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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.

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    $\begingroup$ Have you looked carefully at the product of the first few terms? I think if you use a bit of algebra, you'll see that there is some cancellation between numerators and denominators of adjacent terms. $\endgroup$
    – hardmath
    Jan 2, 2019 at 17:11
  • $\begingroup$ @hardmath That's absolutely true. I also have looked carefully at the product of first few terms, no doubt. $\endgroup$
    – Melz
    Jan 2, 2019 at 17:13
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    $\begingroup$ It would also depend on how you write those terms. If you just write $\left(1 - \frac1{2^2}\right)\left(1 - \frac1{3^2}\right)\left(1 - \frac1{4^2}\right)\cdots$ then I don't think you will see a useful pattern. $\endgroup$
    – David K
    Jan 2, 2019 at 17:16
  • $\begingroup$ I have got $$\frac{8}{15}$$ with my TI calculator $\endgroup$ Jan 2, 2019 at 17:18

2 Answers 2

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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

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  • $\begingroup$ That does not really help. $\endgroup$
    – Melz
    Jan 2, 2019 at 17:16
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    $\begingroup$ @Enzo Perhaps you should try harder to understand it. It is the same hint I would have given; I don't think there is an easier way. $\endgroup$
    – David K
    Jan 2, 2019 at 17:18
  • $\begingroup$ Write as I said and look what will happen $\endgroup$
    – user617243
    Jan 2, 2019 at 17:19
  • $\begingroup$ @Enzo I'll go out on a limb and say that there is no easier way. $\endgroup$
    – John
    Jan 2, 2019 at 17:20
  • $\begingroup$ $$\biggr (\frac{(2-1)(2+1)}{2^2} \biggr ) \biggr ( \frac{(3-1)(3+1)}{3^2} \biggr )$$ No such term cancels out here. $\endgroup$
    – Melz
    Jan 2, 2019 at 17:22
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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 $16!/15!=16$.

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