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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 JuodelÄ—'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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