All Questions
299
questions
8
votes
1
answer
467
views
Identity in Number Theory Paper
In this paper by Jerry Hu, he defines the function
$$f_{s,k,i}\left(u\right)=\prod_{p\mid u}
\left(1-\frac{\sum_{m=i}^{k-1}{s \choose m}\left(p-1\right)^{k-1-m}}{\sum_{m=0}^{k-1}{s \choose m}\left(p-...
8
votes
0
answers
258
views
Help me to get deeper understanding of Euler's proof of his Arithmetical Theorem
With distinct numbers $a_1, a_2, \ldots, a_n$, let's denote the products of the differences of each of these numbers with the each of the rest of them by the following principle:
\begin{align}
(...
7
votes
1
answer
365
views
How to prove the following discovery of Euler?
There exists a series of formulas.
\begin{align*}
\ & \dfrac{1}{(a-b)(a-c)}+\dfrac{1}{(b-a)(b-c)}+\dfrac{1}{(c-a)(c-b)} = 0, \\
\ & \dfrac{a}{(a-b)(a-c)}+\dfrac{b}{(b-a)(b-c)}+\dfrac{c}...
7
votes
3
answers
1k
views
Summation series ($\Sigma$) is to Integral ($\int$)... as Product series ($\Pi$) is to ??
If a Summation series ($\Sigma$) is to an Integral ($\int$)... is there a corresponding concept for a Product series ($\Pi$)?
Summation series ($\Sigma$) is to Integral ($\int$)... as Product series (...
7
votes
2
answers
200
views
Elementary proof of "generalized reverse Bernoulli inequality"
I've stumbled upon the following exercise in an early chapter of an analysis textbook:
Let $a_n$ be a finite, nonnegative sequence such that $\sum_{i=0}^n a_i\le 1$. Prove $$ \prod_{i=1}^n (1 + a_i) \...
6
votes
2
answers
359
views
Simplify $\prod_{k=1}^5\tan\frac{k\pi}{11}$ and $\sum_{k=1}^5\tan^2\frac{k\pi}{11}$
My question is:
If $\tan\frac{\pi}{11}\cdot \tan\frac{2\pi}{11}\cdot \tan\frac{3\pi}{11}\cdot \tan\frac{4\pi}{11}\cdot \tan\frac{5\pi}{11} = X$ and $\tan^2\frac{\pi}{11}+\tan^2\frac{2\pi}{11}+\tan^2\...
6
votes
2
answers
144
views
Combinatorics Problem on proving that a particular sum is 0
I'm having some issues with proving that the following sum is $0$ for any value of $n \geq 2$:
$$
\sum_{j=1}^{n} \frac{1}{\prod_{i=1,i\neq j}^{n}(a_{j}-a_{i})}
$$
where the $a_i$ are non-zero and ...
6
votes
2
answers
199
views
Calculate $\sum_{i = 0}^{n}\ln\binom{n}{i}\Big/n^2$
Calculate $$\sum_{i = 0}^{n}\ln\binom{n}{i}\Big/n^2$$
I can only bound it as follows:
$$\binom{n}{i}<\left(\dfrac{n\cdot e}{k}\right)^k$$
$$\sum_{i = 0}^{n}\ln\binom{n}{i}\Big/n^2<\dfrac{1}{n}\...
6
votes
0
answers
97
views
Prove that if $x_{1}x_{2}...x_{n}=1$ then $\frac{1}{1+x_{1}+x_{1}x_{2}}+...+\frac{1}{1+x_{n-1}+x_{n-1}x_{n}}+\frac{1}{1 +x_{n}+x_{n}x_{1}}\ge 1$ [duplicate]
Prove that if $x_{1}x_{2}...x_{n}=1$ then $\frac{1}{1+x_{1}+x_{1}x_{2}}+...+\frac{1}{1+x_{n-1}+x_{n-1}x_{n}}+\frac{1}{1 +x_{n}+x_{n}x_{1}}\ge 1$.
$x_{1},x_{2},...,x_{n}$ are positive real numbers, and ...
5
votes
4
answers
485
views
Big Greeks and commutation
Does a sum or product symbol, $\Sigma$ or $\Pi$, imply an ordering?
Clearly if $\mathbf{x}_i$ is a matrix then:
$$\prod_{i=0}^{n} \mathbf{x}_i$$
depends on the order of the multiplication. But, ...
5
votes
3
answers
3k
views
swap summation and multiple
In which case can we swap summation and multiple? ie.
$$\sum_{i=1}^n\prod_{j=1}^na_{ij}=\prod_{j=1}^n\sum_{i=1}^na_{ij}$$
if we can't swap like this, please tell me how can we swap them?
5
votes
2
answers
151
views
An equality between a product and a combinatorial sum
I'm trying to prove the following identity (of which I numerically verified the truth) :
$$\text{For every $n\in\mathbb{N}^*$ and $\alpha \in \mathbb{R}\setminus\lbrace-2k\text{ }|\text{ }k\in\mathbb{...
5
votes
3
answers
195
views
Prove that $\frac{\sqrt[n]{\prod_{k = 1}^nx_n}}{m} \ge n - 1$ where $\sum_{k = 1}^n\frac{1}{x_k + m} = \frac{1}{m}$.
Given positives $x_1, x_2, \cdots, x_{n - 1}, x_n$ such that $$\large \sum_{k = 1}^n\frac{1}{x_k + m} = \frac{1}{m}$$. Prove that $$\large \frac{\displaystyle \sqrt[n]{\prod_{k = 1}^nx_n}}{m} \ge n - ...
5
votes
1
answer
1k
views
Sum operator precedence
I'm trying to read some simple equations and in order to interpret them in the right way I need to know $\sum$ and $\prod $ operator range/precedence.
$$ \sum p(s, a) +\gamma $$
is equal to $\sum(p(...
5
votes
1
answer
225
views
Proving $(1+\frac 1n)^{n} = 1 + \sum_{k=1}^n({\frac 1{k!}\prod_{r=0}^{k-1}(1-\frac rn))}$ using the binomial theorem
$$\left(1+\frac 1n\right)^{n} = 1 + \sum\limits_{k=1}^n \left\{\frac 1{k!}\prod_{r=0}^{k-1}\left(1-\frac rn\right)\right\}$$
this exercise is taken from Apostol's Calculus I (page 45) and it's ...