Proving $\prod$ identity

Question

I got stuck working on the following identity. Although I know that both sides of the equation deliver the same result, I'm cannot find the correct steps to convert the right side to look as the left one (or the other way round). Although I know that both sides of the equation give the same result, I can't find the right steps and to convert the right side to look like the left side (or vice versa). What kind of solid mathematical arguments can I use to proof the identity? Could someone give me some advice, please? Many thanks in advance, Phil

Edit:

Maybe I should have provide the origin of these equation. At the beginning I had this assignment: $$\begin{align*} &\forall x \in \mathbb{R}; k \in \mathbb{N}; k \neq 0:&&\\\\ &\begin{pmatrix} x \\ k \end{pmatrix}= \begin{pmatrix} x-1 \\ k-1 \end{pmatrix}+ \begin{pmatrix} x-1 \\ k \end{pmatrix}&& \end{align*}$$ having $$\begin{align*} \begin{pmatrix} x \\ k \end{pmatrix}=\frac{x-j-1}{j}.&& \end{align*}$$

To check the consistence of the last identity I took values for $x=10$, $k=4$ and I could verified that everything is correct. My point is the I would like to bring the $\frac{x}{k}$ into the $\prod$ and this would inplicate a"re-arrangement" of the indices. But as I already mentioned I cannot figure out how.

Of course I could prove it numerically with the example above but, from a purely mathematical point of view, would this method stand as a demonstration? I hope I could provide some additional feedbak.

$$\begin{align*} \prod_{j=1}^{k}\frac{x-j+1}{j}&=\prod_{j=1}^{k-1}\frac{x-j}{j}+\prod_{j=1}^{k}\frac{x-j}{j}\\ \prod_{j=1}^{k}\frac{x-j+1}{j}&=\prod_{j=1}^{k-1}\frac{x-j}{j}+\Big(\frac{x-k}{k}\Big)\cdot\prod_{j=1}^{k-1}\frac{x-j}{j}\\ \prod_{j=1}^{k}\frac{x-j+1}{j}&=\Big(\frac{x-k}{k}+1\Big)\cdot\prod_{j=1}^{k-1}\frac{x-j}{j}\\ \prod_{j=1}^{k}\frac{x-j+1}{j}&=\Big(\frac{x}{k}\Big)\cdot\prod_{j=1}^{k-1}\frac{x-j}{j} \end{align*}$$

Answer 1

We show the following identity is valid \begin{align*} \color{blue}{\prod_{j=1}^k\frac{x-j+1}{j}=\frac{x}{k}\prod_{j=1}^{k-1}\frac{x-j}{j}}\tag{1} \end{align*}

We obtain \begin{align*} \color{blue}{\prod_{j=1}^{k}\frac{x-j+1}{j!}}&=\frac{1}{k!}\prod_{j=1}^{k}\left(x-j+1\right)\tag{2}\\ &=\frac{1}{k!}\prod_{j=0}^{k-1}\left(x-j\right)\tag{3}\\ &=\frac{x}{k!}\prod_{j=1}^{k-1}\left(x-j\right)\tag{4}\\ &\color{blue}{=\frac{x}{k}\prod_{j=1}^{k-1}\frac{x-j}{j}}\tag{5}\\ \end{align*} and the claim (1) follows.

Comment:

• In (2) we use the identity $\prod_{j=1}^{k}j=k!$.

• In (3) we shift the index and start with $j=0$.

• In (4) we factor out $x$ and compensate this by starting with $j=1$.

• In (5) we use $k!=k(k-1)!=k\prod_{j=1}^{k-1}j$ .

Answer 2

Consider the two polynomials of degree $k$ each. \begin{align*} p(x)&=\prod_{j=1}^{k}\frac{x-j+1}{j}\\ q(x)&=\prod_{j=1}^{k-1}\frac{x-j}{j}+\prod_{j=1}^{k}\frac{x-j}{j}=\Big(\frac{x}{k}\Big)\cdot\prod_{j=1}^{k-1}\frac{x-j}{j} \end{align*} The roots of $p(x)$ are $0,1,2,\ldots ,k-1$. These are all the roots of $p(x)$.

Now, we use the second expression for the polynomial $q(x)$ to see that it has the same $k$ roots. Thus, the two polynomial functions must be related as $$p(x)=cq(x), \qquad c \text{ is some constant}.$$ Now choose a value of $x$ (say $x=k$) and compute $p(k)$ and $q(k)$, to determine this $c$.

Answer 3

Here is another way you can try to solve this problem.

We start with the right-hand side and expand the products, trying to match it with the left-hand side product: $$\prod_{j=1}^{k-1}\frac{x-j}{j}+\prod_{j=1}^{k}\frac{x-j}{j}=\frac{x-1}{1}\cdot\frac{x-2}{2}\cdot...\cdot\frac{x-(k-1)}{k-1}+\frac{x-1}{1}\cdot\frac{x-2}{2}\cdot...\cdot\frac{x-k}{k}$$ We can see some common terms in this sum, so we can factor them out and simplify the expression: $$\frac{x-1}{1}\cdot\frac{x-2}{2}\cdot...\cdot\frac{x-(k-1)}{k-1}\cdot\left(1+\frac{x-k}{k}\right)=\frac{x-1}{1}\cdot\frac{x-2}{2}\cdot...\cdot\frac{x-(k-1)}{k-1}\cdot\frac{x}{k}$$ Now, in the left-hand side product $\prod_{i=j}^k\frac{x\color{crimson}{-j+1}}{\color{teal}{j}}$, the term being subtracted in the numerator $(\color{crimson}{-j+1}=\color{crimson}{-(j-1)})$ is just $1$ smaller than the term in the denominator $(\color{teal}{j})$, while in our current expression, these terms are identical.

We can change it by essentially "shifting" each denominator one term left, since our right-hand side expression is a huge product, where every term is multiplied by a fraction (e.g. $\frac{x-2}{2}=(x-2)\cdot\frac{1}{2}$), and we just change the order of multiplication, moving each fraction one term left (Note: the denominator of the first term $\frac{x-1}{1}$ is $1$, so we do not need to do anything with it): $$\frac{x-1}{2}\cdot\frac{x-2}{3}\cdot...\cdot\frac{x-(k-1)}{k}\cdot x$$ In the left-hand side, the term at $j=1$ is just $x$, so we move our $x$ to the first term position and rewrite each numerator to fit the product in the left-hand side, from which the equality follows: $$x\cdot\frac{x-\color{crimson}{2}\color{teal}{+1}}{\color{crimson}{2}}\cdot\frac{x-\color{crimson}{3}\color{teal}{+1}}{\color{crimson}{3}}\cdot...\cdot\frac{x-\color{crimson}{k}\color{teal}{+1}}{\color{crimson}{k}}=\prod_{j=1}^k\frac{x-\color{crimson}{j}\color{teal}{+1}}{\color{crimson}{j}}$$ The right-hand side equals the left-hand side, thus: $$\prod_{i=j}^k\frac{x-j+1}{j}=\prod_{j=1}^{k-1}\frac{x-j}{j}+\prod_{j=1}^{k}\frac{x-j}{j}$$ Hope it helps :)