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March 21[edit]

Fractional factorials[edit]

How are fractional factorials like 3.4! (10.1361019) calculated? My graphing calculator can only do whole numbers and halfs, but the google calculator can do any number I put in. Is there any use for fractional factorials? 68.231.151.161 02:45, 21 March 2007 (UTC)[reply]

See gamma function. --Spoon! 02:56, 21 March 2007 (UTC)[reply]

what is the inverse of x exp(x)?[edit]

if ln(x)is the inverse of exp(x),what would be the inverse of f(x)=(x)*exp(x)?80.255.40.168 10:32, 21 March 2007 (UTC)ARTHER[reply]

See Lambert W function. --Lambiam ^Talk 10:39, 21 March 2007 (UTC)[reply]

is there any general way for this?[edit]

suppose f(x)=f1(x).f2(x)....fn(x) different elementary functions,can we find the inverse of f(x)?in general i mean. for instanse,f(x)=x*sin(x)ln(x) 80.255.40.168 11:35, 21 March 2007 (UTC)ARTHER[reply]

No, there isn't. Consider f1(x)=x and f2(x)=0 (constant), which implies constant function f(x)=0.
Or f1(x)=x and f2(x)=x, which gives f(x)=x². Or f1(x)=sin(x) and f2(x)=1. --CiaPan 14:24, 21 March 2007 (UTC)[reply]

But even when the inverse exists, it's not necessarily expressible in terms of "elementary functions". See the question just above this one, with the special Lambert W function. —The preceding unsigned comment was added by 24.91.135.162 (talk) 15:16, 21 March 2007 (UTC).[reply]

Cardinality[edit]

Is the cardinality of the power set of the natural numbers equal to the cardinality of the set of real numbers? Is the set of real numbers in an interval equal to the set of all real numbers? Black Carrot 18:19, 21 March 2007 (UTC)[reply]

Yes and yes (where I insert "the cardinality of" in the second question). See Cardinality of the continuum. The cardinality of intervals is only mentioned there, but it is trivial to find a bijection from an open interval to

\mathbb {R}

. -- Meni Rosenfeld (talk) 18:46, 21 March 2007 (UTC)[reply]

Meni's comment above about finding a bijection got me thinking that it might not be trivial sounding to everybody. So just in case a reader is having trouble, consider the tangent function tan(x) on the interval

\left({\frac {-\pi }{2}},{\frac {\pi }{2}}\right)

. It is a bijection from an open interval to the reals. If you wanted instead to start with a different interval domain, such as (0,1), try translating the domain as in

\tan \left(\pi x-{\frac {\pi }{2}}\right)

. Dugwiki 20:37, 22 March 2007 (UTC)[reply]

My assumption was that a reader sophisticated enough to follow this discussion, even if not able to think of this example immediately, would benefit from the challenge of trying to find it. While we're at it, I'll mention that finding an explicit bijection between a closed interval and the line is not as easy, but that's where the Cantor–Bernstein–Schroeder theorem comes in handy. -- Meni Rosenfeld (talk) 20:49, 22 March 2007 (UTC)[reply]

Well, the bijection (a,b) → [a,b) is quite obvious — once you know it. Anyway it's much easier to understand than the CBS theorem's proof. You just need to choose some subset, order it into a sequence, and that's (almost) all... ;) --CiaPan 14:44, 26 March 2007 (UTC)[reply]

I agree with the second question being true, but might the first question be equivalent the continuum hypothesis, which can be postulated to be true or false, much like Euclid's parallel postulate, without any inconsistencies arising? Baccyak4H (Yak!) 13:54, 23 March 2007 (UTC)[reply]

Oops, I just read Cardinality of the continuum, which answers the first question in the affirmative. So no continuum hypothesis. Baccyak4H (Yak!) 13:56, 23 March 2007 (UTC)[reply]

Radians and the area of a circle[edit]

If i had a circle of perimeter 1000 how can i work out its area using radians and what formula do i use.

Death of Rats 19:16, 21 March 2007 (UTC)[reply]

What you call "perimeter" is usually called "circumference" in geometry (the length of the perimeter). Using the formulas for circumference and area of a circle given in the respective articles (as well as in our article Circle), you can first, given the circumference, find the radius, and from that the area. If you report the result you find here, we can check if it is right. You don't even need to "use" radians, because the relevant relationships are independent of angular units. --Lambiam ^Talk 19:40, 21 March 2007 (UTC)[reply]

Hang on i messed that question up. Ok i have an equation :

250000 ÷ (n×tan(½(360÷n)×(180÷PI))

How can this be changed to use Radians insted of Degrees? The homework says that a there are two methods for working out the area of a shape using the above formula. One in degrees (above) and one in radians. However i dont have a clue how to convert that to radians. And n is the number of sides in the shape. Here n is 1.

Sorry about Messing question up at the start

Death of Rats 20:06, 21 March 2007 (UTC)[reply]

You have a formula, but not an equation. I can only guess what it represents, and the parentheses do not even match up. Assuming that a final closing parenthesis is missing, it looks rather similar to the formula for the area of a regular polygon with n sides, where the circumference has length 1000, but all together it doesn't work. What is positively weird is that 360 and 180, which both look like angular measures using degrees as the unit, are multiplied together. That does not happen in any decent formula of plane geometry. The factor 180/π is the regular conversion factor from radians to degrees, but 360/n would be an angular measure in degrees. Assuming that the multiplication sign should actually be division, we get as the argument to the tan function ½(360/n)/(180/π) = π/n. Now we get close to the formula for the area of a regular polygon with n sides of length t, which is:

{\frac {nt^{2}}{4\tan(\pi /n)}}~.

Substituting t := 1000/n (since the circumference is nt = 1000), we get

{\frac {1000^{2}}{4n\tan(\pi /n)}}~,

which is the above formula as corrected and simplified. --Lambiam ^Talk 20:48, 21 March 2007 (UTC)[reply]

Also, you wrote "Here n is 1", but that does not make sense; a regular polygon has at least 3 sides. --Lambiam ^Talk 22:10, 21 March 2007 (UTC)[reply]

The area of a disk article may help; a circle is the perimeter (boundary) of a disk. When we are being careful, we say that only the disk has area, not the circle. --KSmrq^T 20:42, 22 March 2007 (UTC)[reply]

Gee Force Formula[edit]

I need to know the mathematical formula to find out how many gees are being applied to a up right person. Can anybody give me a link to a site or if you know the formula, please state it. Thank you. —The preceding unsigned comment was added by 74.132.57.45 (talk) 22:11, 21 March 2007 (UTC).[reply]

I assume you're talking about "gee" as in "multiples of the Earth's gravitational acceleration"? In that case, for a person standing upright, the acceleration is 1g. For any other situation, the formula depends on the situation. --Carnildo 22:35, 21 March 2007 (UTC)[reply]

(after edit conflict) It sounds like a physics question to me. If this is about the force of gravity, the physical orientation or moral standing of the body subject to that force does not matter. If the force is the gravitational force as exerted on a body at rest on the surface of the Earth, then 1 g is a good approximation. --Lambiam ^Talk 22:41, 21 March 2007 (UTC)[reply]

Isn't there some sort of equation where I can plug in weight of the person and how high they are(they are skydiving), to find out how many gees are actually being applied? I'm looking for precise answers with perhaps decimals.

The acceleration is the roughly the same while a person is on earth or skydiving on earth. It is 1 g.

1 g is 9.8 metre/second/second

If you are anal about it, you can calculate it yourself with :

F=G{\frac {m_{1}m_{2}}{r^{2}}}=\left(G{\frac {m_{1}}{r^{2}}}\right)m_{2}

where g is the bracketed factor and thus:

g=G{\frac {m_{1}}{r^{2}}}

in metres/second/second

Where the Big G is the Gravitational constant.

202.168.50.40 00:25, 22 March 2007 (UTC)[reply]

As long as they stay below 10 km height you can use 1.00 g. The gravitational acceleration does not depend on the person's weight and decreases very slowly with height. At 1 km height it is 0.9997 g. At 10 km 0.9969 g. --Lambiam ^Talk 00:34, 22 March 2007 (UTC)[reply]

Beware of mathematicians answering physics questions. This clearly belongs on the science reference desk.

There is, of course, no force of gravity — not as general relativity formalizes it. When standing on the Earth, our bodies are trying to do the natural thing, which is to follow a straightest path through a space curved by mass|energy. The electrostatic interactions between us and the ground prevent that path, and that is what we feel.

If we go to Disneyland and enjoy the Star Tours ride, our bodies are subject to "g-forces" that have nothing to do with gravity. For entertainment purposes, these forces are deliberately limited; however, the pilot of an F-22 Raptor in an air combat situation feels forces so strong they can cause blackout. And if we are lucky enough to get a ride on a Vomit Comet, the plane flies a path carefully chosen to match the natural geodesic path, approximating a "zero-gee" experience.

So, what can we say about skydiving? There is no Earth pressing against our feet, but the air itself presses harder and harder as our speed increases, until finally we reach terminal velocity and can go no faster. Therefore, for most of our descent we feel a force little different from that of standing. However, if our parachutes (we always have two!) fail to open, we will eventually experience large forces of deceleration. :-(

Thus, the answers already given mostly apply, but all the fun is in the details. --KSmrq^T 21:49, 22 March 2007 (UTC)[reply]

Ah.. just to clarify his statement. "For his statement about feeling a force little different from standing" we consider that we reach or are near terminal velocity quite early in the drop. So a graph of the g's felt in the frame of reference of the skydiver would be practically 0 at first (since wind is negligible) but nearing terminal velocity the g's felt would tend toward 1.). I was curious to get an idea how the acceleration slows as the skydiver nears the terminal velocity, but I have yet to find an actual graph. I did find this (The Physics Classroom -- Skydiving) which basically reiterates the same points brought up... however please note the animated GIF at the bottom. The animation would seem to imply that the force caused by the air movement at one point is actually greater than 1 G! This occurs until the forces balance out. This would seem to imply the graph of the velocity as the skydiver reaches terminal velocity is not a gradual asymptotic convergence but a wavy, spring like convergence. (which makes sense to my intuition about it).

Root⁴(one) 06:03, 25 March 2007 (UTC)[reply]

Hmmm I just looked once again upon that animation. The (greater than 1) G's experienced were ACTUALLY DUE to the parachute opening, so don't apply to my arguments about force felt as an object approaches its terminal velocity. I can't prove I'm right, but I can't prove I'm wrong either. So ignore my comment about "wavy, spring like convergence"

Root⁴(one) 15:08, 25 March 2007 (UTC)[reply]