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July 15

Shell/disk volume integrals different?

Something is confusing me. I was trying to remember the volume integrals I learned ages ago, and came across the shell and disk methods at Solid of revolution (hazy memories now). So I thought ah!, I'll do a really easy example: f(x) = x, a=0, b = 1 (a=lower x limit, b=upper x limit). So:

Disk method: $V=\pi \int _{0}^{1}x^{2}dx=\pi [x^{3}/3]_{0}^{1}=\pi /3$ (good, the volume of a cone is correct!).
Shell method: $V=2\pi \int _{0}^{2}x\times xdx=2\pi [x^{3}/3]_{0}^{1}=2\pi /3$ (uh oh).

Clearly I'm making a fully elementary mistake. But surely if f(x) = x, then f(x)^2 === xf(x), and everything is symmetric with this straight line and these limits? Have I even so got the axis of rotation wrong in the shell method?79.76.150.219 (talk) 01:43, 15 July 2008 (UTC)[reply]

I think your problem might be that you aren't actually drawing a cone - that formula is for the volume between f(x) and the x-axis, which is a cylinder with a cone removed (point down), not a cone (draw the graph and shade in the relevant part to check). The volume of that *is* 2pi/3. The question, then, is why did you get the wrong answer with the disk method? That, I'm baffled by... --Tango (talk) 01:59, 15 July 2008 (UTC)[reply]

Oh, and I'm assuming that 2 in the upper limit on the second integral is just a typo, you've switched back to the correct 1 in the next expression. --Tango (talk) 02:01, 15 July 2008 (UTC)[reply]

The correct integral for the shell method is:

\int _{0}^{1}(1-x)\times 2\pi x\,dx.

Drawing a picture might be helpful. siℓℓy rabbit (talk) 02:09, 15 July 2008 (UTC)[reply]

You mean f(x) should be 1-x for a cone, not x? I agree with that, but why does the disk method get the right answer with the wrong function? --Tango (talk) 16:54, 15 July 2008 (UTC)[reply]

Revolution is about the x-axis (at least given the form of integral used for discs), so it is a cone for f(x) = x, and the discs give the right value. The catch is that for shells you still want revolution about the x-axis (while the formula on the Solid of revolution page assumed revolution about the y-axis, giving a cylinder with a cone removed as you suggest) so to construct your shell you need to be careful, since your cylinders lie horizontally, and thus have radius f(x)=x but have height 1-x (distance from function to our upper bound b=1 running horizontally) so we have cylinders of area

2\pi x(1-x)

giving the integral form supplied by Silly rabbit -- Leland McInnes (talk) 20:06, 15 July 2008 (UTC)[reply]

D'oh. Yes, of course, one formula was revolution about the x-axis, the other about the y-axis, leading to different integrands to compute the same volume. Thanks. 79.76.150.219 (talk) 20:49, 15 July 2008 (UTC)[reply]

Well spotted! Thank you! --Tango (talk) 21:17, 15 July 2008 (UTC)[reply]

Happy Numbers

We have an article on happy number and in the article, it describes the "proof" that all integers either end up in a cycle of 1,1,1,1,... or a cycle containing the number 4. Everything is fine until the last line which says that every number above 99 drops below 99 and then an exhaustive search shows us that only these sequences are possible. Without referring to this computer proof, how can we prove that all number between 1 and 99 eventually get into one of the two cycles?--A Real Kaiser...NOT! (talk) 05:39, 15 July 2008 (UTC)[reply]

I wouldn't call that a computer proof. While 99 cases is a bit of a stretch, one could verify it by hand. This isn't the four color theorem. 69.106.57.217 (talk) 06:46, 15 July 2008 (UTC)[reply]

It's less than fifteen minutes' work. I've spent much longer than that trying to understand a proof, let alone produce one. Algebraist 09:59, 15 July 2008 (UTC)[reply]

I suspect the OP actually meant a proof not involving proof by cases, not specifically a computer proof. GromXXVII (talk) 10:55, 15 July 2008 (UTC)[reply]

In that case, I doubt anything satisfactory is possible. You can streamline the case analysis with a little effort, but there doesn't seem to be much point. Algebraist 11:00, 15 July 2008 (UTC)[reply]

Drilling Holes in a Cube

We have a perfect cube with sides 10 units each. A hole has been drilled along its main diagonal (all the way across) with a drill whose radius was one unit. What is the new surface area? I got the answer as $600-2\pi +2\pi {\sqrt {300}}$ . 600 was the original surface area before the drilling. After the drilling, $2\pi$ was lost and then $2\pi {\sqrt {300}}$ (the surface area of the hollow cylinder) was gained. Do you guys agree?--A Real Kaiser...NOT! (talk) 05:49, 15 July 2008 (UTC)[reply]

I think your last two terms are incorrect. The drill was through a diagonal, so it would not have met the faces at a right angle. The hollow shape is also not quite a cylidner as its end shapes are not circles. Maelin (Talk | Contribs) 09:25, 15 July 2008 (UTC)[reply]

Almost agree - you equation is very nearly right - but not perfect as pointed out above.87.102.86.73 (talk) 12:42, 15 July 2008 (UTC)[reply]

To get it absolutely right you'd need to calculate the surface area of the ellipses projected onto the cube corners by the cylinder, as well as the surface area of the trigonal pyramid caps removed from the corners by making the tube..87.102.86.73 (talk) 13:16, 15 July 2008 (UTC)[reply]

My answers: The total "lost" area is

2\pi {\sqrt {3}}

(each face loses

{\frac {\pi }{\sqrt {3}}}

). The area of the inner cylinder-ish surface is

20\pi {\sqrt {3}}+2\pi {\sqrt {2}}-6{\sqrt {6}}

making the total surface area of the object:

600+18\pi {\sqrt {3}}+2\pi {\sqrt {2}}-6{\sqrt {6}}\approx 692.133993088

. How about that, "do you guys agree" now? --tcsetattr (talk / contribs) 22:12, 15 July 2008 (UTC)[reply]

Hemispheres

If I mark three distinct points on a perfect sphere randomly, what are the chances that two of the points are on the same hemisphere?68.126.250.171 (talk) 06:32, 15 July 2008 (UTC)[reply]

There are only two hemispheres in a sphere. Therefore possible distributions for your three points are 3-0, 2-1, 1-2 or 0-3. In all cases one hemisphere has at least 2 points. -- SGBailey (talk) 08:06, 15 July 2008 (UTC)[reply]

This is an instance of the Pigeonhole principle. Maelin (Talk | Contribs) 09:26, 15 July 2008 (UTC)[reply]

Is it possible that the questioner meant 'all three points are on the same hemisphere' which is slightly less trivial?? (are you still there?)87.102.86.73 (talk) 12:40, 15 July 2008 (UTC)[reply]

They may also have meant that the hemispheres are defined before the points are added, such as the Earth's northern and southern hemispheres. In that case, you could have all three points on the border (equator) between the two hemispheres. So, would those be considered to be on the same hemisphere or not ? StuRat (talk) 05:09, 16 July 2008 (UTC)[reply]

The whole sphere/hemispheres part is irrelevant. We can recast this as follows : if a set S is partitioned into two disjoint sub-sets, then for any three members of S, at least two of the three will be members of the same sub-set. As Maelin says, it's a simple application of the pigeonhole principle. As StuRat says, the original formulation unfortunately introduces some ambiguity around the status of points on the boundary between the hemispheres. Gandalf61 (talk) 13:43, 16 July 2008 (UTC)[reply]

Another interpretation is that the hemisphere may not be fixed in advance. In other words, take three points at random on the sphere, and then ask: is there a hemisphere that contains all three points? The answer to this question is yes, almost surely. (The exceptional cases all occur when the three points lie on the same great circle.) siℓℓy rabbit (talk) 14:12, 16 July 2008 (UTC)[reply]

Oh, wait. I was answering 87.102.86.73's modification of the problem (all three points rather than two out of three). siℓℓy rabbit (talk) 14:14, 16 July 2008 (UTC)[reply]

What is this?

Can anyone identify this? I think it might be a cipher or code of some kind. Or, I could be completely wrong :P. VIVID (talk) 13:56, 15 July 2008 (UTC)[reply]

53 6f 6d 65 74 69 6d 65 73 20 61 20 67 69 72 6c 20 63 61 6e 20 67 69 76 65 20 79 6f 75 20 61 20 6c 6f 6f 6b 20 61 6e 64 20 61 20 73 6d 69 6c 65 20 74 68 61 74 20 63 61 6e 20 6d 61 6b 65 20 79 6f 75 20 66 65 65 6c 20 6c 69 6b 65 20 61 20 6d 69 6c 6c 69 6f 6e 20 62 75 63 6b 73 2e 20 20 4f 72 20 69 6e 20 74 68 69 73 20 63 61 73 65 2c 20 61 20 62 69 6c 6c 69 6f 6e 2e

Is is hexadecimal? some context would help -where's it from etc?87.102.86.73 (talk) 14:14, 15 July 2008 (UTC)[reply]

It is simple ascii in hex format. It says "Sometimes a girl can give you a look and a smile that can make you feel like a million bucks. Or in this case, a billion.". -- Q Chris (talk) 14:19, 15 July 2008 (UTC)[reply]

So it was a (simple) cipher of sorts .. see the hex column ofAscii#ASCII_printable_characters which one of you two is the girl?87.102.86.73 (talk) 14:22, 15 July 2008 (UTC)[reply]

Natural oracles giving P≠NP

In Computational Complexity: A Modern Approach, Arora and Barak construct a set B such that P^B≠NP^B. This B is explicitly constructed to ensure that no B-Turing machine decides the set of lengths of strings in B in polynomial time. But is there a more natural set (or one that was not specifically constructed to prove this)? --Taejo|대조 16:01, 15 July 2008 (UTC)[reply]

Sure -- it's the empty set. No one knows how to prove it, of course. --Trovatore (talk) 22:36, 15 July 2008 (UTC)[reply]

Lol, of course. Naturally, I meant a known natural set. --Taejo|대조 21:09, 17 July 2008 (UTC)[reply]

Converting between PPPs of different years

Sigh, OK, help me out on this. Math ain't my strong suit, much less international economics. (And no, this isn't homework. It's work work.)

I'm trying to take a number of different datasets relating to expenditures in terms of Purchasing power parity of a given "current year" and trying to combine them to be in terms of PPP of a fixed year.

So let's say I have spending figures, in "million current PPP$" for the year 2008, that look like this:

	2002	2003	2004
China	1162.4	1362.7	1630.1

And let's say I have spending figures "in millions current PPP$" for the year 2005 that look like this:

	2002	2003	2004
China	71358.6	84646.7	102622.9

OK. Obviously what "current PPP" means has changed a lot in those years. Let's say I have data from the 2005 report that is not in the 2008 report, and I'd like to convert all of the data to "millions of 2005 PPP$". Can I do that with just the information above? Would it be a straightforward conversion factor? I mean, it seems like the 2005 data is around 62 times the 2008 data. Would just using a factor like that produce junk data on the whole, though?

(My problem, if isn't clear, is I have lots of reports, each of which give values I need for different time scales in "current PPP$", which of course changes from year to year. I'm trying to aggregate the data. I'm not having a lot of luck. Or put another way, I am trying to get PPP values that are both currency AND time independent.) --140.247.248.35 (talk) 21:38, 15 July 2008 (UTC)[reply]

With only one datapoint, it could be anything really. It could be a straight conversion factor, but you'd need a 2nd data point to falsify that hypothesis. -mattbuck (Talk) 22:10, 15 July 2008 (UTC)[reply]

Well I'm only converting between these two types of data sets. I've given you three datapoints above—the total datapoints are only about 10 per country (and I think it's pretty clear that these calculations would have be done from country to country separately, as they are based on different national currency changes). So...? --98.217.8.46 (talk) 23:08, 15 July 2008 (UTC)[reply]

Are those genuine figures? A factor of 1/62 in 3 years doesn't sound right. Unless I'm completely misunderstanding this (I'm not an economist), the difference should be purely due to inflation in the US dollar (I'm assuming the dollars in question are US, they usually are for this kind of thing), and that should give a much smaller difference (in the order of 10% rather than 6200%) and it should be in the other direction... --Tango (talk) 23:30, 15 July 2008 (UTC)[reply]

Do I understand the data correctly? Was Chinese expenditure in 2002 equal to 1162 million "2008 US dollars" and 71359 million "2005 US dollars" (where 1162 million "2008 US dollars" represents the value of goods and services that can be purchased with Chinese Yuan equivalent to 1162 million US dollars in 2008 at 2008 purchasing power implied exchange rates)? From this graph and my calc, the Chinese Yuan appreciated by only 13.7% from 2005 to 2008 (in US dollar terms) so something ridiculous must have happened to purchasing power for my interpretation to be correct. Zain Ebrahim (talk) 13:00, 16 July 2008 (UTC)[reply]

I just editted my post above (diff). Zain Ebrahim (talk) 13:39, 16 July 2008 (UTC)[reply]

Oh, haha. I made a gaffe. The 2008 figures are Argentina, whereas the 2005 figures are China. Yeah, OK. Here's the correct 2008 figures:

	2002	2003	2004
China	39444.7	46944.6	57669.6

Which still gives us a change of -55%, which is pretty weird.. --140.247.240.177 (talk) 20:13, 16 July 2008 (UTC)[reply]

That still doesn't make sense... it's closer to the right size change, but it's still in the wrong direction. Due to inflation, a 2008 US dollar is worth less than a 2005 US dollar, so the same quantity measured in 2008 dollars should be larger than when measured in 2005 dollars... --Tango (talk) 22:10, 16 July 2008 (UTC)[reply]

If my definition above is correct then it would be possible (though unlikely) for the number to decrease over time if the purchasing power of 1 yuan increased relative to the purchasing power of one dollar over 2005 to 2008 or if PP exchange rates stayed the same but China had higher inflation. Zain Ebrahim (talk) 11:34, 17 July 2008 (UTC)[reply]

Say China's expenture in year Y was X Yuan (not adjusted for anything). To get this in "2008 US dollars", first you'd convert it into 2008 Yaun (i.e. increase to adjust for inflation (time)) and then convert that into dollars using purchasing power implied exchange rates (to adjust for currency) as of 2008. If I understand you correctly, you have data for ten years at "2005 US dollars" but for fewer years at "2008 US dollars" and you want to convert the data points you don't have in the 2008 report into "2008 US dollars", right? Or is it the other way around?

So the X Yuan spent in year Y is converted into "2005 US dollars". To convert this into "2008 US dollars", all you need are three peices of information:

The 2005 PP implied exchance rate,
Chinese inflation over 2005 to 2008 and
The 2008 PP implied exchange rate

It doesn't depend on Y but it does depend on the country so (if all my assumptions are correct) using the factor seems fine as long as you do it separately for each country. Zain Ebrahim (talk) 11:34, 17 July 2008 (UTC)[reply]

Can't you just use US inflation? I thought the point of PPP was that you didn't need to worry about where it was being spent, that's all taken care of in the exchange rate. --Tango (talk) 16:28, 17 July 2008 (UTC)[reply]

That would work if the expenditures were first adjusted for currency and then for time - I assumed the opposite. I think adjusting for time first is better because it captures the time-value of money in China for China's data. Zain Ebrahim (talk) 19:09, 17 July 2008 (UTC)[reply]

Factorial differentiation

I know that the factorial function can't be differentiated because it is only defined on the (positive?) integers but I don't see why this prevents it from being differentiable. Can someone explain please? Thanks 92.2.122.213 (talk) 22:13, 15 July 2008 (UTC)[reply]

Well, the definition of the derivative involves evaluating the given function arbitrarily close to a given point. For example, to get an approximation of the derivative of the factorial function at 3, you might form the fraction

{\frac {3.000001!-3!}{0.000001}}

But if there's no such thing as 3.000001!, how are you going to do that?

Note that the gamma function is differentiable. --Trovatore (talk) 22:34, 15 July 2008 (UTC)[reply]

(edit conflict) The derivative of a function f at x is (loosely speaking) the limit of the difference quotient

{\frac {f(a)-f(b)}{a-b}}

as the two points a and b both get arbitrarily close to the point x. If a and b can only take integer values, they can't approach each other arbitrarily closely — the smallest possible difference between two different integers is 1. The only way for two integers to get closer than that is for them to be equal — and that won't work, since then the difference quotient would become 0/0, which is undefined. Of course, you can still take finite differences: for example, (x+1)! - x! = x·x!. Also, the factorial function can be extended so that it has values on (almost) all real (and even complex) numbers, and that function can then certainly be differentiated. —Ilmari Karonen (talk) 22:39, 15 July 2008 (UTC)[reply]

You might be interested in Stirling's approximation for factorials which is derived thus:

n! = product(from x=1 to x=n) of x

therefor ln(n!)=sum(from x=1 to x=n) of ln(x) remember ln(abcde)=ln(a)+ln(b)+ln(c)+ln(d)..etc

Therefor ln(n!) can be approximated by the integral of ln(x) between 1 and n

ie ln(n!) ~ xlnx - x + 1 I've ignored the stuff that minimises the error -see Stirling's approximation for that..

Therefor n! ~ e^{xlnx - x + 1} = e^xlnx+e^-x+e¹ = x^x+e^-x+e = e(x/e)^x

So as a rough rule of thumb you can say that the slope of the graph formed by joining the points of n! vs. n is given by d/dx(e(x/e)^x) which you might be able to calculate..

For a better version read the rest of the article!87.102.86.73 (talk) 22:55, 15 July 2008 (UTC)[reply]

OK I see what you're saying. So if you have a function involving the factorial function, eg $f(x)=x!*x^{n}$ , how do you go about differentiating it? Or does this depend on the function? 92.2.122.213 (talk) 18:29, 16 July 2008 (UTC)[reply]

You can't differentiate it, any more than you can differentiate the factorial function itself, and for the same reason. By the way, the answer involving Stirling's approximation is a bit misleading -- the derivative of an approximation of a function doesn't have to be a good approximation to the derivative of the function.

Really, if you want a ("natural") differentiable function that interpolates the factorial function, you have to learn about the Gamma function. It's not that hard, so quit looking for shortcuts. --Trovatore (talk) 18:47, 16 July 2008 (UTC)[reply]

(Yes I gave the stirling approximation 'concept' as a method of just estimating the slope of a graph that almost fits the points given by n! in fact if you read more of that article it too goes onto mention the gamma function)

Also it's worth noting that the gamma function does give a way of getting n! as a continuous function, but that factorial as most people know it just isn't really a curve or line - it's just a set of points - and as such doesn't really differentiate.

As for your function y=x!*x^n then as usual dy/dx = xⁿd/dx(x!) + nx^n-1*(x!) (see ~~[[chain rule]~~ product rule] How you interpret d/dx(x!) depends on what your equation means - it may be meaningful to use a stirling-derived approximation, or not - but the questions seems to be how do I differentiate Gamma(n) with respect to n ?? 87.102.86.73 (talk) 19:05, 16 July 2008 (UTC)[reply]