1
$\begingroup$

It so happens that $\lim_{x \to 0}\frac{e^{\sin{x}} - 1}{x} = 1$

Let $f(x) = e^{\sin{x}} - 1$

$f(x)$ has roots at $n\pi$ where $n\in\mathbb{Z}$

Hence, $$f(x) = A. x(x-\pi)(x+\pi)(x-2\pi)(x+2\pi)\ldots$$ But after solving for $A$ and post simplification, I get:

$$f(x) = x\prod_{n=1}^{\infty}\left(1-\frac{x^2}{n^2\pi^2}\right)$$

But this is the same as the infinite product form of $\sin{x}$. So is this correct?

  • 2
    what about $x$'s where $\sin(x) = 2 \pi i$? – 2017-02-15
  • 0
    Your expansion cannot be correct, as it is odd and $e^{\sin x}-1$ is not. – 2017-02-15
  • 0
    @RossMillikan what is the significance of the oddness? – 2017-02-15
  • 0
    It is a property of the product expansion you offer that is not satisfied by the original function $e^{\sin x}-1$. I have plotted the original function at [http://www.wolframalpha.com/input/?i=expand+exp(sin(x))-1] If one is odd and the other is not, they are not the same function. Your error is to assume that $e^{\sin x}-1$ only has zeros at the integers. It has many more in the complex plane. They are at $m+2n\pi i$ for $m,n \in \Bbb Z$ – 2017-02-15
  • 0
    @james.nixon doesn't that argument hold true for the original basel problem too? – 2017-02-15
  • 0
    @deostroll If $x_0$ satisfies $\sin(x_0) = 2\pi i $ then $e^{\sin(x_0)} - 1 = e^{2 \pi i} - 1 = 0$. Therefore your list of zeroes is incomplete. A complete list of zeroes is given by the preimage $\sin^{-1}(2\pi i k )$ for all $k \in \mathbb{Z}$. – 2017-02-15
  • 0
    @james.nixon I forgot about that identity – 2017-02-15
  • 0
    @james.nixon does it mean the factor $(x-0)$ appears _twice_? – 2017-02-16
  • 0
    @deostroll I'm not sure what you mean. The zero set of your function is $Z = \{x \in \mathbb{C}\,:\,\sin(x) = 2 \pi i k\,\, k \in \mathbb{Z}\}$. – 2017-02-16

0 Answers 0