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In the Wikipedia page for the Basel problem, it says that Euler, in his proof, found that

$$\begin{align*} \frac{\sin(x)}{x} &= \left(1 - \frac{x}{\pi}\right)\left(1 + \frac{x}{\pi}\right)\left(1 - \frac{x}{2\pi}\right)\left(1 + \frac{x}{2\pi}\right)\left(1 - \frac{x}{3\pi}\right)\left(1 + \frac{x}{3\pi}\right) \cdots \end{align*}$$

because the roots are at $\pm\pi, \pm2\pi, \pm3\pi, \cdots$ and finite polynomials are in this form (i.e. $(x-\text{root}_1)(x-\text{root}_2)\cdots$).

How was he able to do this? Why does this not simply make a polynomial function that has the roots same roots of $(\sin x)/x$? Can this method be used to make other trigonometric functions?

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1  
See Sandifer's article Basel Problem with Integrals and Dunham's book p39s. Fine reading, –  Raymond Manzoni Apr 21 '12 at 17:18
    
Precisely what does How was he able to do this? mean? His idea follows from the following algebra: $$ \begin{align} \frac{\sin(x)}{x}&=(x-\pm \pi)(x-\pm 2\pi)\cdots\\ &=\left(\frac{x}{\pi}-\pm 1\right)\left(\frac{x}{2\pi}-\pm1\right)\cdots\\ &=\left(\frac{x}{\pi}-+1\right)\left(\frac{x}{\pi}--1\right)\left(\frac{x}{2\pi}‌​-+1\right)\left(\frac{x}{2\pi}--1\right)\cdots\\ &=\left(\frac{x}{\pi}-1\right)\left(\frac{x}{\pi}+1\right)\left(\frac{x}{2\pi}-1‌​\right)\left(\frac{x}{2\pi}+1\right)\cdots \end{align} $$ This does make a function with the same roots as and yes. (Out of letters) –  000 Apr 21 '12 at 17:20
    
@Limitless Can I use this method on any function? What gave him the ability to do this on sinc? –  Argon Apr 21 '12 at 17:22
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@Argon, yes you can. It is a consequence of algebra. In general, if $y$ is a function in an equation $\text{something}=0$, its roots are the values that satisfy that equation. That, of course, implies that $y(x)=(x-r_1)(x-r_2)\cdots$ where $r_i$ is the $i$th root. –  000 Apr 21 '12 at 17:25
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Limitless, Argon, careful What Limitless is saying is not true. It only works so simply for polynomials. For any function $f(z)$ the function $ e^z f(z) $ has the same poles and zeros as $f(z),$ how would this method distinguish between these function functions? In modern times we have the benefit of the Weierstrass Factorization Theorem which gives us the true result. Euler most likely relied on his incredible intuition for the guess, then his great calculating ability to check it's validity numerically. –  Ragib Zaman Apr 21 '12 at 17:33

2 Answers 2

up vote 3 down vote accepted

To summarize,

How was he able to do this?

He lucked out, really. It was a coincidence that it worked and an intuitive guess. As Ragib Zaman said, "Euler most likely relied on his incredible intuition for the guess, then his great calculating ability to check it's validity numerically."

The derivation (though not accurate for most other functions) is:

\begin{align} \frac{\sin(x)}{x}&=(x-\pm \pi)(x-\pm 2\pi)\cdots\\ &=\left(\frac{x}{\pi}-\pm 1\right)\left(\frac{x}{2\pi}-\pm1\right)\cdots\\ &=\left(\frac{x}{\pi}-+1\right)\left(\frac{x}{\pi}--1\right)\left(\frac{x}{2\pi}-+1\right)\left(\frac{x}{2\pi}--1\right)\cdots\\ &=\left(\frac{x}{\pi}-1\right)\left(\frac{x}{\pi}+1\right)\left(\frac{x}{2\pi}-1\right)\left(\frac{x}{2\pi}+1\right)\cdots\\ &=\prod_{k=1}^{\infty}\left(\frac{x}{k\pi}-1 \right)\left(\frac{x}{k\pi}+1 \right) \end{align}

Why does this not simply make a polynomial function that has the roots same roots of sinx/x?

Note that it does make a function* with the same roots as $\frac{\sin(x)}{x}$. You can see this by equating the first equation with $0$.

Can this method be used to make other trigonometric functions?

I don't think so. From what I understand, there are six basic trigonometric functions and that's all there is. There is no specific limit on the number of trigonometric functions; rather, these six are the only ones that have caught on due to their specialness, power, and use. (You may find it intriguing that there are hyperbolic analogs of these functions: Hyperbolic functions.)

*I do not think this function could be considered a polynomial function.

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7  
Euler was actually much more careful than just "let's see what the roots of $\sin(x)/x$ are and let us pretend that it is a polynomial". In his Introduction to infinitesimal analysis he discussed relatively carefully why it worked for $\sin x=(e^{ix}-e^{-ix})/2i$ and not for $(e^{2\pi i}-1)/2i$ (a function with the same roots). Basically he said $e^{ix}=(1+ix/N)^N$ "for an infinite $N$" and then he checked that some terms were infinitely small. Perhaps not up to today's standards, but much more than nothing. –  user8268 Apr 21 '12 at 19:32
    
@user8268, I do recall there being a particular justification on Euler's part. However, the paper mentioned in Raymond Manzoni's comment gave me the impression that I put in my answer here. I suppose my interpretation was being blatantly simplistic and overly crass, however. On the other hand, I see that wikipedia espouses a very similar train of thought to mine. Oh well. I suppose it's a historically contested issue; not a necessarily mathematical one. :) –  000 Apr 22 '12 at 0:35
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There is an analogous infinite product expansion for cosine. The other four usual trigonometric functions can be obtained by appropriate operations on these infinite products, including the differentiation of the logarithms of these products. –  J. M. May 7 '12 at 2:51

Nearly the same question was posted here recently. I hope this will add a little that is not in the other answers to this present question.

We know that $\dfrac{\sin x}{x}=0$ when $\sin x= 0$ and $x\neq0$, and we know that $\dfrac{\sin x}{x}$ "$=$" $1$ when $x=0$ (I think Euler's way of saying this is that $\sin x = x$ when $x$ is infinitely small). So this function should be $0$ when $x=\pm\pi$ or $\pm2\pi$ or $\pm3\pi$, etc., so it is $$ \begin{align} & \text{constant}\cdot(x-\pi)(x+\pi)(x-2\pi)(x+2\pi)(x-3\pi)(x+3\pi)\cdots \\[8pt] & = \text{constant}\cdot(x^2-\pi^2)(x^2-4\pi^2)(x^2-9\pi^2)\cdots. \end{align} $$ When $x=0$, this is $(-\pi^2)(-4\pi^2)(-9\pi^2)\cdots$. But we saw above that when $x=0$, this is $1$. Hence we have $$ \begin{align} \frac{\sin x}{x} & = \frac{(x^2-\pi^2)(x^2-4\pi^2)(x^2-9\pi^2)\cdots}{(-\pi^2)(-4\pi^2)(-9\pi^2)\cdots} \\[8pt] & = \left(1-\frac{x^2}{\pi^2}\right)\left(1-\frac{x^2}{4\pi^2}\right)\left(1-\frac{x^2}{9\pi^2}\right)\cdots. \end{align} $$

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Excellent second half, I couldn't find this anywhere. –  Alyosha Jan 3 '13 at 18:36

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