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6

We see that $$S_n = 1^{\frac{1}{2}}2^{\frac{1}{4}}3^{\frac{1}{8}}\ldots = \prod_{k=1}^{n}k^{2^{-k}}.$$ Furthermore $$\log S_n = \sum_{k=1}^{n}\log k^{2^{-k}} = \sum_{k=1}^{n}\left(\frac{1}{2}\right)^k\log k.$$ So we see that $$\lim_{n\rightarrow\infty}\log S_n = \sum_{k=1}^{\infty}\left(\frac{1}{2}\right)^k\log k \le ... 5 As Cameron Williams pointed out, it suffices to consider the non-squared version. But note that$$ \prod_{k=0}^{n} (2^{k} + 1) = \prod_{k=0}^{n} 2^{k}(1 + 2^{-k}) = \left( \prod_{k=0}^{n} 2^{k} \right)\left( \prod_{k=0}^{n} (1 + 2^{-k}) \right). $$Taking logarithm, we find that$$ \log \prod_{k=0}^{n} 2^{k} = \sum_{k=0}^{n} k \log 2 = ...

4

In fields (e.g., $\mathbb{Q}$,$\mathbb{R}$,$\mathbb{C}$) and integral domains (e.g. $\mathbb{Z},\mathbb{Z}[x]$), no such pair of "numbers" $n,m$ can exist. For fields, this follows from the invertibility of multiplication. If $nm = 0$ but $n \neq 0$, then $m = n^{-1}0 = 0$, and vice versa. For integral domains, this is an axiom - and it's precisely what ...

4

This is Somos' quadratic recurrence constant $($see also$)$, whose value is about $1.661688^{^-}$ and which is not yet known to possess a closed form. For a similar expression, see the nested radical constant. I think this should help out.

4

Define $a_n:=\prod_{k=1}^n\left(k-\frac 12\right)$. From the relationship $\Gamma(x+1)=x\Gamma(x)$ for $x$ positive, we derive $$k-\frac 12=\frac{\Gamma(k+1-1/2)}{\Gamma(k-1/2)},$$ hence the product which defines $a_n$ is telescopic. We obtain $$a_n=\frac{\Gamma(n-1/2)}{\Gamma(1/2)}.$$ Since $\Gamma(1/2)=\sqrt\pi$, we get the same formula as Wolfram Alpha. ...

4

Use this formula (found here, and mentioned recently on MSE here): $$\prod _{k=1}^{n-1}\,\sin \left({\frac {k\pi }{n}} \right)=\frac{n}{2^{n-1}} .$$ Let $n=13$, which gives $$\left(\sin{\frac{\pi}{13}} \cdot \sin{\frac{2\pi}{13}} \cdot \sin{\frac{3\pi}{13}} \cdots \sin{\frac{6\pi}{13}}\right)\left(\sin{\frac{7\pi}{13}} \cdot \sin{\frac{8\pi}{13}} \cdot ... 3 The factorial has two equivalent definitions. First, the one I expect you have learnt k!=k\cdot(k-1)\ldots2\cdot1 or, "k factorial is the product of all the integers from 1 up to k" This is a good definition in that it's natural - it seems sensible given the situations in which the factorial is helpful (in combinatorics, for instance). On the ... 2 Cauchy product is what you look for. \beta_{n-1} = \sum_{k=0}^{n-1} a_k a_{n-k-1} is the coefficient in the series developpement of$$ \left( \sqrt{1+x} \right)^2 = \sum_{n=0}^\infty \beta_n x^n $$around 0, that is 0 when n>2. 2 Let P the polynomial with degree n defined by$$P(x)=\left(\sum_{i=0}^{n}\prod_{j=0,j\neq i}^{n}\frac{x-x_j}{x_i-x_j}\right)-1$$then we see easily that$$P(x_i)=0,\;\quad\forall i=0,\ldots,n$$so P has n+1 distinct roots hence it's the zero polynomial due to the D'Alembert's theorem. Conclude. 2 So if I understand correctly, what you want is a proof of the Theorem: Let \left\{ f_i \colon X \to Y_i \mid i \in I\right\} a family of maps, where the Y_i are topological spaces, and X is a set. If \tau_1 and \tau_2 are topologies on X with the property that a map g \colon (Z,\tau_Z) \to (X,\tau_k) is continuous if and only if f_i \circ g ... 2 For positive integer n If \sin(2n+1)x=0, (2n+1)x=m\pi\iff x=\frac{m\pi}{2n+1}  where m is any integer From (3) of this, \displaystyle \sin(2n+1)x=2^{2n}s^{2n+1}+\cdots+(2n+1)s=0 where s=\sin\frac{m\pi}{2n+1} So the roots of \displaystyle 2^{2n}s^{2n+1}+\cdots+(2n+1)s=0  are \sin\frac{m\pi}{2n+1}; 0\le m\le2n So the roots of ... 2 No, that is not sufficient. For instance, let \mathbf{B} = \begin{bmatrix}1/2 & 1/2\\1/2 & 1/2 \end{bmatrix} and \hat n = \begin{bmatrix} 1/\sqrt{5} \\ 2/\sqrt{5}\end{bmatrix}. Then \hat n is a unit vector and \mathbf B^T\mathbf B = \mathbf B has the sum of absolute values of the elements in its columns equal to 1. Yet \mathbf B \cdot \hat n ... 2 You demand that (Bn,Bn) = 1 for any unit vector n. From this it necessarily follows that B^TB=I. To see this, first show that B preserves lengths for all vectors. Here is that step: Let v be any non-zero vector. Then (Bv,Bv) = ||v||^2(Bn,Bn) where n is v/||v||. Note that n is a unit vector. By assumption then, (Bn,Bn) = 1 and we have ... 1 For each of the 999 possible first factors, construct an object that will produce the multiples of that factor on demand starting with the highest one. Each object has a primitive to ask it what the next number to produce is (without updating it), and one to move to the next number. Maintain a priority queue of these generator objects, arranged by which one ... 1 As a general advice, you are looking for products of random variables (or, more generally, products of functions of random variables). To determine the distribution (pdf, cdf) of a product of random variables, different techniques may apply. You could look here: What is the distribution of a random variable that is the product of the two normal random ... 1 Let P be the product; then$$\log{P} = \sum_{i=1}^{k-1} \log{\left (1-\frac{i}{2 N} \right )} $$When N is sufficiently large, then we may use the approximation \log{(1-y)} \approx -y and get$$\log{P} \approx -\frac1{2 N} \sum_{i=1}^{k-1} i = -\frac1{2 N} \frac{k (k-1)}{2} = -\frac1{2 N} \binom{k}{2}$$Thus,$$P \approx e^{-\frac1{2 N} ...

1

This is definitely the product rule. Note that $$\frac{1}{r} \frac{\partial}{\partial r} \left(r \; k \frac{\partial T}{\partial r} \right)={1\over r}\left(\frac{\partial}{\partial r}(r)\cdot\left(k \frac{\partial T}{\partial r}\right)+\frac{\partial}{\partial r}\left(k \frac{\partial T}{\partial r}\right)(r)\right)=\frac{\partial}{\partial r} \left(k ... 1 For every k,$$ \binom{k^2+2k}{k^2}=\frac{(k^2+2k)!}{(k^2)!\,(2k)!}=\frac{((k+1)^2)!}{(k^2)!}\,\frac1{(k+1)^2}\,\frac1{(2k)!}, $$hence$$ \prod_{k=1}^n \binom{k^2+2k}{k^2}=((n+1)^2)!\,\left(\prod_{k=1}^n\frac1{k+1}\right)^2\,\left(\prod_{k=1}^n\frac1{(2k)!}\right)=\frac{((n+1)^2)!}{((n+1)!)^2}\prod_{k=1}^n\frac1{(2k)!}. $$This reduces the problem to ... 1 Here is all that is known about this sequence on OEIS. (Or rather, the sequence before squaring it.) I see no formulas that are not recursive or do not involve \sum or \prod. 1 The comma is just a list separator. A listing of events in the probability function is interpreted as the intersection of those events. So P(A,B) is the probability of A and B both occurring. Also expressed as P(A \cap B) The pipes refers to conditional probability. So P(A\mid B) is the probability of A occurring when it is given that B ... 1 The factors 1 - \mathop{\text{cis}} \frac{2k\pi}{n}, k = 1,\ldots, n - 1, are the distinct roots of the polynomial$$(1 - x)^{n - 1} + (1 - x)^{n - 2} + \cdots + (1 - x) + 1. The product of the roots of any polynomial is its constant term, which is $n$ in this case. In case the first statement is not clear: the roots of $y^n - 1$ are ...

1

Suggestion: Start with Karolis Juodelė's suggestion: all $n_i=7$. Keep all $n_i=7$ for $i>2$. Change $n_1$ a little bit and (similar to the implicit function theorem) solve the original equation for $n_2$ as a function of $n_1$. Since you are moving $n_1$ just a bit, $n_2$ will move just a bit as well, so all inequality constraints will hold. After ...

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