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18

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 $$\sqrt[n]P_n\approx\frac{\phi^{(n+1)/2}}{\sqrt5}\approx\frac{F_n}{\sqrt{5\phi}}.$$

7

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:

4

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: $$\sum_{n=1}^{N}\left(\frac{1}{n}-\log\left(1+\frac{1}{n}\right)\right) = H_N-\log(N+1).\tag{B}$$

2

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 $$\dfrac{f(n,k)}... 1 \DeclareMathOperator{\proj}{proj}The term "component" has several different (but partially-overlapping) meanings in linear algebra. These include: Your definition, which is a scalar: If u is a unit vector, then the component of A on u is the scalar c such that$$ cu = (|A| \cos\theta) u = (A \cdot u) u = \proj_{u}A $$is the orthogonal projection ... 1 Basically you want the leading-order digits, starting from the second one, of \sqrt{n!}, where n is a power of ten. Stirling's approximation will give you what you need. Specifically, use$$ n! \sim n^n e^{-n}\sqrt{2\pi n}\left(1 + \frac{1}{12n}\right). $$If n=10^k, then n^n is a large even power of 10 and its square root doesn't affect the ... 1 Right direction. A^2 B-2A -B=0 is equivalent to:$$ \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} $$1 A space X has the "finite-closed" topology, if the only closed sets are X and the finite subsets of X. If X_1,X_2 are two infinite spaces, then X_1 \setminus \{x_1\} is open in X_1 if X_1 has the finite-closed topology. But then (X_1 \setminus \{x_1\}) \times X_2 is also open, so its complement in the product, the set \{x_1\} \times X_2, ... 1 The problem is with the step \mathbb{E}[\exp \sum_{i=1}^M \log \xi_i]\approx \mathbb{E}[\exp M\mathbb{E}[\log\xi_i]]. As M\rightarrow \infty the left and right hand sides of the approximate equality do not converge to each other except in the special case where the expectation and logarithm commute. 1 There is a nice argument based on Weierstrass products. The sine function has its (simple) zeroes at \pi\mathbb{Z} and$$\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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