Sum and product of analytic functions that is not analytic

Question

The function

$$f(x) = \frac{2 + \cos x}{3} (2π - x) + \sin x$$

is the sum/product of analytic functions ($\cos(x)$,$\sin(x)$, linear), but all it's derivatives at $2\pi$ are $0$ ($f^n(2\pi)=0$).

I simply don't understand why.

Thanks,

Answer 1

Says who? $$f^{(5)}(2\pi ) = -2/3.$$

Answer 2

Set $x-2\pi=t$, so $x=t+2\pi$ and $\cos x=\cos t$, $\sin x=\sin t$. Thus you can consider $$ g(t)=-\frac{2}{3}t-\frac{1}{3}t\cos t+\sin t $$ and the Taylor coefficients at $0$ of $g$ are the same as the Taylor coefficients at $2\pi$ of $f$. We have \begin{align} g(t)&=-\frac{2}{3}t-\frac{1}{3}t\left(1-\frac{1}{2}t^2+\frac{1}{24}t^4+o(t^4)\right)+t-\frac{1}{6}t^3+\frac{1}{120}t^5+o(t^5) \\[6px] &=-\frac{2}{3}t-\frac{1}{3}t+\frac{1}{6}t^3-\frac{1}{72}t^5 +t-\frac{1}{6}t^3+\frac{1}{120}t^5+o(t^5) \\[6px] &=-\frac{1}{180}t^5+o(t^5) \end{align} contradicting your statement.

However, even if all derivatives are zero, there would be no problem: the function would be the constant zero function (in the whole circle of convergence of the Taylor series at $2\pi$).

Since the functions involved are entire, the circle of convergence is the whole line (if you think real) or the whole plane (if you think complex). Thus, in order to check that some derivative is nonzero at $2\pi$ you just have to check that $f(x)\ne0$ for some $x$. And $$ f(0)=\frac{2+\cos0}{3}(2\pi-0)+\sin 0=2\pi\ne0 $$