Define the real valued function
$$ \sin:\mathbb{R} \rightarrow \mathbb{R}, \qquad given ~~by \qquad \sin(x) := x-\frac{x^3}{3!} + \frac{x^5}{5!} – \frac{x^7}{7!} + \ldots $$
How does one show $\sin(x)$ is bounded using this definition? Note that you are not allowed to
use the power series of $\cos(x)$ and try to show $\sin^2(x) + \cos^2(x) =1$ and
then prove they are bounded. I want a direct proof using the power series of $\sin(x)$.
Remark: I am looking for a proof that will allow me to modify/mimic the arguments if I am given some DIFFERENT power series that also happens to be bounded (but which doesn't have all the nice properties of sin(x) and cos(x) ). That is the motivation for the question. Take the power series of $\exp(-x^2)$ for example. Why is that bounded?
Best Answer
The coefficient of $x^n$ is $\frac{i^n+(-i)^n}{2n!}$. (We don't really need complex numbers here, but that's a convenient way of explcitly describing the coefficient)
To ban the forbidden tricks, let us simply not use derivatives at all!
We have $$ \sin 1 = 1-\frac1{3!}+\frac1{5!}\mp\ldots>1-\frac1{3!}=\frac56$$ because the summands are decreasing in absolute value. And $$\sin 4 =4-\frac{4^3}{3!}+\frac{4^5}{5!}\mp\ldots<4-\frac{4^3}{3!}+\frac{4^5}{5!}-\frac{4^7}{7!}+\frac{4^9}{9!}= -\frac{268}{405}$$ because all but the first few summands are decreasing in absolute value. By continuity, there exists a number $\pi\in(1,4)$ with $\sin\pi=0$. Then (absolute convergence justifies sum swapping) $$\begin{align}\sin(x+\pi)&=\sum_{n=0}^\infty \frac{i^n+(-i)^n}{2n!}(x+\pi)^n\\&=\sum_{n=0}^\infty\sum_{k=0}^n\frac{i^n+(-i)^n}{2n!}\frac{n!}{k!(n-k)!}x^k\pi^{n-k}\\ &=\sum_{k=0}^\infty \frac{x^k}{k!}\sum_{n=k}^\infty\frac{i^n+(-i)^n}{2(n-k)!}\pi^{n-k}\\ &=\sum_{k=0}^\infty \frac{x^k}{k!}\sum_{m=0}^\infty\frac{i^{m+k}+(-i)^{m+k}}{2m!}\pi^m\\ \end{align}$$ If $k\equiv0\pmod 4$, the inner series is $\sin\pi=0$. If $k\equiv 2\pmod 4$, it is $-\sin\pi=0$. If $k\equiv 1\pmod 4$, it is $c$, and if $k\equiv 3\pmod 4$ it is $-c$, where $c:=\sum_{m=0}^\infty\frac{i^{m+1}+(-i)^{m+1}}{2m!}\pi^m$. We conclude that $$ \sin(x+\pi)=c\sin x.$$ Directly from the series, we see that $\sin$ is odd, i.e. $\sin(-x)=-\sin x$. Hence $$\sin(x-\pi)=-\sin(-x+\pi)=-c \sin(-x)=c\sin x$$ and ultimatley $$\sin x=\sin(x+\pi-\pi)=c^2\sin x$$ for all $x$. Then from $\sin(x+2\pi)=c\sin(x+\pi)=c^2\sin x=\sin x$, we see that $\sin$ is a periodic continuos function, hence bounded.
(Admittedly, this cannot be expanded to $e^{-x^2}$ in any way)