Recent content by pasmith

A generalization is to fix ##g: [0,1] \to \mathbb{R}## and let $$f_n : [0,1] \to \mathbb{R} : x \mapsto \begin{cases} ng(nx) & x < \frac 1n \\ 0 & x \geq \frac 1n \end{cases}.$$ Then $$\int_0^1 f_n(x)\,dx = \int_0^1 g(x)\,dx.$$ The pointwise limit is again zero for ##x > 0##, but at zero we have...

I was going to append this to @chwala's thread here, but thought it deserved a new thread. For ##n \geq 1##, define $$f_n : [0,1] \to \mathbb{R} : x \mapsto \begin{cases} 1 & x = 0, \\ n(1 - nx) & x \in (0, \tfrac 1n], \\ 0 & x \in (\tfrac 1n, 1]. \end{cases}$$ (Note that ##f_n## is...

Consider also ##e^{-n\sin \theta}## for ##\theta \in [0, \pi]##.

Responses, like the original post, should be typed and make use of LaTeX rather than screenshots.

I was working at the UK's Ministry of Defence just over 10 years ago when they rolled out Office 365. This involved the MOD making absolutely certain - no doubt at great expense - that its data remained on secure servers which it controlled and did not leave the United Kingdom.

One can always resort to the Newton-Raphson method to solve ##\mathbf{f}(\mathbf{x}) = \nabla L = 0## by iteration, where each step requires solving the linear system $$ J(\mathbf{x}_n)\mathbf{u} = -\mathbf{f}(\mathbf{x}_n)$$ for ##\mathbf{u} = \mathbf{x}_{n+1} - \mathbf{x}_n## where $$J =...

If $$L = (y_1(x_1) - y_2(x_2))^2 + (x_1 - x_2)^2$$ then the partial derivatives are $$\begin{split} \frac{\partial L}{\partial x_1} &= 2(y_1 - y_2)y_1'(x_1) + 2 (x_1 - x_2) = 0 \\ \frac{\partial L}{\partial x_2} &= -2(y_1 - y_2)y_2'(x_2) - 2 (x_1 - x_2) = 0.\end{split}$$ Adding these gives $$...

This question seems poorly designed, in that part (i) has absolutely nothing to do with part (ii).

Thinking about it more closely, we want ##c_1y_1 + c_2y_2## to be a particular solution. But we're happy if it's only determined up to a complemenary function, because we were going to add one anyway. Two choices of ##(c_1 ,c_2)## cause ##c_1y_1 + c_2y_2## to differ by a complementary function...

I think you have an error in the algebra. Substituting the ansatz and condition into the ODE, we should find $$\begin{split} p(c_1'y_1' + c_2'y_2') &= f - (ph)' \\ c_1'y_1 + c_2'y_2 &= h \end{split}$$ hence leading to $$\begin{split} c_1' &= -\frac{(f - (ph)')y_2}{pW} + \frac{hy_2'}W =...

Any CAS which knows that for ##a \in \mathbb{R}##, ##(a^2)^{1/2} = |a|## should be able to get it right.

Based on the sketch, it seems that ##\theta## is the angle of cylindrical polar coordinates, not spherical polar coordinates. I agree that the base of the cylinder is completely described by ##-\pi/2 \leq \theta \leq \pi/2##. The factor of 2 comes not from rotational symmetry about the origin...

It's a bit confusing to take ##b < a##; usually we have ##a < b##. My approach would be to define $$ g : [0,1]\to\mathbb{R} : x \mapsto f(b) + x(f(a) - f(b)) - f(b+ x(a - b)) \geq 0.$$ Now ##g(0) = g(1) = 0## and the constraint on the sign of ##g## allows you to say something about the signs of...

A Hessian matrix must be symmetric; this matrix is not. This is a consequence of the problem I noted in my earlier post: the vector field specified by the OP is not the gradient of a scalar function.

It may help to note that $$\frac{n^2}{n + \frac15} = n - \frac{1}{5 + \frac 1n}.$$