Articles of products

Continuing direct product on a subcategory

Let $F$ is a full subcategory of a category $G$, both categories having binary direct product. Is it always true that there is such a binary direct product in $G$ that it is a continuation of a binary direct product in $F$? Hm, can it be generalized for infinitary direct products, rather than only binary? […]

Is the product of two discrete $\sigma$-algebras necessarily discrete?

I know that the answer to this question is negative, since proving the opposite is an exercise in Terrance Tao’s Measure Theory book. However, it doesn’t make sense to me. In another part of the same exercise, we show that the product $\sigma$-algebra of two atomic algebras, is again an atomic algebra. As I understand, […]

Proving $\prod((k^2-1)/k^2)=(n+1)/(2n)$ by induction

$$P_n = \prod^n_{k=2} \left(\frac{k^2 – 1}{k^2}\right)$$ Someone already helped me see that $$P_n = \frac{1}{2}.\frac{n + 1}{n} $$ Now I have to prove, by induction, that the formula for $P_n$ is correct. The basis step: $n = 2$ is true, $$P_2 = \frac{3}{4} = \frac{1}{2}.\frac{2+1}{2} $$ Inductive step: Assuming $P_k = \frac12.\frac{k+1}k$ is true for […]

Is there any good approximation for $\prod_{i=3}^k (n-i)$? $(n \gg k)$

Is there any good approximation for $\prod_{i=3}^k (n-i)$? $(n \gg k)$ I know that $\prod_{i=3}^k (n-i) < \prod_{i=3}^k n = n^{k-2}$ Also a tighter upper bound is appreciated.

How to define this pattern as $f(n)$

Given a binary table with n bits as follows: $$\begin{array}{cccc|l} 2^{n-1}…&2^2&2^1&2^0&row\\ \hline \\ &0&0&0&1 \\ &0&0&1&2 \\ &0&1&0&3 \\ &0&1&1&4 \\ &1&0&0&5 \\ &1&0&1&6 \\ &1&1&0&7 \\ &1&1&1&8 \end{array} $$ If I replace each $0$ with a $1$ and each $1$ with $g(n)$ as follows $$\begin{array}{cccc|l} 2^{n-1}…&2^2&2^1&2^0&row\\ \hline \\ &1&1&1&1 \\ &1&1&g(0)&2 \\ &1&g(1)&1&3 \\ […]

How to efficiently compute a*b mod N

I’m trying to solve some problems on interviewstreet. For some problems they mention As the answers can be very big, output them modulo 1000000007. How can I compute a*b mod N where N is a large number like 1000000007. I thought of using (a mod N) * (b mod N) = (a*b mod N) but […]

A product identity involving the gamma function

I have reduced this problem (thanks @Mhenni) to the following (which needs to be proved): $$\prod_{k=1}^n\frac{\Gamma(3k)\Gamma\left(\frac{k}{2}\right)}{2^k\Gamma\left(\frac{3k}{2}\right)\Gamma(2k)}=\prod_{k=1}^n\frac{2^k(1+k)\Gamma(k)\Gamma\left(\frac{3(1+k)}{2}\right)}{(1+3k)\Gamma(2k)\Gamma\left(\frac{3+k}{2}\right)}.$$ As you see it’s quite a mess. Hopefully one can apply some gamma-identities and cancel some stuff out. I have evaluated both products for large numbers and I know that the identity is true, I just need to learn […]

Identity involving a recursive product

Here is yet another problem related to plane partitions. Given the recursive formula $$ \begin{align*} F(0)&=1,\\ F(r)&=\prod_{i=1}^r\frac{i+2r-1}{2i+r-2}F(r-1). \end{align*} $$ How can we prove $$F(n)=\prod_{1\leq i\leq j\leq k\leq n}\frac{i+j+k-1}{i+j+k-2}\ ?$$ EDIT: The solution to this problem can be found in the answer section to this question.

Inequality and Induction: $\prod_{i=1}^n\frac{2i-1}{2i}$ $<$ $\frac{1}{\sqrt{2n+1}}$

This question already has an answer here: Induction and convergence of an inequality: $\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n)}\leq \frac{1}{\sqrt{2n+1}}$ 7 answers

Summation and Product Bounds

If I have a sum or product whose upper index is less than its start index, how is this interpreted? For example: $$\sum_{k=2}^0a_k,\qquad \prod_{k=3}^1b_k$$ I want to say that they are equal to the empty sum and empty product, respectively, but I don’t know. (This question arises from seeking shortened forms for denoting some nested […]