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September 17

Formula for placing things equidistant along a circle circumference

As the subject implies, i need to find a formula to place items an equal distance apart on the circumference of a circle. Specifically i need 24 equidistant locations. I dont know quite yet how i will define the circle, probably with parametrics? But ive looked on google for an answer to this and there doesnt seem to be any. Please help?

Thanks! 137.81.115.58 (talk) 01:26, 17 September 2009 (UTC)[reply]

You can define a circle with radius R in terms of the angle θ around the origin with the parametric equations x = Rcosθ, y = Rsinθ. Evenly spaced points along the circle are at evenly spaced angles, so we divide the full 2π angle of the circle up into 24ths. The points with coordinates (Rcos(2πn/24), Rsin(2πn/24)) with n ranging from 1 to 24 satisfy the conditions you're looking for. Rckrone (talk) 02:44, 17 September 2009 (UTC)[reply]

See Root of unity. Bo Jacoby (talk) 08:58, 17 September 2009 (UTC).[reply]

Note that 24 = 6x2x2 - you can easily divide the circle into 6 arcs, and then half those arcs twice. When you can define how you want to describe the circle it will be easy to give a method for the process.87.102.94.154 (talk) 11:12, 17 September 2009 (UTC)[reply]

help me ...it will safe my future..??

sir/mam....which extra book i adopt ?? i am the student of commerce stream..i choose maths because i want to become good C.A.

you are good to go rambharose..best of luck

You are going to need to be more specific. --Tango (talk) 17:56, 17 September 2009 (UTC)[reply]

Quite so :o) ~~ Dr Dec (Talk) ~~ 18:30, 17 September 2009 (UTC)[reply]

You seem like a nice person, though I'm not sure who 'rambharose' is. I'm going to guess that you need an introductory maths book for a commerse coarse, possibly including accountancy and economic maths? (and maybe statistics?) Maybe someone else would be good enough to give a suggestion?83.100.251.196 (talk) 19:12, 17 September 2009 (UTC)[reply]

Rambharose means in trust of Rama(God) in Hindi:)--Shahab (talk) 18:42, 19 September 2009 (UTC)[reply]

Writing Roman Numerals

Hello

Hope someone can answer this as I can't find it in the system. I understand how roman numerals work but get stuck at large numbers, ie one million is M with a bar, but how do you write 2 million or 10 million etc ?

--Bruceglasgow (talk) 18:42, 17 September 2009 (UTC)[reply]

Two million derives from one million just as two derives from one. If one million is M, two million is easy to generate: MM. Alternately, parentheses may be used to represent "times 1000", such that two million is (MM). Ten million requires some extrapolation from our Roman numerals article, but might be fairly represented as (X), or a longer form. — Lomn 19:21, 17 September 2009 (UTC)[reply]

Equation of oscillator - flat section

A question arises from another persons question about oscillations of things travelling through a spherical body exerting and inverse square force law Wikipedia:Reference_desk/Science#Gravity_at_the_center_of_a_body.

My question begins: If a spherical centre part is removed then within this part the field force is zero. (as per Shell theorem), assuming a body is dropped into the whole hollow sphere (assumming a negligable cylindrical hole for it to travel in or whatever) - then it seems to me that the body will accelerate towards the centre, then when it reaches the empty part no field is experienced and the body will move at constant velocity, until it exits the spherical hole at which point the deacceleration occurs, and so on. Thus it oscillates.. but with a part at which the acceleration is zero...

So the equation of motion (eg acceleration vs time) I would expect to be expressable as a fourier series (ie a series of powers of sines or similar).. Since the function is periodic. But at the times at which the particle is inside the spherical void the acceleration is 0, as are the derivatives of the accelerations.. This seems to contradict the taylor-maclaurin type theorems (ie that a function at any point can be constructed from the function at a given time and all it's derivatives.

My questions are:

Can the function be expressed as a series of sines, ie a fourier series
- If not what sort of function
What goes wrong with the taylor-maclaurin series here? (or what is my error?)83.100.251.196 (talk) 19:22, 17 September 2009 (UTC)[reply]

Taylor-Maclaurin-type results apply to analytic functions. This function is not analytic. The Fourier series of position as a function of time would converge to that function. This is one of the difference between Fourier series and power series. Michael Hardy (talk) 19:57, 17 September 2009 (UTC)[reply]

The page analytic function seems to define itself as "A function is analytic if and only if it is equal to its Taylor series in some neighborhood of every point." (which is somewhat circular in respect to your statement, but I'm not arguing this).

The actual function of this oscillation can be expressed as a fourier series, and so also as a power series (is there an exception here?). I imagine that the square wave is a simple(r) example to discuss.

Is there another way of showing that a given power series (or fourier series) will not be analytic (other than trying the taylor maclaurin series).

Does the fact that eventually a differential of the function includes a discontinuity mean that it will be non-analytic ? - it seems to me that this will be the case, but can't imagine a proof.83.100.251.196 (talk)

You can't have a power series that's not analytic, since a power series is analytic by definition. You're right that a function has to be smooth in order to be analytic. To prove it, any power series is smooth (assuming it converges), so if a function isn't smooth, it's not equal to any power series. The Taylor series article mentions that the converse isn't true: there are smooth functions that aren't analytic.

As for the Fourier series, sine functions are analytic, and so any finite sum of sine functions is analytic, but that doesn't necessarily extend to an infinite sum. So the partial sums of the Fourier series are analytic, but they may converge to something that's not analytic, which is the case with the function being discussed. Rckrone (talk) 21:41, 17 September 2009 (UTC)[reply]

A power series is analytic by definition: not quite so; to be precise, you need to prove that at any point within the convergence disk of your power series, the function represented by the series is also representable by a power series centered at that point: not difficult, but still a computation. --84.221.68.30 (talk) 22:34, 17 September 2009 (UTC)[reply]

ok thanks so far, For the square wave the 'x¹' term would be of the form 1+1/3+1/5+1/7... etc that relates to pi , (I haven't checked the other terms x^3 etc) , but it looks like it can be expressed as a infinite power series (is this right?).

I'm not very familiar with the radius of convergence - ~~though at first sight it looks like the square wave will converge for all values of x ..~~ ah not for the differentials.

I'm fairly familiar with smooth functions though.

It would be great if a few points could be clarified - specifically about the square wave (I'm confused because I think it is non-analytic, yet expressible as a power series):

Can the square wave be expressed in the form a₀ + a₁x¹ + a₂x² + a₃x³ + etc all the way to infinity. (as I suspect) - but this form does not converge - but oscillates and diverges does this relate to it being non-analytic?
- Is this form a power series, or is this the definition of a power series more complex.
~~The radius of convergence of the square wave - infinite ? but only for the fourier series~~ not for the differentials, starting to partially understand.
Apart from lacking smoothness (ignoring smooth non-analytic functions for now) is there another test for lack of analyticalisity in a function? (apart from the obvious taylor maclaurin method)? The test doesn't have to be absolute - just able to reject a set of equations.. Thanks/83.100.251.196 (talk) 23:39, 17 September 2009 (UTC)[reply]

I think you might be confusing approximating the square wave with a sequence of analytic functions (such as with partial Fourier series), and a single power series that converges to the square wave. There's no power series that converges to the square wave, which is to say that it's not analytic. We know this because it's not smooth, and all analytic functions are smooth. We can however approximate the square wave with analytic functions, for example with the Fourier series. The partial sums of the Fourier series give us a sequence of different analytic functions that approach the square wave (at most points).

Besides smoothness I don't know what other criteria there are to decide if a function is analytic besides just comparing it to its Taylor series. Maybe someone else who knows more about this has something. Rckrone (talk) 01:23, 18 September 2009 (UTC)[reply]

Yes exactly that, I know see why attempting to make the corresponding power series for the fourier series fails - because the terms for x are all infinite (or non converging) beyond x¹.

I get the principle now thanks.

Resolved

83.100.251.196 (talk) 11:25, 18 September 2009 (UTC)[reply]

The taylor series for g(z) defines the fourier series for f(x) = g(e^ix) . Bo Jacoby (talk) 10:04, 18 September 2009 (UTC).[reply]

I'll have a look at that too, thanks.83.100.251.196 (talk) 11:25, 18 September 2009 (UTC)[reply]

This corresponds to the rather unusual case when the nth (exponential) Fourier coefficient of f is zero for every negative n. In general, the Laurent series for g defines the Fourier series for f. However, in practice the Laurent series corresponding to the Fourier series of a typical periodic function will have both radii of convergence equal to 1, meaning that it's degenerate (its annulus of convergence is empty). — Emil J. 16:34, 18 September 2009 (UTC)[reply]