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April 15[edit]

roots of an equation[edit]

How many roots does the equation $z^{3}+{\bar {z}}=0$ have?. I guess I could plug in x+iy in place of z, seperate the real and imaginary parts and find out a simultaneous solution but is there a quicker way? Thanks.-Shahab (talk) 05:11, 15 April 2010 (UTC)[reply]

Transform the equation to the form

z^{3}=-{\bar {z}}

, then try substituting

{\bar {z}}=|z|e^{arg(z)\,i}

and consequently

z=|z|e^{-arg(z)\,i}

. --CiaPan (talk) 06:04, 15 April 2010 (UTC)[reply]

It should be clear that any root necessarily has |z| = 1 or |z| = 0. So z = 0 is a root and for |z| = 1 you can multiply through by z to get

z^{4}+|z|^{2}=0

which means z satisfies

z^{4}+1=0

and then you can find the roots of that. Rckrone (talk) 07:52, 15 April 2010 (UTC)[reply]

Notation[edit]

Suppose you are given two functions $f(x)$ and $g(x)$

Given the equation $f(x)=g(x)$ , what is the correct notation for rearranging that equation to express $x$ in terms of $f$ and $g$ ?

$x=...$ (something containing $f$ and $g$ )

--Alphador (talk) 07:02, 15 April 2010 (UTC)[reply]

{x : f(x)=g(x)} is the set of solutions to the equation. x = (f−g)⁻¹(0) is the supposed unique solution. Bo Jacoby (talk) 07:41, 15 April 2010 (UTC).[reply]

How about, where

f(x)=g(x)=y\,\!

, then

x=f^{-1}(y)=g^{-1}(y)\,\!

. ~Kaimbridge~ (talk) 15:54, 15 April 2010 (UTC)[reply]

nature makes its move[edit]

The article probability space says:

Once the probability space is established, it is assumed that “nature” makes its move and selects a single outcome, ω, from the sample space Ω. Then we say that all events from

\scriptstyle {\mathcal {F}}

containing the selected outcome ω (recall that each event is a subset of Ω) “have occurred”. The selection performed by nature is done in such a way that if we were to repeat the experiment an infinite number of times, the relative frequencies of occurrence of each of the events would have coincided with the probabilities prescribed by the function P.

What the heck is that supposed to mean? Is there a more formal description of what happens here? For example, if I repeatedly flip an unbiased coin (infinite Bernoulli process) and "nature" selects the sequence H,T,H,T,H,T... then the frequencies of occurrence coincide with the expected 50% heads, but I wouldn't call it random. Note, I also brought this up on the article talk page. I am basically asking what it means to "observe" a random variable. Thanks. 66.127.52.47 (talk) 08:42, 15 April 2010 (UTC)[reply]

Good question! Probabilities describe unknown or undetermined situations. The result of flipping a coin is unknown until the coin is flipped - then the result is known. Strictly speaking the word 'random' applies not to the result but rather to the situation. Flipping a coin six times, the outcome HTHTHT is not more probable than TTTTTT. Every outcome is unique and no outcome is random! But classifying the possible outcomes according to the number of tails gives different odds to different classes. There is one way to obtain 6 tails, (namely TTTTTT). There are 6 ways to obtain 5 tails, (namely TTTTTH, TTTTHT, TTTHTT, TTHTTT, THTTTT, and HTTTTT). There are 15 ways to obtain 4 tails, (namely TTTTHH, TTTHTH, TTTHHT, TTHTTH, TTHTHT, TTHHTT, THTTTH, THTTHT, THTHTT, THHTTT, HTTTTH, HTTTHT, HTTHTT, HTHTTT, and HHTTTT). There are 20 ways to obtain 3 tails, (namely TTTHHH, TTHTHH, TTHHTH, TTHHHT, THTTHH, THTHTH, THTHHT, THHTTH, THHTHT, THHHTT, HTTTHH, HTTHTH, HTTHHT, HTHTTH, HTHTHT, HTHHTT, HHTTTH, HHTTHT, HHTHTT, and HHHTTT). There are 15 ways to obtain 2 tails, there are 6 ways to obtain 1 tail, and there is only one way to obtain no tails (namely HHHHHH). The total number of outcomes is 64, and the odds for the 7 classes are as 1:6:15:20:15:6:1. Each outcome defines a class. The classes are called events. The idea of a probability space focus on events rather than on outcomes. I hope this helps. Bo Jacoby (talk) 09:50, 15 April 2010 (UTC).[reply]

Thanks for replying, sorry the question wasn't clear. I only meant HTHTHT as the first 6 flips of an infinite sequence that kept strictly alternating between H and T, i.e. a trivial deterministic sequence that still met the defininition, that if you repeat the experiment an infinite number of times, you get 50% probability of H and 50% T. I guess in the case of the Bernoulli process, we can say we're definining an infinite family of random variables r1,r2... so that any finite collection of them have a certain joint probability distribution, and then define a sequence with that property as "iid" (rather than going the other way, starting from a vaguely defined term and calculating the joint distribution). But I don't know what happens when the sample space is more complicated. I'm trying to understand the formalism of probability in terms of sigma-algebras and so forth, not really about flipping coins. But I'm stuck at the part where you actually draw a sample from a distribution and observe it, and how that concept is formalized. 66.127.52.47 (talk) 10:22, 15 April 2010 (UTC)[reply]

If the probability space is finite, (as is the case when flipping a coin 6 times), any subset can be regarded an event having a probability, but if the probability space is infinite, (as is the case when flipping a coin an infinite number of times), individual outcomes (such as your THTHTH...) are not necessarily events. To deal with this complication one must define which subsets are events. The answer is that the set of events must constitute a sigma-algebra. Bo Jacoby (talk) 11:39, 15 April 2010 (UTC).[reply]

Thanks, that's a good point about a single outcome not being an event on the infinite process, but I'm still not doing a good job explaining what my question is trying to get at. Consider the following proposition:

Let U be the uniform probability distribution on the interval [0,1]. Then for any pair of samples x,y drawn from U, we have x+y≤2.

This proposition is obviously true, but I would like to formally prove it as a theorem starting from the definitions of random variables, probability spaces, etc. (and likewise for similar statements). In order to do that, I'm trying to find out how axiomatic probability theory describes the process of drawing a sample from a distribution and inspecting its value. But the closest thing I can find is this thing about "nature making its move", which doesn't sound like a mathematical statement at all. How does "nature" pick ω from Ω, when Ω can be some messy uncountable set? What does it mean to "repeat the experiment" even once, and expect to get a different outcome? I'm still not sure if I'm asking something sensible. 66.127.52.47 (talk) 18:04, 15 April 2010 (UTC)[reply]

In order to prove x+y≤2 you need not describe 'the process of drawing a sample from a distribution and inspecting its value'. The informal talk about "nature making its move" is an attempt to explain the interpretation of the mathematical theory, but is is not part of the mathematical theory. Bo Jacoby (talk) 18:37, 17 April 2010 (UTC).[reply]

The Earth Ellipsoid[edit]

How many points on the Earth need to be precisely defined so as to uniquely define the Ellipsoid of the Earth. —Preceding unsigned comment added by 116.90.224.116 (talk) 10:41, 15 April 2010 (UTC)[reply]

An Earth ellipsoid reference frame model is an oblate spheroid defined by two parameters: its equatorial axis and its polar axis. Adding in three degrees of freedom for the co-ordinates of the centre of the spheroid and two more for the direction of its polar axis, we have seven degrees of freedom altogether. Assuming you can measure the Cartesian co-ordinates of one point relative to another with any required degree of accuracy, then measuring the relative co-ordinates of any seven points in general position on the ellipsoid should give you enough data to uniquely define the ellipsoid ("in general position" means that you do not use points that are in some special relationship to one another, such as seven points all on the equator).

However, note that the Earth ellipsoid is a model, respresenting a standard datum. If you picked seven actual points on the Earth's surface, you could define an ellipsoid that passed exactly through those points - but at other places it would not be a good approximation to the Earth's surface. Earth ellipsoids are usually defined so as to resemble an "average" shape of the Earth in some sense, rather than to pass through any specific points. The more actual points you measure, the closer you can make your datum to resemble the "average". Gandalf61 (talk) 12:06, 15 April 2010 (UTC)[reply]

Excellent answer,Thanks a lot. —Preceding unsigned comment added by Amrahs (talk • contribs) 15:31, 15 April 2010 (UTC)[reply]

The answer is nine. If you have a given ellipsoid and you mark nine points on it then the given ellipsoid is the only one that can contain the nine marked points. Let me explain:

An ellipsoid is an example of a quadric. A general quadric in $xyz$ -space has the equation

a_{1}x^{2}+a_{2}y^{2}+a_{3}z^{2}+a_{4}xy+a_{5}xz+a_{6}yz+a_{7}x+a_{8}y+a_{9}z+a_{10}=0,\,

where each of the $a_{i}$ are real numbers. Thus, for every a ∈ ℝ¹⁰ we get a quadric surface. However, different points of ℝ¹⁰ give the same quadric. We see that a ∈ ℝ¹⁰ and λa ∈ ℝ¹⁰ give the same quadric for any real number λ ≠ 0:

(\lambda a_{1})x^{2}+(\lambda a_{2})y^{2}+(\lambda a_{3})z^{2}+(\lambda a_{4})xy+(\lambda a_{5})xz+(\lambda a_{6})yz+(\lambda a_{7})x+(\lambda a_{8})y+(\lambda a_{9})z+(\lambda a_{10})=0\,

\iff \lambda (a_{1}x^{2}+a_{2}y^{2}+a_{3}z^{2}+a_{4}xy+a_{5}xz+a_{6}yz+a_{7}x+a_{8}y+a_{9}z+a_{10})=0.\,

\iff a_{1}x^{2}+a_{2}y^{2}+a_{3}z^{2}+a_{4}xy+a_{5}xz+a_{6}yz+a_{7}x+a_{8}y+a_{9}z+a_{10}=0.\,

What this means is that the space of quadrics is given by the quotient space ℝ¹⁰/~ where a ~ b if and only if there exists λ ≠ 0 such that a = λb. This quotient space is denoted by ℝℙ⁹, its elements are given in homogeneous coordinates and are of the form (a₁:a₂: … :a₁₀) ∈ ℝℙ⁹. The symbol ℙ is used to denote the projective space. The bottom line here is that ℝℙ⁹ is locally a nine-dimensional manifold. So the space of quadrics has nine degrees of freedom. There are some genericity conditions to worry about in the abstract setting. But in the case of ellipsoids − which are generic quadrics − we need to specify nine points to uniquely define an ellipsoid. So if you have a given ellipsoid and you mark nine points on it then the given ellipsoid is the only one that can contain the nine marked points.

Regarding the genericity conditions. You could have a sphere, which is an example of an ellipsoid, and you could mark your nine points on the equator. Then the only ellipsoid that contains those points would by your sphere. However, the plane (an example of a degenerate, i.e. non-generic, quadric) that contains the equator would also contain these nine points. So the genericity conditions apply to your choice of points. Choosing points on the equator is highly non-generic because they are co-linear. If you perturb any point slightly then you will almost always have nine non-co-linear points. So to satisfy the genericity conditions you need to choose nine points in general position. •• Fly by Night (talk) 15:35, 15 April 2010 (UTC)[reply]

Yes, a general scalene elipsoid has nine degrees of freedom, but an Earth ellipsoid is not a scalene ellipsoid, it is an oblate spheroid - it has rotational symmetry about its polar axis - so it only has seven degrees of freedom. Gandalf61 (talk) 09:54, 16 April 2010 (UTC)[reply]

The earth is approximately spherical. Ellipsoidal is an even closer approximation. The actual shape of the equipotential surface is called the geoid and it is not quite ellipsoidal, because of the unequal distribution of mass of the continents, etc. And even more accurately than the equipotential surface, you really have to describe every single point, to account for every rock, hill, bump on the ground, etc. 66.127.52.47 (talk) 17:41, 15 April 2010 (UTC)[reply]

How does counting degrees of freedom automatically translate into the number of points necessary to define a surface? Doesn't a general line in 3-space have three degrees of freedom? But isn't such a line is uniquely defined by only two points? I'm thinking that perhaps it works for the ellipsoid because it is a n-1 surface in n space (where n=3). What is the name of the general principle here? 124.157.234.136 (talk) 15:11, 16 April 2010 (UTC)[reply]

A general line in 3-space actually has four degrees of freedom, because its equation is

a_{1}x+a_{2}=a_{3}y+a_{4}=z

and each set of four numbers defines a different line. Each given point on the line gives you two equations in the four parameters, so you only need two (non-identical) points to determine all parameters and hence the line uniquely. Whereas with an ellipsoid each point only gives you one equation in its parameters, so to uniquely determine the ellipsoid you need as many points as there are (independent) parameters. Not sure what the name of the general principle is. Gandalf61 (talk) 19:53, 16 April 2010 (UTC)[reply]

In general, does a point on a m-surface in n-space provide

n-m

equations in the parameters of the surface? 124.157.234.136 (talk) 07:43, 17 April 2010 (UTC)[reply]

Price elasticity , quantity demanded[edit]

If the percentagechange in the quantity demanded is equal to the percentage change in the price that cuased the change in the quantity demanded, the elasticity coufficient will be equal to one? —Preceding unsigned comment added by Elista (talk • contribs) 13:07, 15 April 2010 (UTC)[reply]

The instantaneous price elasticity demand coefficient will be negative one, assuming that you meant "increasing the sales price will decrease the sales unit volume by the same percentage". Of course, this only works for small percentage changes, since it would mean increasing price by 100% (doubling the price) would decrease sales unit volume by 100% (so no sales at all). That seems rather odd, so this isn't the best way to represent price elasticity over large changes in price. Since the price-demand curve is nonlinear, the relationship isn't a direct one, but rather one described by a formula. Calculus may be used to derive this formula. StuRat (talk) 16:01, 15 April 2010 (UTC)[reply]

What an amazingly well-written article that is. Maybe we should club together and buy the poor guy that wrote (most of) it a car, or something. Elista, as the article notes, in many real world circumstances positive coefficients are used for ease, so +1 may be the "correct" answer in your context. And yeah, StuRat, you're unlike to have a PED of (-)1 for such an operation. - Jarry1250 ^{[Humorous? Discuss.]} 13:11, 16 April 2010 (UTC)[reply]