Recent content by andrewkirk

There is only one rule. The second one you wrote is just what you get when you take the reciprocals of all the items. Naturally if a=b and neither a nor b is zero then it will also be true that 1/a = 1/b. Whenever you perform the same operation on both sides of the equation, the equality still...

It refers to a semi-ring of sets, defined here, not a semi-ring in the algebraic sense. That is not correct. [0,1] is indeed an element of ##\mathcal A## so this example does not demonstrate non-closure. That example was never going to work because you are subtracting from a set another set...

By that "if necessary" they mean that, if you can't prove $$\mathcal{F}_{\alpha+1}\subset\mathcal{G}$$ by assuming $$\mathcal{F}_{\alpha}\subset\mathcal{G}$$ then try proving it by assuming the stronger precedent: $$\forall \beta \leq \alpha:\ \mathcal{F}_{\beta}\subset\mathcal{G}$$ Since you...

The Hessian approach won't achieve anything if the gradient is zero, as the Hessian will then also be zero. EDIT: That's wrong. Please ignore.o:) We'd need an example of an ##(m-1)##-dimensional submanifold ##S## embedded in ##\mathbb R^m## such that: ##S## is defined by ##f(\mathbf x)=0## for...

A normal vector is defined relative to a surface, not relative to a function. If hypersurface S is a submanifold of differentiable manifold U (the space in which S is embedded) then at point ##p\in S##, the vector ##v## in the ##T_p(U)##, tangent space of U at p, is a normal vector of S iff v...

A derivative calculating by approaching from the right (positive perturbations) is called a right derivative. A derivative calculating by approaching from the left (negative perturbations) is called a left derivative. A function is differentiable at a point only if both of those exist and equal...

Say rather that (5) doesn't say there is an interaction, rather than a collision. We can make physical sense out of any solution to (5) with the following interaction. Mass 1 uses a spring-loaded device to fire a small particle towards Mass 2, which bounces off 2 and returns to 1, reloading the...

In your working you start with the "=54" there. That's never going to work because that's essentially what you have to prove (since 54 is the square of ##3\sqrt 6##), so you can't use it. Instead, define ##x = 3\sqrt{2 - \sqrt 3}+ \frac 3{ \sqrt{2 - \sqrt 3}}## and then try to prove that...

I think the problem lies in an incorrect hidden assumption the OP makes: that the ellipsoid shown is a surface of constant potential, so that the two charges start with the same potential. The electric field outside the conductor is much stronger near the point (on the right) than on the left...

On all the installations of GIT bash I have used, it is a stand-alone program. When you run it, it opens a dedicated terminal window into which you can type commends and see console output. It does not use either CMD or Powershell. The GIT bash shell allows you to use most (not all) Unix Bash...

Doesn't the belt trick demonstrate that the non-nullity of a continuous sequence of rotations between 0 and ##2\pi## (paired with a return to nullity when we rotate through a further ##2\pi## in the same direction) reflects a physical reality? Perhaps the difficulty lies in the statement...

It is not ##w## but ##f## and ##g## that are indicated to be elements of ##L^2## in the OP. ##w## is the measure function used to define the inner product on the space. In the wolfram article I linked they call it ##\mu##. If ##w\equiv 0## then all sets have ##w##-measure zero, so there is only...

According to Wolfram, ##L^2## is actually not a set of functions, but of equivalence classes of functions, where functions that are identical except on sets of measure zero are equivalent. Under that interpretation, we can conclude from ##\langle f, f\rangle = 0## that ##f## is zero except on a...

Very good. makes sense to me.

They must be countable, because the number of digits in a decimal expansion is countable, since we can number (count) each one based on how many places after the decimal point it comes. I intuitively expect each of the ten numerals to occur infinitely often in the expansion. But I can think of...