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 October 27th, 2007, 10:03 AM #1 Newbie   Joined: Oct 2007 Posts: 5 Thanks: 0 Polynomials and Cones Hi all, I'm new both here and to mathematics. I'm working on factoring 3rd and 4th order polynomials and finding it difficult to understand. What I want to avoid is getting into a habit of learning math by memorizing procedures and forumulae, which is exactly how I was originally taught in highschool. Lately I have started to appreciate math conceptually and imaginatively. By that I mean that I can see *why* a simple thing like a cube can be described by mathematics - or is mathematics? (See my other thread...) That kind of learning is dominant in early grades of learning, but eventually gets abandoned - probably due to difficulty - and math becomes memorizing and solving hundreds of problems. Great preparation for a test but I was not learning anything at all. I have got to Polynomials working through Barron's "Algebra" book and hit a wall. I could see how the 2nd order polynomials broke down into factors but I couldn't not really visualize what was going on. Also, it was unhelpful that the book recommended "guessing" as a way of finding the solutions. I won't always be working with problems which I can guess at, and that leaves me incapacitated. The 3rd and 4th order stuff is just beyond me at this point. I can see the solutions there on the page and follow it but I feel like I'm just memorizing procedures again. I want to understand this stuff more deeply, if that makes sense. What I am trying to do is develop a conceptual framework in my mind of mathematics for each unit. I'm not trying to accumulate abilities, I'm trying to get my imagination into math. Does this make sense? Therefore, can anyone recommend to me a resource which will help me in this direction? Very disappointed with my textbooks and the books at the bookstore... and my university library stuff is either too basic or too advanced for me. There seems to be no literature which can teach mathematical *thinking*. Why is that? As an example of what doesn't work for me, any time the author does little more than show the transformation of an equation by steps all I get is memorization or procedure. Math books for example, which have very little written english, are usually quite useless resources for me in this direction. I think that is because I have no yet learned to think mathematically. Without that imaginative aspect, it is all very "surface" level and uninteresting. But when someone shows me how to imagine what I am seeing in the math I get an "Ah ha!"... I'm hoping there is an author here or elsewhere that might be able to do that for the more advanced stuff. A polynomial equivalent of showing how a line becomes a plane, so to speak. Hope I am making sense.
 October 28th, 2007, 01:21 PM #2 Site Founder     Joined: Nov 2006 From: France Posts: 824 Thanks: 7 Very clever point you are discussing now, as it seems that a mathematician reaches its limits once he loses his ability to conceptualize. I myself cannot conceptualize anything in advanced complex analysis, which doesn't prevent me to solve the problems by analogies but I don't really have a logical framework in my mind (for those of you who are somewhat advanced in math, how do you represent yourself an holomorphic function ?). Concerning the 2nd degree polynomials, you can see where the things come from by "completing the square". This is quite a natural approach indeed, as when you have such a polynomial, you may want to write it as a product of two first degree polynomials, and when dealing with squares, things get easier. As for guessing the roots, well, you may want to have a look at the rational root test first. Of course, then, there are a lot of numeric algorithms to find roots but i think this is not the topic of this post. You'll find out that conceptualizing polynomials is pretty easy after a while, but things may become pretty trickier when you start to tackle notions which are generalizations of no real-life ground (like topological spaces), because then you must be able to conceptualize these notions for yourself.
 October 28th, 2007, 03:48 PM #4 Global Moderator   Joined: Dec 2006 Posts: 19,502 Thanks: 1739 Don't worry about 3rd and 4th order polynomials. If you can find one factor, you can simplify the problem by dividing by that factor. Also, a polynomial of odd degree with real coefficients must have at least one real zero. Anything beyond that is just more trouble than it's worth at the stage you're at.
 October 30th, 2007, 02:29 PM #5 Newbie   Joined: Oct 2007 Posts: 5 Thanks: 0 Thanks very much for the help. Good advice too, I fear rationalized frustration creeps in sometimes. cknapp the first section of your example was really helpful, thinking of how the factors lock together. This might be a dumb question, but since the eqation is a square why is it a parabola? Is that because it is a plane? I found a picture here http://en.wikipedia.org/wiki/Image:Conic_sections_2.png that seems to confirm that. So does every value for x and y in the slice of that cone section get described by the equation? Where do the a & b's fit in the plane? And just a curiousity - would we ever have first, (as in order of events), the factors of a polynomial before the equation? What would those be? Getting there I hope... I'm not giving up
October 31st, 2007, 03:18 PM   #6
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Quote:
 Originally Posted by AdamL12 Thanks very much for the help. Good advice too, I fear rationalized frustration creeps in sometimes. cknapp the first section of your example was really helpful, thinking of how the factors lock together. This might be a dumb question, but since the eqation is a square why is it a parabola? Is that because it is a plane?
You're welcome for the help.

This may end up being a bit abstract, but we shall see:
Functions "take in" a value and "output" another value. In general, you take one set (You've probably only dealt with the reals, and integers) and "map" it to another set (Or the same set.)

In the notation you're used to, y is typically a function of x, and both y and x are arbitrary (but related) members of the reals; which is why sometimes you'll see f(x)=2x interchangeable with y=2x.

Functions can be visually represented on a "Cartesian Space", that is, the set made by assigning each pair (member from set one, member from set 2) to a "point" in the plane.

To give an example, most people have only experience with the Cartesian space RxR, which equates to the set of all points (x,y) where x and y are real numbers. With 2 dimensions, this Cartesian space is a plane.

Now, moving back to functions; as far as I would guess, every function you'll be working with in the next few years will be on a cartesian plane between the reals and the reals. Also, y will almost always equal f(x). That is, the x axis represents one set of the reals, the y axis represents the set that the function maps x to.

So, onto our example, with parabolas. The simples tone is y=x^2.
So at every value of x, the function outputs to x^2. Visually, this is the point (x,x^2). What you see is a parabola.

Not sure how well that answered that question.

Quote:
 I found a picture here http://en.wikipedia.org/wiki/Image:Conic_sections_2.png that seems to confirm that. So does every value for x and y in the slice of that cone section get described by the equation?
This is rather difficult to explain, but in short: a plane intersecting a cone will always make a parabola. The math is messy.... yeah. Messy.

Quote:
 Where do the a & b's fit in the plane? And just a curiousity - would we ever have first, (as in order of events), the factors of a polynomial before the equation? What would those be?
Could you try to rephrase this question; I don't understand.

 November 12th, 2007, 06:33 PM #7 Newbie   Joined: Oct 2007 Posts: 5 Thanks: 0 Hi cknapp, Sorry for the delay responding, Halloween and 3 birthdays made things very busy. Thanks for this. When we look at the parabola on the graph (x,y), are we looking at the plane from the side? I'm used to thinking of planes as being set by a boundary and including all the values of x,y in between those boundaries. So, when I look at the parabola as a plane, I tend to think it must be like that. Here I drew a hasty parabola in MSPAINT and also drew some colored lines to show two different areas I want to ask about. The blue area represents what I interpret as the "plane" of the parabola. I'm not sure if those values for x,y in the blue area are actually a part of it or not. IOW, are the points x,y of a parabola *strictly* the parabola line or are the areas within and outside of that line called something particularly? The red area would be the outside of the cone, the "exclusion zone" marked by the cone boundary described by that parabola. Here is a picture from Wikipedia that does the same thing but with better artistic skills It seems that the parabolic line represents the boundary, however, I'm not sure what to call the coloured areas. Also, if you rotate the parabola about the x axis in 3d space, you can see 3 areas, because the parabola is flat. The area on the "plane" of the parabola is green (my guess), but the red area completely surrounds it. In other words, does the parabola extend infinitely in the "Z" axis? I guess I'm getting away from cones there a bit. Interesting stuff anyhow. Thanks for the help.
November 13th, 2007, 08:02 AM   #8
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Quote:
 Originally Posted by AdamL12 Sorry for the delay responding, Halloween and 3 birthdays made things very busy. Thanks for this. When we look at the parabola on the graph (x,y), are we looking at the plane from the side? I'm used to thinking of planes as being set by a boundary and including all the values of x,y in between those boundaries. So, when I look at the parabola as a plane, I tend to think it must be like that.
A plane is an infinitely extended flat surface. So, there are no bounds on x or y within a plane. Obviously, we can't draw a "whole" plane, so we draw the "important" section instead. So, looking at your drawing, the whole picture is a section of the plane, bounded by a rectangle; the blue area is the area above the parabola, and the red area is the area below the parabola.

Quote:
 I'm not sure if those values for x,y in the blue area are actually a part of it or not. IOW, are the points x,y of a parabola *strictly* the parabola line or are the areas within and outside of that line called something particularly?
The line is the parabola, the outside is under the parabola, and the inside is above the parabola. If the parabola were flipped, obviously the area bounded by the parabola would be "under"

Quote:
 Also, if you rotate the parabola about the x axis in 3d space, you can see 3 areas, because the parabola is flat. The area on the "plane" of the parabola is green (my guess), but the red area completely surrounds it. In other words, does the parabola extend infinitely in the "Z" axis?
If you took the plane you drew earlier, and superimposed it onto the 3d graph, so that you could see the plane the parabola is on extending out, you would have 3 shapes: a cone, and a plane, which intersect; and a parabola, which represents the intersection between the two. A parabola is a 2-dimensional shape, so it only falls on that plane. The parabola is infinite, as is the plane, and the cone.
If you bounded either the plane or the cone, that is, took only the region on the object that is above [or below] a certain value of x,y, and or z, the intersection between the two would obviously be bounded. but typically, you don't place artificial bounds on a function.

Cheers

 November 13th, 2007, 09:19 AM #9 Newbie   Joined: Nov 2007 From: East West Kerry Posts: 2 Thanks: 0 For me the best way to solve this problem is with recourse to the factor theorem. If a,b,c,d,.. are roots of the polynomial. the poly factors as (x-a)(x-b)(x-c)... This is easy to visualise if you consider the graph of the poly. At any of the roots, a say, p(a)=0. Hence your factorisation p(x)=(x-a)(x-b)(x-c)... will equal 0 also. Hence to factor a poly plot the poly from [-4,4] at -4,-3,...,4. When the plot crosses the x-axis, the poly is zero there and hence the cross point, x=a say, is a root. So factor out (x-a) with long division of polys - this is intuitive enough. Then work with your new poly; or revert to the normal equations. This works because realistically any poly you will be asked to factor will have an integer root in [-4,4]. This will work for the third degree derived poly also. When you get down to a two-degree poly you have to realise that the roots are going to be often be surds and you won't be able to catch the roots by chance. Loooking at the graph of p(x) and understanding the factor theorem will satisfy the conceptional standards I hope. Let p(x) in P_2[x] have factors (Y-b/2)(Y+b/2). See where that leads. Now a teaser: (x-a)(x-b)(x-c)...(x-z)=?
 November 15th, 2007, 01:44 PM #10 Newbie   Joined: Oct 2007 Posts: 5 Thanks: 0 Got it now cknapp, thanks!

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