HallsofIvy

I have no idea what you mean by "traditional simplification rules". I said "And that reduces to x^4= 0. Can you solve THAT polynomial equation?" Your response was "That is exactly what I am saying: the traditional simplification rules do not solve the equations accurately." ?? That was a "traditional simplification rule" that did solve the equation. What I would mean by "traditional simplification rules" (again, I don't know what you mean by that) solve those equations that can be solved in terms of roots, such as this one, very nicely. Of course, it has been known for a long time that there were polynomial equations that cannot be solved in terms of roots. There are other methods that work, to get at least numerical approximations, to any accuracy, for those

A Delta flyer CAN land on earth can't it?

To Paris, of course! I can't imagine why I would want to go to a place where I can't even breathe!

No. The rate at which a substance cools is proportional to the difference in tempature between the substance and the atmosphere. For one thng, metal (or oliive oil) is not a good insulator. For another, the size of the spoon is very small compared to the surface area of the pot the water is in.

First, what do you mean by "dimension"? You seem to be using the word in a way that make no sense to me!

I have a lot of knowledge that is unvaluable!

You appear to be using "concave" where you mean "convex". A "concave"mirror cannot collect light, it disperses it.

First, as you have already been told this is NOT a "polynomial equation". It can be written x^3= (N^2xc{N^2/x+ x)$. Multiply both sides by that denominator, N^2/x+x. N^2x^2+ x^4= N^2x^2. And that reduces to x^4= 0. Can you solve THAT polynomial equation?

Set up an xy-coordinate system so that (0, 0) is at the lower left. Take the vertex in the middle of the bottom side to be (X, 0). Then the vertex in the middle of the left side is (0, X). The distance between those two points is sqrt{X^2+ X^2}= sqrt{2X^2}= Xsqrt{2} while the distance from the vertex in the middle of the bottom to the vertex on the right end if the bottom is 1- X. Since those sides are the same length we have Xsqrt{2}= 1- X. Xsqrt{2}+ X= (1+\sqrt{2})X= 1. X= \frac{1}{1+ \sqrt{2}. That's not one of the given answers so "rationalize the denominator": X= \frac{1}{1+ \sqrt{2}}\frac{1- \sqrt{2}}{1- \sqrt{2}}= \frac{1- \sqrt{2}}{1- 2}= \sqrt{2}- 1.

Are you sure you are taking this course? Perhaps you accidently went into the wrong class and were given homework for a course you aren't taking! The first problem asks you to integrate [math]\oint_L x^2y da_x- y da_y[/math] around the path consisting of the straight line from (0, 0) to (1, 1), then the straight line from (1, 1) to (2, 0), then the straight line from (2, 0) back to (0, 0). The first line can be written as y= x so dy= dx and [tex]dl= \sqrt{(dx)^2+ (dy)^2}= \sqrt{(dx)^2+ (dx)^2}= \sqrt{2}dx[/tex] with x from 0 to 1. [tex]\v

It's interesting that this test on "Electro-magnetic theory" is pretty much a Calculus test!

The "Gallilean theory" says that velocities add as $v_1+ v_2$. If you see a person driving by at 40 mph, relative to yourself, on the back of a flatbed truck and, just as they pass you, that person throws a ball forward at 30 mph, relative to the truck, Gallilean theor y says that you would see that ball having speed 40+ 30= 70 mph relative to you. Relativity theory says that the velocities add as [tex]\frac{v_1+ v_2}{1+ \frac{v_1v_2}{c^2}}[/tex]. If the velocities are small then [tex]\frac{v_1v_2}{c^2}[/tex] will be negligible and that the two formulas give the same thing to within measurement error. If [tex]v_1[/tex] and [tex]v_2[/tex] are close to c, as [tex]v_1= v_2= 0.6c[/tex] then [tex]\frac{v_1v_2}{c^2}= (0.6)^2= 0.36[/tex] so the velocity of the bullet relative to you is [tex]\frac{0.6c+ 0.6c}{1+ 0.36}= \frac{1.2}{1.36}c= 0.88c[/tex].

The whole point of "irrational numbers", and the set of transcendal numbers is a subset of the irrational numbers, is that they are NOT rational numbers and cannot be written as a fraction with integer numerator and denominator. As Markus Hanke said, 22/7 is an approximation to [math]\pi[/math], it is NOT equal to [tex]\pi[/tex]. (We can "use" rational numbers to define the irrational numbers. We say that two sequences of rational numbers, [tex]\{a_1, a_2, a_3, \cdot\cdot\cdot \}[/tex] and [tex]\{b_1, b_2, ...\}[/tex] are "equivalent" if and only if the sequence [tex]\{a_1- b_1, a_2- b_2, a_3- b_2, \cdot\cdot\cdot\}[/tex] and then define the real numbers to be the set of equivalence classes of such sequences. [So the sequence 3, 3.1, 3.14, 3.141, 3.1415, 3.14156, ... is in the eqivalence class we identify as "[tex]\pi[/tex]]).

You appear to be misunderstanding "inconsistency". It does not mean that you can state an incorrect formula. It means that you can prove, from the basic axioms and definitions, two contradictory statements. In "naive" set theory, a basic axiom is that "if you can define a set rigorously, it exists". Therefore, "the set of all sets that do not contain themselves", since we can, naively, look at the definition of any set and determine whether or not it contains itself, must exist. But once we have decided that it is a set we can ask whether or not it contains itself. It cannot contain itself because that contradicts the fact that it only contains sets that do NOT contain themselves. But if it does not contain itself then it cannot be said that it contains allsets that do not contain themselves.

But no one said "absolutely identical". The question about "morpholgically identical" clones. They will be standing at different positions so I can identify one as "the one on my left" and "the one on my right".

"L: logaritmic" log(x) is continuous for x any positive number, not defined for x non-positive. "A: arc" "arc" is not a function, it is a geometric object. "P: polynomic" Polnomials are continuous for all x. "T: trygnometric " sine and cosine are conntinuous for all x. tangent is continuous for all x except multiples of pi, cotangent is continuous for all odd multiples of pi/2. secant is cotinuous for all x except odd multiples of pi/2, and cosecant is continuous for all x except multiples of pi. "E: exponential." e^x and the general a^x, for positive a, is continuous for all x.

You appear to be confusing "mathemtatics" with "applictions of mathematics". In mathematics, we have abstract functions, such as "y= bsin(ax)", "y= be^(ax)", etc. There are NO "units" associated with x or y. IF we want to apply those to specific problems, then you need to choose units for x and y that correspond to those applications. For example, if you have a problem in which you need need to calulate an ocean wave you might well choose a sine or cosine function, the "ideal" periodic functions, then "x" might be measured in "meters". In order to be able to use bsin(ax) or be^(ax) we must have "a" having units of "1/meter" as swansont and others said. And if the function is to return "meters" then "b" must have units of "meters". (Of course, it might occure that a and/or b have numerical value of "1" so they are obscured but units are still there.)

Let me add this: one of the things Sir Isaac Newton was concerned with was the force that kept the planets in motion about the sun. He knew "mass times accelertion equals force" from Gallileo and knew, from Copernicus' measurements, that planets closer to the sun move faster than when they were farther from the sun. From that he believed the force, and so the acceleration depended upon the distance from the sun. But "acceleration" is "change in speed divided by change in time" just as "speed" is "change in position divided by change in time" while "distance" is given at a specific time, not over a change in time! That was why Newton had to invent the Calculus- in order to be able to define "instantaneous speed", "instantaneous acceleration" or, generally, the rate of change at a specific time rather than over a span of time.

We are, at any given time, moving in one direction. That direction vector might be the sum of many different vectors but it is still one direction. (If we were moving in many different directions at the same time we would be torn apart!)

I didn't think Richard was that bad!

4500,000 years will reduce to 30% with e^{500,000*\alpha}= 0.3. 500,000*\alpha= ln(0.3) so \alpha= \frac{ln(0.3)}{500,000}= ln\left(3^{1/500000}\right). And then to find the half life, set e^{\alpha x}= e^{ln\left(3^{1/500000}\right)t}= 3^{(1/500000)t= 0.5 and solve for t.

The fact that you posted this question at "3 A.M in the morning" means you need to go to sleep!

First, in mathematics a "series" is an infinite sum of numbers of functions. A "power series" is an infinite sum of powers of the variable, usually "x". An example is sum_{n=0 to infinity} a_n(x- x_0)^n= a_0+ a_1(x- x_0)+ a_2(x- x_0)^2+ \cdot\cdot\cdot. A "Taylor's series" is a power series "representing" a given function derived in a particular way- a_n, the coefficient of x^n is the nth derivative of f evaluated at x_0 divided by n!. (I say "representing" because the Taylor's series of a function is not necessarily equal to that function.) An simple example is f(x)= e^x. All derivatives of e^x are e^x again and its value at x_0= 0 is 1. The Taylor's series for e^x about x0= 0 is Sum x^n/n!. It can be shown that this series does converge to e^x for all x. In fact it converges fast enough that "truncating" it (cutting it short) at, say n= 2, gives a reasonable approximation. That is e^x is approximately 1+ x+ x^2/2 for small x. For example, if x= 0.1, e^x= e^0.1= 1.1051709180756476248117078264902... while 1+ 0.1+ 0.1^2= 1.11. If x= 0.01, e^x= e^0.01= 1.0100501670841680575421654569029 while 1+ .01+ 0.01^2= 1.0101

I am going to jump in here to point out that "the square root of i" is an ambiguous phrase. Every non-zero complex number has two square roots. In the real numbers, one is positive and the other negative so we can agree that "the square root of a", if it is a real number, means the positive square root. Since the complex numbers are not an "ordered field" we cannot make that distinction. We cannot, in general, distinguish "the square root" of a complex number.