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## Hot answers tagged products

49

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 + ... 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. $$Now, if f(n)=1 then the product is 0. I assume you know how to handle "divergence to 0"? To simplify, let's assume f(n)\ne 1 for all n. Then we have that A_k>A_{k+1} for all k, since ... 23 To elaborate, and extend on GEdgar's answer: there is what is called the q-Pochhammer symbol$$(a;q)_n=\prod_{k=0}^{n-1} (1-aq^k)$$and (a;q)_\infty is interpreted straightforwardly. The product you are interested in is equivalent to \left(\frac1{10};\frac1{10}\right)_\infty\approx0.8900100999989990000001. One can also express the q-Pochhammer ... 21 Note that$$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}.$$19 First of all, it is almost impossible to have zero hairs. Even bald people presumably have some amount of hair on their chest, their arms, or their legs. I shudder to think what kind of chemicals you'd need to douse on yourself to get rid of all of your hair. A much more reasonable answer is something like (3 \times 10^5)^{7 \times 10^9}: I happen to know ... 18 If you take any sequence a_1,a_2,a_3,\ldots whose sum is \log_b (1/2), then b^{a_1}, b^{a_2}, b^{a_3},\ldots is a sequence whose product is 1/2. Later note: Notice that \frac 1 2 + \frac 1 4 + \frac 1 8 + \frac 1 {16} + \cdots = 1. If you multiply every term by \log_b \frac 1 2 then you get a series whose sum is \log_b \frac 1 2. Still later ... 18 Your friend is right: the partial products are clearly \le 3/4, and since the infinite product converges (by the test you showed), its value must also be \le 3/4. To find the actual value, note that 1 - 1/n^2 = (n-1)(n+1)/n^2, and so the infinite product is$$ \prod_{n=2}^{\infty}\left(1 - \frac{1}{n^2}\right) = ...

17

$$\prod_{k=1}^{n-1}\sin\left(\frac{k\pi}{n}\right)=\frac{n}{2^{n-1}}$$ 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 ...

16

I will answer your question "Most importantly I'd like to know why $$\prod (1+|a_n|) \to a < \infty \quad \Longrightarrow \quad \prod (1+ a_n) \to b \neq 0. "$$ We will first prove that if $\sum \lvert a_n \rvert < \infty$, then the product $\prod_{n=1}^{\infty} (1+a_n)$ converges. Note that the condition you have $\prod (1+|a_n|) \to a < ... 15 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 Use the formula$\sin(x) = \frac{1}{2i}(e^{ix}-e^{-ix})$to get $$\prod_{k=1}^{n-1} \sin(k\pi/n) = \left(\frac{1}{2i}\right)^{n-1}\prod_{k=1}^{n-1} \left(e^{k\pi i/n} - e^{-k\pi i/n}\right) = \left(\frac{1}{2i}\right)^{n-1}\left(\prod_{k=1}^{n-1} e^{k\pi i/n} \right) \prod_{k=1}^{n-1} \left(1-e^{-2k\pi i/n} \right).$$ The first product simplifies to ... 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_ndenote 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 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)\\ ... 12 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).$$You can do the same thing with the generalized mean, replacing \log and ... 11 The identity in the question is wrong. The correct one is$$ \prod_{i=1}^n\prod_{j=1}^na_ia_j=\left(\prod_{i=1}^na_i\right)^{2n}. $$As a quick `sanity check' you can try counting the number of a's in the product on each side. The expression in the question had 2n^2 on the left but only 2n on the right, so couldn't possibly be correct. One possible ... 10 The idea of using \cos x=\frac{1}{2}\frac{\sin 2x}{\sin x} is a nice one, and works quickly. We should not use this when x=\frac{n\pi}{n},because of the \frac{0}{0} issue. Also, as you observe, the product is 0 if n is even, so we needn't bother. So let n be odd. Look at the product from k=1 to k=n-1. As k ranges over these values, the ... 10 Try to find a sequence such that \frac{n+1}{2n}=\prod_{j=2}^nx_j (it will do the job). We have x_2=3/4 and$$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. 9 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 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]$$we find that ... 9 This is an example of a Multiple Zeta Value, namely \zeta (2,2,2,\cdots,2). On the page http://www.usna.edu/Users/math/meh/mult.html there are several relations satisfied by such MZVs. For example, \zeta (2,2,2,2, \cdots) = (2n + 1) \zeta (3,1,3,1,\cdots) where there are 2n copies of 2 on the left, and n blocks of (3,1) on the right. So for ... 8 You can start a summation or a product anywhere, not just at 1. There is no law mandating that they must start at 1. In this case, if you were to allow i=1, then the first factor of the product would be$$\left(1 - \frac{1}{1^2}\right) = 0$$and so the entire product would be zero, which would make everything rather silly. So you start at 2 instead ... 8 Simplest way to establish this is by induction on n. The case n=1 is immediate; the case n=2 is the usual product rule. Assuming you have established the desired formula$$\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)$$for n, then to get the n+1 case we have: ... 8 There is a term for this (actually, more than one). It's called the product integral or multiplicative integral, and together with the corresponding derivative you get what's called product calculus or multiplicative calculus or non-Newtonian calculus. In addition, your idea of obtaining different versions of calculus by considering the generalized mean is ... 7 Yes \boxed{\displaystyle e^{-\zeta'(2)}} I think. To prove it start with :$$\zeta(2-x)=\sum_{k=1}^\infty \frac {k^x}{k^2}$$and compute the derivative! The trick is that the derivation will create a \ln(k) term at the numerator. At the end take the limit as x\to 0. 7 The monic Chebyshev polynomial of the second kind,$$\hat{U}_n(x)=\frac{\sin((n+1)\arccos\,x)}{2^n \sqrt{1-x^2}}$$can be easily seen to have the roots x_k=\cos\dfrac{\pi k}{n+1} for k=1,\dots,n. By Vieta, the constant term of \hat{U}_n(x) is equal to$$\hat{U}_n(0)=\prod_{k=1}^n \cos\dfrac{\pi k}{n+1}$$and thus$$\prod_{k=1}^n \cos\dfrac{\pi ... 7 By looking at the decimal representation, it appears that: $$\prod_{i=1}^\infty\left(1-\frac1{10^i}\right)= \sum_{i=1}^\infty \frac{8 + \frac{10^{2^i-1}-1}{10^{2i-1}} + \frac1{10^{6i-2}} + \frac{10^{4i}-1}{10^{12i-2}} }{10^{(2i-1)(3i-2)}}$$ I don't have a proof, but the pattern is so regular that I'm confident. 7 Your argument about the product starts well, but then it goes off the rails. Why do you say that "it doesn't account for functions that are not surjective"? Or "functions with vertical asymptotes"? It certainly doesn't include functions that are not defined on all of\mathbb{R}$, but then neither does$\mathbb{R}^{\mathbb{R}}$. As for surjectivity, nothing ... 7 We begin with a simple inequality. For every$x$in$[0,\frac12]$,$\mathrm e^x\leqslant2$hence$\frac12\leqslant\mathrm e^{-x}$. Integrating this on the interval$[0,x]$yields$\frac12x\leqslant1-\mathrm e^{-x}$, that is,$1-\frac12x\geqslant\mathrm e^{-x}$. Applying this to$x=\frac1{2^{k-1}}$for each$k$between$2$and$n$, one gets$\$ ...

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