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By the Binet formula, $$F_n\approx\frac{\phi^n}{\sqrt5}.$$ Then multiplying the $n$ first estimates $$P_n\approx a\frac{\phi^{n(n+1)/2}}{\sqrt5^n}.$$ By numerical computation, $a\approx 1.22674201072$. We can deduce an expression for the geometric average $$\sqrt[n]P_n\approx\frac{\phi^{(n+1)/2}}{\sqrt5}\approx\frac{F_n}{\sqrt{5\phi}}.$$


This is known as a product integral (Wikipedia). There seem to be competing definitions, but the one that makes the most sense to me is "Type I", $$\prod_{a}^{b}f(x)^{dx}=\lim_{\Delta x\to 0}\prod f(x_i)^{\Delta x}=\exp\left(\int_a^b\ln f(x)\,dx\right)$$ Also from that page:


In order to prove that $C=1$, you need to prove that: $$ \gamma = \sum_{n\geq 1}\left(\frac{1}{n}-\log\left(1+\frac{1}{n}\right)\right) \tag{A}$$ but that is exactly the usual definition of the Euler-Mascheroni costant, since: $$ \sum_{n=1}^{N}\left(\frac{1}{n}-\log\left(1+\frac{1}{n}\right)\right) = H_N-\log(N+1).\tag{B}$$


Let $m = \pi(k)$. Then for $n > m$, $$\log f(n,k) - \log f(n,0) = \sum_{i=m+1}^n \log(1 - k/p_i) - \sum_{i=1}^m \log(p_i) $$ As $i \to \infty$, $\log(1 - k/p_i) \sim \dfrac{-k}{p_i} \sim \dfrac{-k}{i \log i}$, and $\int_{2}^n 1/(t \log t)\; dt \sim \log \log n$, so we should get $$ \log f(n,k) - \log f(n,0) \sim -k \log \log n$$ and thus $$\dfrac{f(n,k)}...


$\DeclareMathOperator{\proj}{proj}$The term "component" has several different (but partially-overlapping) meanings in linear algebra. These include: Your definition, which is a scalar: If $u$ is a unit vector, then the component of $A$ on $u$ is the scalar $c$ such that $$ cu = (|A| \cos\theta) u = (A \cdot u) u = \proj_{u}A $$ is the orthogonal projection ...


Basically you want the leading-order digits, starting from the second one, of $\sqrt{n!}$, where $n$ is a power of ten. Stirling's approximation will give you what you need. Specifically, use $$ n! \sim n^n e^{-n}\sqrt{2\pi n}\left(1 + \frac{1}{12n}\right). $$ If $n=10^k$, then $n^n$ is a large even power of $10$ and its square root doesn't affect the ...


Right direction. $A^2 B-2A -B=0$ is equivalent to: $$ \frac{2A}{B} = (A^2-1) \tag{1}$$ that in our case follow from the identity: $$ c^{2^n}-1 = \left(c^{2^{n-1}}+1\right)\cdot \left(c^{2^{n-2}}+1\right) \cdot \ldots \cdot\left(c^{2}+1\right)\cdot(c+1)\cdot(c-1).\tag{2} $$


A space $X$ has the "finite-closed" topology, if the only closed sets are $X$ and the finite subsets of $X$. If $X_1,X_2$ are two infinite spaces, then $X_1 \setminus \{x_1\}$ is open in $X_1$ if $X_1$ has the finite-closed topology. But then $(X_1 \setminus \{x_1\}) \times X_2$ is also open, so its complement in the product, the set $\{x_1\} \times X_2$, ...


The problem is with the step $\mathbb{E}[\exp \sum_{i=1}^M \log \xi_i]\approx \mathbb{E}[\exp M\mathbb{E}[\log\xi_i]]$. As $M\rightarrow \infty$ the left and right hand sides of the approximate equality do not converge to each other except in the special case where the expectation and logarithm commute.


There is a nice argument based on Weierstrass products. The sine function has its (simple) zeroes at $\pi\mathbb{Z}$ and $$\frac{\sin x}{x}=\prod_{m\geq 1}\left(1-\frac{x^2}{n^2\pi^2}\right)\tag{1}$$ holds. The $\sin(n x)$ function has its zeroes at $\frac{\pi}{n}\mathbb{Z}=\left(\pi Z\right)\cup\left(\pi \mathbb{Z}+\frac{1}{n}\right)\cup\ldots\cup\left(\...

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