Is it possible to create a cubic function that coincides with a trigonometric function in the interval between two local extremums?

Question

Is it possible to create a cubic function with the roots $a$, $b$ and $c$, $f(x) = K(x-a)(x-b)(x-c)$, that coincides with a sine or cosine function, $g(x)$, in the interval $d \le x \le e$, $g'(d) = g'(e) = 0$?

Answer 1

It is not possible, because $f^{(4)}\equiv0$, and $g^{(4)}\not\equiv0$.

Answer 2

It's not possible (assuming $d\neq e$). I am not sure what the quickest way to prove this is, but using complex analysis you can just say the following: if this is true, then $f-g$ is a function that can be extended to all of $\Bbb{C}$ which is $0$ on all of $[d,e]$. Since $[d,e]$ has a limit point, this impies $f-g$ is constantly zero, i.e. $f=g$. But this can't be true (for instance because $g$ has infinitely many zeroes while $f$ is only zero at $a$, $b$ and $c$).

If you don't know/don't like complex analysis though, there's probably some other way to see this.

Answer 3

It is not possible for a cubic function to coincide perfectly with a trigonometric function over any interval. If $f(x)$ and $t(x)$ were such cubic and trig functions, then their difference $g(x)=f(x)-t(x)$, would be an entire function that is identically zero on an interval. Therefore, by the identity principle from complex analysis, we would have $g(x)=0$ everywhere.

That said, we can get a cubic function that is pretty close to a trigonometric function. One standard way to do that is by writing the $3$rd degree Taylor polynomial approximating the trig function at a point in the interval of interest.

Answer 4

No. One argument could rely on the algebraic independence of various values of the trigonometric functions. But it's probably easier to note that for $g(x)=\cos{(\alpha x+\beta)}$ with $\alpha \neq 0$ (notice this includes sine functions too), $$ \frac{g''(x)}{g(x)} = -\alpha^2 $$ is independent of $x$ and not zero (apart from finitely many removable singularities), whereas for any cubic $f(x)=ax^3+bx^2+cx+d$ with $a\neq 0$, $$ \frac{f''(x)}{f(x)} = \frac{6ax+2b}{ax^3+bx^2+cx+d} $$ is not constant.