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This question is off an exam I took. Naturally I set out to first to create a Maclaurin power series.

$f(x) = \frac{1}{\cos(x)}$ $\qquad \quad \quad \quad \quad f(0)= 1$

$f'(x)= \frac{\sin(x)}{\cos^2(x)}$ $\quad \quad \quad \quad \quad f'(0)= 0$

$f''(x) = 2\frac{\sin^2(x)}{\cos^3(x)} +\frac{1}{\cos(x)}$ $\quad f''(0)= 1$

Power series is then:

$$1 + \frac{x^2}{2!} + \frac{x^4}{4!}+...=\sum \frac{x^{2n}}{(2n)!}$$

I performed the ratio test on this power series and found that it converges everywhere. However, my professor wrote that:

The function $1/\cos x$ does not have a value at $x = \pi/2$; therefore the inside of the circle or interval of convergence cannot include $x = \pi/2.$ ” This point is a “singularity”, and no other singularity lies closer to the origin.

This should be taken very seriously.*

What is this singularity he is referring to and why does it make my ratio test answer incorrect?

Thomas Andrews
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stack ex
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1 Answers1

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The pattern you noticed in the derivatives does not continue at the 3rd derivative. The series for $ \frac{1}{\cos x} = \sec x $ is

$$ \sum_{n=0}^\infty \frac{(-1)^n x^{2n} E_{2n}}{(2n)!} $$

where $E_n$ is the $n$-th Euler number. The radius of convergence is $(-\frac{\pi}{2}, \frac{\pi}{2})$.

Your professor's note is probably meant to be interpreted as, "The fact that $f$ is undefined at $x=\frac{\pi}{2}$, but your series is should be a hint that you have the wrong series," or as @MiloBrandt pointed out below "the radius of convergence is the the distance to the nearest singularity in the complex plane, which is at $\frac{\pi}{2}$."

Dando18
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  • The professor's hint might also be noting radius of convergence equals the distance to the nearest singularity in the complex plane - that gives a much easier method for finding the radius of convergence. (At least, if this was in the context of complex analysis; it's not so clear from the question though) – Milo Brandt May 11 '17 at 01:30
  • @MiloBrandt yeah that makes more sense. For some reason I interpreted this as a Calc 2 question and responded as such. – Dando18 May 11 '17 at 01:32