### Show Posts

This section allows you to view all posts made by this member. Note that you can only see posts made in areas you currently have access to.

### Messages - Ge Shi

Pages: [1]
1
##### Quiz-2 / Re: Q2 TUT 5201
« 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)

2
##### Quiz-1 / Re: Q1: TUT 0203
« on: September 28, 2018, 05:11:53 PM »
Since the circle through 0, 2+2i, 2-2i,
It means that the circle through (0,0), (2,2) and (2,-2)
thus, the equation of this circle in complex form is  |z-2|=2

3
##### Quiz-6 / Re: Q6--T0401
« on: March 17, 2018, 01:50:00 PM »
(a)

4
##### Quiz-6 / Re: Q6--T0201
« on: March 17, 2018, 12:11:47 AM »
(a)
In the attachement

(b)
When t approaches to infinity, the solution is approaches to zero

Since $\lambda_1=-3$ , $\lambda_2=-1$
Eigenvalues are real but unequal and have the same sign, x=0 is a node and asymptotically stable.

5
##### Quiz-6 / Re: Q6--T0401
« on: March 16, 2018, 11:48:09 PM »
(a)
https://imgur.com/a/W9njS

(b)

When t approaches to infinity:
if C2 is not equal to zero ,the solution is unbounded.
if C2 is equal to zero, the solution approaches to zero.

Since $\lambda_1=-1$ , $\lambda_2=2$
Eigenvalues are real but unequal and have the opposite signs, x=0 is a saddle point and unstable.
I've attached the graph.

Pages: [1]