Recent content by Nono713

Your bound for y is wrong, draw a diagram with your bounds (e.g. y as a function of x) and you'll see that's not the bounds of a circle quadrant. Hint: the relation is x^2 + y^2 = 3^2

Thanks again for your help! One last question: I assume the N to N and S to S magnet layout is required in order to get the correct field configuration for the desired eddy currents and/or diamagnetism effects, however is it possible to space the magnets a little bit with something nonmagnetic...

An irrational \xi admits a finite or periodic continued fraction representation if and only if there exist integers a, b, c, d with c \geq 1 and d nonzero such that \displaystyle \xi = \frac{a + b \sqrt{c}}{d} So in short the answer to your question is no; the only classes of numbers which...

The unit step function is zero if its argument is negative, and t - n will eventually become negative for sufficiently large n.

Thanks a lot for the thorough answer! I hadn't realized that the field would be zero inside the cylinder; naive me had expected it to kind of work like a normal magnet lining the inner cylinder walls :D good thing you mentioned the axial force as that could seriously hurt someone. The design...

Hi! Suppose I have a cylinder of pyrolytic carbon graphite (strong diamagnet) that's vertically placed inside a hollow cylinder made out of approximately radially magnetized neodymium. Assume the graphite is frictionlessly held into place vertically somehow (so gravity can be ignored) but can...

Do you know about Euler's theorem, or Fermat's little theorem? Powers of 5 are periodic modulo 503, so your expression can be simplified if you can find what that big exponent is modulo that period. Euler's theorem tells us that the period is divisible by divides 503 - 1 = 502 (since 503 is...

Hint: multiply through by $\log(x)$ then substitute $u = \log(x)$, you are left with a quadratic equation in $u$.

Show that this series diverges: $$\sum_{n = 0}^\infty \cos \left ( n^2 \right )$$ (in the sense that it takes arbitrarily large values as $n \to \infty$)

Re: number of digit

Well, you just have to multiply 17 with itself twice, that is, 17^3 = 17 * 17 * 17... then you can use whatever method you want to do it "easily"/quickly. For instance if I did not have access to a calculator what I would do is start with one multiplication like this: 17 * 17 = (20 - 3) * (20 -...

That's right :) using this technique you can easily construct a group element that has the desired prime order, as you have shown. In general, if $g$ has order $nm$ then $g^n$ has order $m$ and symmetrically $g^m$ has order $n$.