Question

In Problem 21, what do you think is the interval of convergence for the Maclaurin series of $\mathit{s}\mathit{e}\mathit{c}\mathbf{}\mathit{x}$?

Short answer

The interval of convergence, around $x=0$ is$\left(\frac{-\pi }{2},\frac{\pi }{2}\right)$

Step-by-step solution

Given Information

The given value

$secx$

To obtain the Macluarin series

We found the Maclaurin Series, centered at$x=0$of the sec$x$by using the Trigonometric Property

$\sec x=\frac{1}{\cos x}$

Our aim is to find the interval of convergence of the Maclaurin Series of the sec function.

We do that, by equating the denominator by zero, yields

$\cos x=0$

$x=\frac{2n+1}{2}\pi$

Where$n=\dots ,-1,0,1,\dots$

Around $x=0$

The radius of convergence is$\frac{\pi }{2}$and the interval of convergence is$\left(\frac{-\pi }{2},\frac{\pi }{2}\right).$From the graph below, you can deduce that the sec function diverges at$x=\frac{2n+1}{2}\pi .$

The interval of convergence,$x=0$around is $\left(\frac{-\pi }{2},\frac{\pi }{2}\right)$.