That's very interesting. What I was thinking was that you could take any integer, and multiply it by two, then by pi, then subject that result to the floor function, and then find 2k(pi)-floor(2k(pi)) and it could always be lower and it all depends on the digits in pi, it's sort of like asking...
I sort of feel we are as well, let's hope someone will come in and say something more constructive then either of us have had to contribute. By the way, since |sin(n)|>0, the openess of the set means that there is always a neighbourhood (i.e., another open set) smaller than |sin(n)| but bigger...
Well, for one thing, the sequence |sin(n)| does not converge, when you start taking limits to infinity, you can't apply the axiom since infinity is not a number. It's the same as asking why some infinite series of rational numbers converge to irrational numbers. We're not taking limits here. If...
I don't really see what's wrong with my argument, so you would probably have to point it out to me. If I missed anything in the statement I first gave, it was the citing of the axiom of the completeness of the reals (and the absolute sign, which I later added), once applied, the proof is complete.
The maximum lower bound, as defined by the axiom of the completeness of the reals, is zero in this case, hence the lower bound is not negative. The lower bound is zero, but because the set is opened, the function is never zero, given the arguments.
This is because the the argument of the sine function in question can never be multiples of two pi, which are the only arguments for which the function can be zero, and it can't because all natural numbers are integers and pi is an irrational number, and any integer times two (which is an...
I got 60*9.81*cos(11)= 577.79 N, keep a few more digits in calculation, by the way, and round at the end, especially since you are looking at small numerical answers.
Sorry for leaving out the absolute signs. Added, it does, by the axiom of the completeness of the reals, if a set has a lower bound, it has a maximum lower bound which is real.
No, there isn't a particular "formal" way to do proofs, (excluding those types already formalised, such as mathematical induction proofs,) other than transforming most of the words into symbols, which you seem quite adept at doing already and adding "Q.E.D." at the end.
Factor \frac{1}{30} out of every term first, i.e., multply the (k+1)^4 term by thirty and put it atop the fraction as well, then expand and factorise, using the remainder theorem, of course, since you know what factors to expect.
If he is sliding down with constant velocity, he is not accelerating as he would be if there were no friction, hence the force of friction is equal in magnitude and opposite in direction to the component of gravitational force parallel to the incline.
If the Natural Numbers are defined as the set of positive integers, as opposed to the definition where the Natural Numbers are the set of non-negative integers, since no natural number is a integer multiple of 2\pi, and |sin(n)| only equals to zero when n= 2k\pi, and is between in the range (0...
I think the second statement is meant to be an argument based on the theory of motion of Aristotle's. And as for number six, I don't think you determined who was at fault, nor can you.
The acceleration would also be the angular acceleration of the pulley times its diameter, and that times its mass would be the force on the pulley, I think. So calculate the vertical and horizontal components of the acceleration vector on the pulley, remembering all the while that only the...
For any function f(x), f(x-a) moves the graph right by a units, a*f(x) stretches the graph by a factor of "a", f(ax) increases the frequency of the graph "a" times, f(x)-a moves a graph down by "a" units.
Since a and b are both composite, let ab= mnjk, where two factors of mnjk is a and the other two is b. If p is prime and it divides ab = mnjk, then taking any combination of two of m,n,j, or k to be one factor and two to be the other, it is evident that one of the two factors must be divisible...
The initial vertical component of the velocity of the released item is not zero, due to the vis inertiae, since the aeroplane was travelling at two hundred and fifty-five kilometres per hour times the sine of thirty degrees vertically.
Assuming that there is no friction, since the skier is not accelerating, there is no unbalanced force. The normal force is balanced by the component of the gravitational fforce perpendicular to the incline, the pulling force of the rope is balanced by the component of the gravitational force...
Say, for the train, which is initially (i.e., at time t=0) 32 metres ahead of the car, and travels with a constant velocity, say v_{train}. Then we have for the train s=32+v_{train}t.
Let the point when the measurement starts be s=0, then you have all the initial conditions. Now consider the distance s as a function of time t, so s(0)=0 for the car and s(0)=32 for the train. Write two such formulae, one to describe each object. At the point where the car passes the train, the...
The assumption of translational symmetry is one of the most basic assumptions of physics without which physics is generally meaningless. If this assumption is challenged, other physical laws should not be invoked to further an argument. And Newton's theory of universal gravitation follows from F=ma.
I already gave the formulae above, and by those (i.e., by equating the ss, as I mentioned), I came to your conclusion that "
I have figured that the time to catch up = Da/(Vb-Va)".