0
$\begingroup$

Given $$ f(x) = \frac{1}{x^2+\cos^2(x)} $$ We can use the binomial expansion to rewrite $f(x)$ as $$ f(x) = \sum_{k=0}^\infty {-1\choose{k}}\frac{\cos^{2k}x}{x^{2k+2}} $$ Or, in other terms $$ f(x) = \sum_{k=0}^\infty \frac{(-1)^k \cos^{2k}x}{x^{2k+2}} $$ If I remember correctly the formula for R.O.C is: $$ (x+y)^n \Longrightarrow \, -1<\frac{x}{y} < 1 $$ How would I go about solving this for: $$ -x^2 < \cos^2x < x^2 $$ Thanks.

$\endgroup$
1
  • 4
    $\begingroup$ ROC is for power series and your series is not a power series. $\endgroup$ Commented Feb 15, 2021 at 8:43

1 Answer 1

1
$\begingroup$

The term radius of convergence is reserved for power series, such as $\sum_n a_nx^n$. Your series does not fit this format. However, it is a geometric series in $\cos^2 x/x^2$. It will converge if and only if $$ \left\lvert \frac {\cos^2 x}{x^2} \right \rvert < 1.$$ At a minimum this will be the case whenever $|x| > 1$ since (for real values of $x$) $|\cos x| < 1$. But in fact, still concentrating on real $x$, for $0 < x < 1$, the ratio $\cos^2x/x^2$ is decreasing and greater than $1$ for small $x$. Therefore the inequality is true for all larger $x$ once $x > \theta$ where $\theta$ is the solution to $\cos \theta = \theta$, about $0.7$. Because each term is an even function, the sign of $x$ is unimportant and convergence of your sum arises for all $|x|> \theta$, which is the solution to the inequality you have written.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .