How does one evaluate the multiplication $f(2)\cdot f(3)\cdot f(4)\cdots f(15)$ by formulating?

Question

Suppose

$$f : \mathbb{Z}^+ \rightarrow \mathbb{R}$$

$$f(x) = 1-\dfrac{1}{x^2}$$

How does one evaluate the multiplication $\prod_{i=2}^{15} f(i)=f(2)\cdot f(3)\cdot f(4)\cdots f(15)$?

Here I have to see the trick that directly yields the calculation.

How come that we write this using $\Pi$ (product) notation? I'll be glad to hear your dear thoughts.

Answer 1

Hint: $f(x) = \frac{x^2-1}{x^2} = \frac{(x-1)(x+1)}{x^2}$
and take product of $f(2),f(3),...,f(n)$

Look: $$ \frac{(1)(3)}{2^2} \cdot \frac{(2)(4)}{3^2} \cdot \frac{(3)(5)}{4^2} \cdot ... \frac{(n-3)(n-1)}{(n-2)^2} \cdot \frac{(n-2)n}{(n-1)^2} \cdot \frac{(n-1)(n+1)}{n^2} = \frac{1}{2} \cdot \frac{n+1}{n} $$ and take this for your task

Answer 2

It is also convenient to use the factorial notation $n!=1\cdot2\cdots n$. We obtain \begin{align*} \color{blue}{\prod_{i=2}^{15}f(i)}&=\prod_{i=2}^{15}\left(1-\frac{1}{i^2}\right)\\ &=\prod_{i=2}^{15}\frac{i^2-1}{i^2}\\ &=\prod_{i=2}^{15}\frac{(i-1)(i+1)}{i^2}\\ &=\frac{14!\cdot 16!/2}{\left(15!\right)^2}\tag{1}\\ &\,\,\color{blue}{=\frac{8}{15}}\tag{2} \end{align*}

Comment:

• In (1) we use $\prod_{i=2}^{15}(i-1)=1\cdot 2\cdots 14=14!$ and $\prod_{i=2}^{15}(i+1)=3\cdot4\cdots 16=\frac{1}{2}\cdot 16!$.

• In (2) we see that $14!/15!=1/15$ and $16!/15!=16$.