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1 vote
1 answer
89 views

Asymtotic of some binomial sum

Assume $n$ is a positive odd integer, I need to find the asymptotic as $n$ goes to infinity of the sum $$s(n,x)=\frac1x\sum_{k=0}^n (-1)^k\binom{-x-\frac12}{k}\binom{x-\frac12}{n-k},$$ where the ...
lomarkly's user avatar
  • 134
3 votes
1 answer
162 views

Prove combination of polynomials must be odd polynomials with positive coefficients.

Let $m,n\geq0$ are integers, show that $$p_{m,n}(x)=\sum_{k=0}^{m} \binom{2x+2k}{2k+1} \binom{n+m-k-x-\frac{1}{2}}{m-k}$$ must be an odd polynomial(all coefficients of even power of $x$ is $0$) with ...
lomarkly's user avatar
  • 134
2 votes
0 answers
56 views

$f_n(ab) = f_n(a)f_n(b) + f_n(a-1)f_n(b-1) + f_n(a-2)f_n(b-2)+ ...+f_n(a-n)f_n(b-n)$

I was toying around when I noticed for $a,b > 0$: $$f(ab) = f(a)f(b) + f(a-1)f(b-1)$$ is satisfied by $f(n) = T_n$ ; the triangular numbers $n(n+1)/2$. This equation is not an addition formula, not ...
mick's user avatar
  • 16.4k
2 votes
0 answers
70 views

Expectation of a certain polynomial expression in Rademacher random variables.

Let $N_1,k \ge 1$ be integers and let $N = N_1 k$. Let $G_1,...,G_k$ be an equi-partition of $[N] := \{1,2,\ldots,N\}$. Thus, $|G_j| = N_1$ for all $i$. Let $\mathcal S$ be the transversal of this ...
dohmatob's user avatar
  • 9,575
6 votes
1 answer
194 views

Conjecture: Shallow diagonals in Pascal's triangle form polynomials whose roots are all real, distinct, and in $(-2,2)$.

Here are the first few "shallow diagonals" in Pascal's triangle. We can use these shallow diagonals to make polynomials. Note that the even degree polynomials have an extra $\pm x$ at the ...
Dan's user avatar
  • 25.7k
5 votes
1 answer
215 views

Series of polynomials very nearly follows binomial coefficients but doesn't quite

I'm modelling a system using a Markov chain and by a few iterations of the transition matrix I can see a pattern emerging in the resulting polynomial that really looks like Pascal's triangle, but isn'...
Justin Booth's user avatar
1 vote
2 answers
153 views

Find the coefficients of product

Given the following product, $$(1+ax)(1+a^2x)(1+a^3x)\cdots (1+a^mx) $$ where $a$ is some real number which will be taken to be unity in the end. I want to know the coefficient of general term of ...
Craig Thone's user avatar
5 votes
2 answers
169 views

Prove that $\sum_{k=1}^n\frac{\prod_{1\leq r\leq n, r\neq m}(x+k-r)}{\prod_{1\leq r\leq n, r\neq k}(k-r)}=1$

For arbitrary $x$ and $1\leqslant m\leqslant n$, prove the following: $$\sum_{k=1}^n\frac{\prod_{1\leq r\leq n, r\neq m}(x+k-r)}{\prod_{1\leq r\leq n, r\neq k}(k-r)}=1$$ I'm looking for a proof that ...
Rusurano's user avatar
  • 848
1 vote
1 answer
97 views

Sum with binomial coefficient using identity

I want to prove: $\displaystyle \sum_{k=0}^n (-1)^k \binom{x}{k} = (-1)^n \binom{x-1}{n}$ using: $(1-z)^x \cdot \frac{1}{1-z} = (1-z)^{x-1}$ I know how to do it with induction but i somehow can't ...
elson1608's user avatar
  • 121
0 votes
0 answers
42 views

Expressing $j$-nomial coefficients in terms of binomial coefficients

Call the $k$th coefficient of $\left(\sum_{i=0}^jx^i\right)^n$ the $j$-nomial coefficient $C(n,k,j)$, so that the numbers $C(n,k,2)$ are just the binomial coefficients $\binom{n}{k}$. From the OEIS ...
Thomas Anton's user avatar
  • 2,346
0 votes
1 answer
51 views

Show with the help of binomial theorem that these two equations are equal?

Show with the help of binomial theorem that these two expression are equal for $n\ge 0$ then this $$ \sum_{k=0}^n \binom n k x^k (2+x)^k = \sum_{k=0}^{2n} \binom {2n} k x^k $$ I don’t know how to do ...
First_1st's user avatar
7 votes
3 answers
275 views

Proof of a neat pattern in polynomials

Let $f_1:\mathbb{R}\to\mathbb{R}$ such that $$f(x) = ax + b\space\space \forall\space x \in \mathbb{R}$$ It can be easily verified that $$f(x)-2f(x-1)+f(x-2)=0 \space \forall \space x \in \mathbb{R}--...
neofyt's user avatar
  • 271
3 votes
1 answer
889 views

What is the most efficient algorithm to evaluate a polynomial of n degree at K points?

A brute force approach would be to evaluate each point for each of the terms of a polynomial, which will be $O(Kn^2)$. If we use logarithmic exponentiation to find each $x^i$ then it becomes $O(Kn \...
ishandutta2007's user avatar
1 vote
0 answers
50 views

Nonnegative integer combinations of binomial coefficients

A classical result of Polya and Ostrowski states that integral linear combinations of binomial coefficients ${x\choose k}$ is exactly the set of all polynomials $f(x)\in\mathbb{Q}[x]$ such that $f(\...
QMath's user avatar
  • 427
2 votes
1 answer
129 views

Concrete Mathematics: What is polynomial argument

In the book Concrete Mathematics: A foundation for Computer Science of the chapter Binomial Coefficients, It gives an identity $$ (r - k) {r \choose k} = r {r - 1 \choose k} $$ for positive integers r....
Abhisek's user avatar
  • 465

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