Country Boy

Not in terms of elementary functions. It might be possible to solve in terms of the "Lambert w-function" which is defined as the inverse function to f(x)= xe^x.

I would recommend, instead, subtracting the two functions and integrating ln(x)- (x-2)= ln(x)- x+ 2. That way you don't have to worry about where the functions are positive or negative. Within the limits of integration ln(x)- x+ 2 is always positive.

More specifically "density" is "weight divided by volume". In order to determine the density- to see if it matched that of gold, Archimedes had to determin BOTH.

Perhaps it would help if you defined exactly what YOU mean by "universe". In its general sense, it is defined as encompassing everything and so there cannot, by definition, be more than one universe. And what do you mean by "budding"?

As insane alien said, "speed" is not the right word. By any reasonable definition of "speed", the "speed" of time is "one second per second" and cant change!

To reiterate, yes, he did weigh the crown. But you can make two crowns weighing exactly the same thing from different percentages of gold and silver just by having the one with more silver slightly larger. Archimedes did not use "displacement" to measure weight, he used it to measure the volume of the crown so that he could calculate density = weight/volume.

What do you mean by "using" imaginary numbers? If you just mean that you don't want imaginary numbers in the final result, Here's how I would do it: the equation [math]x^4+ 1= 0[/math] has roots [math]\frac{\sqrt{2}}{2}+ i\frac{\sqrt{2}}{2}[/math], [math]\frac{\sqrt{2}}{2}- i\frac{\sqrt{2}}{2}[/math], [math]-\frac{\sqrt{2}}{2}+ i\frac{\sqrt{2}}{2}[/math], and [math]-\frac{\sqrt{2}}{2}- i\frac{\sqrt{2}}{2}[/math] so [math]x^4+ 1= (x-(\frac{\sqrt{2}}{2}+ i\frac{\sqrt{2}}{2}))(x- \frac{\sqrt{2}}{2}- i\frac{\sqrt{2}}2})}(x- (-\frac{\sqrt{2}}{2}+ i\frac{\sqrt{2}}{2}x-(\frac{\sqrt{2}}{2}+ i\frac{\sqrt{2}}{2}[/math] Now write those factors as [math](x-\frac{\sqrt{2}}{2}- i\frac{\sqrt{2}}{2})(x-\frac{\sqrt{2}}{2}+ i\frac{\sqrt{2}}{2}= (x-\frac{\sqrt{2}}{2})^2+ \frac{1}{2}[/math] and [math](x+\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2})(x+ \frac{\sqrt{2}}{2}+ i\frac{\sqrt{2}}{2})= (x+ \frac{\sqrt{2}}{2})^2+ \frac{1}{2}[/math] which have only real coefficients.

A method I prefer is "row reduction". You are probably familiar with the concept but I will review it here: There are basically three kinds of "row operations" you can do to a matrix: 1) Multiply every number in a row by the same number. 2) Swap two rows. 3) Add a multiple of one row to another. The crucial point is that every row operation corresponds to multiplication by a specific matrix: in fact, if you apply a row operation to the identity matrix, you get an "elementary" matrix- multiplying any matrix by that elementary matrix is the same as applying the corresponding row operation to it. Write your matrix "A" and the identity matrix side by side. Apply row operations, one after another, to reduce A to the identity matrix. This can be done fairly mechanically. (If you get a row consisting entirely of 0s, you can't do this. That is exactly the situtation in which A does not have an inverse.) As you apply each row operation to A, apply it also to the identity matrix beside. By the time you have reduced A to the identity matrix, you will have changed the identity matrix to A-1. This is because applying the sequence of row operations to A is the same as multiplying A by their corresponding elementary matrices. Multiplying all of those elementary matrices would give the one matrix that, multiplying A gives the identity matrix: A-1. Applying the row operations to the identity matrix is the same as multiplying them all by the identity matrix, a quick way of "multiplying" those elementary matrices without having to write them down.

K!! is right, that equation is separable and integrates to give y= Cx. But that doesn't mean you can't look for an integrating factor! There is no general method of finding integrating factors. One thing you can do is assume some particular "type" of function and just try that. For example, if I look for an integrating factor that is a function of x only, say v(x), multiplying by v(x) we have v(x)ydx- v(x)xdy. In order that this be "exact", we must have (v(x)y)_y= (-v(x)x)_x. On the left, we have (v(x)y)_y= v(x), on the right, (-v(x)x)_x= -v'x- v. Setting those equal, v= -xv'- v or xv'= -2v. That is, again, a separable equation: dv/v= -2dx/x which integrates to ln(v)= 2ln(x)= ln(x^(-2)) so v= x^(-2)= 1/x^2. How did I know to look for an integrating factor that was a function of x only? Lucky guess. (Helped by the fact that you had already told me that 1/x^2 was an integrating factor!) From looking at ydx- xdy, you might just as well guess a function of y only: v(y). Then we would have v(y)ydx- v(y)xdy and would have to have yv'+ v= -v so that yv'= -2v. Yes, v(y)= 1/y^2 is also a perfectly good integrating factor. The form of -1/(x^2+ y^2) makes me suspect that they changed to polar coordinates and found the integrating factor from that. If x= r cos(t), y= r sin(t), then dx= cos(t)dr- r sin(t)dt and dy= sin(t)dr+ r cos(t)dt. ydx= (r sin(t))(cos(t)dr)- (r sin(t))(r sin(t)dt) = r sin(t)cos(t)dr- r^2 sin^2(t)dt - xdy= -(r cos(t))(sin9t)dr)- (r cos(t))(r cos(t)dt) = -r sin(t)cos(t)dr- r^2 cos^2(t)dt so ydx- xdy= -r^2 dt. Since the coefficient of dr is 0, the derivative of that multiplied by anything, with respect to t, will be 0. In order to have this exact, we must also have the derivative of the coefficient of dt be 0 which is true, of course, if that coefficient is a constant: multiplying by 1/r^2 give -1 dt. An integrating factor is 1/r^2= 1/(x^2+ y^2). (The "-1" in the numerator of what you give is irrelevant). Notice that there are many possible integrating factors- and no general method for finding them.

?? What you are describing is exacty what we mean by "light moving"! Just as the individual molecules of water in a water wave move up and down, it is the WAVE that moves across the water, not any physical object. What you are talking about is precisely what a "wave" is.

Are you clear on what "Chaos Theory" is? It has nothing to do with uncertainty or chance. In a sense, Chaos Theory is the opposite of probability. In probability, we have individual events that happen purely arbitrarily but "in the long" run, may turn out to be very close to determined- as in the famous demonstration where you drop beads on to nails so that each on bounce left or right at random but the mass of beads come right up to the "bell curve" already drawn on the glass. In Chaos Theory, every step is given by a specific algorithm- but the algorithm is so complex, that "in the long run", very very tiny differences are expanded to where, unless you are absolutely perfect in your initial conditions, you can have no idea what the result will be.

You said The one that "moves away and comes back" is the one accelerating. While velocity is "relative", in special relativity, acceleration is not. In general relativity, acceleration can be interpreted as a "gravitational" force. The one accelerating feels a greater "gravitational" force which affects his clock. Notice that the twin paradox depends on one of the twins coming back. That cannot be done without acceleration. If you assume that two people, with different motions, are at one point at time t= 0, then move away, each sees the other as in motion and each sees the other's time ellapsing more slowly than his own. There is no paradox in that.

You are following a formula? surely it doesn't say "replace m/1 with (m/2)^2! They aren't equal and you can't do that. I imagine what you are doing is dividing the coefficient of m by 2 and squaring that: (1/2)^2= 1/4, not (m/2)^2 and it doesn't go into the middle of the equation. Okay, now you are back on track! Since you don't say what is in the back of the book that is "way off", I don't know what you did. You have completed the square when you write m^2+ m+ 0.25= 3.25 or (m+ .5)^2= 3.25. But what was the problem really? To solve the equation? In that case, you need to recall that there are two numbers whose square is 3.25: m+ .5= +/- sqrt{3.25} and the solutions to the equation are m= -.5 +/- sqrt{3.25}.

First of all, state the problem accurately! The projection of a helix onto a plane normal to the axis is a circle. Apparently you want a helix projected onto a plane parallel to the axis. That projection will be a cycloid. http://mathworld.wolfram.com/Cycloid.html

Well, that depends upon how you define 'physical curvature' doesn't it? You could certainly define it as "giving a specific result on this experiment". Are you saying that you reject the very concept of measuring angles? If you have two straight lines intersecting at a point, the "measure of the angle between them" has very specific definition which works as well for angles on a sphere as on a plane. (Oh, and "straight line" can be given a specific definition that would include great circles on a sphere.) Now, we come to your: a being in the surface of the sphere would not see any difference between his "lines" and your "lines"- and would measure angles in exactly the same way. The being would however find that angle sum in a triangle depends upon the size of the triangle. Whether he would then come to the conclusion that his space was "curved" I can answer to- but certainly we can distinguish between one "curvature" and another through intrinsic measurements. That was a theorem of differential geometry due to Gauss. On what basis do you contend that? Indeed, what definitions are you giving to "angle", "line", etc.? Those are all carefully defined in differential geometry- as well as the notion of "intrinsic measurement". No, we wouldn't observe anything of the sort- and you can't contend that we would because you haven't done the experiement! The "experiment" can be done, mathematically, in differential geometry and gives the results that you disagree with. And you don't recognise from that that they would have to agree that there are trangles in which the angle sum is not 180 degrees? Choose a point on this line which returns to its starting point, A, and move at right angles to it. Since the first line meets itself, this second line will have to meet it in a second point, B- it's not to difficult to show that it must also be at right angles. Go back to any third point, C, on the new line, turn at right angles and go until you meet the first line, which must happen, at D. You now have two triangles in which two of the angles are right angles- and so the angle sum must be greater than 180 degrees. Sorry I was late in reading this. You might want to look for a basic book on "Differential Geometry". You should find that much of it is devoted to "intrinsic properties" and especially the "Gaussian" or "intrinsic" curvature which is defined through "intrinsic" measurements.

Your job involves solving classical high school algebra problems?

Assuming that axiom is equivalent to the continuum hypothesis, why would you consider it "more obvious" than the continuum hypothesis? It looks much more complicated to me.

Are you under the impression that Gummi Berries have some botanical reason for including or not including something? (I hate to be the one to break it to you but there are NO bears in gummi bears!) (I'm not so sure about the gummi worms.)

If it is the integral itself that is bothering you, u= 1-x^2 is an obvious substitution.

?? The set of nxn matrices, indeed, even of general n by m matrices, forms a vector space with the usual addition of matrices and multiplication by numbers.

Irrelevant to what I said. "Applying" the negation of the axiom of choice to physics does not prove that negation is true. One can apply an incorrect idea to a physical situation and get a true result. And you have only a very vague argument that your result is true. I might also point out that your physics is wrong. "The mass must go somewhere" is untrue. There is no reason to believe the laws of "conservation of mass" or even "conservation of mass/energy" apply either before the "big-bang" or after the "big-crunch".

You do understand, don't you, that once you have introduced the " physical space of elementary particles" you are no longer talking about mathematics? In any case, I cannot see how this has anything to do with the axiom of choice.

I think what you are asking is not "how do I solve for t", but "what does t mean". Let me point out that it is not enough to just learn "formula" if you don't learn, at the same time, exactly what each of the symbols represents! Here, "t" represents the TIME. Exactly what number that is depends on on where YOU decide t= 0 to be. Normally, it is best to choose t= 0 at the beginning- when the object is let fall or thrown. By the way, I really hate it when people say something like "g= gravity". "Gravity" (and THAT really should be "the force of gravity) is a force while g is acceleration. That is "shorthand" for "g is the acceleration of a body due to gravity, at the surface of the earth". And shorthand can be very confusing. Cap'h Refsmmat gave an excellent analysis of your problem. Go back and look at that.

"DivideByZero" seems to think it amusing to give a question so vaguely phrased that he can then declare that no one has answered it correctly.