Recent content by pasmith

Alternatively, we can also consider that the general solution of ##F_{n+2} = F_{n+1} + F_n## is $$\begin{split} F_n &= A \left( \frac{1 + \sqrt{5}}{2}\right)^n + B\left( \frac{1 - \sqrt{5}}{2}\right)^n \\ &= A \varphi^n + B (-\varphi^{-1})^n \\ &= \varphi^n \left(A + B (-\varphi^{-2})^{n}\right)...

A slight correction is indeed required. From ##a_n \geq 1## we have ##a_n \varphi \geq \varphi ## so that $$ |a_{n+1} - \varphi| \leq \frac{1}{\varphi} |a_n - \varphi|$$ which together with ##\varphi > 1## is sufficient for ##a_n \to \varphi##.

There is a much easier way to arrive at these results. From ##a_{n+1} = 1 + a_n^{-1}## and the definition of ##\varphi## we can obtain $$ a_{n+1} - \varphi = (-1)\frac{a_n - \varphi}{a_n\varphi}.$$ Assuming ##a_n \geq 1## we have ##a_n \varphi \geq \varphi > 1## so that $$ |a_{n+1} - \varphi| <...

So that we understand the proofs, and can satisfy ourselves that they are indeed correct proofs. We don't take other people's word for it in mathematics.

It is a fact of basic group theory that conjugation preserves the order of an element (because conjugation is an isomorphism from a group to itself). The order of a cycle is the same as its length: the identity (generally written as a single 1-cycle) has order 1. a transposition (1 2) has order...

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 =...