0

Can we write the following product in terms of binomial coefficients ? $(x-1)\cdot(x-3)\cdots (x-(n-2))$. i.e the the product take up odd numbers.

wkmath
  • 3
  • Using only binomial coefficients? Why would you want to restrict yourself like that? Using any simplification? Yes, of course. See double factorial. Remember that you can "multiply by one" whenever you like and it doesn't change the values, and that a value divided by itself equals $1$... So you have for instance $(x-1)(x-3)(x-5)\cdots = \frac{(x-1)(x-2)(x-3)(x-4)(x-5)\cdots}{(x-2)(x-4)\cdots}$. – JMoravitz Apr 25 '22 at 12:54
  • Are you assuming $n$ is odd? [otherwise last term doesn't match pattern] – coffeemath Apr 25 '22 at 12:55
  • As for your specific expression, it is unclear whether $x$ is even or odd. It is unclear if $n$ is even or odd. It is unclear whether you allow things like $2^k$ or factorials in your simplification, and so on... – JMoravitz Apr 25 '22 at 12:55
  • Is there any notation for the product $(x-1)\cdot(x-3)\cdots(x-(n-2))$ over certain factorial ? (here $x$ is a complex number) – wkmath Apr 25 '22 at 12:57

1 Answers1

0

Rewriting your question... you ask to simplify $\underbrace{(x-1)(x-3)(x-5)\cdots (x-(2k-1))}_{k~\text{terms in the product}}$ where $x$ is any complex number.

I reintroduce the notation for the falling factorial $z\frac{n}{~}=\underbrace{z(z-1)(z-2)\cdots(z-n+1)}_{n~\text{terms in the product}}$

I will assume that $x$ is not an integer from $\{0,1,2,3,\dots,k\}$ to avoid possible division by zero errors. Following from the approach used for simplifying double factorials and using the more generic falling factorials rather than factorials, we have:

$(x-1)(x-3)(x-5)\cdots(x-(2k-1)) = \dfrac{x(x-1)(x-2)(x-3)\cdots(x-(2k-1))}{x(x-2)(x-4)\cdots (x-(2k-2))}$

$ = \dfrac{x\frac{2k}{~}}{x(x-2)(x-4)\cdots(x-(2k-2))} = \dfrac{x\frac{2k}{~}}{2^k(\frac{x}{2})(\frac{x}{2}-1)(\frac{x}{2}-2)\cdots(\frac{x}{2}-k+1)}$

$$=\dfrac{x\frac{2k}{~}}{2^k(\frac{x}{2})\frac{k}{~}}$$

In the case that $x$ was among the numbers $1,3,5,\dots,2k-1$ then the product is simply zero, and in the case that $x$ was among the numbers $0,2,4,6,\dots,2k-2$ then it is as above but having skipped the zero term from numerator and denominator.

JMoravitz
  • 79,518