[t] is the floor function, and t just represents a generic argument. So for example [0.5]=0, [0.9]=0, [1.01]=1, [1]=1, [23.567]=23, and so on. You simply ignore whats written after the ...

The answer is already on the other page, basically, but here we go. For every t\geqslant0, let \Lambda(t)=\int\limits_0^t\lambda(s)\mathrm ds, hence \frac{\mathrm d}{\mathrm dt}\Lambda(t)=\lambda(t) ...

There’s a small error in your proof that x_n\ge 0 for all n: it’s not true that \frac{x_n}{x_n+1}>\frac1{x_n+1}\;, since in fact it turns out that x_n\le 1 for all n. However, given the ...

Consider the part P_N = \{ x : 1 \leqslant \lvert x\rvert \leqslant 2, \lvert x_1\rvert \leqslant 1/N\} of the annulus. On P_N, the integrand is bounded below by \frac{1}{2^{10}\cdot 3^{10}}, ...

For x=0 to be the coordinate of the local maximum, there must exist a neighborhood of 0 such that f(0) is the maximum. Since cosine attains 1 when x=0, a sufficiently small right neighborhood of ...

You can write x_{n+1}\geq x_n and x_{n+1}\leq x_n like so: x_n\leq x_{n+1}\leq x_n so x_n=x_{n+1}.