Maclaurin series for sin(x) representation
Olivia Zamora 
The Maclaurin series for $\sin(x)$ is:
$$ \sin(x) = x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \dfrac{x^7}{7!} ... $$
Which according to wikipedia is:
$$ \displaystyle \sum_{n=0}^{\infty} \dfrac{(-1)^n}{(2n+1)!} x^{2n+1}$$
I however don't understand why this notation is correct at all. It only seems to work if you consider that $x$ is the zeroth term. Why is it the zeroth term?
$\endgroup$ 61 Answer
$\begingroup$What do you mean by the zeroth term? Do you mean the term given by $n = 0$? In this case simple evaluation shows this term to be $$\frac{(-1)^{0}}{(2(0)+1)!}x^{2(0)+1} = \frac{1}{1!}x^{1} = x$$
If, however, you are wondering why the first non-zero term of the Maclaurin series is $0$, then it is as simple as observing the following:
Let $f(x) = \sin(x)$.
Then $f'(x) = \cos(x)$
... $f''(x) = -\sin(x)$
... $f'''(x) = -\cos(x)$
... $f^{(4)}(x) = \sin(x)$
and so on. It is easy to note that for $f^{(k)}(0) = 0$ for $k$ even, since for $k$ even, $f^{(k)}(x) = c\sin x$, where $c$ is $1$ or $-1$, and $\sin(0) = 0$.
On the other hand, this gives us an expression for odd $k$. Writing $k$ odd as $2n+1$ for some $n \in \mathbb{N}$, we have $f^{(2n+1)}(x) = (-1)^{n}\cos(x)$, and so $f^{(2n+1)}(0) = (-1)^{n}\cos(0) = (-1)^{n}$. The rest of the expression given by wikipedia, then, follows from the definition of a Maclaurin series.
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