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This follows from the following characterisation of the eigenvalues of symmetric matrices.

\[\lambda_k(A) = \min_{\dim(U) = n-k+1} \max_{0 \neq x \in U} \frac{x^TAx}{\|x\|_2}\]

Here the minimum is taken over all subspaces $U$ of $\mathbb{R}^n$ of dimension $n-k+1$. (I chose to write $\lambda_K(A ...

It remained unanswered for quite some time. Since I have been asked for it, here is a solution:

The answer is affirmative.

Given a sequence $a = (a_1,a_2,a_3,\ldots)$, where $a_i \in \{1,2\}$ for each $i$, we construct the subset $S_a$ of $(0,1)$ as follows:

Every element of $S_a$ is of the form ...

We have also seen this here :

The basic steps are as follows: We consider independent Poisson distributions \(X_1,X_2,\ldots\) with parameter 1. We know that \(Y_n = X_1 + \cdots + X_n\) is Poisson with parameter \(n\). From the central limit theorem, \((Y_n-n)/\sqrt{n}\) converges in distribution ...

I presume you can show that \(f'(1) = 1\) showing that \(x=1\) is a stationary point. Now observe that for \(0 < x < 1\) we have \(x^x < x^1 = x\). So \(x^{x^x} < x^x\), i.e. \(f(x) < 0 = f(1).\) Also, for \(x > 1\) we have \(x^x > x\) giving \(f(x) > f(1)\). So \(x=1\) is a point of inflection.

To use Stirling's approximation observe that \[ \prod_{r=1}^n \left(1 + \frac{r}{n}\right)^{1/n} = \left( \frac{(n+1)(n+2)\cdots (n+n)}{n^n}\right)^{1/n} = \left(\frac{(2n)!}{n^n(n!)}\right)^{1/n}.\] By Stirling's approximation we have \[ \frac{(2n)!}{n^n(n!)} \sim \frac{\sqrt{4\pi n}(2n/e)^{2n}}{n ...

Indeed as Apostolos says, the integral diverges. What jacks's proof shows is that for every \(0 < \varepsilon < \pi/2\) we have that \(\int_{\varepsilon}^{\pi-\varepsilon} \lfloor \cot{x} \rfloor \; dx = -\pi + 2\varepsilon.\) However this does not mean that the value of the integral converges to ...

If this is a Riemann integral, we use the fact that any Riemann integrable function has a point of continuity. (In fact the set of point of discontinuity has measure zero - this is Lebesgue's criterion.) So there is an interval \(I \subseteq [0,x]\) and a positive real number \(a\) such that \(f(t ...

No it is not. For example \(z\) is analytic but \(\overline{z}\) is not. (As can be checked e.g. by the Cauchy-Riemann equations.)

By the way, "analytic" and "locally represented by a convergent power series" mean the same thing.

Hi Apostole,

It is a nice proof! I have not seen it anywhere.

Given two points \(x,y \in \mathcal{D}\) we define \[\displaystyle{D(x,y) = \frac{f(x)-f(y)}{x-y}}.\] Note that for \(x < y < z\) we have \[ D(x,y) = \alpha D(x,z) + (1-\alpha)D(z,y)\] where \(\alpha = \alpha(x,y,z) = \displaystyle{\frac{z-x}{x-y} \in (0,1)}.\) It follows that either \(D(x,z ...