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A rescaling is needed for a nontrivial limit. As discussed in Iteration of Sine and Related Power Series by C. Towse (2014), denoting the $n$-th iterate by $\sin^{\circ n}x$, one has the limit $$\lim_ …

The OP asks where this recursion relation might appear in a research context. It appears as a discretization of the Emden–Fowler nonlinear differential equation, $$f''(t)=t^{p}[f(t)]^q,$$ for $p=1$, $ …

UPDATE I tried to evaluate the sum numerically for large $n$ and what I find does not support the conclusion I give below, that the large-$n$ limit equals 1 independent of $c$. Here is a plot for $c= …

The sum $\sum _{n=0}^{\infty}t^{n^2}$ evaluates for $t<1$ to an elliptic theta function, and then taking the limit $t\rightarrow 1$ from below gives $$\lim_{t\nearrow 1}\sqrt{1-t}\sum _{n=0}^{\infty}t …

Mathematica can actually solve the recursion relation in closed form, $$F_n(n)=-\tfrac{1}{2}(n^2-1)^{-1}\left(\frac{n-1}{n}\right)^n\left[n \left(\frac{n}{n-1}\right)^n \Phi \left(\frac{n}{n-1},1,n+1\ …

This integral can actually be evaluated in closed form, from which the large-$n$ asymptotics follows readily: $$\int_0^{1} dt\, I_{2 t - t^2}(a,b)= \frac{1}{B(a,b)}\int_0^1 dt\,\int_0^{2t-t^2} ds\,s^{ …

A series with precise error estimates is derived in Error bounds for a uniform asymptotic expansion of the Legendre function: $$P_n(\cosh x)=\left(\frac{x}{\sinh x}\right)^{1/2}\sum_{\nu=0}^{\infty}c …

Q: Is $\lim_{x \to \infty}\frac{h(x)}{r(x)} = \frac{\pi}{2+\pi} $ ? A: use that $r(x)=h(x)+2a(x)$, hence $$\frac{h(x)}{r(x)} = \frac{h(x)/a(x)}{2+h(x)/a(x)}$$ and $$\lim_{x\rightarrow\infty}\frac{h(x) …

A general approach to obtain upper and lower bounds on $P(|X_1|\leq M_1, |X_2|\leq M_2,\dots, |X_d|\leq M_d)$ for a singular multivariate Gaussian, with a noninvertible covariance matrix, is developed …

Expansion of the binomial ${n\choose k}$ around the maximum $k=n/2$ gives a narrowly peaked Gaussian for large $n$, $$\binom{n}k=\frac{2^n}{\sqrt{n\pi/2}} \exp\left(-\frac{(k-n/2)^2}{n/2}\right) \left …

You can reduce the evaluation of $S_n$ to a quadrature by means of the Abel-Plana formula, $$S_n=\sum _{s=1}^{n-1} g(s)=\int_1^{n-1}g(s)\,ds+\tfrac{1}{2}g(1)+\tfrac{1}{2}g(n-1)$$ $$\qquad\qquad -\,2\o …

For any $\delta,x$ such that $0<\delta\leq |x|<\pi$ one has $|F_N(x)|\leq[2\pi(N+1)\sin^2(\delta/2)]^{-1}$, so the error in the delta-function approximation is of order $1/N$. Two explicit examples, i …

Use the expansion $$\zeta(s)=\frac{1}{s-1}+\gamma+{\cal O}(s-1),$$ hence, since $\zeta(x)=1+2^{-x}+3^{-x}+4^{-x}+5^{-x}+\cdots$, you have $$\zeta(\zeta(x))=\frac{2^x}{1+(2/3)^x+(2/4)^x+(2/5)^x+\cdots} …

The sum is only convergent for $p>1/2$, so for $p=1/2$ you could regularize it by adding a small positive increment: $p=1/2+\epsilon$, $\epsilon>0$. For $a\gg 1$ you can then approximate the sum by …

I find it useful to represent the Legendre function in terms of a hypergeometric function, using a formula from Wikipedia, $$f(\theta)=\frac{ (1+\cos \theta)^{\mu/2} \, _2F_1\left(-\nu,\nu+1;1-\mu;\f …