### Author Topic: Q2 TUT 5201  (Read 3162 times)

#### Victor Ivrii ##### Q2 TUT 5201
« on: October 05, 2018, 06:15:41 PM »
Show that each of the following series converges for all $z$:
\begin{align*}
&\sum_{n=0}^\infty \frac{z^n}{n!}, && \sum_{n=0}^\infty (-1)^n \frac{z^{2n}}{(2n)!}.
\end{align*}

#### Ge Shi

• Jr. Member
•  • Posts: 8
• Karma: 3 ##### Re: Q2 TUT 5201
« Reply #1 on: October 06, 2018, 01:51:09 AM »
(1)
Apply ratio test:
|[Z^n+1 / (n+1)!] / [Z^n / n!]|=|Z| / n+1
Limit |Z| / n+1= 0 < 1 as n approaches infinity
thus it converges for all z

(2)
Apply ratio test:
|[(-1)^n+1*Z^2(n+1) / (2n+1)!]/ [(-1)^n*Z^2n / (2n)!]| = |Z^2| / 2n+1
Limit |Z^2| / 2n+1 = 0 < 1 as n approaches infinity.
thus it converges for all z

Beyond readability (and sanity)
« Last Edit: October 06, 2018, 05:52:30 AM by Victor Ivrii »

#### Jeffery Mcbride

• Full Member
•   • Posts: 24
• Karma: 19 ##### Re: Q2 TUT 5201
« Reply #2 on: October 08, 2018, 04:33:05 PM »
Every power series has a convergence radius R, where Sum[anxn] converges if |x| < R.

The first summation is the power series equal to ez

and we have an = 1/n!

lim |an+1| / |an| = 1 / (n + 1)

= 0. So our convergence radius R is infinity and the power series converges for all z.

The second summation is the power series equal to cos(z) and we know:

cos(z) = (1/2)*(eiz + e-iz)

From the first series, we know the ez is convergent on all z, so cos(z) is also convergent on all z.

#### hanyu Qi

• Jr. Member
•  • Posts: 12
• Karma: 4 ##### Re: Q2 TUT 5201
« Reply #3 on: October 08, 2018, 09:26:37 PM »
In attachment.

#### Victor Ivrii ##### Re: Q2 TUT 5201
« Reply #4 on: October 08, 2018, 09:58:59 PM »
Jeff, learn a bit of LaTeX, since without it anything but the most simple math expressions will be out of your reach
http://forum.math.toronto.edu/index.php?topic=610.0
Alex, learn how to scan properly