2
$\begingroup$

Show that the power series of $\cos(x)$ converges uniformly to $\cos(x)$ on every bounded interval.

My attempt: The power series for $\cos(x)$ is $$\sum_{n=0}^{\infty} \frac{{(-1)^{n}}{x^{2n}}}{(2n)!}$$ A sequence of functions $f_n: X \rightarrow Y$ converges uniformly if there exists a function $f: X \rightarrow Y$ such that for every $\epsilon > 0$ there is an $N$ such that $d_Y(f_n(x),f(x)) < \epsilon$ for every $ x \in X$ and every $ n \geq N$. In our problem, $$f_n = \sum_{j=0}^{n} \frac{{(-1)^{j}}{x^{2j}}}{(2j)!}, \qquad f = \cos(x).$$ Using the ratio test I got the radius of convergence is equal to infinity. Can I then conclude that the power series converges uniformly to $\cos(x)$ on every bounded interval because the radius of convergence is $(-\infty,\infty)$?

$\endgroup$
1
  • 3
    $\begingroup$ That depends on what you know about power series. If you know a power series converges compactly on its (open) disk of convergence, you're done. If not, you need to prove it at least in the special case of $\cos$. $\endgroup$ – Daniel Fischer Mar 16 '14 at 15:00
1
$\begingroup$

Hints:

  • For every $x$ such that $|x|\leqslant K$ and every $N\geqslant1$, $$\left|\sum_{n\geqslant N} \frac{(-1)^nx^{2n}}{(2n)!}\right|\leqslant\sum_{n\geqslant N}\frac{K^{2n}}{(2n)!}\leqslant\frac{K^{2N}}{(2N)!}\sum_{i\geqslant0} \frac{K^{2i}}{(2N)^{2i}}.$$
  • For every $N\geqslant2K$, $$ \frac{K^{2N}}{(2N)!}\leqslant\left(\frac{\mathrm e}4\right)^{2N},\qquad\sum_{i\geqslant0} \frac{K^{2i}}{(2N)^{2i}}\leqslant2.$$
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.