All Questions
61
questions
3
votes
5
answers
218
views
Coefficient of $x^{21}$ in $(1+x+x^2+\dots+x^{10})^4$
Find the coefficient of $x^{21}$ in $(1+x+x^2+\dots+x^{10})^4$
I tried splitting the terms inside the bracket into two parts $1+x+\dots+x^9$ and $x^{10}$, and then tried binomial theorem, but that ...
- 565
2
votes
0
answers
77
views
Closed expression for a combinatorial sum
The following equality is true for every positive integer $n$ :
$$\sum_{k=0}^n {n \choose k} = 2^n $$
It is a special case ($p = 0$) of the sequence :
$$S_{p, n}=\sum_{k=0}^n k^p {n \choose k} $$
For ...
- 389
2
votes
4
answers
258
views
Coefficient of $x^k$ in polynomial
Let $k, n, m \in \mathbb{N}, k \le n.$ Find the formula for coefficient of $x^k$ in $(x^n + x^{(n-1)} + ... + x^2 + x + 1)^m$.
answer is in this question: faster-way-to-find-coefficient-of-xn-in-1-x-...
- 311
0
votes
1
answer
87
views
Simplifying a sum of binomials, is there a closed form?
I am struggling with a formula that I derived, which (I believe) can be simplified further.
In principle, I want to determine the coefficients of the following polynomial:
\begin{align}
p(x) = (1+x+.....
- 31
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 ...
- 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 ...
- 134
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 ...
- 9,575
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 ...
- 505
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 ...
- 2,346
1
vote
1
answer
33
views
Simplify $\sum_{t=k}^{n} (\binom{n}{t} \cdot a^{t-1} \cdot (1 - a)^{n - t - 1} \cdot (t - n \cdot a))$
I was working on my probability theory homework and I found probability density function that looks as following
$$\sum_{t=k}^{n} \binom{n}{t} \cdot a^{t-1} \cdot (1 - a)^{n - t - 1} \cdot (t - n \...
- 983
3
votes
1
answer
135
views
Reciprocal binomial coefficient polynomial evaluation
The conventional binomial coefficient can be obtained via
$$
f(x, n) = (1+x)^n = \sum_{i=0}^n { n \choose i} x^i
$$
And the function $f$ can be every efficiently performed on evaluation.
I'm ...
- 1,271
6
votes
2
answers
121
views
Is the sum $\sum_{k=0}^n{n \choose k}\frac{(-1)^{n-k}}{n+k+1}$ always the reciprocal of an integer $\big(\frac{(2n+1)!}{(n!)^2}\big)$?
Denote the sum $$S_n := \sum_{k=0}^n{n \choose k}\frac{(-1)^{n-k}}{n+k+1}$$
This value arose in some calculations of polynomial coefficients. I'm not used to dealing with expressions of this sort. ...
- 1,044
1
vote
0
answers
120
views
Proving this equation involving derivative of a rook polynomial.
So, there is an exercise in Section 8.3 in Alan Tucker's Applied Combinatorics regarding rook polynomials. The problem is as follow.
Let $R_{n, m}(x)$ be the rook polynomial for an $n \times m$ ...
- 11
6
votes
4
answers
258
views
If $P(x)$ is any polynomial of degree less than $n$, show that $\sum_{j=0}^n (-1)^j\binom{n}{j}P(j)=0$. [duplicate]
If $P(x)$ is any polynomial of degree less than $n$, then prove that
$$\sum_{j=0}^n (-1)^j\binom{n}{j}P(j)=0$$
My approach was to try and prove this separately for $j^k\ \ \forall\ \ k<n$, instead ...
- 641
2
votes
0
answers
55
views
Formula for $\sum\limits_{k=1}^m\binom{N}{k}k^n$ [closed]
For given natural numbers $m, n$ and $N$, is there a compact formula for the expression
$$\sum\limits_{k=1}^m\binom{N}{k}k^n\; ?$$
We can assume that $N>m$.
- 43