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I was working on a formula, and along the way, I came across this product:$$(x-1)(x-2)(x-3)(x-4)\ldots(x-k+1)\tag1$$ This is just a quick question, but

Question: Is there a shorthand notation for $(1)$? I mean, besides writing all of the terms out.

An example would be something like the sigma notation $\sum$ meaning adding, or $\prod$ meaning the product of the terms.

I feel like there is a notation for $(1)$, but I just don't know it. Any suggestions?

pseudoeuclidean
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Frank
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  • This Wikipedia page may be useful. – angryavian Dec 30 '16 at 05:32
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    Your title does not match the expression in the body of your question. Additionally, the "$+1$" in the expression does not make sense to me. I think you mean to write $$(x-1)(x-2)(x-3)...(x-k)$$ – pseudoeuclidean Dec 30 '16 at 05:32
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    @pseudoeuclidean No, I meant $(x-k+1)$. For example, let's let $c_{\text{something}}:=(x-1)(x-2)(x-3)\ldots$. If that 'something'$=3$, then it would be$$c_3=(x-1)(x-2)$$If it was $4$, then the product would be$$c_4=(x-1)(x-2)(x-3)$$ – Frank Dec 30 '16 at 05:36
  • @angryavian So close! We're missing the $x$ term... Although, we could have$$\dfrac {(x)_k}x$$ – Frank Dec 30 '16 at 05:41

4 Answers4

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If falling powers aren't common enough, you could write it fairly economically as $$\frac{(x-1)!}{(x-k)!}$$

Tad
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this notion in Mathematics is called a falling factorial (more info from MathWorld). The general notation is $$ (x)_n = x(x-1) \ldots (x-(n-1)), $$ so your expression is $$ (x-1) \times \ldots \times (x-k+1) = (x-1) \times \ldots \times (x-(k-1)) = (x-1)_{k-1} $$

gt6989b
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    $(x)_n$ is usually denoted by $x^{\underline n}$ to avoid ambiguity. – Henricus V. Dec 30 '16 at 06:00
  • @HenryW. i now recall this notation from my combinatorics class, but can you please refresh my memory -- what is the possible ambiguity with $(x)_n$? Sequences are denoted $(x_n)$, is there something else? – gt6989b Jan 03 '17 at 15:55
  • I think if the expression in the bracket is simple (e.g. a single number) there is no confusion, but for more complex equations it can be misinterpreted as index notation. – Henricus V. Jan 04 '17 at 17:23
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Yes there is. We can write it as $$\prod_{i=1}^{k-1}(x-i).$$

Xam
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If we multiply the expression by $x$ and divide by $k!$ we have,

$$\frac{x(x-1)....(x-k+1)}{k!}$$

This is precisely ${x \choose k}$, so your product can be rewritten as,

$$k! {x \choose k} \frac{1}{x}$$

After reversing the original operations performed.

If you don't like the $x$ in the denominator here is another solution.

Replace $x$ by $x+1$ in the expression to get,

$$x(x-1).....(x-(k-1)+1)$$

Again we recognize this as,

$$(k-1)! {x+1-1 \choose k-1}$$

Then we replace $x+1$ by $x$ to reverse what we just did,

So the expression is

$$=(k-1)!{x-1 \choose k-1}$$

Ahmed S. Attaalla
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