Wikipedia:Reference desk/Archives/Mathematics/2007 March 27

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[edit] March 27

[edit] Completeness

I ask this question so that we can probably add the reasoning to the article on completeness.

Is a scalar valued random variable indexed by the mean always complete?

In other words, if X is a random variable with E(X) = 0, then is it true that

E(f(X+θ)) = 0 for all θ => f = 0 almost everywhere?

--Hirak 99 05:54, 27 March 2007 (UTC)

Well, it's not true. I've constructed this counterexample, please notify me if you think it is wrong.
Let X follow Uniform[0,1]. Let g(a)=sin(2πa). Then E(g(X+θ))=0, no matter what θ is.
--Hirak 99 13:37, 27 March 2007 (UTC)
Isn't E(X) = 1/2 in your example? -- Jokes Free4Me 10:49, 28 March 2007 (UTC)
You're right. Sorry, I meant Uniform[-1/2,1/2] :) I have updated the article on completeness also with this example. Cheers --Hirak 99 12:03, 28 March 2007 (UTC)
Why is E(X)=\frac12? \forall\theta\,\int_\theta^{\theta+1} \sin 2\pi x\,dx=0... you'd need \left[-\frac12,\frac12\right] for g(a) = sinπa, but not with the 2. Right? --Tardis 16:49, 28 March 2007 (UTC)
For X ~ Uniform[0,1], E(X)=½. So we define X ~ Uniform[-½,½] so that E(X)=0. Also then \forall\theta\,\operatorname{E}(f(X+\theta)) =\int_{-\frac{1}{2}}^{\frac{1}{2}} \sin 2\pi (x+\theta)\,dx =-\frac{1}{2\pi} \left[ \cos(2\pi (x+\theta)) \right]_{-\frac{1}{2}}^{\frac{1}{2}}=0. Without the factor of 2 however, it may not be 0 for all θ. --Hirak 99 20:08, 28 March 2007 (UTC)

[edit] Linear beam theory

I've been trying to find references on the dynamics of a cantilever beam (assume Euler-Bernoulli beam equation) coupled to a forced mass-spring-damper system, but unfortunately I haven't found any that use a partial differential equation formulation rather than a Lagrangian approach. I'm trying to model this as m\ddot{u}(x,t)+EIu''''(x,t)=0 with the BCs u(0,t) = 0, u'(0,t) = 0 (fixed beam BCs) and u''(1,t) = 0 (zero bending moment) and M\ddot{u}(1,t) + 2\zeta\dot{u}(1,t) + \kappa^2u(1,t) + EIu'''(1,t) = A\sin(\omega t).

Does this look reasonable to anyone? Or should there be a m\ddot{u}(1,t) term in the BCs as well as the M\ddot{u}(1,t) one?

Also, if anyone has any suggestions on analytical solutions to this that would be great!

Thanks 137.222.183.82 17:58, 27 March 2007 (UTC)

[edit] Ode to a differential equation

In mathematics, an ordinary differential equation (or ODE) is a relation that contains functions of only one independent variable, and one or more of its derivatives with respect to that variable.

A simple example is Newton's second law of motion, which leads to the differential equation

m \frac{d^2 x}{dt^2} = f(x),\,

Which one is the independent variable? Is it x or t ? 202.168.50.40 21:17, 27 March 2007 (UTC)

To someone not familiar with the conventions (most readers who need the article?) I can see how this might be confusing. The intent is that x (position) is a function of t (time), with f (force) a fixed function that depends on x. We would use a law of motion to solve for motion; that is, to find x at each instant in time. The big hint here is that we see a derivative of x with respect to t, not the other way around. --KSmrqT 22:22, 27 March 2007 (UTC)
Sounds like some conventions need to be explained in some article. Which article? and why the little "f"? Shouldn't it be "F(x)", or wouldn't it be better described as "F(t)", since x=g(t) for some function g of t? In any case maybe I need to learn something. Root4(one) 02:58, 28 March 2007 (UTC)
The forcing function can be thought of as a function of x (consider a ball rolling around in a weirdly-shaped bowl) or as a function of t (consider a piece of paper blowing in gusty winds). Of course, all such functions can be written as f(t): = f(x(t),t) because all forces can depend at most on position and time, and we're only interested in one position per time. But to actually do that transformation requires that we know x(t), which we don't. So we write f(x) or f(x,t) whenever there's a spatial variation in the force. (Bonus question: why can't we write f(x): = f(x,t(x))?) --Tardis 17:14, 29 March 2007 (UTC)

[edit] Volume of a sphere

I know that the volume of a sphere is given by the formula: V = \frac{4}{3}\pi r^3., but I do not understand how someone managed to get to that equation. Can someone explain this to me please, considering that my mathematical knowledge is limited? Thank you. --80.229.152.246 21:25, 27 March 2007 (UTC)

A simple way using calculus is to calculate the volume of the solid of revolution formed by revolving a half-circle around its diameter using disk integration. So let us revolve the region between y=\sqrt{r^2-x^2} and y=0 around y=0. We get
\pi\int_{-r}^r \sqrt{r^2-x^2}^2 \,dx which is easy to evaluate to \frac{4}{3}\pi r^3. --Spoon! 22:07, 27 March 2007 (UTC)
(after edit conflict) Do you know any calculus? If you have a sphere of radius r, and it expands to a radius r+ε, where the increase in radius ε is very small, the volume of the thin shell by which the sphere grows is very close to the thickness ε of that shell times the original surface area A(r) = 4πr2 of the sphere, or εA(r). So the derivative of V(r) with respect to r, which is the limit of (V(r+ε) − V(r))/ε, is equal to A(r). V(r) can then be found by finding the antiderivative of A(r), with the boundary condition that V(0) = 0. Indeed, if V(r) = 4/3πr3, d/dr V(r) = 4πr2 = A(r). Note that there is a similar relationship between the area enclosed by a circle and its circumference: d/dr πr2 = 2πr. The above is not the only way of finding the answer, but it is the one I like best.  --LambiamTalk 22:20, 27 March 2007 (UTC)

The volume of a sphere is twice the volume of a hemisphere.

The volume of a hemisphere can be made by stacking discs of deminishing radius on top of each other.

The volume of a disc is (area of disc * thickness of disc) which is ( Pi*r_disc^2 * thickness ).

If you sum up all the volumes of the discs, you would get the volume of the hemisphere.

If you multiple the volume of a hemisphere by two, you get the volume of a sphere.

Now the tricky bit is working out how many discs, the radius of each individual disc and the uniform thickness of each disc. Ultimately, you need an infinite number of discs with zero thickness. At this point you would need to learn about the mathematical concept of limit.

I would advise you to work out the volume of a hemisphere with

(a) 1 disc

(b) 2 discs

(c) 4 discs

And get a feel of the concept of limits as the number of discs kept on doubling each time.

PS: Do not confuse the radius of each individual disc ( r_disc1 , r_disc2 , ... ) with the radius of the sphere ( r ) .

202.168.50.40 22:41, 27 March 2007 (UTC)

Mathematical discovery can happen in many ways, and a given fact can be proved in many ways. Suppose we have a sphere with a given radius, say a radius of 1. If we expand it in all three independent directions in space so that it becomes a sphere of radius r, then we can easily convince ourselves that the volume is multiplied by r3 (that is, a factor of r for each direction). So originally, someone might make a really big ball and fill it with water, then drain and measure the contents. That would empirically suggest the 4⁄3π constant. Of course, π cannot be expressed as a ratio of integers, and our measurements will always have a little error, a little uncertainty, so this could only help us guess. But most mathematical theorems begin with guesses derived from observations. (Usually the observations are more abstract than this and the guesses more difficult to explain.)
This particular relationship holds an interesting position in the history of mathematics, because Archimedes proved it in antiquity, a very long time before calculus was invented, which itself was long before calculus was placed on a solid formal foundation using "limits". He was more proud of this accomplishment than of any other, and I can't say I blame him! You can find his description of his method in his own words (translated into English) here, starting at page 191.
He states his discovery in a slightly different form, saying that if we fit a cylinder tightly around the sphere (so that its base has the same radius and its height is the diameter), then the volume of the cylinder is half again as large as that of the sphere. He assumes we already know that the volume of a cylinder is the height times the area of the base, and that the area of a circle of radius r is πr2. So the cylinder volume is (2r)(πr2), and a little algebra gives the formula we want.
You may enjoy reading Archimedes first. I think it will give you a greater appreciation for modern methods, since this is a problem a first-year calculus student should be able to solve quickly and easily. Then when someone suggests calculus is hard, you can reply "Compared to what?"!
For an easier warm-up exercise, read our "area of a disk" article, which also explains both ancient and modern methods. --KSmrqT 23:14, 27 March 2007 (UTC)

Thanks for all the answers! I thought that it could be done reasonably easily with calculus (but I don't know any so I couldn't work it out), but I had read that it had been found by Archimedes, and I know that he lived a long time before calculus was discovered and so I wanted to find out what his method was. Thanks for all the info --80.229.152.246 16:15, 28 March 2007 (UTC)