November 29[edit]

Ohh, Ohh, I'm the first person to use this[edit]

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[edit]

Very large numbers[edit]

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]

I think by deduction "octilliion" would be more wider understood than you give credit for. It is one of the prefixed used more often such as "octopus" or "octagon" being associated with eight (therfore another 6 lots of 3 zeros from billion). This might be argued of course, but I vehmently disagee with using analogies such as you give. The perception of how far the moon is, appart from being big or a long way to drive! would be farily abstract and would vary between people even if the distance were given. Dainamo 15:23, 8 December 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]

In an English speaking encyclopedia, the use of billion, trillion etc. are fairly well understood in the English speaking world. The use of, for example, 10^12 as a billion in the UK and commonwealth is generally understood to be old fashioned and more or less abondoned in favour of 10^9 in every day use, the media and also government reporting.Dainamo 15:27, 8 December 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[edit]

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[edit]

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]