Country Boy

Of course I do! I'm not talking about solving equations, I am talking about the inverse function! You, in your first post, referred to "the inverse function of quintic polynomial" You are aware, are you not, that f(x)= x^2, because it is not a one-to-one function, does NOT HAVE an inverse function.

Then "continous" is the wrong word. Functions are continous or discontinuous. Sets (domains) are "connected" or "non-connected". You may me translating from another language in which the same terms are used but, again, in English, "continuous" only applies to functions, "connected" to sets. I don't see what that has to do with the problem.

No, in general, a quintic polynomial function does NOT have an inverse function. Just as a quadratic polynomial does NOT have an inverse function. (Since the quintic is of odd degree it, unlike a quadratic, MIGHT have an inverse. That why I said "in general".)

No, I don't think it MUST be done that way. In fact, the solution I THINK you are giving, x> 1, is clearly incorrect. There are many different ways to do a problem like that. What I would have done was immediately combine the logarithms: log(x-1)- log(x+2)>0 is the same as log((x-1)/(x+2))> 0 and that means, of course that (x-1)/(x+2)> 1. Now use the advice you were given originally: think it this as an equation: (x-1)/(x+2)= 1 is the same as x-1= x+ 2 which has no solution. The point of "treat it as an equation" is that the only place a CONTINUOUS function can change from "< 1" to ">1" is where it is EQUAL to 1. Since x-1= x+2 is impossible that never happens. However, the function (x-1)/(x+2) is NOT continuous at x= -2. Of course, the orginal function log(x-1) is only defined for x> 1. If x= 2, say, then log(x-1)-log(x+2) becomes log(1)- log(3) which is negative. log(x-1)- log(x+2)< 0 for all x> 1 and is undefined for any other x. The original inequality, log(x-1)- log(x+2)> 0 is NEVER true.

The law of conservation of energy does not say that MECHANICAL energy is conserved. Here you have lost energy through the resistance of the wires and friction in the parts, without which you could not have turned the blades of the fan. In net, the temperature of the fan will go up.

??? The general quintic does NOT have an inverse function.

Then GRAPH the function! It should look awfully familiar. Probably, if you use a graphing calculator, it will completely ignore the "difficulty". Just check what value the graph gives at the "difficult" point. I'm sorry but this makes no sense to me. You seem to be defining "continuous" or "discontinuous" to be a property of the DOMAIN of a function rather than the function itself- which it definitely is not. The function f(x)= 0 is x<= 0, 1 if x> 1 has domain all real numbers but is discontinuous at x= 0. The function f(x)= 0 if x is rational, 1 if x is irrational, has domain all real numbers but is discontinuous for all x. The function f(x)= 1/n if x is rational and x= m/n reduced to lowest terms, 0 if x is irrational, has domain all real numbers but is discontinuous for all x EXCEPT x= 0.

That's certainly plausible but I was specifically asking Intothevoidx what definition of continuous HE was using since he is asking about continous functions but tells us he hasn't "gone over limits". What is his DEFINITION of "continuous" if it does not involve limits?

That measures volume. a cubic centimeter of WATER is approximately a gram. It doesn't follow that the mass of the object you were using the water to determine the volume of has that same mass.

I wondering exactly what DEFINITION of "continuous" you are using?

The nice thing about arithmetic sequences is that the "average" of all the numbers in the sequence is just the average of the first and last numbers. That is, if the sequence {a1, a2, ..., an} is an arithmetic sequence, then (a1+ a2+ ... + an)/n= (a1+ an)/2. And so the sum, a1+ a2+ ... an is just that average (a1+a2)/2 times n.

I'm not familiar with this. "ABC" is a real number. What number? The cosine of the angle between sides? I would be more inclined to use modulus than arg: |A- C|= |-2- 4j|= [math]\sqrt{4+ 16}= \sqrt{20}[/math]. Since this is an equilateral triangle, we must have also |z- A|= |z- C|=[math]\sqrt{20}[/math] also. Letting z= x+ yj, [math]\sqrt{(x-2)^2+ (y-3)^2}= \sqrt{20}[/math] and [math]\sqrt{(x-4)^2+ (y-7)^2}= \sqrt{20}[/math]. Of course, those immediately give [math](x- 2)^2+ (y-3)^2= 20[/math] and [math](x- 4)^2+ (y- 7)^2= 20[/math]. You should be able to see that, after you multiply those out, the [math]x^2[/math] and [math]y^2[/math] terms cancel and you have two linear equations to solve for x and y.

HOW did you get that? You seem to just have 0.5 times the MacLaurin coefficient of [math]e^x[/math] when n is odd, 1 times if n is even. If f(x)= [math]e^{-x/2}[/math] then f(0)= 1 as you have. f'x)= [math]-1/2 e^{-x/2}[/math] so f'(0)= -1/2 also as you have. f"(x)= [math]1/4 e^{-x/2}[/math] so f"(0)= 1/4 and the MacLaurin coefficient is (1/4)/2= 1/8, not at all what you have! In general the nth derivative of [math]e^{-x/2}[/math] is [math](-1)^nx^n/2^n[/math] and so the nth coefficient of the MacLaurin series is [math](-1)^n/(2^n n!)[/math]. You could also do that just by replacing x in the MacLaurin series for [math]e^x[/math] by -x/2: [math]x^n[/math] becomes [math](-1)^n x^n/2^n[/math]. In particular, taking x= 1, you should get [math]1/\sqrt{e}[/math] is approximately 1- 1/2+ 1/8- 1/48+ 1/384+ 1/3840= 0.6044, a pretty good approximation to [math]1/\sqrt{e}[/math] which is about 0.6053

And in what system is that not DEFINED as 2?

In your first example, since the total force due to the spherical shell on the mass insde it is nil, yes, at a given instant and assuming non-zero relative speeds to begin with, u1= u2. More correctly, a1= a2 In your second example, under the same conditions as before, Yes, again, u1= u2. As for your third question, I do not know what a "white mass" is but clearly u1 and v will be equal and opposite while massm2, the "feather mass", will have a much higher acceleration and will accelerate toward either m1 or m2 depending upon which is closer. All that is pretty "Newton"- there is no reason to appeal to Relativity. Since you kindly awaited my answer I will await your response!

I also dislike the notion of "speed of time". Not that we don't move through time- we all do- at one second per second! it's just that our usual notion of "speed" is "rate of change compared to time"- which makes the "rate of change of time compared with time" seem redudant! However, that is not what is meant in relativity. If I am moving at 99% the speed of light compared to you, I would observe time changing very slowly for YOU compared to ME- YOUR second would be much longer than MINE.. Of course, since speed is "relative", you would observe my time changing very slowly compared to yours! It is even possible to calculate the precise change and that calculation has been experimentally verified- Cosmic ray, moving at high speed relative to us have a much shorter "lifespan" that suc cosmic rays should have by our calculation.

As long as you don't mind if WE have no idea what you are talking about!

No, it's not. A "3D curve" would be a one-dimensional object in three dimensions: a line in space or a spiral, for example. A surface, lke this paraboloid, is a 2-dimensional object.

1 5 a -4 -5a -16 a 20 16 I assume you added four times the first row to the second row and subtracted a times the first row to the third row to get 1 5 a 0 20-5a 4a-16 0 20-5a 16-a^2 Do you notice that the numbers in the second and third rows of the second column are the same? I would go ahead and subtract the second row from the third: 1 5 a 0 20-5a 4a-16 0 0 32-4a- a^2 Now 20- 5a= -5(a- 4) and 4a-16= 4(a-4) so dividing the second row by 20-5a give 0 1 -4/5. You have now 1 5 a 0 1 -4/5 0 0 32-4a- a^2 Of course, you can subract 5 times the second row now from the first row: 1 0 a+ 4 0 1 -4/5 0 0 32- 4a- a^2 Can you finish?

No, you are correct- the "factorial" is only defined for positive itegers. What others are saying here is that the "gamma function" has the property that it is identical to the factorial for positive integer values of the argument but is defined for other numbers as well. (It is NOT defined for negative integers because the integral does not exist in that case.)

First, what kind calculator are you talking about and, second, what do you mean by "tackle" an element? The Texas Instruments calculators do, I believe, keep x and y values of graphed functions in memory.

The real problem is that we were given a function and asked how to "solve it" but were never told what we were to solve! know0algebra, are you still with us? What, exactly, is your question?

No, a Taylor series is NOT a polynomial it is an infinite series. A polynomial, by definition, has a "highest power" (its degree) and is a finite sum. If you "chop" of a Taylor series at, say, the nth power, you get a "Taylor polynomial" which is the approximation Klynos was talking about. For an analytic function, its Taylor series is exactly equal to it, not an approximation. There are, however, "smooth" (infinitely differentiable) functions that are NOT analytic and not equal to their Taylor's series. One such is f(x)= e^{-1/x^2} if x is not 0 and f(0)= 0. All derivatives exist and area 0 at x= 0 so the Taylor series for it around x= 0 is identically 0. That, of course, converges for all x but converges to f only for x= 0.

Not restricting to square roots but roots in general- there exist polynomial equations whose solutions are not "surds"- they cannot be written in terms of any nth roots.

So "thin slab" perpendicular to the x-axis would be a square with base running from y= x^2 to x= y^2 (y= sqrt(x)). It's "thickness" would be dx so it would have volume x^2(sqrt(x))dx. To find the volume of the whole figure, integrate that with x going from 0 to 1. Do you see why x is from 0 to 1? Differentiating x^2/a^2+ y^2/b^2= 1 with respect to x gives 2x/a^2+ 2y/b^2 y'= 0 so y'= -b^2x/a^2. Differentiating x^2+ y^2= a^2 with repspect to x gives 2x+ 2y y'=0 so y'= -x/y.