Recent content by pasmith

Note that \mathbf{J} is zero everywhere except on r = a where it is not defined, but causes a jump in \mathbf{B} across r = a. Therefore you need to take the general solutions in a vacuum which are proportional to \sin \theta or \cos \theta, which are different for r < a and r > a, and adjust...

In fact we have uniform continuity, since \delta = \epsilon works independently of x.

We are not even given that X is a vector space, so we don't have a concept of linearity. @Lambda96: All you have available to you are the properties of an arbitrary metric and the definition of continuity with respect to that metric. Consider the triangle rule.

You should start a new thread. Note that this new question is a generalization of the question you posted in this thread.

|\phi(f) - \phi(g)| = \left| \int_a^b f(x)\,dx - \int_a^b g(x)\,dx \right| = \left| \int_a^b f(x) - g(x)\,dx\right| etc.

Define "hollow". Do you mean a circular annulus lying between 0 < a \leq r \leq b? In that case yes, both J and Y are admissible solutions. For orthogonality relations, see section 11.4 of Abramowitz & Stegun, available at...

Hint: By basic properties of the Riemann integral, \left| \int_a^b f(x)\,dx \right| \leq \int_a^b |f(x)|\,dx \leq (b- a) \sup |f|.

You know from the Cayley-Hamilton theorem that A = \begin{pmatrix}2 & - 5 \\ 4 & - 2 \end{pmatrix} satisfies its own characteristic polynomial, so that A^2 = -4^2I. It follows that you can compute \Phi(t) = \exp(At) directly: \begin{split} \exp(At) &= \sum_{n=0}^\infty \frac{A^{2n}t^{2n}}{(2n)!}...

Note that sapping a and b swaps 3a^2 + 2b and 3b^2 + 2a, leaving the maximum invariant. Note also that 3a^2 + 2b is strictly increasing in the direction of increasing b, and 3b^2 + 2a is strictly increasing in the direction of increasing a. These observations suggest that \min_{(a,b)} \max...

No. The indefinite integral already includes an arbitrary constant; you don't need to add an e^{At}x(0) term in this case. That leaves e^{At} \int e^{-At}f(t)\,dt.

They are evidently using e^{At}\int^t e^{-As}f(s)\,ds with an indefinite integral, so that an arbitrary constant of integration must be included. If instead the integral is made definite with a lower limit of t_0, then the arbitrary constant becomes \vec{x}(t_0).

You need to end up with \sinh^2 \frac x2, so start from \cosh 2x - 1 = 2\sinh^2 x. But using these identities comes very close to assuming exactly what the question is asking you to show. In any event, the simplest approach is the straight forward \begin{split} (e^{x/2} - e^{-x/2})^2 &=...

It i a counter example to the assertion f''(x) > 0 \Rightarrow f(x) > 0.

What do you get if you differentiate x(t) = \Phi(t)x_0 for constant x_0?

I think they have seen that the hypoteneuse is 5km and have therefore assumed, entirely incorrectly, that the triangle is the well-known right-angled triangle with sides 3 km, 4 km and 5km.