Wikipedia:Reference desk/Archives/Mathematics/2010 April 4

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April 4

When will things in the U.S. seem to have been improving the most? (on Humanities)

Just after I posted Wikipedia:Reference desk/Humanities#When will things in the U.S. seem to have been improving the most? I realized that it may be more of a pure math question than a humanities (economics) question because of the way the assumptions are stated, so I am also posting a cross-disciplinary pointer to it here. 99.25.114.221 (talk) 00:03, 4 April 2010 (UTC)

Product of two differentials

Is the product of two differentials zero? Yakeyglee (talk) 05:27, 4 April 2010 (UTC)

No. What makes you think it would be? Theresa Knott | token threats 07:56, 4 April 2010 (UTC)

Depends on which sense you mean. Formally, product of differentials is the wedge product (search for the article differential form). They are in general not zero. If you mean the product of two infinitesmals is so small it must be zero, you should take some look at non-standard analysis. Money is tight (talk) 10:04, 4 April 2010 (UTC)

There are some versions of calculus with infinitesimals in which the square of any of the infinitesimals used is zero. This gives a simple way of finding certain derivatives:

${\frac {(x+dx)^{3}-x^{3}}{dx}}={\frac {3x^{2}\,dx+3x\,dx^{2}+dx^{3}}{dx}}={\frac {3x^{2}\,dx}{dx}}=3x^{2}$

That is of course not how it works in Robinson's system. I know how to construct Robinson's nonstandard reals; I don't know the specifics of systems like this one. Michael Hardy (talk) 17:24, 4 April 2010 (UTC)

Even in unformalized infinitesimal calculus, the product of differentials figures into higher-order derivatives. Consider the second derivative $d^{2}y \over dx^{2}$ or the mixed partial derivative, $\partial ^{2}f \over \partial x\partial y$. I'd simplify ${\frac {3x^{2}\,dx+3x\,dx^{2}+dx^{3}}{dx}}$ as = $3x^{2}+3x\,dx+dx^{2}$ before taking dx→0. Note, I fixed a typo in your equation, you had an exponent 3 instead of 2 in the second fraction. 66.127.52.47 (talk) 18:26, 4 April 2010 (UTC)

Yeah, but in $d^{2}y \over dx^{2}$ aren't you technically not squaring anything? Isn't it just so you don't have to write really small if you have numerous derivatives of derivatives (i.e.- so you don't have to write ${\frac {d\left({\frac {d\left({\frac {dy}{dx}}\right)}{dx}}\right)}{dx}}$ for the third derivative of $y$) Yakeyglee (talk) 00:22, 6 April 2010 (UTC)

...which formally can be noted as if it was multiplication, and then, yes, shortened as raising to powers:
${\frac {d\left({\frac {d\left({\frac {dy}{dx}}\right)}{dx}}\right)}{dx}}={\frac {d}{dx}}\,{\frac {d}{dx}}\,{\frac {d}{dx}}\,y={\frac {d\,d\,d}{dx\,dx\,dx}}\,y={\frac {d^{3}}{dx^{3}}}\,y={\frac {d^{3}y}{dx^{3}}}$
CiaPan (talk) 14:49, 8 April 2010 (UTC)

One-way repeated measures ANOVA or something better?

So I have a dataset comprising of cell counts of a specific type of brain cells. I've counted these cells in various locations of the brain using serial sectioning, to determine a location gradient. I have done this for two different age-groups. Basically I want to see if there are any differences in number of cells per location and overall gradient between the two age groups. Would I be right in analysing this with a reapeated measures one-way ANOVA with the different locations being the repeated measurements, or is there a better way to do this?

Kind regards and thanks in advance! 82.21.153.228 (talk) 11:34, 4 April 2010 (UTC)

ANOVA assumes normality, while cell counts have a poisson distribution. Bo Jacoby (talk) 19:56, 4 April 2010 (UTC).

Countable closed partition of R

I'd like to gather different proofs of the following well-known fact (I guess there are lot of them):

• There is no partition of R into countably many closed sets.

How do you prove it? The way I do it is as follows: start from such a partition $\{C_{i}\}_{\in \mathbb {N} }$, reach a contradiction by exhibiting a continuous non-constant function $f:\mathbb {R} \to \mathbb {Q} .$ This function is defined inductively taking it identically 0 on $C_{0},$ and identically 1 on $C_{1};$ and for any k≥2, and any connected component J of $\mathbb {R} \setminus (C_{0}\cup ..\cup C_{k-1}),$ f is chosen to be constant on $J\cap C_{k},$ equal to the arithmetic mean of the (already defined) values of f on the boundary of J (which consists of one or two points of $C_{0}\cup ..\cup C_{k-1}$). I have in mind anoter couple of possibile proofs (e.g. exploiting the Baire category theorem), but they seem to be delicate. --pma 15:13, 4 April 2010 (UTC) PS: To me countable means countably infinite.

You may wish to add extra conditions to exclude (R,Ø,Ø,Ø,...) as a solution. –Henning Makholm (talk) 23:50, 4 April 2010 (UTC)

A partition of a set X is a set of nonempty subsets of X such that every element x in X is in exactly one of these subsets, so Henning's example is not a partition. It furthermore contains two sets, R and Ø. Is two considered countably many here? If so, then {R} is partition of R into countably many, namely 1, closed sets. Bo Jacoby (talk) 13:05, 5 April 2010 (UTC).

The word "countable" is used by some people to mean countably infinite and by others to mean finite or countably infinite. If it matters which you mean and isn't clear, you need to specify. The OP specifies that he means countably infinite. Henning is defining a partition as an n-tuple of subsets, rather than a set of subsets, which I think is non-standard. --Tango (talk) 15:24, 5 April 2010 (UTC)

Just add the word nontrivial before partition, where I'd say the one set partition is pretty clearly trivial. It doesn't matter if he means countably infinite or if countable means countable infinite or finite as long as you get rid of the trivial partition. The fact is, you can't partition the reals into finitely many closed sets either. And, this is somewhat trivial because the real numbers are connected. Any finite (> 1) partition into closed sets would contradict this. If I have n closed sets, then the union of n - 1 of them is closed and the remaining one is closed and those 2 sets (the union and then the remaining set) are a partition of the reals into nonempty closed sets, which implies the reals are not connected. So, the interesting case is the countably infinite case because a union of countably many closed sets is not necessarily closed. StatisticsMan (talk) 17:24, 5 April 2010 (UTC)

thanks to everybody --pma 16:37, 7 April 2010 (UTC)

Well, we didn't really give you anything. I'd ask on sci.math or something. I have asked questions there as well. There are lots of research mathematicians and such there and I'm sure if there are other proofs that someone there will know one or two. Basically, there are just a lot more question answerers there so a lot better chance of getting an answer to such a question. StatisticsMan (talk) 19:33, 7 April 2010 (UTC)

yes, but it is not urgent, and what is important after all is talking together ;) Thanks for the link also--pma 20:47, 7 April 2010 (UTC)

Invalid Triangle?

Why is wolfram alpha drawing an invalid triangle with this query? http://www.wolframalpha.com/input/?i=triangle+%7C+vertex+coordinates%28A{-28%2C24}%2C+B{-44%2C14}%2C+C{84%2C-38}%29

without letters preceding co ordinates, it's fine. Can I rearrange the namings so that it works?

PerfectProposal 16:59, 4 April 2010 (UTC)

I don't understand the syntax used, but it appears to me that Alpha is interpreting your input as six points, not three. —Bkell (talk) 17:18, 4 April 2010 (UTC)

It doesn't understand that you are labeling the points. So, delete the A, B, C

http://www.wolframalpha.com/input/?i=triangle+%7C+vertex+coordinates({-28%2C24}%2C+{-44%2C14}%2C+{84%2C-38})

StatisticsMan (talk) 18:09, 4 April 2010 (UTC)

Why do you want to name the points? There doesn't seem to be a way of doing that in Wolfram Alpha. --Tango (talk) 18:49, 4 April 2010 (UTC)

Interesting Calculus Examples (integrals)

So, I am teaching Calc 1 and I have just been thinking about interesting examples to give my students for integrals specifically. At this point, they have not done the chapter on "Techniques of Integration" so they only know substitution really and no other techniques. Do you know any interesting examples that could be enlightening to students? This depends on what you consider interesting and that is fine. One example I think of as interesting is $\int \sec ^{8}x\tan x\,dx$ and the reason is you can use more than one choice for your substitution. $u=\sec x$ is obvious but $u=\sec ^{2}x$ also works. This is interesting to me because it shows students that the $u$ does not have to be one specific thing. This sort of backs up that when you pick a $u$, you are basically just guessing and seeing if it works out. StatisticsMan (talk) 18:20, 4 April 2010 (UTC)

There being a choice of substitution is very common. In fact, there is always a choice, just sometimes only one of them is sensible. If u=f(x) works then so will u=f(x)+1 (at least it should, it haven't tried to prove it in full generality). One choice of substitution I quite like is for integrals like $\int {\sqrt {x+1}}\,\mathrm {d} x$. There is the obvious substitution, u=x+1, but you can also go with u²=x+1 and get rid of ugly square root (the same idea applies to any integral with an n-th root in it, although obviously it is more complicate is the derivative of the radicand isn't constant, so you can make the example as complicated as you wish). Proving the formulae with integrals being inverse trig functions is quite interesting too (integral of 1/sqrt(a²-x²), etc.). --Tango (talk) 00:22, 5 April 2010 (UTC)

I remember being surprised in integration by parts, to find that sometimes you end up with what you started with plus some constant terms, and then have an equation you can solve. The example

$\int \sin(x)e^{x}\,dx$

from Calculus/Integration techniques/Integration by Parts shows what I mean. 66.127.52.47 (talk) 04:47, 5 April 2010 (UTC)

That's not much good before they've learned integration by parts, though. --Tango (talk) 15:26, 5 April 2010 (UTC)

In the original example, $u=\tan x$ also works and it's a bit of a challenge to show, as predicted by theory, that the antiderivatives generated by the two methods differ by a constant. In my experience though, students have a very different definition of 'interesting' than math professors.--RDBury (talk) 08:42, 5 April 2010 (UTC)

Oh yea, good point, by converting the extra secants into tangents. That makes it even interestinger, or more interesting if you want. StatisticsMan (talk) 15:58, 5 April 2010 (UTC)

I found it interesting that $u=\sin x$ as well as $u=\cos x$ works for

$\int \sin x\cos x\,dx$

or $u=\tan x$ as well as $u=\sec x$ work in

$\int \tan x\sec ^{2}x\,dx$

In the latter example, students are puzzled for a while how two different substitutions lead to "different" answers,

${\frac {1}{2}}\tan ^{2}x+C$ and ${\frac {1}{2}}\sec ^{2}x+C$

(Igny (talk) 03:27, 7 April 2010 (UTC))