All Questions
7
questions
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If $p$ is a prime and $m,n\in\mathbb{N}$, prove that $\binom{pm}{pn}\equiv\binom{m}{n}$ mod $p$.
Hint: Compare the binomial expansions of $(1+x)^{pm}$ and of $(1+x^m)^p$ in $\mathbb{F}_p[x]$.
If $R$ is a commutative ring, then the set of all polynomials with coefficients in $R$ is denoted by $R[x]...
1
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0
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71
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Why do two ways of expanding the same formal polynomial lead to matching coefficients?
There are a few proofs in which the technique is to expand the product of some formal polynomials in $\mathbb{R}[x_1,x_2,\ldots,x_k]$ in more than one distinct way and then we can match up the ...
1
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1
answer
34
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how to reduce $(1-\alpha)^{T-i}$ into a sum
I'm given the following proof:
\begin{align}
& \sum^T_{i=k} \alpha^i(1-\alpha)^{T-i} \binom{T}{i}
\\&=\sum^T_{i=k} \alpha^i \binom{T}{i} \sum_{j=0}^{T-i}(-\alpha)^j\binom{T-i}{j}
\\&=\...
1
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1
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49
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Binomial expansion of $(1+X)^b$ of the form $\sum\limits_k b^kf_k(X)$ with $f_k(X)$ polynomial
For each $b \in \mathbb{N}, (1+X)^b=\sum\limits_{n=0}^b{{b}\choose {n}}X^{n}$ is a polynomial function of $X$. How to write $(1+X)^b$ in terms of $b$ as a closed expression of the form $$(1+X)^b=\sum\...
3
votes
1
answer
463
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Gaussian polynomial identities
I'd appreciate any hints for showing that these identities are true for Gaussian polynomials. I've tried to approach the problem using basic algebra but it gets messy very quickly and I've gotten ...
0
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0
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44
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How to calculate the $k$-dimension of a subspace of a polynomial ring? [duplicate]
Let $k$ be an infinite field and $R:=k[x_1,...,x_n]$ the polynomial ring in $n$ indeterminates.
Why is the $k$-dimension of $U$ given by $\begin{pmatrix} n+m-1 \\ m\end{pmatrix}$, when $U$ is the ...
10
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1
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Vandermonde identity in a ring
Let $R$ be a commutative $\mathbb{Q}$-algebra. For $r \in R$ and $n \in \mathbb{N}$ we can define the binomial coefficient $\binom{r}{n}$ as usual by $\binom{r}{0}=1$ and $\binom{r}{n+1}=\frac{r-n}{n+...