Wikipedia:Reference desk/Archives/Mathematics/2008 December 13

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December 13

Is PlanetMath a valid source?

Does PlanetMath (including papers and expositions) count as a valid source, in terms of not being considered original research?
-Dee ♥ Dee   74.10.197.201 (talk) 17:13, 3 December 2008 (UTC)

We do have quite a few articles that use material from PlanetMath, for example with the {{planetmath}} template. But in general PlanetMath is not a good reference for a Wikipedia article, since the articles there are so terse and often don't include references to the broader literature. A good textbook or research paper is a much better reference for a Wikipedia article. — Carl (CBM · talk) 18:28, 3 December 2008 (UTC)

Also take note of the PlanetMath Exchange GromXXVII (talk) 16:26, 6 December 2008 (UTC)

I don't mean create a Wikipedia article from a PlanetMath article or research paper, I just mean use it as a reference for a nugget of information. For instance, say it is accepted that "A+B=C" and an article or paper in PlanetMath claims that also "A+B=J*K" and shows how it is so. That would be considered unacceptable as "original research" if it was claimed by an author on Wikipedia with no verifying outside source. What I want to know is if I can present "A+B=J*K" as a piece of information in a Wikipedia article (not (necessarily) that the whole article is about it), with the PlanetMath paper being a valid source?
-Dee ♥ Dee  17:23, 13 December 2008 (UTC)

I guess so if you're very certain about the result but I'd be very careful because people can write all sorts of rubbish there and be very sure of themselves even though it's nonsense they're spouting. A bit like Wikipedia really. Really you should try and find some journal or book but it might be a good place to just get some ideas. Try out Google scholar for instance instead, it might find you a better place to look for a citation. Dmcq (talk) 17:44, 13 December 2008 (UTC)

Continued fraction

Hi. I'm learning a little bit about continued fractions, and I worked through a theorem that says that the continued fraction $\left[a_{0};a_{1},a_{2},\ldots \right]$ converges if and only if the sum $\sum _{n=0}^{\infty }a_{n}$ diverges. This leads me to ask: to which real number does $\left[1;{\frac {1}{2}},{\frac {1}{3}},{\frac {1}{4}},\ldots \right]$ converge? I've googled a bit, but I can only find treatments of continued fractions where the elements are integers. I ran the fraction in a spreadsheet, and it converges to somewhere between 1.75191 and 1.75197.

Any ideas? -GTBacchus^(talk) 18:43, 13 December 2008 (UTC)

Computing the first few convergents and looking up the numerator (1, 3, 9, 45, 75, 175, 1225, 11025...) and denominator (1, 1, 7, 19, 53, 81, 823, 5359, ...) sequence in OEIS gives this and this. Fredrik Johansson 20:57, 13 December 2008 (UTC)

The answer is ${\frac {2}{\pi -2}}$. I found this by computing a 50-digit approximation of the limit and running PSLQ. If anyone is curious, I used gmpy, mpmath and essentially the following code:

from mpmath import limit, identify, mp, mpf, inf
from gmpy import mpq

# from http://lists.canonical.org/pipermail/kragen-hacks/2000-March/000235.html
def ratf(frac):
	if len(frac) == 0:
		return (1, 0)
	elif len(frac) == 1:
		return (frac[0], 1)
	else:
		remainder = frac[1:len(frac)]
		(num, denom) = ratf(remainder)
		return (frac[0] * num + denom, num)

def f(n):
    p, q = ratf([mpq(1,k) for k in range(1,n+1)])
    r = p/q
    return (mpf(int(r.numer())) / int(r.denom()))

mp.dps = 50
a = limit(lambda n: f(n)+f(n+1), inf, method='richardson')/2
identify(a, ['pi'])

Now, a proof would be nice... Fredrik Johansson 21:11, 13 December 2008 (UTC)

Wow, that's cool. You can use that answer - $\left[1;{\frac {1}{2}},{\frac {1}{3}},{\frac {1}{4}},\ldots \right]={\frac {2}{\pi -2}}$, and then calculate remainders obtained by leaving off inital elements:

$\left[{\frac {1}{2}};{\frac {1}{3}},{\frac {1}{4}},{\frac {1}{5}},\ldots \right]={\frac {\pi -2}{4-\pi }}$,

$\left[{\frac {1}{3}};{\frac {1}{4}},{\frac {1}{5}},{\frac {1}{6}},\ldots \right]={\frac {8-2\pi }{3\pi -8}}$,

$\left[{\frac {1}{4}};{\frac {1}{5}},{\frac {1}{6}},{\frac {1}{7}},\ldots \right]={\frac {9\pi -24}{32-9\pi }}$,

$\cdots$

I don't know what any of that means, but there are some appealing patterns in there... -GTBacchus^(talk) 21:48, 13 December 2008 (UTC)

Working backwards, the problem gets simpler if you add a leading 1 so as to consider [1; 1, 1/2, 1/3...]. Then the answer is $\pi /2$. This is now exactly the value of the Wallis product; it should be possible to show that the convergents are equal. Fredrik Johansson 21:59, 13 December 2008 (UTC)

Cool. Thank you for your replies. That continued fraction [1;1,1/2,1/3,...] is kind of a nice expression for π/2, although it doesn't converge quickly. I'm also kind of wondering about the continued fraction $\left[1;{\frac {1}{p_{1}}},{\frac {1}{p_{2}}},\ldots \right]$ where p_n is the n^th prime. I'll bet the convergence is slow for that one, because $\sum _{n=1}^{\infty }{\frac {1}{p_{n}}}$ goes to infinity like ln(ln(n)). -GTBacchus^(talk) 00:41, 15 December 2008 (UTC)

While the original continued fraction doesn't converge quickly, the error is very predictable and that can be exploited. In particular, Richardson extrapolation (as used above) practically converts logarithmic convergence to linear (every term roughly adds a digit). The prime continued fraction is harder to compute because the convergence is erratic, so standard extrapolation methods don't work. Fredrik Johansson 14:14, 15 December 2008 (UTC)

For a proof: as far as I know the main source of computated continued fractions, including this one, are suitable orthogonal polynomials. For the three terms recursive relation produces a continued fraction in the same way that the Fibonacci sequence gives the continued fraction for the golden number. I should think a little to give you a precise derivation for your CF this way, but as you are interested in the topic, you can have a look to Akhiezer's book on the classical moment problem, where everything is very clearly explained. PMajer --84.221.209.229 (talk) 19:02, 20 December 2008 (UTC)

Geometric shapes

A perfectly round object, in two-dimensions, is a "circle" ... in three dimensions, a "sphere". What is the correct math/geometry term for a three-dimensional object (say, an aspirin or a drink-coaster or a coin) ... where you essentially have a "two-dimensional circle" with additional depth? Thanks. (Joseph A. Spadaro (talk) 18:55, 13 December 2008 (UTC))

A (solid) cylinder. --Tango (talk) 18:58, 13 December 2008 (UTC)

Yes, and more specifically, a right circular cylinder (as opposed to an oblique cylinder or an elliptical, parabolic, or hyperbolic cylinder). StuRat (talk) 01:29, 14 December 2008 (UTC)

Of dimension 3 (and no, it won't be solid) because a 'two-dimensional circle' is a sphere in Euclidean 3-space and its product with a one-dimensional object is 3 dimensional (imbedded in Euclidean 4-space).

Topology Expert (talk) 22:05, 13 December 2008 (UTC)

The OP gives real world examples - a coin is quite definitely a solid cylinder. Yes, the OP is using technical terminology a little incorrectly (just replace "circle" by "disc" and "sphere" by "ball" throughout), that isn't a reason to intentionally misinterpret them. --Tango (talk) 23:10, 13 December 2008 (UTC)

I did not intentionally misinterpret him. I knew what he was referring to (your answer) but I gave the mathematically correct (based on his language) answer anyway.

Also, note that the OP (i.e., I) used quotation marks in describing the shapes ... conceding that these were not correct and accurate mathematical terms, but rather layman's terms. (Joseph A. Spadaro (talk) 23:47, 13 December 2008 (UTC))

Note that an aspirin isn't quite flat on the top and bottom, so isn't a cylinder. I'm not exactly sure what you'd call that shape, though. It's a solid of revolution, but that's still rather vague. StuRat (talk) 01:36, 14 December 2008 (UTC)

Thanks to all ... this was helpful. (Joseph A. Spadaro (talk) 00:16, 15 December 2008 (UTC))

Measure Theory questions

Hello, I am currently studying for my final in the first semester of a measure theory class. So, I have several easy questions which I maybe already know the answer to but want to make sure my understanding is correct since analysis is a subject where even when I think I am right, I am sometimes wrong. If someone smarter than me says I am right or explains it better, it will help my understanding.

1. Implicit in the definition of a Riemann integral is that the value of the integral is finite since the function must be a bounded function and the integral must be on a closed interval?

2. A proposition in Royden says "A bounded function f on [a, b] is Riemann integrable if and only if the set of points at which f is discontinuous has measure zero." But, doesn't f have to be bounded to be Riemann integrable? So, could this also be stated, "A function f on [a, b] is Riemann integrable if and only if f is bounded and the set of points at which f is discontinuous has measure zero."?

3. Would any integral (other than improper) learned in undergraduate calculus automatically be Riemann integrable by the analysis definition? I guess in calculus you define a Riemann integral as a limit of a sum of areas of rectangles each with the same width, whereas in the analysis definition, the widths also have to go to 0 (sort of since this is defined using sups and infs over finite partitions) but do not necessarily have to be the same width.

4. Is a continuous function on a closed interval automatically bounded? If not, it would have to be infinity at some point by the extreme value theorem.

5. Proposition 4.15 in Royden's book partially says if f and g are integrable over E and f is less than or equal to g a.e., then the integral of f is less than or equal to the integral of g over E. It does not say anything about if f = g a.e., but implied by this is the integrals are equal since if f = g a.e., f is less than or equal to g and greater than or equal to g a.e.

6. Same propsition also says "If A and B are disjoint measurable sets contained in E, then $\int _{A\cup B}f=\int _{A}f+\int _{B}f.$" Implicit in this statement is that if f is integrable on a measurable set E, then f is integrable over any measurable subset of E, correct?

Thanks for any help. StatisticsMan (talk) 19:33, 13 December 2008 (UTC)

I can answer #4. Any continuous function on a closed interval satisfies the conditions of the extreme value theorem, so it attains a maximum and minimum value, therefore it's bounded.

For #5, I'm not exactly sure what your question is, but if f=g a.e., then f-g=0 a.e., so the integral of f-g is 0, so the two integrals are equal. Also, your reasoning above seems right.

Your #1 sounds right to me, unless there's something I'm missing. -GTBacchus^(talk) 20:11, 13 December 2008 (UTC)

Question 1: Remember, intervals such as [0, +infinity) are also called closed intervals (link to to the article because there are different definitions of 'closed interval'). To avoid any confusion, any bounded function on a compact set in R should be Riemann integratable (note that this applies to any finite set for instance). A very important thing to remember here lies in your fifth question.

Question 2: OK. Any continuous function on [a,b] must be bounded because [a,b] is compact. This is a very important theorem that might be helpful to remember. In fact, if the set of points at which f is discontinuous, has measure 0, then f must be bounded (since ‘almost continuous’ implies bounded for a function on [a,b]; try to prove this yourself!).

Question 3: The purpose of mathematics (the first purpose at least) is to generalize ‘simple concepts’. For instance, as you will know, the Lebesgue measure generalizes the concept of volume. Therefore, the Lebesgue measure of a closed unit disk should be pi. Since integrals in analysis are generalizations of integrals in calculus, the definitions should agree at their intersection; that is, on the real line, the definitions should agree.

Question 4: As I have mentioned, a closed interval can be [0, +infinity) and in that case the answer is no. I encourage you to see compact space and note that any continuous real-valued function on such a space must be bounded.

Question 5: Correct! A good (an easy exercise) would be to verify (yourself) that if a function from a measure space to the real line is almost everywhere 0, then its Lebesgue integral must be 0 (hint: the countable union of sets of measure 0 is measure 0). I am sure that they will ask something similar in the final exam (these are standard).

Question 6: You don't really need this preposition to conclude that but it is true (remember 'measurable' is important here).

Hope this helps.

Topology Expert (talk) 22:00, 13 December 2008 (UTC)

TE, on your response to question 1, no, I don't agree that [0,+∞) is standardly considered a closed interval. It is a closed set, and also an interval, but it is not a closed interval, which ordinarily means an interval of the form [a,b].

On an unrelated note, please stop de-indenting all your comments to the far left. In your response above, it was fine, because it stretched over several long paragraphs, but the way you do it habitually is quite irritating and makes the discussions harder to read. --Trovatore (talk) 22:06, 13 December 2008 (UTC)

That is what I thought too. But in modern day mathematics a 'closed interval' is an interval that is closed (think about it: if an interval is closed (equivalently a closed interval), you are referring to a topology so?). Better (notice the indent)?Topology Expert (talk) 22:15, 13 December 2008 (UTC)

It's not a given that you are referring to a topology, nor that a closed interval is the same as an interval that's closed (closed has lots and lots of meanings). But, assuming you are using closed in a topological sense, given that you're referring to +∞ at all, you might well have in mind a topology in which +∞ is an accumulation point of [a,+∞), and in that case [a,+∞) is not in fact closed. --Trovatore (talk) 22:28, 13 December 2008 (UTC)

Let us just not argue about it. I am stubborn about this and will stick to my definition. Topology Expert (talk) 22:30, 13 December 2008 (UTC)

Well, you had better explain it when you use it this way, as you will not be understood otherwise. In standard usage [a,+∞) is not a closed interval. --Trovatore (talk) 22:39, 13 December 2008 (UTC)

Thanks for the responses everyone. Speaking of topology, I'm in that class as well and we are using Munkres book, Topology. He calls what I meant by closed intervals, closed intervals, which are [a, b]. And, he calls $[a,\infty )$ closed rays, noting that in the order topology they may end up being closed intervals if the topology has a maximum element but otherwise it means everything greater than or equal to a (and similar definitions for open interval and open rays and such). This distinction is important the way he uses them in that the open intervals form a basis and the open rays form a subbasis for the order topology.

But, speaking of my number 6, if a function is measurable on a set E, is it also measurable on any measurable subset of E? Thanks again for your help. StatisticsMan (talk) 22:49, 13 December 2008

(UTC)

I disagree with your answer to Q2 - the function f(x)=1/x with f(0)=0 is unbounded on any interval which includes 0, yet is only discontinuous at one point (which clearly has zero measure). --Tango (talk) 22:58, 13 December 2008 (UTC)

I accept that I was incorrect for Q2. Thankyou, Tango for spotting this.

Yes. If a function is measurable on a set E. then it must be measurable on any subset of E (formally speaking, its restriction to any measurable subset of E is measurable). This is very similar to the analagous case of a continuous function between topological spaces but in this case, it does not matter what subspace of E you choose.

Topology Expert (talk) 09:47, 14 December 2008 (UTC)

About closed intervals; that exactly how I called them too until Oded told me otherwise (and gave a reference). I think that the definition of a closed interval is not exactly agreed upon in the mathematical community. Topology Expert (talk) 09:52, 14 December 2008 (UTC)

Statistics Man: Could you please tell us what particular concepts are taught in that course? I am happy to help (even in topology if you have any questions) (I am a bit bored with the reference desk questions nowadays, you see so I am happy to answer any question that are a little less 'trivial' (even if it is basic topology or measure theory)). Topology Expert (talk) 09:56, 14 December 2008 (UTC)

Sure, this is the first semester of a two semester course and I believe we are just going straight through Royden's Real Analysis. So far, we did the review chapters, Chapter 3 is on measure theory where we only did Lebesgue measure (I think next semester we will deal with other measures). Chapter 4 is on defining the Lebesgue integral and talking a bit about how the Riemann integral has the same value if it is defined and that the improper Riemann integral may exist if it's not integrable but if it is integrable, then they also agree. Chapter 5 is all about the fundamental theorems of calculus and figuring out when the derivative and integral are inverses of each other so we also talk about bounded variation and absolute continuity and such. Chapter 6 is on L^p spaces including Minkowski's inequality, Holder's inequality, completeness of the L^p spaces and Riesz representation theorem. We also had exercises on l^p spaces and proved most of those for l^p as well. Chapter 7 is on metric spaces, which we have not quite finished. So, that's what we did this semester. We have not done abstract measure spaces. Next will be topology and later on we get to abstract measure spaces and such. I don't know much about it yet.

It just occurred that maybe you meant the topology course. I am not sure, I'll just tell you both. In topology, we are using Munkres' Topology and we went through Chapters 2-4 mostly. 2 is all the basics of topology, defining topology, basis, different types of topologies like order, product, metric, subspace, quotient. Chapter 3 is on connectedness, compactness, different types of compactness and local of most things. Chapter 4 is on the separation and countability axioms. So, we talked about Hausdorff, regular, and normal where our definition is that 1 point sets are closed in all of these. And, we talked about first and second countability. And, we also mentioned a couple other of each type of axiom but didn't spend as much time on these.

Don't worry, I'm sure I'll be back with more questions. StatisticsMan (talk) 15:11, 14 December 2008 (UTC)

Thanks for the info. And just a tip for the final exam: I am assuming that it is a 'take-home' one and in that case they generally ask you to prove theorems that have similar proofs to those in the text. So it would be good to understand the idea of each proof very thoroughly. Topology Expert (talk) 15:44, 14 December 2008 (UTC)

Actually, the final is in-class, 2 hours. That's actually nice I believe. Our homework takes so long every time that I feel I don't have time to study. A take-home final would do the same. Now, I have time to read through every section and think about things and ask questions and try to understand everything well. StatisticsMan (talk) 22:50, 14 December 2008 (UTC)