We have for n > N that (dropping the constant multiple of \epsilon) |f_n(x) - f(x)| < \epsilon. We can write this in the following two ways:
\begin{split}
f_n(x) - \epsilon &< f(x) < f_n(x) + \epsilon \\
f(x) - \epsilon &< f_n(x) < f(x) + \epsilon. \end{split} Ignoring the lower bounds and...
E(2) also includes reflections. The rotations and translations generate a subgroup SE(2) of this group.
E(2) is a semidirect product of O(2) and \mathbb{R}^2 under vector addition. It acts on \mathbb{R}^2 as
(A, b) : x \mapsto Ax + b and its composition law is
(A, b)(C, d) = (AC, Ad + b)...
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...
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.
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...
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 &=...