Take: $$a_0=x,~~~~b_0=y$$ $$a_{n+1}=\frac{a_n+\sqrt{a_nb_n}}{2},~~~~b_{n+1}=\frac{b_n+\sqrt{a_nb_n}}{2}$$ Then we obtain as a limit the logarithmic mean of $x,y$: $$\lim_{n \to \infty} a_n=\lim_{n \to \infty} b_n=\frac{x-y}{\ln x-\ln y}$$ I don’t know how to prove this. But I do know that numerically it fits really well. In fact, the best approximation is obtained if we take geometric mean of $a_n,b_n$: $$x=5,~~~~y=3$$ […]

In answering calculate the mean and variance of the highest number drawn on lottery based on the lowest number drawn, I couldn’t find the mean and variance of the order statistics of a discrete uniform sample without replacement anywhere, so I figured I’d derive them here for future reference. Draw $k$ distinct numbers uniformly from […]

This question already has an answer here: If $\sigma_n=\frac{s_1+s_2+\cdots+s_n}{n}$ then $\operatorname{{lim sup}}\sigma_n \leq \operatorname{lim sup} s_n$ 2 answers

We know that if we iterate arithmetic and harmonic means of two numbers, we get their geometric mean. So, basically if we need to compute the square root of $x$: $$\sqrt{x}=\sqrt{1 \cdot x}=AHM(1,x)$$ $$a_0=1,~~~~b_0=x$$ $$a_{n+1}=\frac{a_n+b_n}{2},~~~~~b_{n+1}=\frac{2a_nb_n}{a_n+b_n}=\frac{a_nb_n}{a_{n+1}}$$ As far as I know, this expression will converge for any real positive $x$. See this and this question for […]

I want to create an ordered sequence of various ‘three-number means‘ with as many different elements in it as possible. So far I’ve got $12$ ($8$ unusual ones are highlighted): $$\sqrt{\frac{x^2+y^2+z^2}{3}} \geq \color{blue}{ \frac{\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2}}{3 \sqrt{2}}} \geq $$ $$\geq \color{blue}{\frac{\sqrt{(x+y)^2+(y+z)^2+(z+x)^2}}{2 \sqrt{3}} } \geq \frac{x+y+z}{3} \geq $$ $$ \geq \color{blue}{ \frac{\sqrt[3]{(x+y)(y+z)(z+x)}}{2} } \geq \color{blue}{\sqrt{\frac{xy+yz+zx}{3}}} \geq $$ $$\geq […]

Let’s consider the function defined by the integral: $$R(a,b,c,d)=\int_0^\infty \frac{dx}{\sqrt{(x+a)(x+b)(x+c)(x+d)}}$$ I’m interested in the case $a,b,c,d \in \mathbb{R}^+$. Obviously, the function is symmetric in all four parameters. This function has some really nice properties. $$R(ka,kb,kc,kd)=\frac{1}{k} R(a,b,c,d)$$ Thus: $$R(a,a,a,a)=\frac{1}{a}$$ Moreover: $$R(a,a,b,b)=\frac{\ln a-\ln b}{a-b}$$ This is the reciprocal of the logarithmic mean of the numbers $a$ and […]

Is there a general closed form or the integral representation for the limit of the sequence: $$a_{n+1}=\frac{\sqrt{(a_n+b_n)(a_n+c_n)}}{2} \\ b_{n+1}=\frac{\sqrt{(b_n+a_n)(b_n+c_n)}}{2} \\ c_{n+1}=\frac{\sqrt{(c_n+a_n)(c_n+b_n)}}{2}$$ in terms of $a_0,b_0,c_0$? $$L(a_0,b_0,c_0)=\lim_{n \to \infty}a_n=\lim_{n \to \infty}b_n=\lim_{n \to \infty}c_n$$ For the most simple case $a_0=b_0$ we have some interesting closed forms in terms of inverse hyperbolic or trigomonetric functions: $$L(1,1,\sqrt{2})=\frac{1}{\ln(1+\sqrt{2})}$$ $$L(1,1,1/\sqrt{2})=\frac{2 \sqrt{2}}{\pi}$$ […]

Define our Harmonic sequence for two numbers such that \begin{equation} a_{n+1} = \frac{2a_nb_n}{a_n + b_n} \end{equation} and our geometric sequence \begin{equation}b_{n+1} = \sqrt{a_nb_n} \end{equation} such that as $n \rightarrow \infty$ we tend towards the Geometric-Harmonic Mean. The arithmetic-geometric mean can be defined by the following two sequences. First compute the arithmetic mean of two numbers […]

Let $A$, $G$ and $H$ denote the arithmetic, geometric and harmonic means of a set of $n$ values. It is well-known that $A$, $G$, and $H$ satisfy $$ A \ge G \ge H$$ regardless of the value $n$. Furthermore, for $n=2$ we have $$G^2 = AH$$ By coincidence I found the following result for $n=3$ […]

For every nonnegative $(a_0,b_0,c_0)$, consider $$a_{n+1}=\sqrt{a_n \frac{b_n+c_n}{2}},\quad b_{n+1}=\sqrt{b_n \frac{c_n+a_n}{2}},\quad c_{n+1}=\sqrt{c_n \frac{a_n+b_n}{2}}$$ $$M(a_0,b_0,c_0)=\lim_{n \to \infty} a_n=\lim_{n \to \infty} b_n=\lim_{n \to \infty} c_n$$ This simple three term iterated mean gives several interesting closed forms for particular cases (confirmed to $14$ digits so far): $$M(1,1,2)=\frac{3^{3/4}}{\sqrt{\pi}}=1.28607413715749$$ $$M(1,1,\sqrt{2})=\frac{2}{\sqrt{\pi}}=1.12837916709551$$ $$M(1,1,\sqrt{3})=\frac{2^{3/4}}{\sqrt{\arccos(-1/3)}}=1.21670090936316$$ $$M(1,1,\sqrt{3}/2)=\frac{1}{\sqrt{\ln 3}}=0.95406458200000$$ These particular closed forms (if true) imply that the […]

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