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3 votes
2 answers
533 views

Is knowing the Sum and Product of k different natural numbers enough to find them?

Can we uniquely identify the set of k different natural numbers (no two are the same) by knowing only their sum and product (and the value of k itself)?
PM 77-1's user avatar
  • 133
3 votes
1 answer
923 views

Sum over subsets of $\{1,2,\ldots,n\}$ of terms involving a product over that subset

I'm attempting to perform a sum, using products, using all possible combinations, in a function. How would I go about doing this? (I really need to find something that works.) For example, say I ...
Plasmarob's user avatar
  • 133
3 votes
2 answers
2k views

How to go from a sum to a product and a product to a sum?

I have read here (third post down) that exponentials turn sums into products and logarithms turn products into sums. Can someone please further explain this?
zerosofthezeta's user avatar
3 votes
2 answers
178 views

Prove that $\frac1{a(1+b)}+\frac1{b(1+c)}+\frac1{c(1+a)}\ge\frac3{1+abc}$

I tried doing it with CS-Engel to get $$ \frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)} \geq \frac{9}{a+b+c+ a b+b c+a c} $$ I thought that maybe proof that $$ \frac{1}{a+b+c+a b+b c+a c} \geq \...
Albert Wijaya's user avatar
3 votes
4 answers
80 views

Sum and product of integers conserving order

I have the feeling this is true, but can't prove it: $$\sum_n^An\lt\sum_n^Bn\implies\prod_n^An\lt\prod_n^Bn$$ Where $A\subset\mathbb N-\{0, 1\},B\subset\mathbb N-\{0, 1\}$ Example: $$3+4\lt5+6\...
throwaway's user avatar
3 votes
2 answers
153 views

How to show this identity $\prod_{q=1}^k\frac{1}{1-qz}=\sum_{j=1}^{k}jz\prod_{q=1}^j\frac{1}{1-qz}+1$ avoiding a proof by induction

When looking at a nice problem regarding Stirling numbers of the second kind a challenge was to show the validity of \begin{align*} \color{blue}{\prod_{q=1}^k\frac{1}{1-qz}=\sum_{j=1}^kjz\prod_{q=1}^j\...
Markus Scheuer's user avatar
3 votes
2 answers
81 views

$AM-GM$-ish inequality

Suppose $x_0, \cdots, x_n$ are positive reals. Suppose that: $$\sum_{i = 0}^n \frac{1}{1+x_i} \leq 1$$ Then show that: $$\prod_{i=0}^{n} x_i \geq n^{n+1} $$ I got to this problem by rewriting problem ...
mtheorylord's user avatar
  • 4,284
3 votes
1 answer
146 views

Upper bound on product of finitely many non negative terms

Let $a_i, 1 \leq i \leq n$ be non-negative real numbers. Let $S$ denote their sum. To prove $$\prod_{k=1}^n(1+a_k) \leq 1 + \frac{S}{1}+\frac{S^2}{2!} + \ldots + \frac{S^n}{n!}$$ Let $S = \sum_{k=1}^...
PAMG's user avatar
  • 4,500
3 votes
1 answer
1k views

Triples of natural numbers with same sum and product

Im looking at pairs of triples of natural numbers without repititions such that the sums of the two triples are equal and the products of the two triples are equal. To be precise: Let $x<y<z$ ...
Uli Fahrenberg's user avatar
3 votes
1 answer
197 views

Partition Proof

Let $\lambda$ be a partition of $N$ of rank $r$. How can I show that: $$\sum_wx^\lambda(w)=f^\lambda(-1)^{t(\lambda)}\prod^r_{i=1}(\lambda_i-1)!(\lambda'_i-1)!$$ where $w$ ranges over all ...
user avatar
3 votes
1 answer
307 views

Infinite Sum of Infinite Product

I've got an expression, $$ \left( \prod^{n}_{i = 1} \mu a_i \right) \left(\sum^{n}_{j=1} \frac{1 - a_j}{\prod^{j}_{k=1} \mu a_k} \right) $$ Where each $0 < a_j < 1 / \mu$ and $\mu > 1$, so ...
abnowack's user avatar
  • 155
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 ...
Erik Olesen's user avatar
  • 1,008
3 votes
1 answer
383 views

Nested operation notation convention for evaluation (particularly for Pi and Sigma)

I have a question on the notation for two products. For Pi: Is this statement true? $$\prod_{k=1}^{l}a_k\prod_{m=1}^{n}b_m = \prod_{k=1}^{l}\prod_{m=1}^{n}a_kb_m \neq a_1a_2\dots a_lb_1b_2\dots b_n = \...
Superman's user avatar
  • 255
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+...
user avatar
3 votes
0 answers
68 views

Evaluating a summation of product

Show that for any integer $k>1$ $$ \sum_{\substack{i_j \in \{0, 1\} \forall j < k, i_k = 0}} \prod_{j = 1}^{k} \left(i_j + (-1)^{i_j} \frac{a+ (j - 1) c - c \sum_{\lambda = 0}^{j - 1} i_\lambda}{...
ParaN3xus's user avatar

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