Wikipedia:Reference desk/Archives/Mathematics/2006 November 3

= November 3 =

Polynomials
Hi, I was wondering what some aplications of polynomials are in real life, besides of the standard "throwing of a ball," or "calculating compound interest" kind of stuff. Thanks! --novama


 * Polynomials are heavily used "behind the scenes". For example, the shape of this glyph for the letter “g”:
 * g
 * is described by polynomial curves, called Bézier curves. Surfaces like the bodies of airplanes and automobiles are designed using polynomials, via CAD software. Structural properties, like the strength of a bridge, are analyzed using polynomials, via FEM software. The elaborate computer models used to predict the weather also depend on polynomials. While the most common tool of applied mathematics is the line or plane (linear algebra), the next most common is likely the polynomial. --KSmrqT 05:52, 3 November 2006 (UTC)
 * is described by polynomial curves, called Bézier curves. Surfaces like the bodies of airplanes and automobiles are designed using polynomials, via CAD software. Structural properties, like the strength of a bridge, are analyzed using polynomials, via FEM software. The elaborate computer models used to predict the weather also depend on polynomials. While the most common tool of applied mathematics is the line or plane (linear algebra), the next most common is likely the polynomial. --KSmrqT 05:52, 3 November 2006 (UTC)


 * Another use of polynomials is in Taylor polynomials. Polynomials are very "well-behaved" functions, so if we have a less cooperative function, we can approximate it using a Taylor polynomial and then we can analyse the Taylor polynomial instead of the original function. Maelin (Talk | Contribs) 10:15, 5 November 2006 (UTC)

Optimal angle for a thrown projectile
It's obvious that when throwing a projectile on level ground for the longest distance the best angle is 45. But If you're throwing a projectile from a given height above the ground, how do you determine the optimal angle to throw it at (given a velocity). I've thought about this problem for a while, and I don't even know where to start. I want to know how I would go about solving it (would I use differential equations? Optimizations?) any help would be appreciated.


 * This can be solved by more elementary means. Let h be the given height, v the initial velocity with which the projectile is thrown, and α the optimal angle to be determined. Express x (horizontal position) and y (vertical position) as functions of t (time), using g for the gravitational acceleration:
 * x(t) = vt cos α
 * y(t) = vt sin α − &#x00bd;gt2 + h
 * Here we keep the direction of the throw in the first quadrant, where both the sin and cos are positive. Let T be the time when the projectile hits the ground. So then y(T) = 0, which gives us the equation
 * vT sin α − &#x00bd;gT2 + h = 0.
 * Put X = x(T). We want to find α maximizing X. Equivalently, we can maximize
 * X2 = v2T2 cos2 α = v2T2 (1 − sin2 α).
 * Using the equation resulting from y(T) = 0, you can now eliminate sin α, resulting in an equation expressing X2 as a function of T2. So find the (positive) value of T maximizing X2. Substitute this in the equation from y(T) = 0, and solve for sin α. (Leaving T unexpanded as long as you can should simplify the manipulations.) Then α is the arcsin of the result.
 * If you introduce the ratio between initial potential energy and kinetic energy as a parameter:
 * ρ = Epot / Ekin = 2gh / v2,
 * you can express α as a function of ρ, eliminating h, v and g in one fell swoop. --Lambiam Talk  06:19, 3 November 2006 (UTC)

Are these properties of lattices easy to prove?
Hello,

I'm taking my first steps into the world of lattices. Unfortunately my syllabus gives a lot of properties without proofs, so they could either be really trivial or just too hard to prove in this course. It could be either but I've been thinking about it for some time now.

1. a lattice $$(P,\leq )$$ is sais to be distributive if

$$a\vee(b\wedge c)=(a\vee b)\wedge (a\vee c)$$

Now my syllabus doesn't even give the following equivalent property but the article does: it's equivalent to : $$ a\wedge(b\vee c)=(a\wedge b)\vee (a\wedge c)$$

Looks very dual, but is it trivial?!

2. In a bounded lattice with smallest element 0 and biggest element 1, a complement of an element a is an element x such that $$a\vee x=0$$ and$$ a\wedge x=0$$ Now if we also assume distributivity, which we will assume, one can prove there can be at most one complement of an element, which we will denote as $$a'$$

Now I should accept these properties : $$(a\vee b)'=(a')\wedge (b')$$

$$(a\wedge b)'=(a')\vee (b')$$

Now I am convinced that I need BOTH of the (dual) properties in my first question to solve this. Is it trivial then?

Thank you very much, Evilbu 12:45, 3 November 2006 (UTC)


 * I don't know if it is trivial, but it is not particularly hard. You will need the absorptive properties
 * $$a\vee(a\wedge b)=a$$
 * $$a\wedge(a\vee b)=a$$
 * Then use the given distributive property with the substitution
 * $$(a,b,c):=(a\wedge b,a,c)$$
 * and simplify. --Lambiam Talk  13:30, 3 November 2006 (UTC)


 * Okay thanks, but let us be clear : you are giving me a hint for the first one right?
 * It looks very good but then I got this : $$(x\wedge y)\vee (x\wedge z)=x \wedge ((x\wedge y)\vee z)$$. That's one x in the RHS too many :(.  What did I do wrong?Evilbu 13:51, 3 November 2006 (UTC)


 * You did nothing wrong; you just are not finished. If you follow Lambian's suggestion, you continue by applying apply the first distributive law once more on the expression $$(x\wedge y)\vee z$$, too.
 * As for your second exercise, I hope that $$ a\vee x=0$$ is just a typo. If you correct it, you should be able to prove something about 0' and 1' immediately, and then see that already the property $$ (a\vee b)'=(a')\wedge (b')$$ alone is meaningless if a' isn't well-defined.  (Test b=1.)  It is also quite possible to prove the unicity directly from the distributivity, without accepting either of the extra conditions.


 * A general remark to your first question: What makes it non-intuitive is that it in general is 'locally false'. In other words, if you just know that $$a\vee(b\wedge c)=(a\vee b)\wedge (a\vee c)$$ for three specific elements a, b, and c, then it does not follow that $$a\wedge(b\vee c)=(a\wedge b)\vee (a\wedge c)$$. In fact, consider the lattice $$N_5 = \{0,1,a,b,c\}\,$$, with the partial order given by $$0 < a < b < 1\,$$ and $$0 < c < 1\,$$, but no other relations (so that $$a \wedge c = b \wedge c = 0$$ and $$a \vee c = b \vee c = 1$$), then
 * $$a\vee(b\wedge c) = a = (a\vee b)\wedge (a\vee c)$$ but $$a\wedge(b\vee c) = a \neq b = (a\wedge b)\vee (a\wedge c)$$.
 * Thus, in general, you employ that the first distributive law holds for all triples $$(a,b,c)\,$$ of elements, in order to deduce the second distributive law even for a particular triple.
 * (On the other hand, if there is no sublattice isomorphic to $$N_5\,$$ of your lattice, then the lattice is modular; and if you are given three elements in a modular lattice, then either none or all six of the possible distributivity relations among these three elements hold. This is an elementary but slightly harder exercise to prove.) JoergenB 16:48, 3 November 2006 (UTC)


 * Thanks, so now (for the first question) I got that I have to prove $$x\wedge ((x\vee z)\wedge (y\vee z))=x\wedge(y\vee z)$$

Now I guess the important things to note is that $$(x\vee z)\geq x\geq (x\wedge(y\vee z))$$?Evilbu 22:19, 3 November 2006 (UTC)

expected value
Hello I'm having trouble with my maths homework. I've read the textbook chapter, Expected value and some googling results and I still can't figure this out. COuld somoene point me in the right direction or give me a hint?

Suppose that we roll a die until a 6 comes up. a. what is the probability that we roll the die n times? b. what is the expected number of times we roll the die?

Thanks, -DAN


 * Dear Dan, you are in luck. I have notified the Koreans and they have agreed to solve your problem and share the answer with the world.


 * Please see: ai.dgu.ac.kr/class/pds/2006-DiscreteMath_HW2_Sol.pdf


 * (edit conflict) Your answer to a will give you the means to calculate b. To examine a, it may be easiest to start with some special cases of n:  What is the probability of n = 1 (alternatively stated, what is the probability of getting a 6 on the first roll)?  If n = 2, then you rolled a non-six, then a six.  What is the probability of that occuring?  Can you generalize this probability to n > 2?


 * Once you have a function pn which describes the probability at each n, use the expected value formula to answer part b
 * $$\sum_{i=1}^{\infty}n*p_n$$
 * Write again if you need help evaluating this sum. --TeaDrinker 22:28, 3 November 2006 (UTC)


 * You might also be interested in geometric distribution (where rolling a 6 is a "success"), but be careful which version you use. – AlbinoMonkey (Talk) 23:46, 3 November 2006 (UTC)