# Reducing $\prod \limits_{0 \le j \ne i \le n} \frac{n+1-j}{i-j}$ to $\frac{(n+1)!}{(n+1-i)\cdot i! \cdot (n-i)!}(-1)^{(n-i)}$

How could we show that: $$\prod_{0 \le j \ne i \le n} \frac{n+1-j}{i-j} = \frac{(n+1)!}{(n+1-i)\cdot i! \cdot (n-i)!}(-1)^{(n-i)} .$$

The module suggest we could reduce it by simply writing $$\prod_{0 \le j \ne i \le n} \frac{n+1-j}{i-j} = \frac{(n + 1)n \cdots (n + 1 − (i − 1))(n + 1 − (i + 1)) \cdots 1}{i(i − 1) \cdots 1 · (−1) \cdots (−(n − i))} ,$$ but I am not able to figure out the result from here.

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Why the formulas are so large? –  Fabian May 30 '11 at 19:27
$\quad$Reduced. –  Quixotic May 30 '11 at 19:30
Your fomula doesn't make sense unless $i$ is a fixed number. Where does it come from? –  N. S. May 30 '11 at 19:31
$i$ is fixed it's a part of the problem that involves operation with la-granges interpolation formula. –  Quixotic May 30 '11 at 19:38
That formula seems to evaluate to $$\frac{(n+1)!}{(n+1-i)\cdot i! \cdot (n-i)!}(-1)^{(n-i)}$$. Are you sure the n+1-i term is a factorial? –  EuYu May 30 '11 at 19:52

The numerator is

$$\prod_{0 \le j \ne i \le n} (n+1-j) \,.$$

By writing it term by term you get

$$(n+1)n(n-1)....(n+1-(i-1))(n+1-(i+1))...1 = \frac{(n+1)!}{n+1-i} \,.$$

The denominator is

$$\prod_{0 \le j \ne i \le n} (i-j)= [i(i-1)(i-2)...(i-(i-1))][(i-(i+1))(i-(i+1))...(i-n)] \,.$$

The first bracket is exactly $i!$, while the second is $(-1)(-2)...(-(n-i))$. By taking a minus out of each bracket you get $(-1)^{n-i}(n-i)!$.

Thus the denominator is $(-1)^{n-i}(n-i)!i!$.

Combining those you get exactly the formula Yuqing posted.

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This might be silly but when $(-1)^{n-i}$ goes in the numerator won't it would be $(-1)^{-n+i}$? what I am missing? –  Quixotic May 30 '11 at 20:09
$\frac{1}{(-1)^{n-i}}=\frac{[(-1)^{n-i}]^2 }{(-1)^{n-i}}$ ;) –  N. S. May 30 '11 at 20:12
Lol...clever ;) –  Quixotic May 30 '11 at 20:13

$$\prod_{0 \le j \ne i \le n} \frac{n+1-j}{i-j} = \frac{\prod_{0 \le j \ne i \le n}(n+1-j)}{\prod_{0 \le j \ne i \le n}(i-j)}$$

we can rewrite this as follows

$$\frac{\prod_{0 \le j \ne i \le n}(n+1-j)}{\prod_{0 \le j \ne i \le n}(i-j)} = \frac{\prod_{0 \le j \le n}(n+1-j)}{(n+1-i) \times \prod_{0 \le j < i}(i-j) \times \prod_{i+1 \le j \le n}-(j-i)}$$

You can simplify the above expression to get an expression similar to what you have written. However, there is only $(n+1-i)$ instead of of the $(n+1-i)!$ in the denominator on the right hand side.

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what I am more interested to know, how are you rewriting? :)Precisely the denominator. –  Quixotic May 30 '11 at 20:00
@debanjan, I'm simply splitting the product into two terms, one where $j<i$, where the individual terms of the product $\prod_{0\le j <i}(i-j)$ are all positive and another where the product is of negative terms. Try it with a small example (say $n=3$ and $i=1$) and it should be clear. –  svenkatr May 30 '11 at 20:04

I assume only $j$ is the running index. Typically the way you have written the product on the left side it means that $i$ and $j$ are both running indices. You could mention this explicitly by using the following notation. $$\prod_{\substack{j=0\\ j \neq i}}^{n} \frac{n+1-j}{i-j}$$ Split the product into two $$\prod_{\substack{j=0\\ j \neq i}}^{n} \frac{n+1-j}{i-j} = \prod_{j=0}^{i-1} \frac{n+1-j}{i-j} \times \prod_{j=i+1}^{n} \frac{n+1-j}{i-j}$$ $$\prod_{j=0}^{i-1} \frac{n+1-j}{i-j} = \frac{(n+1)n(n-1)(n-2) \cdots (n-i+2)}{(i)!} = \frac{(n+1)!}{i! (n-i+1)!}$$ $$\prod_{j=i+1}^{n} \frac{n+1-j}{i-j} = (-1)^{n-i} \frac{(n-i)!}{(n-i)!} = (-1)^{n-i}$$ Hence, we get $$(-1)^{n-i} \frac{(n+1)!}{i! (n-i+1)!}$$

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Actually both i and j are running index in the actual problem,but you are right this is the better way to explain for this subproblem. –  Quixotic May 30 '11 at 20:19