# Interpretations of $\sin(x)$

As sin(x) has an infinite number of maximum and minimums, I wondered if $\sin(x)$ could be interpreted in such a way as: $$ax^\infty+bx^{\infty-1}\cdots zx$$Or something. Am I talking nonsense here or is there actually an interpretation of sine that involves polynomials?

• Things like "$\infty - 1$" don't make sense, but Euler's solution of the Basel problem (finding the value of $\sum_{n = 1}^\infty \tfrac{1}{n^2}$) involves looking at $\sin x$ as an infinite product $x \prod_{n \in \Bbb Z \setminus \{0\}} (1 - \tfrac{x}{n\pi})$ (analogous to a polynomial being a product $\prod (x - \lambda)$, where $\lambda$ ranges over the roots). As the answers that have appeared below suggest, you want to read about Taylor polynomials to make this rigorous. Mar 14, 2013 at 20:52

Really good question!

You would be right on that there exists such an interpretation for sine. It is called sin's Maclaurin representation and looks like this:

$$\sin x=x- \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \frac{x^9}{9!}+\dots$$

• This is also known as a Mauclarin or Taylor Series. There exists a distinction between the two but it would require a knowledge of calculus to understand. Mar 14, 2013 at 20:49
• There exists a distinction between the two [Maclaurin or Taylor Series] but it would require a knowledge of calculus to understand. It seems to me that the difference between the two can be understood using the precalculus idea of a horizontal shift. Mar 14, 2013 at 21:02

You're close, though what you wrote is kind of nonsense because there's no such thing as having $x^{\infty}$.

The Taylor expansion of sine is $\sin(x)=\displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!}$, so sine is actually given by an infinite polynomial which is equal to it everywhere. You can prove this using calculus.

There are many ways of unterstanding $\sin(x)$ for example the one as a taylor series $$\sin(x)=\sum_{k=0}^\infty (-1)^k \frac{x^{2k+1}}{(2k+1)!}$$ or as $$\sin(x)= \frac{e^{ix} - e^{-ix}}{2i}$$ Those gives you the geometric interpretation as you are going in the complex plane and seeing the so called euler formula: $$e^{ix} = \cos(x) + i \sin(x)$$ Well both of them really don't point that stuff out with the infinit many maximums, there are other definition like $$\sin(x)=x \prod_{k=1}^\infty \left( 1- \frac{x^2}{k^2 \pi^2}\right)$$ which gives you an idea that the $\sin(x)$ function really infinitly many zeroes and this one can be interpreted really nice as a polynomial (written as a product of it's zeroes).

Another way to define the $\sin(x)$ function is over a ordinary differential equation: We call $\sin(x)$ the solution of $$\begin{cases} y''=-y\\ y'(0)=1\\ y(0)=0 \end{cases}$$ Here one can "see" that this functions has infinitly many zeroes, when we start at $0$ we know the function is increasing (as $f'>0$) so the function will increase, but the second derivative is lower than $0$ when $x$ is a bit bigger than $0$ so the function is increasing slower till the maximum, than it is decreasing, it will become smaller $0$ and then the idea goes on like that.