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52

Using $$1+\tan x = \frac{\sin x + \cos x}{\cos x} = \frac{\sqrt{2} \cos (45^{\circ} - x)}{\cos x},$$ the product can be written as: \prod_{x=1}^{45}(1+\tan x^\circ) = 2^{45/2} \prod_{x=1}^{45} \frac{\cos (45 - x)^{\circ}}{\cos x^{\circ}} \stackrel{(1)}{=} 2^{45/2} \cdot \frac{\prod\limits_{x=0}^{44} \cos x^{\circ}}{\prod\limits_{x=1}^{45} \cos ... 35 We have \begin{align*} p_{k} &= \prod_{n = 2}^{k} \left( 1 - \frac{1}{n^{2}} \right) = \prod_{n=2}^{k} \frac{(n-1)(n+1)}{n^{2}} \\ & = \frac{1 \cdot 3}{2 \cdot 2} \cdot \frac{2\cdot 4}{3 \cdot 3} \cdot \frac{3\cdot 5}{4 \cdot 4} \cdot \cdots \cdot \frac{(k-2)\cdot k}{(k-1)\cdot(k-1)} \cdot \frac{(k-1)(k+1)}{k\cdot k}\\ & = \frac{1}{2}\left(1 + ... 26\prod_{k=1}^{n-1}\sin\left(\frac{k\pi}{n}\right)=\frac{n}{2^{n-1}}$$Here, let$$x=\prod_{k=1}^{999}\sin\left(\frac{k\pi}{1999}\right) \tag{1}$$Since \sin t=\sin (\pi-t), therefore,$$x=\prod_{k=1}^{999}\sin\left(\frac{(1999-k)\pi}{1999}\right)=\prod_{k=1000}^{1998}\sin\left(\frac{k\pi}{1999}\right) \tag{2}$$Multiplying equation (1) by equation ... 23 The product can never be 1. Recall that the product is defined as the limit of the partial products$$ \prod_{n\le k}\left(1-\frac1{f(n)}\right)=A_k. $$Now, if f(n)=1 then the product is 0. I assume you know how to handle "divergence to 0"? To simplify, let's assume f(n)\ne 1 for all n. Then we have that A_k>A_{k+1} for all k, since ... 22 Note that$$1-\frac1{k^2}=\left(1-\frac1k\right)\left(1+\frac1k\right)=\frac{k-1}{k}\frac{k+1}{k}=\frac{a_k}{a_{k-1}}$$with a_k= \frac{k+1}k, hence this is a telescoping product, i.e.$$ \prod_{k=2}^n\left(1-\frac1{k^2}\right)=\frac{a_2}{a_1}\frac{a_3}{a_2}\cdots\frac{a_n}{a_{n-1}}=\frac{a_n}{a_1}=\frac{n+1}{2n}.$$19$$P=\prod_{k=1}^{n-1}\sin(k\pi/n)=(2i)^{1-n}\prod_{k=1}^{n-1}(e^{ik\pi/n}-e^{-ik\pi/n})=(2i)^{1-n}e^{-i\pi n(n-1)/(2n)}\prod_{k=1}^{n-1}(e^{2ik\pi/n}-1)=(-2)^{1-n}\prod_{k=1}^{n-1}(\xi^k-1)=2^{1-n}\prod_{k=1}^{n-1}(1-\xi^k),$$where \xi=e^{2i\pi/n}. Now note, that x^n-1=(x-1)\sum_{k=0}^{n-1}x^k and x^n-1=\prod_{k=0}^{n-1} (x-\xi^k), thus ... 19 First of all, it is almost impossible to have zero hairs. Even bald people presumably have some amount of hair on their chest, their arms, or their legs. I shudder to think what kind of chemicals you'd need to douse on yourself to get rid of all of your hair. A much more reasonable answer is something like (3 \times 10^5)^{7 \times 10^9}: I happen to know ... 18 Your friend is right: the partial products are clearly \le 3/4, and since the infinite product converges (by the test you showed), its value must also be \le 3/4. To find the actual value, note that 1 - 1/n^2 = (n-1)(n+1)/n^2, and so the infinite product is$$ \prod_{n=2}^{\infty}\left(1 - \frac{1}{n^2}\right) = ...

18

If you take any sequence $a_1,a_2,a_3,\ldots$ whose sum is $\log_b (1/2)$, then $b^{a_1}, b^{a_2}, b^{a_3},\ldots$ is a sequence whose product is $1/2$. Later note: Notice that $\frac 1 2 + \frac 1 4 + \frac 1 8 + \frac 1 {16} + \cdots = 1$. If you multiply every term by $\log_b \frac 1 2$ then you get a series whose sum is $\log_b \frac 1 2$. Still later ...

15

Use the formula $\sin(x) = \frac{1}{2i}(e^{ix}-e^{-ix})$ to get \begin{align*} \prod_{k=1}^{n-1} \sin(k\pi/n) &= \left(\frac{1}{2i}\right)^{n-1}\prod_{k=1}^{n-1} \left(e^{k\pi i/n} - e^{-k\pi i/n}\right) \\ &= \left(\frac{1}{2i}\right)^{n-1}\left(\prod_{k=1}^{n-1} e^{k\pi i/n} \right) \prod_{k=1}^{n-1} \left(1-e^{-2k\pi i/n} \right). \end{align*} The ...

14

Since $$1-\frac1{k^3}=\frac{(k-1)(k+\frac12+\frac{\sqrt3}2i)(k+\frac12-\frac{\sqrt3}2i)}{k^3}$$ and $$k+a=\frac{\Gamma(k+a+1)}{\Gamma(k+a)},$$ every term in the product is a ratio of the Gamma functions. Also there is a formula $$\Gamma \left(\frac{1}{2}-i y\right) \Gamma \left(\frac{1}{2}+i y\right)= \pi \text{sech}\pi y.$$ In particular for the end ...

14

You don't need to introduce a new concept for this. The geometric mean of $y_i$ is nothing but $\exp$ of the arithmetic mean of $\log y_i$, and this generalizes in the straightforward way to integration: $$\exp\left(\frac{\int_{x_0}^{x_1} \log f(x)dx}{\int_{x_0}^{x_1} dx}\right).$$ You can do the same thing with the generalized mean, replacing $\log$ and ...

14

The roots of the polynomial $X^{2n}-1$ are $\omega_j:=\exp\left(\mathrm i\frac{2j\pi}{2n}\right)$, $0\leq j\leq 2n-1$. We can write \begin{align} X^{2n}-1&=(X^2-1)\prod_{j=1}^{n-1}\left(X-\exp\left(\mathrm i\frac{2j\pi}{2n}\right)\right)\left(X-\exp\left(-\mathrm i\frac{2j\pi}{2n}\right)\right)\\ ...

13

Intuitively, we have $$\log\left( 1 + \frac{k}{n^2} \right) = \frac{k}{n^2} + O\left(\frac{1}{n^2}\right) \quad \Longrightarrow \quad \log \prod_{k=1}^{n} \left( 1 + \frac{k}{n^2} \right) = \frac{1}{2} + O\left(\frac{1}{n}\right)$$ and therefore the log-limit is $\frac{1}{2}$. Here is a more elementary approach: Let $P_n$ denote the sequence inside the ...

13

HINT: It never hurts to gather some data by doing some actual computation: $$\begin{array}{c|l} n&\prod_{k=2}^n\frac{k^2+k+1}{k^2-k+1}\\ \hline 2&\frac73\\ 3&\frac{\color{red}7}3\cdot\frac{13}{\color{red}7}\\ 4&\frac{\color{red}7}3\cdot\frac{\color{blue}{13}}{\color{red}7}\cdot\frac{21}{\color{blue}{13}}\\ ... 13 This is going to be a little out of the blue, but here goes. Consider the function$$f(x) = \frac{\arcsin{x}}{\sqrt{1-x^2}}$$f(x) has a Maclurin expansion as follows:$$f(x) = \sum_{n=0}^{\infty} \frac{2^{2 n}}{\displaystyle (2 n+1) \binom{2 n}{n}} x^{2 n+1}$$Differentiating, we get$$f'(x) = \frac{x \, \arcsin{x}}{(1-x^2)^{3/2}} + \frac{1}{1-x^2} ...

12

The identity in the question is wrong. The correct one is $$\prod_{i=1}^n\prod_{j=1}^na_ia_j=\left(\prod_{i=1}^na_i\right)^{2n}.$$ As a quick `sanity check' you can try counting the number of a's in the product on each side. The expression in the question had $2n^2$ on the left but only $2n$ on the right, so couldn't possibly be correct. One possible ...

11

Your first term is $r^2$, where $$r=\frac{2\cdot 4\cdots (2n)}{1\cdot 3\cdot 5\cdots (2n-1)}=\frac{(2\cdot 4\cdots (2n))^2}{(2n)!}=\frac{4^nn!n!}{(2n)!}=\frac{4^n}{{2n\choose n}}$$ Now, ${2n \choose n}$ is the central binomial coefficient, which is known to be approximately $\large \frac{4^n}{\sqrt{\pi n}}$ for large $n$. Hence $r^2\approx \pi n$. You are ...

10

Try to find a sequence such that $\frac{n+1}{2n}=\prod_{j=2}^nx_j$ (it will do the job). We have $x_2=3/4$ and $$x_{n+1}=\frac{\prod_{j=2}^{n+1}x_j}{\prod_{j=2}^nx_j}=\frac{n+2}{2(n+1)}\frac{2n}{n+1}=\frac{n(n+2)}{(n+1)^2}=\frac{n^2+2n}{(n+1)^2}<\frac{n^2+2n\color{red}{+1}}{(n+1)^2}=1.$$ So $x_n=\frac{n^2-1}{n^2}$ does the job.

10

The idea of using $\cos x=\frac{1}{2}\frac{\sin 2x}{\sin x}$ is a nice one, and works quickly. We should not use this when $x=\frac{n\pi}{n}$,because of the $\frac{0}{0}$ issue. Also, as you observe, the product is $0$ if $n$ is even, so we needn't bother. So let $n$ be odd. Look at the product from $k=1$ to $k=n-1$. As $k$ ranges over these values, the ...

10

As Dylan has already mentioned, the category-theoretic product in $\textbf{Rel}$ is the disjoint union of sets. We can verify this by hand: \begin{align} \textbf{Rel}(X, Y \amalg Z) & = \mathscr{P}(X \times (Y \amalg Z)) \\ & \cong \mathscr{P}((X \times Y) \amalg (X \times Z)) \\ & \cong \mathscr{P}(X \times Y) \times \mathscr{P}(X \times Z) = ...

9

(The following is a joke.) Put $a_n:={1\over 6}n(n+1)(n+2)$ and $b_n:={1\over2}n(n+1)$, and define the magic constant $$\xi:=\sum_{n=1}^\infty {n!\over 2^{a_n}}\ =\ 0.630882266676063396815526621896\ldots\quad .$$ Then $$n!=\bigl\lfloor\> 2^{a_n}\xi\>\bigr\rfloor\quad \bigl({\rm mod}\ \ 2^{b_n}\bigr)\qquad(n\geq1)\ .$$ Try it out!

9

Looking at the bigger picture a little, the idea behind most answers given works not just for product but also for sum, difference, quotient, exponent, and most other binary operations. The common idea is that for many ways that f can vary, g can also vary in some way that cancels it out, making the product (or sum, etc.) constant. For instance, as long as ...

9

This is an example of a Multiple Zeta Value, namely $\zeta (2,2,2,\cdots,2)$. On the page http://www.usna.edu/Users/math/meh/mult.html there are several relations satisfied by such MZVs. For example, $\zeta (2,2,2,2, \cdots) = (2n + 1) \zeta (3,1,3,1,\cdots)$ where there are $2n$ copies of $2$ on the left, and $n$ blocks of $(3,1)$ on the right. So for ...

9

We have the following infinite product representation for $\sinh\,z$: $$\frac{\sinh\,z}{z}=\prod_{k=1}^\infty\left(1+\frac{z^2}{k^2\pi^2}\right)$$ Comparing this with $$e^{ax}-e^{bx} = x(a-b)\exp\left[\frac{a+b}{2}x\right]\prod_{i=1}^{\infty}\left[1+\frac{(a-b)^2x^2}{2k^2\pi^{2}}\right]$$ we find that ...

8

You can start a summation or a product anywhere, not just at $1$. There is no law mandating that they must start at $1$. In this case, if you were to allow $i=1$, then the first factor of the product would be $$\left(1 - \frac{1}{1^2}\right) = 0$$ and so the entire product would be zero, which would make everything rather silly. So you start at $2$ instead ...

8

There is a term for this (actually, more than one). It's called the product integral or multiplicative integral, and together with the corresponding derivative you get what's called product calculus or multiplicative calculus or non-Newtonian calculus. In addition, your idea of obtaining different versions of calculus by considering the generalized mean is ...

8

Since nobody seems to be willing to answer, let me try to summarize the hints from the comments. First, we can write factorial as $$n!=\int_0^{\infty}x^n e^{-x}dx.$$ One of the consequences of this formula is Stirling's approximation: as $n\rightarrow\infty$, $$n!\sim \left(\frac{n}{e}\right)^n\sqrt{2\pi n}.\tag{1}$$ At first sight, if there was an ...

8

Simplest way to establish this is by induction on $n$. The case $n=1$ is immediate; the case $n=2$ is the usual product rule. Assuming you have established the desired formula $$\left(\prod_{i=1}^n f_i(x)\right)' = \sum_{i=1}^n \left(f_i'(x)\prod_{\stackrel{1\leq j\leq n}{i\neq j}}f_j(x)\right)$$ for $n$, then to get the $n+1$ case we have: ...

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