I'm reading Introduction to the Theory of Statistics by Mood and Graybill. One of the exercises in the early chapters is to show the following identity:

$$\prod^{2n+1}_{i=1}(x+n+1-i) = x\prod^{n}_{i=1}(x^2-i^2)\tag{1}$$

Attempting this exercise is the first time I've needed to show something truly complex. I've never before needed to attempt so much creativity and I find it very enjoyable. To me, it is my initiation ceremony into the world of real mathematical problem solving (not the trivial kinds of problems you encounter in high school and intro semester college courses). But as the three methods I tried on my own were unfruitful, I come here for help.

My initial observation was that the $2n+1$ on the left could be made into an $n$ as it is on the right by removing one single iteration and cleverly grouping the multiplications so that you can split the remaining $2n$ iterations into $n$ iterations.

Approach (1)

My first approach naively removed the $+1$ term in the $2n+1$ on the left, hoping that there would be some grouping for which I would be able to remove an $x$ from the entire multiplication and end up with the term on the right. I started this:

$$ \begin{align} \prod^{2n+1}_{i=1}(x+n+1-i) &= (x+n+1-(2n+1))\prod^{2n}_{i=1}(x+n+1-i) \\ &= (x-n)\prod^{2n}_{i=1}(x+n+1-i) \tag{2} \end{align} $$

Then attempting to group the multiplication by $i$ and $i+1$, we get

$$ \begin{align} (x-n)\prod^{2n}_{i=1}(x+n+1-i) &= (x-n)\prod^{n}_{i=1}(x+n+1-i)(x+n+1-i-1) \\ &= (x-n)\prod^{n}_{i=1}(x+n+1-i)(x+n-i) \\ \tag{3} \end{align} $$

Nothing spectacularly nice came out of this so far, and there are two more approaches, so I decided to pursue them before going deeper into (1).

Approach 2

This time I tried to use a spinoff of the binomial coefficient.

I can rewrite the left of $(1)$, substituting $u=x+n+1$


$C_i$ is an interesting number. Take $\binom{2n+1}{i}$ sets of every number from $1$ to $2n+1$, resulting in the set $Q$. For each of the elements of $Q$, uniquely remove $i$ numbers, that is, all elements of $Q$ should be a unique combination of length $n-i$ of all the numbers from $1$ to $2n+1$. Take the product of each of interval-sets in $Q$ and sum them. This results in the coefficient $C_i$.

In the same vein, we can rewrite the right of $(1)$ as


Here $G_i$ is another coefficient function, the same thing as $C_i$ but using $i^2$ instead of $i$.

I would also try rewriting equation $(3)$, already simplified so its limits of multiplication are $1$ and $n$, as a binomial type sum.

However, I realized that this approach could get very complex, and a third approach looked promising so I decided to pursue that next.

Approach (3) MOST PROMISING This is similar to my first approach but less naive. I tried to remove an $x$ term from the product and somehow group the product to get the bounds from $2n$ to $n$.

Using $i=n+1$ as the index of removal, I get out $(x+n+1-(n+1))=x$. Now my intervals of multiplication are $[1,n] \bigcup [n+2, 2n+1]$. Then, grouping $(u-i)$ and $(u-(2n+2-i))$ (using $2n+2$ because $i$ starts at $1$ and not $0$),

$$ \begin{align} x\prod^{2n}_{i=1}(x+n+1-i) \\ &= x\prod^{n}_{i=1}(x^2-i^2+2in+2i-n^2-2n-1) \end{align} $$

I have the $x^2-i^2$ but also a whole lot of junk.

Perhaps an alternate pairing would do better?

Now I'm pairing $(u-i)$ with $(u-(n+1+i))$ to obtain

$$ \begin{align} x\prod^{2n}_{i=1}(x+n+1-i) \\ & = x\prod^{n}_{i=1}(x+n+1-i)(x+n+1-(n+1+i)) \\ &= x\prod^{n}_{i=1}(x^2-i^2+in+i+nx+x) \end{align} $$

This has the $x^2-i^2$ term but also a ton of junk. Another seeming dead end.

I've done my own exploration and now I need some guidance. What is the best path to pursue? Where is my mindset wrong? What should my initial attack have been?


After some valuable discussion in the answer below I realized I had neglected an essential criterion for pairing factors, namely that the $n+1$ cancel out for both factors. If we pair $x+n+(i-n)$ and $(x+n-(i-n))$ then the RHS of $(1)$ immediately follows. The criterion for the solution

  1. Have an $x$ factor
  2. Limits of multiplication upper bounded at $n$
  3. Product term is $(x+1)(x-1)$, i.e. the pairing scheme must cancel out the $n+1$ for both factors.

$$\begin{align*} x \prod_{i=1}^n (x^2 - i^2) &= x \prod_{i=1}^n (x-i)(x+i) \quad \text{factor the difference of squares}\\ & = x \prod_{i=1}^n (x-i) \prod_{i=1}^n (x+i) \quad \text{split into two products} \\ &= (x-0) \prod_{i=1}^n (x-i) \prod_{i=-n}^{-1} (x-i) \quad \text{flip sign of second product's index} \\ &= \prod_{i=-n}^n (x-i) \quad \text{Recombine into single product} \\ &= \prod_{j=1}^{2n+1} (x-(j-(n+1))) \quad \text{let } i = j-(n+1) \iff j = (n+1)+i \\ &= \prod_{j=1}^{2n+1} (x+n+1-j). \end{align*}$$


If you fully factor the RHS (including the factor of $x$), and sort the resulting factors in ascending order, the factors on the LHS are exactly the same as those on the RHS, except in reverse order. If there's still some doubt, try an example with a small value of $n$, say $n = 3$.

  • $\begingroup$ Perhaps, but I don't see the point. If the linear factors of the LHS are identically equal, apart from order, to the factors of the RHS, then it's immediate that the equation is an identity. $\endgroup$ – quasi Jan 3 '17 at 6:13
  • $\begingroup$ Of course, I'm sure you know that $(x^2 - i^2) = (x+i)(x-i)$. $\endgroup$ – quasi Jan 3 '17 at 6:15
  • $\begingroup$ Yes you are making a good point. I know that the general case follows directly. However, I'm more interested in seeing what such a proof would look like as a learning exercise. $\endgroup$ – theideasmith Jan 3 '17 at 6:15
  • $\begingroup$ But that is the point -- to recognize a one-to-one correspondence. $\endgroup$ – quasi Jan 3 '17 at 6:16
  • $\begingroup$ I know this is all probably so clear to you. The approaches are clear to me, but some of these truths that jump out to you don't immediately jump out to me. To be truthful I did consider expanding the LHS. I didn't go through with it. $\endgroup$ – theideasmith Jan 3 '17 at 6:17

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