All Questions
29
questions
26
votes
1
answer
860
views
Is this algebraic identity obvious? $\sum_{i=1}^n \prod_{j\neq i} {\lambda_j\over \lambda_j-\lambda_i}=1$
If $\lambda_1,\dots,\lambda_n$ are distinct positive real numbers, then
$$\sum_{i=1}^n \prod_{j\neq i} {\lambda_j\over \lambda_j-\lambda_i}=1.$$
This identity follows from a probability calculation ...
user940
17
votes
2
answers
862
views
proof of $\sum\nolimits_{i = 1}^{n } {\prod\nolimits_{\substack{j = 1\\j \ne i}}^{n } {\frac{{x_i }}{{x_i - x_j }}} } = 1$ [duplicate]
i found a equation that holds for any natural number of n and any $x_i \ne x_j$ as follows:
$$\sum\limits_{i = 1}^{n } {\prod\limits_{\substack{j = 1\\j \ne i}}^{n } {\frac{{x_i }}{{x_i - x_j }}} } ...
- 173
10
votes
1
answer
717
views
Product of Sines and Sums of Squares of Tangents
There is a nice formula for products of cosines, found by multiplying by the complementary products of sines and using the double angle sine formula (as I asked in my question here): $$\prod_{k=1}^n \...
- 8,945
8
votes
4
answers
21k
views
Change from product to sum
We know that : $$a \times b = \underbrace{a + a + a + ... + a}_{\text{b times}}$$
That's how we convert from a product to a sum.
So what happens if we go a little further?
That is : $$\prod\limits_{a}^...
- 2,221
4
votes
3
answers
868
views
Finding $\frac{\sum_{r=1}^8 \tan^2(r\pi/17)}{\prod_{r=1}^8 \tan^2(r\pi/17)}$
I have tried to wrap my head around this for some time now, and quite frankly I am stuck.
Given is that :
$$a=\sum_{r=1}^8 \tan^2\left(\frac{r\pi}{17}\right) \qquad\qquad b=\prod_{r=1}^8 \tan^2\left(\...
- 183
4
votes
2
answers
82
views
How to define this pattern as $f(n)$
Given a binary table with n bits as follows:
$$\begin{array}{cccc|l}
2^{n-1}...&2^2&2^1&2^0&row\\ \hline \\ &0&0&0&1 \\ &0&0&1&2 \\ &0&1&0&...
- 265
3
votes
1
answer
64
views
What is condition that the sum of $n$ complex numbers eaquals their product
Let $n\geq2$ and let $\{z_1,\dots,z_n\}$ be a set of complex numbers.
Is there a condition on the $z_i$'s such that
$$\sum_{i=1}^n z_i=\prod_{i=1}^n z_i$$
is identically true?
For $n=2$ the ...
- 1,008
3
votes
1
answer
92
views
Proving that, if $f(k)=\prod_{i=1}^ka_i+\sum_{b=1}^{k-1}(1-a_{k-b})\prod_{i=1}^ba_{k-b+i}$, then $f(k+1)=f(k)\cdot a_{k+1}+(1 - a_k)a_{k+1}$
Given that $$f(k) = \prod_{i=1}^k a_i + \sum_{b=1}^{k-1} (1-a_{k-b}) \prod_{i=1}^b a_{k-b+i}$$ for all $k$ where $(a_1, a_2, a_3, \ldots )$ are random constants, prove that: $$f(k+1) = f(k) \cdot a_{k+...
user851578
2
votes
3
answers
68
views
Is there an easy way of seeing that this sum is $0$?
Let $a,b,c,d$ be real positive numbers. Is there an easy way of seeing that
$$(c-d)(b-d)(b-c)-(c-d)(a-d)(a-c)+(b-d)(a-d)(a-b)-(b-c)(a-c)(a-b)=0 $$
without doing the whole multiplication ?
- 2,294
2
votes
2
answers
147
views
Find the value of $a_0a_1a_2\cdots a_n\left(\frac{1}{a_0}+\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}\right)$
Given that the sequence $\left\{a_n\right\}$ satisfies $a_0 \ne 0,1$ and $$a_{n+1}=1-a_n(1-a_n)$$ $$a_1=1-a_0$$
Find the value of $$a_0a_1a_2\cdots a_n\left(\frac{1}{a_0}+\frac{1}{a_1}+\frac{1}{a_2}+...
- 13.2k
2
votes
2
answers
120
views
Product of $n$ terms of sequence where the $n^{th}$ term is of the form $(x^{a^n}+1)$
While practicing from a book I found a product in the form $$(x^{a^1}+1)\cdot(x^{a^2}+1)\cdot(x^{a^3}+1)\cdot(x^{a^4}+1)$$ and was immediately curious if I could a formula to solve the product for $n$ ...
- 121
2
votes
3
answers
92
views
Expansion of $\prod_{j=1}^{n} \left( \sum_{i = 1}^{m} x_{i, j} \right)$
I would like to know if there is a sum-of-products expansion for the following product-of-sums. A special case is given here for the difference of two entries.
$$\prod_{j=1}^{n} \left( \sum_{i= 1}^{m} ...
- 1,876
2
votes
1
answer
2k
views
Product of two sums, one finite and one infinite
I'm working on a problem and I'm not sure how to find the product of these two sums:
$\left(\sum_{k=0}^{\infty}\text{something}\right)\left(\sum_{k=n}^{n}\text{something else}\right)$
The "something"...
- 371
2
votes
1
answer
47
views
Double Sum to Product Derivation
The function after the double-sigma sign can be separated into the
product of two terms, the first of which does not depend on $s$ and
the second of which does not depend on $r$. Source
Is the ...
- 1,834
2
votes
1
answer
129
views
Let $A_{k}=\{0,... ,n\}\setminus\{k\}.$ How to prove $\sum_{k=0}^{n}\left[(-1)^{k+1}\prod_{\substack{i,j\in A_{k}\\i<j}}(a_{i}-a_{j})\right]=0$?
Let $A_{k}=\{0,1,\ldots,n\}\setminus\{k\}$ for each $k=0,1,\ldots ,n$.
I think that the following equality is true for all $n\in\mathbb{N}, n\geq 2$ :
\begin{align}
\sum_{k=0}^{n}\left[(-1)^{k+1}\...
- 23