of sin(x/2) using the appropriate half-angle formula and use the fact that sin(y)/y -> 1 as y -> 0. AndrewWTaylor (talk) 18:37, 5 July 2014 (UTC)
Was that theorem obtained "without invoking the heavy machinery"? —Tamfang (talk) 09:14, 6 July 2014 (UTC)
You mean the theorem that sin(y)/y -> 1 as y -> 0? Yes, it can be proved by a fairly straightforward geometrical method; e.g. see this video. (Using Taylor or L'Hôpital is no good, as they both assume you know the derivatives of the trig functions, and that depends on this limit.) AndrewWTaylor (talk) 11:55, 6 July 2014 (UTC)
Plotting a graph may be a good starting point to get a hint about a possible value of the limit, and could also be a good way to cheek a calculation, but just this. --pma 16:52, 8 July 2014 (UTC)
'to cheek a calculation' I like it! CiaPan (talk) 08:09, 9 July 2014 (UTC)

# July 5

## How to categorize these images?

I recently uploaded these two images to supplement the four referenced in the Fundamental polygon article. What categories should I put them in? In particular, I don't know if they're actually fundamental polygons or not: it's been a while since I've proven that something is a closed surface. Jbeyerl (talk) 17:37, 5 July 2014 (UTC)

It's not definite, but I lean against those ending up being manifolds. The group representation for the first would be A2 = 1, and the second would be A2B−2 = 1. — Arthur Rubin (talk) 01:17, 8 July 2014 (UTC)
A2 = 1 is the projective plane. For the second, by cutting diagonally from the top left to the bottom right, then gluing along A, you get the standard presentation of the Klein bottle.--Antendren (talk) 04:42, 8 July 2014 (UTC)

# July 8

## How does this graph work? How is it possible?

This is the weirdest real-life graph I've ever seen: [[1]]
Source:
IMF Staff Discussions Note: Emerging Markets in Transition: Growth Prospects and challenges.
[2] (bottom of page 4)
— Preceding unsigned comment added by 14.200.130.252 (talk) 11:32, 8 July 2014 (UTC)

Graphical representations of parametric equations often look like that. Dolphin (t) 12:22, 8 July 2014 (UTC)
Thank you for that. It did help shed a bit of light and introduced me to paramtric vs non-parametric equations :) I guess there wasn't a proper function in this graph, it was just simply data points plotted and therefore the line we see in the graph is a bit misleading because we could join the 'dots' in any way we like. 14.200.130.252 (talk) 14:22, 8 July 2014 (UTC)
The dots are joined in chronological order. You can see the years marked next to each location in the graph. -- Meni Rosenfeld (talk) 15:45, 8 July 2014 (UTC)
That graph doesn't seem to show a case where one axis variable is strongly dependent on the other, which is what is most familiar (for example, population is strongly dependent on time). If you plot two largely independent values, the graph will take on a rather random look (for example, sound volume versus frequency in a piece of music). Such graphs seem less useful, to me. I'm not sure how you are supposed to use it to predict anything. StuRat (talk) 14:30, 8 July 2014 (UTC)
Such a graph can help you tell whether the variables are independent. —Tamfang (talk) 04:52, 9 July 2014 (UTC)
It isn't intended to predict anything; it simply provides an easily read display of how the two variables related to each other at different times. It's no different in principle than plotting the path followed by a vehicle or a person over time, like this map or this map... except that with a map, latitude and longitude are obviously related, whereas with a graph showing two functions over time, their relationship may not be obvious. — Preceding unsigned comment added by 50.100.189.160 (talk) 06:22, 9 July 2014 (UTC)

# July 10

## Lie algebras of U(p, q)

Hi!

Which, if any, is correct?

$\mathfrak{u}(p, q) = \left\{\left .\left(\begin{matrix}X_{p \times p} & Z_{p \times q} \\ -\overline{Z}^{\rm{T}} & Y_{q \times q}\end{matrix}\right)\right| \overline{X}^{\rm T} = -X,\quad \overline{Y}^{\rm T} = -Y\right\}.$
$\mathfrak{u}(p, q) = \left\{\left .\left(\begin{matrix}X_{p \times p} & Z_{p \times q} \\ +\overline{Z}^{\rm{T}} & Y_{q \times q}\end{matrix}\right)\right| \overline{X}^{\rm T} = -X,\quad \overline{Y}^{\rm T} = -Y\right\}.$

I have one opinion and my reference another about the +/- sign in the lower left. YohanN7 (talk) 21:29, 10 July 2014 (UTC)