Show that $\sin x$ cannot be polynomial using the concept of boundedness Use the concept of boundedness of a function to show that the functions $\sin x$ and $\cos x$ cannot be polynomials.
This is an exercise in my textbook from the section discussing the Weierstrass M-test.
My thought, since it is from the section discussing Weierstrass M-test, is to express $sinx$ using Taylor expansion. We can then show that the series $\displaystyle\sum_{j=1}^{\infty}{f_j}$ converges uniformly; I don't know what to do next with the series being uniformly convergent.
Any idea would be appreciated.
 A: Recall the Taylor series for the cosine function:
$$\cos x = \sum_{k=0}^\infty(-1)^n \frac{x^{2k}}{2k!}. $$
For $L>0$ we have
$$\left|(-1)^k \frac{x^{2k}}{2k!}\right|\leqslant \frac{L^{2k}}{2k!}=:M_k.  $$
Since
$$\sum_{k=0}^\infty M_k =\sum_{k=0}^\infty \frac{L^{2k}}{(2k)!}\leqslant \sum_{k=0}^\infty \frac{L^k}{k!}=e^L<\infty, $$
it follows that $\cos$ converges uniformly on $[-L,L]$. Taking $L=\pi$ is sufficient to show that $\cos$ converges uniformly on $\mathbb R$, since $\cos$ is periodic ($\cos(x+2\pi)=\cos x$). Since $\cos\in\mathcal C^\infty([-\pi,\pi])$, it is continuous, and is therefore bounded (in particular, $|\cos x|\leqslant 1$ for all $x$).
Let $f(x)=\sum_{k=0}^n a_kx^k$ be a polynomial with $n\geqslant 1$; assume without loss of generality that $a_n>0$. Then $$f(x) = x^n\sum_{k=0}^n a_k x^{n-k} $$
and
$$\lim_{x\to\infty}\sum_{k=0}^n a_k x^{n-k} = a_n, $$
so as $x^n\stackrel{n\to\infty}\longrightarrow\infty$, we see that $$\lim_{n\to\infty}f(x)=\infty,$$
so that $f$ is unbounded. Hence, $\cos$ cannot be a polynomial.
A: I can't guess what the author's intended solution is, but my first thought was that since $\sin(x)$ is bounded, and since any non-constant polynomial is unbounded, $\sin(x)$ cannot be written as a polynomial.
You are left with proving that any non-constant polynomial is unbounded.
A: Assume on the contrary that $\sin x=a_{n}x^{n}+...+a_{0}$ is a polynomial.  ...(1)
Since $\sin x \leq 1$ for all real $x$, thus we must have $\sin x=a_{0}$ since the RHS in (1) is greater than 1 for any real $x>1$.
This means that $\sin x$ is a constant for all $x$, absurd.  
