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Truncated exponential series equation

$\textbf{Problem}.$ For a positive integer $n$ let $x_n$ be the unique positive real root of the equation $$\sum_{k=0}^n \frac{x^k}{k!}=\frac{e^x}{2}.$$ Prove that $\displaystyle\lim_{n\to\infty}(x_n-n)=\frac{2}{3}.$ $\textbf{Comment.}$ The limit $\displaystyle\lim_{n\to\infty} e^{-n}\sum_{k=0}^n \frac{n^k}{k!}$ is well known to be $\displaystyle\frac{1}{2}$. It could be derived via nice probabilistic structure related to Poisson distribution. This limit hints that the value of the root $x_n$ should be close to $n$.  $\textbf{Sketch of a solution of the problem}$. Defining  $$g(x)=\frac{1}{n!}\int_0^x s^ne^{-s}\ ds-\frac{1}{2}$$  the original equation could be rewritten in the form $g(x)=0.$ One should observe that $g$ is monotonically increasing and the unique root is in the interval $(n,n+1)$. Then one uses one iteration of the Newton method initialized at $n$ to obtain an approximation of the solution $\tilde x_n$. Using the final asymptotic obtained at the end of the this answer a

Proof of Gelfand-Phillips Theorem

$\textbf{Theorem.}$ Let $X$ be a Banach space and $A\subseteq X$. Prove that $A$ is precompact (in the norm) if and only if for every $w^*$-convergent to $\textbf{0}$ sequence $\{x_n^*\}_{n\ge 1}\subseteq X^*$ it holds that $\{x_n^*\}_{n\ge 1}$ converges uniformly to $\textbf{0}$ on  $A.$   $\textbf{Proof.}$ If $A$ is precompact the result follows from Arzela-Ascoli theorem (combined with the fact that $w^*$-convergent sequences are norm bounded). Now we prove the reverse direction.  First we proof the following characterization of compactness in normed spaces: $A$ is compact if and only if $A$ is bounded and for any $\varepsilon>0$ there exists a finite-dimensional space $F$ such that $A\subseteq F+\varepsilon\mathbf{B}.$ $\textit{Proof}.$  If $A$ is precompact, then for any $\varepsilon>0$ there exists $\{a_i\}_{i=1}^n\subseteq A$ such that $A\subseteq\bigcup_{i=1}^n\mathbf{B}_{\varepsilon}(a_i)$. In particular $A$ is bounded and \[A\subseteq \text{span}(\{a_i\}_{i=1}^n)+\var

Closed orbits are finite

 Let $f:[a,b]\to [a,b]$ be continuous function and $x_0\in [a,b]$. Consider the sequence $X=\{x_n\}_{n\ge 0}$ defined by $x_n=f(x_{n-1})$ for $n\ge 1$. Assume that $X$ is a closed set. Prove that it consists of finitely many elements. (The problem is taken from IMC 2002. The set $X$ is called the orbit of $f$ starting from $x_0$) $\textbf{Proof.}$ Let us assume that $X$ is infinite. So it has a limit point, which should belong to the set (since it is closed). Let $$n_0:=\min\{n\ | \ x_n \text{ is a limit point of } X \}.$$ Then define $\hat X=\{x_n\}_{n\ge n_0+1}$. Thus $\hat X\cup \{x_{n_0}\}$ is compact and $f:\hat X\cup\{x_{n_0}\}\to \hat X$ is a continuous function. Since the image of compact sets under continuous functions are compact set, we must have that $\hat X$ is compact as well. But it is clearly not closed, since $x_{n_0}$ is a limit point $\hat X$ and does not belong to $\hat X$.

There is no lowest rate of convergence

 Let $\{a_n\}_{n\ge 1}$ be a sequence of positive numbers such that  $$\sum_{n=1}^{\infty} a_n$$ is convergent. Then there exists a sequence $\{b_n\}_{n\ge 1}$ with $\displaystyle \lim_{n\to \infty}\frac{b_n}{a_n}=\infty$ and such that  $$\sum_{n=1}^{\infty} b_n$$ is also convergent. First proof (a la Calculus I, based on ideas of Zhivko Petrov ). Consider $$S_n=\sum_{k=n}^{\infty}a_k.$$ Clearly $S_n\to 0$. Define $$b_n=\sqrt{S_n}-\sqrt{S_{n+1}}.$$ Clearly $b_n>0$ for all $n$ and $b_n\to 0$. Moreover  $$\sum_{n=1}^\infty b_n=\sqrt{S_1},$$ hence is convergent. On the other hand $$\lim_{n\to\infty}\frac{b_n}{a_n}=\lim_{n\to\infty}\frac{1}{\sqrt{S_n}+\sqrt{S_{n+1}}}=\infty.$$   Second proof . (Based on Folland 's book) Assume on the contrary that there is lowest decaying sequence, i.e.  for some sequence $a=\{a_n\}_{n\ge 1}$ with convergent series, holds that for any sequence of positive numbers $\{b_n\}_{n\ge 1}$, the series $$\sum_{n=1}^\infty a_nb_n$$ is convergent if and only

Two functional inequalities

$\textbf{Problem 1.}$ Find all functions $f:\mathbb{R}\to\mathbb{R}$ such that for all $x$ and $y$ holds $$f(x)f(x+y)\ge f(x)^2+xy.$$ $\textbf{Problem 2.}$ Find all differentiable functions $f:\mathbb{R}\to\mathbb{R}$ with $f(0)=0,\ f(1)=1$ and such that for all $x$ and $y$ holds $$f(x+y)\ge 2022^xf(x)+f(y).$$

Haudorff dimension and Liouville numbers

In this post we present a proof of the fact the the set of Liouville numbers contained in $[0,1]$ has Hausdorff dimension $0$.  Why is this result interesting, apart from finding explicitly the Hausdorff dimension of a particular set? Regarding Lebesgue measure, it is not obvious whether there exists an uncountable set with measure zero. Lioville numbers serve as such example. However, Liouville numbers even serve as an example of an uncountable set with Hausdorff dimension zero, and this is much stronger. That is because every set of dimension zero has measure zero, whereas for example, another common reference of an uncountable set with measure $0$, the Cantor set, has positive Hausdorff dimension ($\ln 2/\ln 3$). Thus the set of Lioville numbers, is in a sense, way smaller than the Cantor set. One could also say that Liouville numbers are example of the smallest possible uncountable sets among the sets with measure zero. You can look here, for another construction of uncou

Problem proposed by prof. Gadjev

$\textbf{Problem.}$  Let $f:[0,1]\to\mathbb{R}$ be a continuous function. Find the limit  $$\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n f\left(\frac{\ln k}{\ln n }\right).$$   $\textbf{Solution.}$  Fix $\varepsilon>0$. We aim to show that for large enough $n$   $$\left|\frac{1}{n}\sum_{k=1}^n f\left(\frac{\ln k}{\ln n }\right)-f(1)\right|<3\varepsilon$$  which would mean that the limit is $f(1).$ Let $M$ be an upper bound for $|f|$ on $[0,1]$, i.e. $|f(x)|\le M$ for all $x\in [0,1].$  For all $n\ge 1$ define $$k(n)=\left\lfloor\frac{n}{n^{1/\sqrt{\ln n}}}\right\rfloor.$$ The number is chosen in such a way that $$\lim_{n\to\infty}\frac{k(n)}{n}=0 \ \ \& \ \  \lim_{n\to\infty}\frac{\ln(k(n))}{\ln n}=1.$$  Split the sum in two: $$\left|\frac{1}{n}\sum_{k=1}^n f\left(\frac{\ln k}{\ln n }\right)-f(1)\right|\le \left|\frac{1}{n}\sum_{k=1}^{k(n)} f\left(\frac{\ln k}{\ln n }\right)\right|+\left|\frac{1}{n}\sum_{k=k(n)+1}^{n} f\left(\frac{\ln k}{\ln n }\right)-f(1)\right| \ \ \ (*)$$  T