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Multiplication and addition should really be thought of as different operations on real numbers: for example, how do you interpret $$\sqrt{2} \cdot \sqrt{2} = 2$$ by using addition? That $a \cdot b$ is expressible in terms of addition is really only a consequence of the distributive property, $$x(y+z) = xy+xz:$$ you can write $b= \underbrace{1 + \dotsb + ... 5 Your first formula works well for a pair of numbers$a,b$such that$b$is an integer. But when$b$is not, the product/sum equivalence is much less obvious. If$b$is a rational, say$\dfrac pq$, you can still do with $$x=a\times\frac pq\iff x\times q=a\times p\iff\underbrace{x + x + x + ... + x}_{\text{q times}}=\underbrace{a + a + a + ... + a}_{\text{p ... 5 In this answer, I will provide a method that uses iterated partial summation on the alternating Harmonic numbers that will allow you to explicitly evaluate either of these integrals for any n. Start by expanding the series$$\frac{1}{1-\prod_{k}x_{k}}=1+\prod_{k}x_{k}+\prod_{k}x_{k}^{2}+\cdots$$so that ... 5 Notice that you can write your polynomial as x^n((y/x)^n-1), so we can look for factorizations of t^n-1 and then substitute. Over the integers or the rational numbers, we have a complete answer. They are certain polynomials \Phi_k, called cyclotomic polynomials, one for eachtive integer k. There are irreducible and have the property that ... 4$$\prod_{k=2}^{K}\frac{k^2+k-2}{k^2+k}=\prod_{k=2}^{K}\frac{k-1}{k}\prod_{k=2}^{K}\frac{k+2}{k+1}=\frac{1}{K}\cdot\frac{K+2}{3}$$hence the limit equals \color{red}{\displaystyle\frac{1}{3}}. 3 We can write, from the difference of squares, \begin{gather*} \left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\dots\left(1-\frac{1}{n^2}\right)\\ =\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\dots\left(1-\frac{1}{n}\right)\cdot \left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\dots\left(1+\frac{1}{n}\right)\\ ... 3 In my opinion, these are easiest to understand using the Binet formula F_n = \frac{1}{s-t}(s^n - t^n), where s,t are the positive and negative roots of X^2 - X - 1 = 0. This reduces these equations to facts about polynomials in s and t. For example:$$(s^{m+n}-t^{m+n})^2 - (s^{2m} - t^{2m})(s^{2n} - t^{2n})  = s^{2m}t^{2n}+s^{2n}t^{2m} - ... 2 Brute force: let$S = UV. Then \begin{align*} \Pr[S \le s] &= \Pr[UV \le s] = \Pr[V \le s/U] = \int_{u=0}^1 \Pr[V \le s/u] f_U(u) \, du \\ &= \int_{u=0}^1 \int_{v = 0}^{s/u} \frac{b^a v^{a-1} e^{-bv}}{\Gamma(a)} \, dv \, du \\ &= \frac{b^a}{\Gamma(a)} \left( \int_{v=s}^{\infty} \int_{u=0}^{s/v} v^{a-1} e^{-bv} \, du \, dv + \int_{v=0}^s ... 2 Put P(x)=\prod_{\substack{k=1}}^n (a_k - x) . Then A_i=-P’(a_i). 2 We have that \Pr\{A\cup B\}=\Pr\{A\}+\Pr\{B\}-\Pr\{A\cap B\}=\Pr\{A\}+\Pr\{B\}-\Pr\{A\}\Pr\{B\} if A and B are independent events. For example, if n=2, \begin{align*} \Pr\{\psi=0\}&=\Pr(\{\xi_1=0\}\cup\{\xi_2=0\})\\ &=\Pr\{\xi_1=0\}+\Pr\{\xi_2=0\}-\Pr\{\xi_1=0\}\Pr\{\xi_2=0\}\\ &=1-p^{2}. \end{align*} In general, we need to use the ... 2 Let us show by induction you tagged that for n\ge 2,\prod_{k=2}^{n}\left(1-\frac{1}{k^2}\right)=\frac{n+1}{2n}.$$For n=2, it is true since 1-\frac{1}{2^2}=\frac{3}{4}=\frac{2+1}{2\cdot 2}. Supposing that \prod_{k=2}^{n}\left(1-\frac{1}{k^2}\right)=\frac{n+1}{2n} gives you ... 1 It follows from theorem 40.1(b) on p. 360 (the product rule) and from corollary 39.7 on p. 351 (the differential in terms of the partial differentials) of Robert G. Bartle's "The Elements of Real Analysis", 2nd edition (John Wiley & Sons, 1976), that if n \in \{1, 2, \dots\}, if f, g : \mathbb{R}^n \rightarrow \mathbb{R} and if c \in \mathbb{R}^n ... 1 As anakhronizein wrote product sin (k), k=0..10 The trick lies in the main page, where you find on the right side the bar "Example". If you click it, you find pretty much everything. For example. the answer to your question can be found by then choosing "Mathematics", and finally "Calculus & Analysis". I hope it helps for the future! 1 For example: product (1)/n, n=1..10 Computation here. 1 It can be instructive to look at the set-theoretic product here. Indeed, when the natural numbers are formalised as sets, it's true that the product of natural numbers is (the isomorphic to) the product of the corresponding sets; and multiplication in a field, commutative ring, algebra or whatever, can be considered to generalise that of natural numbers. So ... 1 Note that the multiplication in a ring is a binary operation and we want to define an n-ary operation where n is allowed to range over ... whatever we want and as far as we can consistantly go. The usual recursive definiion works for all finite n from a starting point of our choice. Maybe the most natural chioce for a start would be n=2, where the ... 1 While the product operation is defined by the axioms of a field or ring, nothing in those axiom defines what is meant by the operation \prod which acts on an ordered subset S of the field or ring. So the meaning of the product over zero elements (or for that matter over one element) is a matter of how definition of the operation \prod (S) deals with ... 1 Can you show \displaystyle \prod_{k=2}^n\frac{k^2+k-2}{k^2+k}=\frac{n+2}{3n} The result follows easily from there. 1 HINT: We have$$\begin{align} \prod_{k=2}^n\frac{k^2+k-2}{k^2+k}&=\prod_{k=2}^{n} \left(\frac{k+2}{k+1}\frac{k-1}{k}\right)\\\\ &=\prod_{k=2}^{n} \left(\frac{k+2}{k+1}\right)\prod_{k=2}^{n} \left(\frac{k-1}{k}\right) \end{align}$$1$$\begin{align}\lim_{n\to\infty}\prod_{k=2}^{n}\frac{k^2+k-2}{k^2+k}&=\lim_{n\to\infty}\prod_{k=2}^{n}\frac{(k+2)(k-1)}{k(k+1)}\\&=\lim_{n\to\infty}\frac{\prod_{k=2}^{n}(k+2)\prod_{k=2}^{n}(k-1)}{\prod_{k=2}^{n}k\prod_{k=2}^{n}(k+1)}\\&=\lim_{n\to\infty}\frac{\frac{(n+2)!}{3!}\times (n-1)!}{n!\times ... 1 This is going to be a very primitive attempt to answer this question. Yet I think it is worth posting since this kind of calculations appear many times in calculus. We take the log of the limit and then we Taylor expand the following function: \log\left(2 - e^{\frac{1}{k}}\right) = \log\left(1 - \sum\limits_{p=1}^\infty \frac{1}{p!} ... 1 A simple answer that covers most of the purely algebraic examples is that any class of structures that can be axiomatised by Horn clauses gives rise to a concrete category whose forgetful functor commutes with products. (In another question Concrete category with non-standard products, I use the term concrete category with standard products for this ... 1 You have the right joint density function provided you get the right boundaries: $$f_{S,T}(s,t) = \begin{cases} e^{-t} & \text{if }0<s<t, \\ 0 & \text{otherwise.} \end{cases}$$ (You can draw a picture and see that this is half of the first quadrant.) Now suppose we want the marginal distributions ofS$and$T$.$$f_S(s) = \int_s^\infty ... 1 When you multiply a matrix by a scalar, every entry in the matrix gets multiplied by that amount. E.g.:$2\cdot \left[\begin{smallmatrix} a&b\\c&d\end{smallmatrix}\right]=\left[\begin{smallmatrix} 2a & 2b\\ 2c&2d\end{smallmatrix}\right]\$ Wolfram seems to have interpreted your input as something similar, which may have precedence in coding, ...