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Maple says $$\sum _{j=0}^{n}{\frac {{n\choose j} \left( -z \right) ^{j}}{j!}}={L}_n \left(z \right)$$ Laguerre polynomials. See https://en.wikipedia.org/wiki/Laguerre_polynomials and go down to "clos …

Off-the-wall suggestion... Take $n$ even, I call it $2n$ now. Then asymptotically as $n \to \infty$ $$ \binom{2n}{2n-2j-1}^{-1/(2n-2j-1)} - \binom{2n}{2n-2j}^{-1/(2n-2j)} \sim \frac{1}{2n}\log \frac{ …

half-discrete analytic First do the formal calculation, then discuss its validity. $$\begin{align} &\sum_{m = 0}^{\infty} \binom{x}{m} \sum_{k = 0}^{m} \binom{m}{k} (-1)^{(m - k)} \operatorname{sin} …

Take $x_0=0$, then the integral $$ \int_{-\infty}^\infty \sin\left(\frac{x}{\varepsilon}\right)dx $$ diverges for every $\varepsilon>0$. You can try a principal value for this. There is a whole bran …

What happens with this way of doing it? A rational function is a map from the Riemann sphere $\mathbb C \cup \{\infty\}$ to itself. The Riemann sphere is compact, so has a unique uniform structur …

The sigma algebra generated by $X$ is countably generated. Thus $X$ is measurable for $$ \sigma\left(\bigcup_{k=1}^\infty \mathcal{B}_{i_k}\right) $$ for some increasing sequence $i_1 \le i_2 \le \do …

An example. $X = [0,1]$ with the usual metric. $\mathcal C[0,1] = \mathcal C_b[0,1] = \mathcal C_0[0,1]$. $\mathcal C[0,1]^* = \mathcal M[0,1]$. Let $\mu_n$ be the unit point-mass at $1/n$ and $\mu …

A popular English-language treatment de Boor, Carl, Cutting corners always works, Comput. Aided Geom. Des. 4, 125-131 (1987). ZBL0637.41014. For this subsequent de Rham paper MR0095227 de Rham, George …

weak discrete-analytic Closed form here involves some ${}_2F_1$ functions, so I cannot provide proofs. Numerically, though, it seems that $\sin(ax)$ is weak discrete-analytic for $a$ up to some valu …

The answer is "no" ... also see a Usenet sci.math discussion in July, 2009.

For question 3. Using Maple's gfun library, I get this $4$-term recurrence for $t_{7A}$: \begin{align} 0 &=14\, \left( 2\,n+3 \right) \left( n+2 \right) \left( n+1 \right) u \left( n \right) \\ & - …

The reason in ordinary calculus is: if the function is Riemann integrable, then any tags can be chosen in any partitions (maximum length going to zero) and it converges to the integral. And of cours …

Comment. I think it should work like this. Let $\alpha = \sum_{i=0}^\infty\sum_{j=0}^\infty a_{ij}, \beta = \sum_{j=0}^\infty\sum_{i=0}^\infty a_{ij}$ both exist. Assume $\alpha < \beta$. Let $\va …

Not an answer Merely remarks. Let me use superscript $[k]$ for $k$-fold composition. $\log^{[3]} n$ means $\log\log\log n$. As I remarked on the other question, for fixed $a$ and $n$, the value $f(k, …

Your condition 2 adds nothing. Given any $\text{Lim}$ you can write it as $\lim \circ f$ where $f$ maps each sequence $\mathbf x$ to the constant sequence with value $\text{Lim}(\mathbf{x})$.