I am trying to find the radius of convergence for the following function
$$ f(x)=\sin(\pi x/4)$$
I already found the Maclaurin series of the function and applied the ratio test but seems I cant get the radius of convergence right. I find the radius
$R = 4/\pi$
Best Answer
You have that the derivatives around $x=0$ of $y=\sin x$ are
$$\{ y^{(n)}(0)\}=\{0,1,0,-1 ,\dots \}$$
Since you have a multiplicative factor of $\dfrac{\pi }{4}$ this changes to
$$\{ y^{(n)}(0)\}=\left\{0,\dfrac{\pi }{4},0,-\dfrac{\pi^3 }{4^3} ,\dots \right\}$$
As a consequence you have that the coefficients are
$$c_n=\frac{(-1)^n}{(2n+1)!} \left(\frac{\pi}{4}\right)^{2n+1}$$
You can readily check that
$$\lim \frac{c_{n+1}}{c_{n}}=0$$
from which the radius is $\infty$, i.e., the whole extended real line.
As a general result, the convergence radius of the series for
$$y=\sin(ax+b)$$
is the whole real line.