Country Boy

As pointed out, the limit is obviously 1. I really hate the phrasing "whats the last number in the infinitely long pattern?" Obvously an "infinitely long" sequence doesn't HAVE a "last number" and it is very misleading to refer to the limit of a sequence that way.

Since x is not 0, from the first equation y= 1/x. Then the second equation becomes zx= 1/x so z= 1/x^2. I'm not sure why you wrote "there is a solution". Generally, if you have 2 equations with 3 unknown numbers, there are an infinite number of solutions: choose x to be anything other than 0 or +/- 1 and solve for y and z. For example, x= 2, y= 1/2, z= 1/4 so x^2y= (4)(1/2)= 2= x, zx= (1/4)(2)= 1/2= y. Another, x=3, y= 1/3, z= 1/9 so that x^2y= 9(1/3)= 3= x, zx=(3)(1/9)= 1/3= y. Etc.

There are many different ways of doing a differentiation on a computer, including some sybolic differentiations. I don't know what you mean by "what method"- there are many methods and many programs that will do that.

My apologies, I didn't read it correctly! (Not all that uncommon, unfortunately.) Yes, I read it as asking if every pseudo-vector could be represented as a cross product of vector- which you and he said was true. You read his question correctly.

I think Bignose misunderstood your question. He is saying that "not every vector is a pseudo-vector". It is true that every pseudo-vector can be represented as the cross product of two vectors. It is, perhaps, more accurate to represent a pseudo-vector as an anti-symmetric 3 by 3 matrix.

No, you did hnot get it right! I think your error was that when you "completed the square" for the y terms you added "4" on the right hand side but forgot that it should be multiplied by the 4 multiplying (x+1)^2. You should have had 16 on the right hand side, not 4, and then, dividing the entire equation by 16, (x+1)^2/16+ (y+2)^2/4= 1. By the way, why did you title this "standard form for a circle"?

[math] x = 3 \sqrt{(x+3)^3 - (x+1)^3 + (x+8)^3)} [/math] Now. that one's easy! Square both side and multiply out the cubes to get a cubic equation- then use the cubic formula. The way Atheist interpreted it (you didn't have enough parentheses to make it clear what the square root applied to), squaring gives a 6th degree polynomial equation- and there is no general formula for the solution of 6th degree equations.

Is us learning grammar?

The "real" equation? What you originally said was Now which equation did you ask your teacher about? The real inequality you originally gave us was log(x-1)- log(x+2)> 0 The inequality log((x-1)/(x+2))> 0 is satisfied as long as (x-1)/(x+2)> 1. That is the same as (x-1)/(x+2)-(x+2)/(x+2)= -3/(x+2)> 0, which, in turn, is true as long as x+2 is negative: x+ 2< 0 or x< -2. However, the original inequality, as you give above, was log(x-1)- log(x+2)> 0. It can only have, as solutions, the solutions to the "equivalent" log((x-1)/(x+2))> 0 and those only if they satisfy the original inequality. Since any x< -2 makes the arguments of both logarithms negative, there is NO solution to your equation. I am sure your teacher is a brilliant man. Unfortunately, he can only answer the question you ask, which may not be the correct one!

I have no idea what you are talking about. In any case, various mathematical system model parts of the real world to a lesser or greater extent but are not the real world themselves. Making up an equation tells us NOTHING about reality. "Or what else is so fundamental about A+B=B+A that it can pertain fully or correctly to physical reality?" The only thing "fundamental" about it is that it is very simple. What reason do you have to believe that it pertains "fully or correctly to physical reality"?

The standard way of writing a linear transformation from space U to space V, with U having basis {u1, u2, ..., um} and V having basis {v1, v2, ..., vn} is to apply the linear transformation to each of u1, u2, ..., um in order. Write the resulting vector in terms of v1, v2, ..., mn. The coefficients in that linear combination are the columns of the matrix. Here, U= V= R3 and the "canonical basis" is <1, 0, 0>, <0, 1, 0>, <0, 0, 1>. Project <1, 0, 0> into this plane, and write that as <a, b, c> . Then "a b c" will be the first column. Do the same with <0, 1, 0> and <0, 0, 1> to get the second and third columns. Those, the projections of <1, 0, 0>, <0, 1, 0> and <0, 0, 1> in the subspace perpendicular to <1, 1, 1>, are the "3 vectors" Cpl. Luke is talking about. Notice that a projection operator is clearly not invertible (it is not one-to-one) so your matrix will not be invertible and may well have two or three identical columns.

Sorry, there was a typo in my example: g should be g: {(0, 3), (1, 2), (2, 1), (3, 0)}. Now, the ranges are {0, 1, 2, 3} and {3, 2, 1, 0}, which are, of course the same set- order is not relevant in a set. "All that's needed" for WHAT? The sets of ordered pairs show exactly what f and g "do with the numbers": f(0)= 3, f(1)= 1, f(2)= 2, f(3)= 3 (f is the "identity" function on its domain) while g(0)= 3, g(1)= 2, g(2)= 1, and g(3)= 0 (after my correction). If you do not understand in what sense these sets of ordered pairs ARE functions, then you seem to have trouble understanding the basic definition of "function" (and at least a little difficulty believing what I am saying). Please take the example I gave you and to your calculus lecturer, who, I am sure, is quite good, and ask him/her to explain the example to you.

The derivative essentially makes precise the notion of "rate of change" at a particular instant. It is very easy to define "average rate of change" but that necessarily involves a change in the independent variable. You don't have any change in the independent variable at a particular instant! Consider the problem of planetary motion. Newton, and many others, believed there was a force on planets that depended on their distance from the son. But imagine that you are in a rocket ship, above the plane of the planets and you take a "snapshot" of the planets. You could measure the distance from the sun to each planet and (if you knew how force depended on that distance) calculate the force at that instant. Then, since F= ma, you could calculate the acceleration of the planet at that instant. But does that make sense? If acceleration involves "change in velocity over a change in time" and even velocity involve "change in position over a change in time" and there is no "change in time", what possible sense could it make? Newton (and Leibniz) developed calculus, and the derivative, specifically to be able to make sense of that.

You keep asserting that "the domain of one function can be the range of of another". I've never said they can't! I have agreed with that several times now. I also have never suggested that a function cannot have range and domain identical- f(x)= x is a good example. I have been refusing to accept your assertion that "a function is its output". They are not at all the same thing! I notice you appended ", and depends on its input". That's precisely my whole point- the function is exactly HOW the "output" depends on the "input"! Then I don't know what more to say. That set of duples does a lot more than "represent the domain and range". The functions f: {(0,1), (1,1), (2, 2), (3, 3)} and g: {(0, 3), (1, 2), (2, 1), (3, 1)} have exactly the same Domain: {0, 1, 2, 3} and exactly the same range, {0, 1, 2, 3}, which are themselves identical, but are very different functions!

You are making a gorilla your ideal of "masculinity"? You are aware that there are FEMALE gorrillas, are you not?

I assume that you were bowing deeply when you said that!

Well, it's practically a quote from any basic textbook. I have been known to glance at one occassionally!

Oh, yes. We can always restrict the domain so that the function becomes one-to-one, and so has an inverse function on that domain. I have no problem with that. I was only objecting to talking about the inverse function of a quintic polynomial.

I never said the domain of some function, f, cannot be the range of another function- since both domain and range are sets. that's quite possible. My point was that a function is NOT its domain or range. Earlier, when I said that I did not see what your comment about the domain of one function being the range of another had to do with the question of whether the word "continuous" applied to the function or domain, you said, "If not, then you should consider the possibility that you still don't understand what a mathematical (or any other kind of) function is, possibly. I recommend a chat with a math lecturer (one with a degree, maybe)." Actually, I talk to people with math degrees every day- I have a Ph.D. myself and have been teaching college math for about 30 years. I know very well what a function is as well as the concepts of "domain" and "range" of a function: A "function" is a set of ordered pairs with the property that no two pairs have the same first member. The "domain" of a function is the set of all first members in those pairs, the "range" is the set of all second members (x and y if you like). It is common to specify the domain and give some rule by which the "y" corresponding to a given "x" in the domain. It is also common to just give the "rule" with the understanding that the domain is the set of values to which the rule can be applied. However, the "domain" and "range" of a function are distinct from the function itself- that has been my point all along.

Actually almost all integrable functions have do not have anti-derivatives that can be written in a simple form.

Not since Euclid, a couple thousand years ago! Suppose there were a largest prime number. p. Multiply all the prime numbers up to p together and add 1. That number is not divisible by any prime number less than or equal to p since you would have a remainder of 1. That means that either this new number is prime itself or there is some other prime number, larger than p. Either way, there cannot be a largest prime number.

No, that's using language correctly. Something you have to be very careful about in mathematics. I really can make no sense out of that sentence. "function (i.e. the range)"? No, a "function" is distinct from its range. Yes, of course. ANY set can be the domain or the range of a function. In any case, in mathematical analyis, the word "continuous" is defined only for functions, not for sets.

You have symbols that my internet reader can't interpret. However, yes, though I am no expert, "naive set theory" has been replaced by "classes" which are essentially hierarchies of "sets". At the base, the things that are still called "sets", cannot have sets as members. Then we have a "class" that can contain sets, a class that can contain THOSE things, etc.

Yes, it does matter but you should have been able to see what to do. You have h=(h+0.5gt1^2)/t1)t2 - (0.5gt2^2) which can't be correct- there are more ")" than "(". Assuming you meant h=((h+0.5gt1^2)/t1)t2 - (0.5gt2^2), multiplying out that first term on the right, is h= h(t2/t1)+ 0.5gt1t2- 0.5gt2^2. Now subtract that first term from both sides h- h(t2/t1)= 0.5gt1t2- 0.5gt2^2 factor h out on the left: h- h(t2/t1)= h(1- t2/t1)= h(t1/t1- t2/t1)= h(t1-t2)/t1. (That's only the left side!) On the right, we can factor out 0.5, g, and t2: 0.5gt1t2- 0.5gt2^2= 0.5gt2(t1- t2). (That's only the right side.) Going back to the entire equation we now have h(t1-t2)/t1= 0.5gt2(t1- t2). Finally, multiply both sides by t2/(t2-t1). Of course, (t1-t2)/(t2-t1)= 1 so we have h= 0.5gt1t2.

To answer the question implied by the title, the only "massless particle" is the photon.