0
$\begingroup$

Find the Maclaurin series for $f(x) = e^{-2x}$, and find the interval of convergence for the series.

I got the maclaurin series to be this

$e^x = 1 - 2\frac{x}{1!} + 2^2\frac{x^2}{2!} + \cdots$

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

Using the ratio test to find the interval of convergence:

$$\lim_{n\to\infty} \left| \frac{(-1)^{n+1}2^{n+1}x^{n+1}}{(n+1)!} \frac{n!}{(-1)^n 2^n x^n} \right| = 2|x| \lim_{n\to\infty} \frac{1}{n+1} = 2|x|(0) < 1$$

Therefore for any value of x is in the interval of convergence. Is this right?

$\endgroup$
  • 1
    $\begingroup$ Yes, you're right $\endgroup$ – caverac Apr 19 '17 at 18:57
  • $\begingroup$ Do I need to check the endpoints at x = 0, or no? Im a bit confused when the ratio test gives $\infty$ or zero. If its $\infty$ instead of 0, so like $2|x|\infty > 1$ Can I say there is no point in the interval that is convergence? $\endgroup$ – Tinler Apr 19 '17 at 18:57
  • $\begingroup$ The limit is 0 regardless of the value of $x$, so no need to check anything else $\endgroup$ – caverac Apr 19 '17 at 18:58
  • $\begingroup$ There are no endpoints to check. The interval of convergence is $(-\infty,\infty)$. $\endgroup$ – John Wayland Bales Apr 19 '17 at 18:59
  • $\begingroup$ What happens if its $2|x|\infty > 1$ ?. As in instead of 0 i put infinity. $\endgroup$ – Tinler Apr 19 '17 at 19:05
0
$\begingroup$

The series representation for the exponential function converges everywhere (uniformly as well), as so any related sum will also converge everywhere.

$\endgroup$
0
$\begingroup$

This Maclaurin series is valid for $|x|<\infty$.

You know this because the series is valid over that domain. There are no points where the function "blows up".

By convention, the Maclauren series in centered about $x=0$. A power series allows three possibilities:

  1. Series converges only at $x=0$.
  2. Series converges on an interval centered at $x=0$.
  3. Series converges for all values of $x\in\mathbb{R}$.
$\endgroup$

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.