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November 29

Ohh, Ohh, I'm the first person to use this

What would happen if you tried to take (d²fn(H,T))/(dT²) and expand it all the way to (d^$\infty$fn(H,T))/(dT^$\infty$), thanx --Aolanonawanabe 23:45, 29 November 2005 (UTC)[reply]

Can you give an example of the kind of function "fn" you have in mind? Or am I misunderstanding the question? Dmharvey 23:54, 29 November 2005 (UTC)[reply]

Conceptual, the identity of fn is just an arbitrary function--Aolanonawanabe 23:55, 29 November 2005 (UTC)[reply]

Well, it's a bit odd, because it looks like you're differentiating a function of x and y with respect to T. Dmharvey 00:33, 30 November 2005 (UTC)[reply]

Yes, yes, that wasn't well thought out, fixed now--Aolanonawanabe 00:39, 30 November 2005 (UTC)[reply]

We can take derivatives to any natural number order we like: first, second (as shown), third, and so on. We can also improvise some fractional derivatives. But even though it is quite common to speak of a function as being infinitely differentiable (also known as "smooth"), I don't recall seeing infinity itself (for some specific choice of infinity, such as ω) used as you proposed. Reasonable definitions could surely be given, but for a general function the result may not be well-defined. (And presumably you mean x and y to be functions of T?) --KSmrq^T 00:37, 30 November 2005 (UTC)[reply]

This seems like an interesting question. Why not define d^ωf/dx^ω? And why stop there, keep going to other ordinals! The problem as I see it is that such a thing probably wouldn't exist in most cases. A possible definition would be lim dⁿf/dxⁿ. Ignoring for the moment the question of which topological space to take that limit in, we can still see that D^ωe^x should equal e^x, and D^ωp(x) for any polynomial p should be 0 (something that wouldn't happen with any finite derivative). On the other hand, even D^ωe^2x would probably have to be undefined, and I think most other (even smooth) functions would be undefined as well (remember that you're raising to an infinite power an operator that is already unbounded in the usual Banach norm, for example). Anyway, I've never seen such a thing done, I just thought I would add my 2¢ anyway, for whatever their worth. Thanks for stopping by! -lethe ^talk 01:07, 30 November 2005 (UTC)[reply]

November 30

Very large numbers

i would like to know if their is a specific name for 13,700,000,000,000,000,000,000,000,000 or in scientific notation 1.37*10^28. yes, that is an absurdly large number, i know. That is the number of CO2 atoms in a metric ton. i know the number line runs quintillion,quadrillion,trillion,billion,million,thousand, hundreds. all i really need to know is what to call the 13.7________. i would like only serious answers, and i would prefer to have them soon. NO CRAP SHOTS, because you will be cited for a research project for my project. —preceding unsigned comment by 68.116.142.153 (talk • contribs) 01:28, 30 November 2005

Maybe you didn't see the warning about posting e-mail addresses here. Anyway, the answer depends on what system you're using. Under the system used in the United States, the answer would be "13.7 octillion". In the system traditionally used in the UK, it would be "13,700 quadrillion". However tendency worldwide nowdays seems to be moving towards the US system. --Trovatore 01:36, 30 November 2005 (UTC)[reply]

68.116.142.153, you seem to be new here. Except in rare circumstances, you're not supposed to take people's remarks off a discussion page (which this is all but technically). If you want to change your question, you should add new remarks saying "Oh, I really meant....", rather than trying to revise history. --Trovatore 01:48, 30 November 2005 (UTC)[reply]

Although formally it is true that according to US conventions the number could be named thirteen point seven octillion, in fact few people would be sure what you meant. If the purpose of language is effective communication, then you should stick with the true convention for this situation, scientific notation. If you want to convey to readers a sense of how large 10²⁸ is, then a common ploy is to use a counting or measuring analogy, such as "would reach to the moon and back 42 times" (or whatever). --KSmrq^T 01:58, 30 November 2005 (UTC)[reply]

well, i'll be using a few different ways to convey it, and unfortunatley, that wasn't the correct number, i would like to know any of the possible names for 2.07*10^31. i'm trying to convey how freakisly huge this number is to a class of moderatley low intelligence.

A number of 31 orders of magnitude larger than 2.07, how about that?--Aolanonawanabe 02:10, 30 November 2005 (UTC)[reply]

If this is for a research project, you should use it as an opportunity to hone your research skills. Instead of asking here, try a web search. Then use what you learn to obtain a definitive reference you can cite. If I tell you that in Germany the term for 10³⁰ is "Quintillion", how reliable is my information? Did I just hear from someone who heard it from someone who is guessing? Is there an offical document, some kind of German or European standard? Do all parts of the government, industry, and society follow the same convention? Also, an audience may be uneducated or misinformed or opinionated, yet still intelligent and interested and capable of learning. People tend to notice if you don't respect them; they return the favor. Is that what you want? Some extremely bright and educated people contribute to Wikipedia; should they regard you as of "moderatley [sic] low intelligence"? --KSmrq^T 03:38, 30 November 2005 (UTC)[reply]

Why don't you read Names of large numbers and do your own research? ;-) --hydnjo talk 04:34, 30 November 2005 (UTC)[reply]

well, when i did my searching intially, i was only able to find up to 10^18. thank you for being helpful, and giving me an answer to my question, and of telling me where to look, instead of just pointing out what i should do, with no real advice. thnx hydnjo.

Fractional brownian motion

Is there a method of determining whether a sample sequence of data is fractal brownian motion? --HappyCamper 01:32, 30 November 2005 (UTC)[reply]

Ooh ooh, I know this one. You take the Fourier transform and then take its absolute square to get the spectral density. If the spectral density is proportional to 1/f^2 (f is the frequency), it's brown noise, if it's proportional to 1/f it's pink noise, and so on with the other colors of noise. It's only a true fractal if the spectrum follows that function from 0 all the way to infinite frequency, which can't happen with digital data, but if it follows it pretty well out to the Nyquist frequency that's good evidence. —Keenan Pepper 14:27, 30 November 2005 (UTC)[reply]

Objects over the horizon

Okay, maybe this is the wrong page to ask, but seeming as I am virtually innumerate, I thought that this new Math Refdesk would be a good place to start. The question: How tall would an object need to be to be visible from 200 miles away, presuming a flat landscape, say, Nebraska? I know there's an equation to figure this out, but I don't trust myself to do it right. Thanks, gang. 205.145.64.64 21:04, 30 November 2005 (UTC)(actually Brian Schlosser42 21:07, 30 November 2005 (UTC), I forgot to log in.)[reply]

A quick geometry calculation makes it seem to me that the top of an object of height h is just visible distance s away, then

\cos {\frac {s}{R}}={\frac {R}{R+h}}={\frac {1}{1+{\frac {h}{R}}}}.

Solving for h gives

h=R\left({\frac {1}{\cos {\frac {R}{s}}}}-1\right)

so if the radius of the Earth is R =6 370 km, and s=200 miles=321 km, the we get about 8.1 km=5 miles.

I arrived at this formula like this: the arc length along a circle (which is the distance along the Earth's surface in this case) is given by s=Rθ, where θ is the angle subtended by the arc. The radius to the viewing point, the line of sight, and the radial length to the top of the tower make a right triangle with near leg R and hypotenuse h+R, from which I know that cos θ = R/(R+h). Put the two equations together, and I get my formula. -lethe ^talk 22:49, 30 November 2005 (UTC)[reply]

I've looked at the wikipedia page for horizon, and it contains a different formula

\ell ={\sqrt {2Rh+h^{2}}}.

This formula is for the distance of the line of sight, while the formula I gave is for the distance along the Earth. Take your pick about whether you wanted to ask about the distance along the Earth's surface or the straight line of sight. I suppose that unless the height of tower is on the order of the radius of the planet, then the two answers will be about equal. In that case √13h is a good approximation with h given in kilometers. In your case, the answer is 5 miles from all three formulas. -lethe ^talk 23:08, 30 November 2005 (UTC)[reply]

There is a conflict here, because "flat as Nebraska" could imply either a plane or a smooth sphere. On a plane, we find ourselves having to consider visual acuity plus simple trigonometry. --KSmrq^T 23:13, 30 November 2005 (UTC)[reply]

Thanks for the info. I appreciate it. I was thinking of a roughly smooth sphere, like the midwestern US. Five miles high is suitably impressive. Thanks again!Brian Schlosser42 11:59, 1 December 2005 (UTC)[reply]

December 1

Burning Energy in Transportation

Is it true that you burn roughly the same amount of calories traversing a mile long track with a bike, walking or running?

No. From our article on bicycle:

In both biological and mechanical terms, the bicycle is extraordinarily efficient. In terms of the amount of energy a person must expend to travel a given distance, investigators have calculated it to be the most efficient self-powered means of transportation. From a mechanical viewpoint, up to 99% of the energy delivered by the rider into the pedals is transmitted to the wheels, although the use of gearing mechanisms may reduce this by 10-15%. In terms of the ratio of cargo weight a bicycle can carry to total weight, it is also a most efficient means of cargo transportation.

On firm, flat, ground, a 70 kg man requires about 100 watts to walk at 5 km/h. That same man on a bicycle, on the same ground, with the same power output, can average 25 km/h, so energy expenditure in terms of kcal/kg/km is roughly one-fifth as much. Generally used figures are

1.62 kJ/(km∙kg) or 0.28 kcal/(mile∙lb) for cycling,
3.78 kJ/(km∙kg) or 0.653 kcal/(mile∙lb) for walking/running,
16.96 kJ/(km∙kg) or 2.93 kcal/(mile∙lb) for swimming.

Hope that helps. - Fredrik | t c 04:57, 1 December 2005 (UTC)[reply]

I hope the answer is appreciated; but this question is not a mathematics question. It would be better to ask it at the Science reference desk. Butchers and plumbers are both professionals, but you wouldn't want to ask one to do the other's job. --KSmrq^T 06:45, 1 December 2005 (UTC)[reply]

Twin prime number

Can someone tell mathematically why the middle number between twin prime number is always divisible by 6? Eg: between the twin prime number pair 17 and 19, 18 is divisible by 6.Similarly between the twin prime number pair 29 and 31, 30 is divisible by 6. Can someone give a mathematical proof of why this occurs?

Regards, Vijai12:13, 1 December 2005 (UTC)

If your first prime number is p, then p cannot be divisible by 2 or 3. Therefore it must be either 1 or 5 modulo 6. (See modular arithmetic). Similarly, the second prime number p + 2, since it cannot be divisible by 2 or 3, must be either 1 or 5 modulo 6. Putting this together, the only possiblity left over is that p is 5 mod 6, and p+2 is 1 mod 6. Therefore the number in the middle, p + 1, is 0 mod 6; that is, divisible by six. Dmharvey 12:45, 1 December 2005 (UTC)[reply]

Or, equivalently: Both primes are odd. Therefore, the number in between is even. One of every consecutive n integers is divisible by n and neither of the primes are divisible by 3 so the number in between is. If it's divisible by three and even, it's divisible by 6. Superm401 | Talk 20:42, 1 December 2005 (UTC)[reply]

Seems to me that 3 and 5 are considered twin primes. Both proofs must be amended to exclude this case. --KSmrq^T 21:53, 1 December 2005 (UTC)[reply]

Fair enough. It's not so much that the proofs need to be amended, but that the premise needs to be changed. The integer between two twin primes that are both greater than 3 is always divisible by 6. Superm401 | Talk 02:57, 3 December 2005 (UTC)[reply]

December 2

(no questions today)

December 3

proof that evaluation at a point is continuous

Time to try out the math reference desk for one of my own questions.

So a textbook I'm reading says that evaluation of maps between topological spaces is only a continuous operation on the space of maps (carrying the compact-open topology) if the domain is locally compact and Hausdorff (our page on compact-open topology suggests that this can be weakened to regularity). It's stated without proof, so I suppose it should be easy. I've tried my hand at the proof, but my proof doesn't seem to rely on Hausdorff, so it's either wrong, or I've got some unstated assumption.

Here it is:

I'm looking to show that the evaluation map e: X×Top(X,Y) → Y given by (x,f) ↦ f(x) is continuous if X is a locally compact Hausdorff space (this can apparently be weakened to regularity) and Top(X,Y) (the hom-set between X and Y, that is, the continuous maps between X and Y) carries the compact open topology.

Let V be an open set in Y, and (x₀,f₀) be an element of e⁻¹(V). We want so show that e⁻¹(V) is open, i.e. that every pair (x₀,f₀) is contained in a neighborhood in e⁻¹(V) open in Top(X,Y).

Since X is locally compact, every open neighborhood of x₀ contains a compact subset. f₀(x₀) ∈ V, therefore U=f₀⁻¹(V) contains x₀. It is open since f₀ is continuous. Let K be a compact neighborhood of x₀ in U, and M an open neighborhood of x₀ in K. K⊆U, so f₀(K) ⊆ V, and so f₀∈W(K,V), the set of continuous maps φ∈Top(X,Y) such that φ(K)⊆V, which is open in the compact-open topology by definition.

Thus (x₀,f₀)∈M×W(K,V), and it remains to show that M×W(K,V) is contained in e⁻¹(V). For any (ξ,φ)∈ M×W(K,V), since ξ∈K, then we have e(ξ,φ)=φ(ξ)∈V since φ(K)⊆V.

Therefore every point in M×W(K,V) is also in e⁻¹(V), i.e. M×W(K,V)⊆e⁻¹(V). Since every point (x₀,f₀) of e⁻¹(V) has an open neighborhood contained in e⁻¹(V), e⁻¹(V) is open, and e is continuous.

Somehow, I don't seem to have used that X is Hausdorff anywhere in this proof. So either my proof is wrong, or I've used it somewhere implicitly without knowing it.

So can anyone offer any comments? Thanks. -lethe ^talk 00:11, 3 December 2005 (UTC)[reply]

The only place I can see anything possibly going wrong is when you assert that x₀ is contained in a compact set K which itself contains an open neighbourhood of x₀. If X is Hausdorff then certainly this is what locally compact means. But if X is not Hausdorff, I'm not so sure. The article locally compact suggests that there are several definitions of "locally compact" in use for non-Hausdorff spaces. Dmharvey 01:25, 3 December 2005 (UTC)[reply]

Oh, I see. In my proof, I used the third definition of local compactness (every point has a local base of compact neighborhoods) listed at the bottom of that article. If I had taken the first definition (every point has a compact neighborhood) then my proof wouldn't have gone through, since I would have no guarantee that K⊆U. Then requiring that the space be Hausdorff is just to ensure that I'm allowed to use the stronger version of local compactness with impunity. That's a satisfactory answer to me. Thanks Dmharvey. -lethe ^talk 02:26, 3 December 2005 (UTC)[reply]

Right. In fact, I think (assuming I haven't screwed up) I can give you a specific example of the result breaking down when you use that weaker notion of locally compact. Take a nice metrisable but not locally compact space (e.g. any infinite dimensional hilbert space will do), call it H. Take X = H union {P} where P is an extra point, and take the topology of X to be all the open sets of H plus one extra open set -- the whole of X. This is non-Hausdorff. It is however locally compact according to the weak definition, since X itself is compact (in a stupid sort of way). Now take S = {0,1}, with the weird topology where {1} is open but {0} isn't. Then I believe that the evaluation map on X \times Hom(X, S) is not continuous. In particular, if you take the inverse image of {1}, then this set is not open. To show this, you could take a point (Q,f), where Q is a point of H and f is a characteristic function of some small open set U containing Q, and then check that (Q,f) cannot be contained in any rectangular basic open set contained in X \times Hom(X, S). Exercise: check this :-) I think it comes down to the fact that in H, every compact set is the intersection of all open sets that contain it. (Warning: this entire paragraph may be garbage.) Dmharvey 04:00, 3 December 2005 (UTC)[reply]

OK, let's see. The continuous maps here are the characteristic maps of open sets in X. It seems like the only compact neighborhood in X is X itself (if H has no compact neighborhoods, that is.) The singleton containing the characteristic function of X (which is just a constant function) is open in Hom(X,S). Thus it seems to me thatU×1 is an open neighborhood of Q contained in e⁻¹, and therefore evaluation is continuous for this space, unless I'm mistaken. Hmm... -lethe ^talk 17:23, 4 December 2005 (UTC)[reply]

No, U \times 1 does not contain (Q, f), since f is not equal to the constant 1 function on X. Dmharvey 20:30, 4 December 2005 (UTC)[reply]

Finding B*B

Since there is a Math reference desk, I thought I'd throw out a problem I tried to answer, but gave up on. I don't think it is quickly solvable. I was working on a project in college and I thought I was doing great by reducing a formula to C*C-A = B. B had to be a square value of an integer. Since I could just assume A was a square of something, I thought this could be the classic AA+BB=CC sort of equation. C is not a single number. It is a formula itself that produces a set of integers. Given C and A, it appears to me to impossible to quickly come up with a B that is the square of an integer (or even find a B that is an integer itself). Was I missing something? --Kainaw ^(talk) 05:05, 3 December 2005 (UTC)[reply]

I'm really confused. You're trying to find a possible value of C that makes C² − A the square of an integer? What is the formula for C? Is A an integer? You really need to give more explicit information. —Keenan Pepper 06:55, 3 December 2005 (UTC)[reply]

How can I say this diplomatically…? If this is the best you can do in describing the question, it is no surprise you couldn't answer it; nor can we. It sounds like perhaps there is a Diophantine equation in there somewhere, but you give far too little information. Such equations can be easy or hopeless, depending on the details. (Fermat's Last Theorem is such an integer equation.) Care to try again? --KSmrq^T 07:54, 3 December 2005 (UTC)[reply]

C, in the program, is not a single integer. It is an infinite set of integers (the program can go up to 256 digits). Given A, which is also an integer because that is all the program can produce, each C returns an integer B. However, most of them are not true - they are truncated because the program is unable to represent anything but an integer. So, the goal was to quickly find at least one C that would return a true integer B. Looking back at my notes, I rewrote the problem as: Given an integer, find two squares of integers that are that distance apart. IE: given 1328, find B and C such that BB-CC=1328. I can graph the answers in the real number set - which does hit an integer now and then. But quickly finding only integers is, in my opinion, impossible. --Kainaw ^(talk) 18:12, 3 December 2005 (UTC)[reply]

Ah, now I understand the problem. It depends on the factorization of A. B^2 − C^2 factors as (B − C)(B + C). B − C and B + C are either both odd or both even (just try all the possibilities). So, A must be either the product of two odd factors (so it's odd) or the product of two even factors (so it's a multiple of 4). If A is odd, then B = (A + 1)/2 and C = (A − 1)/2 is a solution. If A is a multiple of 4, then B = A/4 + 1 and C = A/4 − 1 is a solution. If A is even but not a multiple of 4, there is no solution.

In the case A = 1328, 1328 = 4*332 so B = 333 and C = 331 works. 333^2 − 331^2 = 1328. —Keenan Pepper 19:05, 3 December 2005 (UTC)[reply]

Is that, in fact, what you were asking? —Keenan Pepper 19:06, 3 December 2005 (UTC)[reply]

Well, a very simple case is A = 2C - 1. I'll leave the proof to the reader. --jpgordon∇∆∇∆ 18:45, 3 December 2005 (UTC)[reply]
- And an equally simple case is A = 4C - 4; and, if you wish to continue, A = 6C - 9. Oh, what the heck, for any integer N, A = 2NC - N^2. That's probably enough hints. --jpgordon∇∆∇∆ 18:55, 3 December 2005 (UTC)[reply]
  - (slight correction: for any integer N < C.) --jpgordon ∇∆∇∆ 20:27, 3 December 2005 (UTC)[reply]

Thanks. That was pretty much where I ended. From memory, I think I even ended up twisting it into the quadratic formula, but with the 4AC part being the incoming series of integers. In the end, stepping through all possibilities of one formula was no better than stepping through all possibilities of another one. It still ran too slow and I quit and went to a different project. This little formula just stuck in my head because I wasn't sure that it didn't have a discrete solution. --Kainaw ^(talk) 23:20, 3 December 2005 (UTC)[reply]

Banded operators

My turn! I'm looking for references and reading suggestions, since I suspect the problem is hard or at least general.

Never mind, the question was worded poorly.

Suppose one has a bi-infinite sequence $\{a_{n}\}=...,a_{-1},a_{0},a_{1},a_{2},...$ of real or complex numbers. Define an operator M with matrix elements $M_{mn}=a_{n-m}$ so that each row of the matrix is just the $\{a_{n}\}$ shifted over by one. I want to find the eigenvalues and eigenvectors for this M. I believe this is more or less solvable when only a finite number of the $a_{n}$ aren't zero; e.g. the discrete Laplace operator is a famous special case. I'm interested in the general case, where the $\{a_{n}\}$ may be or may contain convergent subsequences; where the subsequences converge to zero, or not, or have convergents that are dense in an interval/disk. The simplest non-trivial case is probably the one with the $a_{n}\to 0$ as $n\to \pm \infty$ . Theorems, special cases, equivalent problems, etc. are invited. linas 19:30, 3 December 2005 (UTC)[reply]

Could you give a little more context? I'm not a functional analyst and "banded operator" gets a few hits, but few enough that I wonder if it might be a typo for "bounded operator". In the context of vector spaces in general, you've described a vector space and a class of linear operators on it, and certainly it has some nontrivial eigenvectors. For example the sequence {...,4,2,1,1/2,1/4,...} is an eigenvector of eigenvalue 2, for the operator that shifts everything 1 space to the right.

If you want a normed linear space or something more (Banach space, Hilbert space) then maybe you should tell us what the extra structure is. There's no obvious "canonical" norm to put on this thing. --Trovatore 20:39, 3 December 2005 (UTC)[reply]

Oops, sorry, I misread your question. --Trovatore 20:42, 3 December 2005 (UTC)[reply]

Thanks, but I'm asking the wrong question. Yes, $x^{n}$ is a formal eigenvector. I'm trying to figure out what I want ask, and this wasn't quite it. linas 23:20, 3 December 2005 (UTC)[reply]

December 4

help?

0 3 1
2 2 −1
1 −3 2

I have to expand the given terminate along the third colum, and I've no idea how 2 do that. Any chance that someone could explain it to me withou using rocket science? — Preceding unsigned comment added by 129.108.96.87 (talk • contribs)

See the article, Determinant. This section illustrates how to expand a determinant along a row or column. —Keenan Pepper 00:31, 4 December 2005 (UTC)[reply]

Requesting math logic statements

The soon-to-be-featured article Intentionally blank page discusses the usage of "This page is intentionally left blank", which is a self-refuting meta-reference, in that it falsifies itself by its very existence on the page in question. Can someone construct equivalent mathematical statements to be used in this article? Thanks! — BRIAN0918 • 2005-12-4 15:21

P\to \neg P

?

No, that's equivalent to

\neg P

. I think a good answer to this would have to use Gödel numbering in order to make statements about other statements.

This is very similar to the kind of statement Gödel used to prove his incompleteness theorem. Actually, in a way, it's the opposite. The blank page notice says, "there exists no statement on this page", but it itself is a statement on the page so it must be false. Gödel's undecidable proposition says, "no statement of a certain type can be proven within this formal system", but it itself is a statement of that type, so it cannot be proven, although it is true. (The alternative, that it could be proven although it were false, would mean the system were inconsistent.) —Keenan Pepper 16:31, 4 December 2005 (UTC)[reply]

So is it possible for a mathematical logic statement to be constructed that is equivalent to what I'm asking for? — BRIAN0918 • 2005-12-4 16:38

To quote from Foundations of Higher Mathematics: "Some sentences that are not propositions are ... and "This sentence is false". Image the last sentence as a true-false question. The question would be unfair exactly because the sentence is not a proposition." E.g. paradoxes cannot be captured in propositional logic. Also note that the proposition "This page is intentionally left blank." does not make it untrue. It is the act of printing that proposition on the page that makes it false. In this sense the proposition is not a paradox at all, it is simply false. —R. Koot 21:21, 4 December 2005 (UTC)[reply]

But it sounds like Gödel numbering could work in this case. I don't know much about number theory, but it seems like the phrase could be represented by a relation that defines a set as being empty while being a part of that set. -- BRIAN0918 21:28, 4 December 2005 (UTC)[reply]

The "self-referential propositional calculus" of Yiannis N. Moschovakis is expressive enough to capture the liar. (Note that Gödel sentences do not capture the liar; they assert their own unprovability, not falsehood.) Moschovakis gives SRP a semantics using least-fixed-point recursion. The liar comes out neither true nor false using that semantics. --Trovatore 21:28, 4 December 2005 (UTC)[reply]

What kind of series is this?

Let $P(n)$ be a polynomial in n. What is the series

\sum _{n=1}^{\infty }{\frac {z^{n}}{P(n)}}

If $P(n)$ is a monomial, then this is the polylogarithm. If all of the roots of the polynomial are degenerate, then this is the Lerch transcendant. But what about the general case? Can it be solved or re-expressed in terms of known special functions? linas 16:00, 4 December 2005 (UTC)[reply]

Perhaps a Hypergeometric function? I'm not sure if its completely general. PAR 16:49, 4 December 2005 (UTC)[reply]

Yes, its a special case

\,_{p}F_{p-1}

w/ degenerate args. I guess I'm fishing for insights from number theory or something. linas 19:40, 4 December 2005 (UTC)[reply]

December 5

Compact Operators

Let $X$ , $Y$ , and $Z$ be Banach spaces. Assume that $K$ is a compact operator from $X$ to $Y$ , and $T$ is a linear operator from $Z$ to $Y$ . I am looking for a proof (or a counterexample) that if the image of $T$ is in the image of $K$ ( $\operatorname {im} \,T\subseteq \operatorname {im} \,K$ ) then $T$ is also compact. Thanks! (Igny 03:55, 5 December 2005 (UTC))[reply]

Assume

Y

is not a compact space; why would you expect

T

to be a compact operator? --KSmrq^T 19:04, 5 December 2005 (UTC)[reply]

Precisely because its image is in the image of another compact operator,

\operatorname {im} \,K\subseteq Y

. Also, it may be important, that

T

is assumed to be a bounded operator. (Igny 21:40, 5 December 2005 (UTC))[reply]

Sorry, that's a non-answer; you merely stated the conjecture again. Why do you think

K

is relevant? --KSmrq^T 03:14, 6 December 2005 (UTC)[reply]

Continuity of Derivatives

Hello! I need a little help with this question. I tried searching for it but was unable to find it on the web. Hypothesis: Let F(x) be a real valued differentiable function on the interval [a,b]. Let f(x) be its derivative. Conclusion f(x) is continuous on the same interval.

I wanted to know how to go about proving or disproving it.

Thank You,

Regards, Abhinav Mehta.

This is false. Try to find an example of a function whose derivative exists but is not continuous. Hint: One way this can happen is if the function oscillates faster and faster as it approaches a point, but it's contained within an envelope such that its derivative exists at that point, but is not continuous there. —Keenan Pepper 17:33, 5 December 2005 (UTC)[reply]

Correction: if F(x) is differentiable (that is its derivative, f(x), exists) at every point of an open interval (a,b), then f(x) is continuous on (a,b). Example: F(x)=xsin(1/x) is continuous but not differentiable at x=0, F'(0) does not exist, and f(x) is not continuous at x=0. A rather loose proof is based on definition of the derivative,

\lim _{x\to x_{0}}f(x)-f(x_{0})=\lim _{x\to x_{0}}\left(\lim _{h\to 0}{\frac {F(x+h)-F(x)}{h}}-\lim _{h\to 0}{\frac {F(x_{0}+h)-F(x_{0})}{h}}\right)=0

You can switch the limits, since all limits exist (F(x) is continuous and differentiable). (Igny 18:55, 5 December 2005 (UTC))[reply]

I disagree with Igny. Take

F(x)=x^{2}\sin(1/x)

(extended to zero by F(0) = 0). Then F'(x) exists for all x, and you can check that F'(0) = 0, even though F'(x) is very discontinuous at x = 0. Dmharvey 19:20, 5 December 2005 (UTC)[reply]

Yes you are right. Indeed the convergence in lim_h is not uniform over x, so you can not easily switch limits.(Igny 21:29, 5 December 2005 (UTC))[reply]

December 6

Precision and Tolerance

PLEASE HELP ME!!

CAN SOMEONE PLEASE DO THE FIRST ONE I REALLY DONT UNDERSTAND!! CAN YOU SHOW ME HOW IT IS DONT PLEASE!!What is the precision, tolerance, accuracy for the following:

4.7 m

5.21 ft

6.03 s

39 2/3 in.

It is impossible to know the accuracy without knowing the true value. Other than that, see Accuracy and precision and do your own homework. —Keenan Pepper 00:13, 6 December 2005 (UTC)[reply]

Thank you so much, but this isn't my homework!! but I thank you for all the help but if i may ask one more question: What do you mean by the true value? I really dont understand im sorry!

If these are exact values, then their accuracy is perfect (0). If instead they are measurements of physical things, the accuracy is the difference between the real value and the measured value. Say the 4.7m figure is the length of a rope. If the rope is actually 4.85m, then the measurement is accurate to +/- 0.15m, because that's how much it's off by. If you only have the measurement, but not the true value, then you have no way of knowing how accurate it is. It could be dead on (high accuracy), or it could be a really bad measurement (low accuracy). —Keenan Pepper 02:17, 6 December 2005 (UTC)[reply]

HELP ME

WHAT IS THE PRECISION ACCURACY AND TOLERANCE FOR 4.3 M AND EXPLAIN PLEASE I DONT UNDERSTAND THIS AT ALL. CAN YOU GIVE DETAILED ANSWER AS TO WHY?

Where did you get "4.3 M" from? The answer is going to depend on where that measurement came from. What type of instrument was used to measure it? A ruler? If so, how was the ruler marked--just with marks every meter, every 10 cm, every cm, every mm? Is 4.3m a single measurement, or is it the average of several measurements? The answer to your questions depends on where your number came from. Chuck 02:56, 6 December 2005 (UTC)[reply]