The following is a problem from Ch. 15 of Spivak's Calculus
- (a) After all the work involved in the definition of $\sin$, it would be disconcerting to find that $\sin$ is actually a rational function. Prove that it isn't. (There is a simple property of $\sin$ which a rational function cannot possibly have.).
(b) Prove that $\sin$ isn't even defined implicitly by an algebraic equation. That is, there do not exist rational functions $f_0,...,f_{n-1}$ such that
$$(\sin x)^n+f_{n-1}(x)(\sin x)^{n-1}+...+f_0(x)=0,\text{ for all } > x\tag{1}$$
Hint: Prove that $f_0=0$, so that $\sin{x}$ can be factored out. The remaining factor is $0$ except perhaps at multiples of $\pi$. But this implies that it is $0$ for all $x$ (Why?) You are now set up for a proof by induction.
My question is precisely the one in bold above.
It has been asked before here, but the answer there is but a hint that is already contained in the solution manual.
Here is how the question arises in the context of a proof of part $(b)$.
$(a)$
A rational function has a finite number of roots (unless it is zero everywhere). $\sin$ has infinite roots $$\sin{k\pi}=0,k\in\mathbb{Z}$$
and $\sin$ is not $0$ everywhere. Thus $\sin$ is not a rational function.
$(b)$
$$x=k\pi\implies\sin(k\pi)=0$$
So the entire expression $(1)$ becomes just $f_0(k\pi)=0$ for all $k\in\mathbb{Z}$. But $f_0$ is a rational function so it must be $0$, otherwise it'd have infinite roots.
Then $(1)$ becomes
$$\sin{x}[ (\sin{x})^{n-1}+f_{n-1}(x)(\sin{x})^{n-2}+...+f_1(x) ]=0\tag{2}$$
Now $x\neq k\pi, k\in\mathbb{Z}\implies\sin{x}\neq 0$, so the expression in brackets must equal zero.
So here is the part that my question is about.
We have
$$(\sin{x})^{n-1}+f_{n-1}(x)(\sin{x})^{n-2}+...+f_1(x)=0\tag{3}$$
Here is what the solution manual says at this point
The term in brackets is continuous and $0$ except perhaps at multiples of $2\pi$, so it is $0$ everywhere.
This doesn't explain why.
$(3)$ is true for all $x\neq k\pi$, but possibly not at $x=k\pi$.
Rational functions can also be undefined at a finite number of points, but I believe this case is ruled out for any points because $(1)$ is true for all $x$, correct?
Ie, even at points $x=k\pi$, the lefthand expression in $(3)$ must at least be defined for $(2)$ and thus $(1)$ to be true, correct?
Therefore, the rational functions in question are continuous everywhere, $\sin$ and powers of $\sin$ are continuous everywhere, and so the expression in $(3)$ is continuous everywhere.
But if we assume that the lefthand expression in $(3)$ is anything other than $0$ at $k\pi$, then it is discontinuous at such points, which is a contradiction.
I think the general idea is on the right track, but I don't feel very comfortable with the details of this last part.
So, going back to original the problem statement: Why?
