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### Messages - Ge Shi

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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

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##### 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

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##### 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.

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##### 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.

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##### MAT244--Misc / Re: week 7 quiz?
« on: February 11, 2018, 04:07:34 PM »
No, since we have our midterm, we don't have a quiz next week but we still have the tutorial.
You can check the website
http://www.math.toronto.edu/courses/mat244h1/20181/homeassignments.html

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##### Web Bonus Problems / Re: Web bonus problem--Week 3
« on: January 22, 2018, 11:49:46 PM »