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Messages - RunboZhang

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Chapter 2 / LEC 0201 Question
« on: September 18, 2020, 02:20:45 PM »
Hi, we just had a lecture on homogenous ODE this morning, and when I went through the slides afterwards, I have found a confusion in the slide P6(pic is also attached below). I am wondering why the domain is strictly greater than minus root two? Why can't it be smaller than minus root two? Is it because the initial condition of this question is greater than 0 so that our answer needs to be greater than 0? (But why?)

Chapter 1 / Re: Section 1.2 questions - using the definition?
« on: September 18, 2020, 12:40:03 PM »
Hi, below this comment I have attached my answer to the question. Basically we can use either of these two methods to solve for this kind of problem. So the answer for your first question is yes. For your second question, I think a general solution to this one should definitely include the locus of z, which is the function of the circle. It can be either in the form of
Code: [Select]
(x-x_0)^2 +(y-y_0)^2 = R^2 or in the form of
Code: [Select]

Chapter 2 / Textbook Section2.1 Example5
« on: September 16, 2020, 09:26:38 PM »
Hi guys, I have been stuck at example 5 of section2.1(page30) for quite a while. In particular, I do not understand why the lower bound of the integral is the initial point t=0. Why can't it be the upper bound?
(example screenshot is attached below)

Chapter 2 / Re: Boyce-DiPrima Section 2.1 Example 1
« on: September 16, 2020, 09:21:19 PM »
I think Suheng Yao has pointed out a crucial point: product rule. Indeed, if you observe LHS=(4+t^2)(dy/dt) + 2ty, you may find the pattern of u'v+uv' where u=y and v=4+t^2. By applying product rule here, we can make LHS become a single expression and thus make it a separable differential equation (which is already discussed in the lecture).

I think a general method of solving this kind of problem is introduced later in the textbook and it is quite useful in solving inseparable first order differential equation.

Chapter 1 / Question about Inversion
« on: September 16, 2020, 04:11:52 PM »
Hi guys, we talked about inversion during the previous lecture and I am a bit confused by the last slide. So by definition, we have
 z-> w=z^-1, and then we can calculate the inversion of any point either by its inverse or its polar form. We have also proved that the inversion of a circle is a vertical straight line in the same slide. But I am a bit confused by the red highlighted part, "inversion is self-inverse". My thought is that an inversion of a circle is a straight line and correspondingly the inversion of that straight line is the original circle. And this property thus makes it self-inverse. I don't know if my understanding is correct so I am writing this post to look for some help. And one more brief question, do we have any restriction on inversion/self-inversion?

(btw slide pic is attached below)

Chapter 1 / Re: More on inversion
« on: September 16, 2020, 03:46:45 PM »
Hi Zhekai, thank you for your sharing. I have two questions regarding your notes. Firstly, how did you derive A prime and B prime? Did you convert A and B in polar form and calculate their inverse, and then represent on the graph? Secondly, how did you know that z*zbar = r^2?

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