Toronto Math Forum

MAT334-2018F => MAT334--Tests => Quiz-2 => Topic started by: Victor Ivrii on October 05, 2018, 06:15:41 PM

Title: Q2 TUT 5201
Post by: Victor Ivrii 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*}

Title: Re: Q2 TUT 5201
Post by: Ge Shi 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)

Title: Re: Q2 TUT 5201
Post by: Jeffery Mcbride on October 08, 2018, 04:33:05 PM

Every power series has a convergence radius R, where Sum[a_nxⁿ] converges if |x| < R.

The first summation is the power series equal to e^z

and we have a_n = 1/n!

lim |a_n+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)*(e^iz + e^-iz)

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

Title: Re: Q2 TUT 5201
Post by: hanyu Qi on October 08, 2018, 09:26:37 PM

In attachment.

Title: Re: Q2 TUT 5201
Post by: Victor Ivrii 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 (http://forum.math.toronto.edu/index.php?topic=610.0)
Alex, learn how to scan properly