Recent content by bel

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    Proving an inequality - |sin n|>c

    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...
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    Proving an inequality - |sin n|>c

    Hahah, I can't say that was all too constructive =), but I'm glad it was interesting for you all the same.
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    Proving an inequality - |sin n|>c

    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...
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    Proving an inequality - |sin n|>c

    The maximum lower bound is zero, that was guarenteed by the axiom aforementioned, but the values are never zero, hence all values are positive.
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    Proving an inequality - |sin n|>c

    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...
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    Factor out the x

    Try long division and then express the remainder term as a rational function, and then apply L'hopital's rule to it.
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    Proving an inequality - |sin n|>c

    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.
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    Proving an inequality - |sin n|>c

    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.
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    Proving an inequality - |sin n|>c

    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...
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    Kinetic friction.

    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.
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    Proving an inequality - |sin n|>c

    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.
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    Kinetic friction.

    Calculate the magnitude of your normal force again, I got a different answer.
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    Proving an inequality - |sin n|>c

    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.
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    Mathematica Mathematical Induction

    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.
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    Kinetic friction.

    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.
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