Country Boy

Is there going to be an "answer to the twin paradox" any time soon?

If |x|= x^2+ x- 3, then either x>= 0 or x< 0. if x>= 0 then |x|= x= x^2+ x- 3 so that x^2= 3. That has two solutions but only one is non-negative. If x< 0, the |x|= -x= x^2+ x- 3 so that x^2+ 2x+ 3= (x+ 2))(x+ 1)= 0. That has two negative solutions.

You can also think of this as a parallelogram with sides of length 3 and 5. |a+ b| is the length of one diagonal, |a- b| is the length of the other. In particular, since |a+ b|= 7, one half of that parallelogram, the one with the larger angle between the sides, is a triangle with sides of length 3, 5, and 7. Taking [tex]\theta[/tex] to be the angle between the two sides, by the cosine law, [tex]7^2= 3^2+ 5^2- 2(3)(5)cos(\theta)[/tex]. You can use that to solve for [itex]\theta[\itex]. then the other angle, [tex]\phi[/tex], is the supplement of [tex]\theta[/tex]. And, letting x= |a- b|, [tex]x^2= 3^2+ 5^2- 2(3)(5)cos(\phi)[/tex].

How could you possibly be asking how to find the first and second derivatives if you are such a "beginner" that you do not know what "domain" means?

In this case you are wrong and your boss is right. (Don't your just hate that?) On many metals, aluminum, say, or copper, surface corrosion (or "patina") forms a protective barrier, protecting the rest of the metal. But iron doesn't work that way. Rust actually encourages more rust and should be removed as soon as possible.

Well, I can't speak for your mental state! 0.9(recurring) MEANS that the string of 9's has no end so how can you have a 5 AFTER it? Since 0.9(recurring) can be interpreted as the limit of the sequence 0.9, 0.99, 0.999, 0.9999, ..., you might try to interpret this, as I did in an earlier post, as the limit of the sequence 0.95, 0.995, 0.9995, 0.99995,... But it is easy to show that the limits of 0.9, 0.99, 0.999, ..., 0.91, 0.991, 0.9991, ..., 0.95, 0.995, 0.9996,... or, indeed, 0.9a, 0.99a, 0.999a, ... where a is any digit, are all the same: 1. And as far as "i" is concerned, far from being "impossible" or even "improbable", it certainly does exist and is used regularly in physics or even engineering.

A list of such symbols, together with the ascii and html representations can be found at http://www.barzilai.org/math_sym.htm

??? Actually, it would be illogical to say you DO understand something while claiming it is meaningless.

0.99999999... is, by definition, the limit of the sequence 0.9, 0.99, 0.999, 0.999, etc. That is a geometric sequence and it is easy to show that its limit (and so the value of 0.9999... ) is exactly 0.11/(1- 0.9)= 0.1/0.1= 1.

You are asking for a whole course! Post some specific problems that you have and we may be able to help you.

"but if it did have an end"? And you just said "unending string of 0s"? Surely you see the logical falacy! The nearest I can come to matching that is to consider the sequence of finite decimals: 0.1, 0.01, 0.001, 0.0001, 0.00001, etc. Its limit could be considered to be "an unending string of 0s with a 1 on the end". Of course, it is easy to see that the limit is simply 0. In fact, the limit of the sequence 0.a, 0.0a, 0.00a, 0.000a, 0.0000a, etc. is 0 no matter what digit 0 is. It really is just an unending string of 0s!

I don't know what you mean by "inverse function (range)" since "inverse function" and "range" are very different things. In general, functions of two variables do not have inverse functions since they are seldom "one to one". Finding the range of a function of two variables depends strongly on exactly what the function is.

By the way, the Hewlet-Packard calculators used to (I don't know if they still do) use "reverse Polish notation" which uses a completely different "precedence" convention. It was supposed to be easier to enter complicated problem using "reverse Polish" but I always found it difficult to not unconsciously convert to "ordinary" notation (thus getting the wrong answer). "Reverse Polish notation" requires that you work through a formula from left to right, each time you reach a binary operator, apply it to the last two numbers. For example, (1+ 4)*6+ 3 would be entered as 1, 4, +, 6, *, 3, +. (Notice that you need a "enter" key to separate numbers. As you entered that, the calculator would save 1 and 4, then seeing the "+" add those, saving 1+4= 5, then save 6, then seeing the "*" do 5*6= 30, saving 30. Then save 3 and, when you enter the last "+", do 30+ 3= 33.

The nth Taylor polynomial requires the nth derivative. The "Taylor expansion" or "Taylor series" requires all of them. One answer to your question is that if we define sine and cosine by their Taylor series, we can get the nth derivative by differentiating the series term by term: If sin(x) is defined by sin(x)= x- x^3/3!+ x^5/5!+ ..., then it is obvious that sin(0)= 0, sin'(x)= 1- 3x^2/3!+ 5x^4/5!+...= 1- x^2/2!+ x^4/4!= cos(x), sin"(x)= -2x/2!+ 4x^3/4!+ ...= -(x- x^3/3!+...)= -sin(x) etc. You could then show that y= sin(x) satisfies the differential equation y"= -y as well as the initial condition sin(0)= 0, sin'(0)= 1 and so, by the existance and uniqueness" theorem for initial value problems, is identical to the "traditional" sine function. Given any infinitely differentiable function we can define its Taylor's series at, say, 0 (the "McLaurin series"). It does NOT follow that the series converges to that function. For example, the function [math]f(x)= e^{-\frac{1}{x^2}}[/math] for [math]x\ne 0[/math], f(0)= 0 is infinitely differentiable at 0. All derivatives are equal to 0 at x= 0 so its Taylor's series about x= 0 is identically equal to 0. That converges, of course, for all x but is equal to f(x) only for x= 0.

Those are triangular numbers. n is a triangular number if n objects can be placed in an equilateral triangle form. 3 objects is obvious. 6 objects form a triangle with rows containing 3, 2, and 1. 10 objects would be 4, 3, 2, 1 and is the standard form for bowling "ten-pens". If n objects are placed in a triangle in which the longest row contains k objects, to make a larger triangle, you need to add a row of k+1 objects. That way, N_(n+1)= N_n+ n. A number is triangular if and only if it is of the form n(n+1)/2 for some positive integer n. That is because the triangular numbers are those that can be written 1+ 2+ 3+ 4+ ...+ n. If you reverse the order and write n+ ...+ 4+ 3+ 2+ 1 and add first numbers of each sum, second numbers of each sum, etc., you will find that every sum is n+1 (top number increases by 1, bottom number decreases by 1 so the sum is always the same). Since there are n such sums, that total will be n(n+1). Since we got that by doing the sum twice the value of just one sum is n(n+1)/2.

I'm sorry, I really am not sure what you are saying. You understand that there is no way to write M= AC+ CA as a single multiplication. Are you asking if E as a single "formula"? Certainly that can be done with a little algebra. You have defined M= AC+ CA and then E= BM+MB. Replace M by AC+ CA and you have E= B(AC+CA)+ (AC+ CA)B= BAC+ BCA+ ACB+ CAB. Since matrix multiplication is not, in general, commutative, you will need all four multiplications.

Looking at the other topics here, I'm not clear how serious one is supposed to be! Eratosthenes, a Greek living in Egypt about 200 BC accurately calculated the radius of the earth (and so the curvature:1 divided by the radius). He had heard that, in a certain town on the Nile River, south of where he lived, on a certain day of the year, the noon sun shone straight down a well without touching the walls of the well (i.e. the town was on the Tropic of Cancer and the day was midsummer). Eratosthenes hired a man to walk from his village to that southern town, counting his steps and thus finding the distance between the two towns. He also measured the angle the sun made on that day in his own town. Assuming the sun was far enough away that the two light paths could be taken to be parallel, the ratio of that angle to 360 degrees must be the same as the ratio of the distance between the two towns divided by the circumference of the earth. Notice, by the way, that this was a good 1700 years before Columbus! The usual idea that every one except Columbus believed the earth was flat is simply not true. All educated people of the time new about Eratosthenes and knew the earth was a globe. Columbus was one of a small group who believed that Eratosthens calculation was wrong and that the earth was much smaller than he had calculated. Apparently their argument was a purely "stylistic" one- they did not believe that the earth could have all of it land (the "known" land of the day, Europe, Africa, and Asia) on one side of it and only empty ocean on the other. Apparently it didn't occur to them that there might be unknown land on the other side of the earth!

You put this in the "evolution" area but you don't appear to be talking about evolution but about "training" carnivores to be herbivores. If you are talking about evolution, I believe most of the evidence is that it went the other way- herbivores evolved into carnivores.

How is that against the law of conservation of momentum? The "center of system changes" because, if you were not including the "other objects" initially you have changed the system. If you include any objects the bullet or gun will touch, any objects these new objects touch, etc. in your system initially, the center of momentum does not change and momentum is conserved. On the contrary, a patent examiner in Switzerland pointed out that the "conservation of energy" was wrong over a hundred years ago and that was immediately accepted! Your reference appears to be saying that the earth does NOT go in an ellipse because it actually rotates around the earth-moon center of mass. I thought that was well known. I think I was about 7 when I read that "the earth rotates in an ellipse" but was 13 when I was taught that that was only approximate- for exactly the reason you state. Have you calculated exactly how far from the center of the earth the earth-moon center of mass is and so exactly how far off a true ellipse the earth's orbit is? Just how MUCH effect on weather, etc. does that have?

] Well, I DON'T agree with your "axiom 1". In fact, since [math]\infty[/math] is NOT a real number, I have no idea what you could mean by [math]1/\infty[/math]. Assuming that you are working in some extended that allows arithmetic with [math]\infty[/math], do you have any evidence that (x/y)/(p/m()= (x/y)(m/p) is true in that system?

And "learning formulas" is NOT learning mathematics!

Your question is, in a sense, answered by the title of Halmo's classic book on linear algebra: "Finite Dimensional Vector Spaces". Linear algebra is basically the study of finite dimensional vector spaces. You can use "vectors" to represent the solutions to systemes of linear equations and so that becomes a small part of Linear Algebra. Solutions to linear differential equations can be represented as vector spaces as well. But that's only a very small part of linear algebra. You will find "eigenvalues" and "eigenvectors" (defined in linear algebra) in all parts of mathematics. Essentially, "vector spaces" encapsulate our basic concept of "linear" behavior and so almost anything involving some concept of "linearity" might come under the heading of linear algebra. And, as Bignose said, you linearize, that is, approximate by a linear function, any "smooth" function locally so linear algebra techniques can be used to get approximate solutions to problems that, themselves, do not fall under "linear algebra". Some linear problems, involving PARTIAL differential equations or spaces of functions, involve infinite dimensional vector spaces and so come under "Functional Analysis" rather than "Linear Algebra".

It's easier use "conservation" of energy.All the way from the surface of the earth to the center, your speed, and so kinetic energy is increasing and potential energy decreasing. From the center of the earth on, you speed is decreasing. It will be 0 (and your kinetic energy will be 0) when your potential energy is exactly what it was to begin with- and so you will be the same distance from the center of the earth. (That is ignoring "technicalities" like air resistance or friction against the side of the hole as well as the fact that the earth is not a perfect sphere. Not to mention what would happen if the other end of hole was under water!)

The components of v, (1,-2,1), sum to 1-2+1= 0. The compnents of w, (0,1,-1), sum to 0+ 1- 1= 0. If C and D are any constants, then Cv+ Dw= (C, -2C+ D, C- D) and the components sum to C+ (-2C+D)+ (C- D)= C(1- 2+ 1)+ D(0+ 1- 1)= 0.

Not "a couple of inches over night" certainly but most boys, around the age of 11 or 12 will grow several inches over a few months. Yes, it is fairly typical of living creatures to grow at different rates at different places in their life span- what you are calling growth spurts.