All Questions
298
questions
29
votes
1
answer
60k
views
Can we express sum of products as product of sums?
I've got an expression which is sum of products like:
$$a_1 a_2 + b_1 b_2 + c_1 c_2 + \cdots,$$
but the real problem I'm solving could be easily solved if I could convert this expression into ...
- 453
28
votes
2
answers
15k
views
How to interchange a sum and a product?
I have this expression:
$$\sum_{\{\vec{S}\}}\prod_{i=1}^{N}e^{\beta HS_{i}}=\prod_{i=1}^{N}\sum_{S_{i}\in\{-1,1\}}e^{\beta HS_{i}} \qquad (1)$$
Where $\sum_{\{\vec{S}\}}$ means a sum over all possible ...
- 943
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:
$$\sum\limits_{i = 1}^{n } {\prod\limits_{\substack{j = 1\\j \ne i}}^{n } {\frac{{x_i }}{{x_i - x_j }}} } ...
- 173
12
votes
1
answer
588
views
New Year Maths 2015
In the spirit of the festive period and in appreciation of the encouraging response to my Xmas Combinatorics 2014 problem posted recently, here's one for the New Year!
Express the following as a ...
- 22.8k
12
votes
1
answer
882
views
Showing $\sum\limits^N_{n=1}\left(\prod\limits_{i=1}^n b_i \right)^\frac1{n}\le\sum\limits^N_{n=1}\left(\prod\limits_{i=1}^n a_i \right)^\frac1{n}$?
If $a_1\ge a_2 \ge a_3 \ldots $ and if $b_1,b_2,b_3\ldots$ is any rearrangement of the sequence $a_1,a_2,a_3\ldots$ then for each $N=1,2,3\ldots$ one has
$$\sum^N_{n=1}\left(\prod_{i=1}^n b_i \right)^...
- 2,038
11
votes
2
answers
387
views
Prove $\sum_{k=0}^n \binom{n}{k}(-1)^k \frac{x}{x+k} = \prod_{k=1}^n \frac{k}{x+k}$ and more
The current issue (vol. 120, no. 6)
of the American Mathematical Monthly
has a proof by probabilistic means
that
$$\sum_{k=0}^n \binom{n}{k}(-1)^k \frac{x}{x+k} = \prod_{k=1}^n \frac{k}{x+k}
$$
for ...
- 108k
11
votes
0
answers
241
views
Are $(2,28)$ and $(5,3207)$ the only solutions $(m,n)\in\mathbb{N}^2$?
I noticed something as I was playing around with prime numbers. By denoting $p_i$ the $i^{\text{th}}$ prime number, I discovered the following:
$$
\begin{align}\prod_{i=1}^2\left(p_i^{ \ 2}+i\right)&...
- 9,487
10
votes
1
answer
713
views
Product of Sines and Sums of Squares of Tangents
There is a nice formula for products of cosines, found by multiplying by the complementary products of sines and using the double angle sine formula (as I asked in my question here): $$\prod_{k=1}^n \...
- 8,885
10
votes
4
answers
369
views
Geometry problem boils down to finding a closed form for $\sum_{n=1}^{k}{\arctan{\left(\frac{1}{n}\right)}}$
I was solving the following problem:
"Find $\angle A + \angle B + \angle C$ in the figure below, assuming the three shapes are squares."
And I found a beautiful one-liner using complex numbers:
$(1+...
- 596
10
votes
1
answer
507
views
Why is this sum equal to $0$?
While solving a differential equation problem involving power series, I stumbled upon a sum (below) that seemed to be always equal to $0$, for any positive integer $s$.
$$
\sum_{k=0}^s \left( \frac{ \...
- 365
10
votes
1
answer
472
views
If integration is a continuous analog of summation (Addition), what is the continuous analog of multiplication (Product)?
One definition of integration over a continuous interval [a,b] into n subintervals with equal width $\Delta x$, and from each interval choose a point $x_i^*$. Then the definite integral of $f(x)$ ...
- 101
9
votes
3
answers
487
views
Question about Euler's approach to find $\sum_{n=1}^\infty\frac1{n^2}=\frac{\pi^2}6$
For a freshman calculus project, I used Euler's approach to find $\sum_{n=1}^\infty\frac1{n^2}=\frac{\pi^2}6$, and noted from Wikipedia's explanation that the infinite product representation of $\frac{...
- 265
8
votes
4
answers
21k
views
Change from product to sum
We know that : $$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}^...
- 2,211
8
votes
2
answers
373
views
Showing an indentity with a cyclic sum
Let $n\geqslant2$, and $k\in \mathbb{N}$
Let $z_1,z_2,..,z_n$ be distinct complex numbers
Prove that
$$ \sum_{i=1}^{n}\frac {{z}_{i}^{n-1+k}} { \prod \limits_{\substack{j = 1\\j \ne i}}^{ n }{ (z_i-...
- 35.9k