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Messages - Victor Ivrii

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Chapter 2 / Re: Lec5101 Question Mock Quiz 1
« on: September 20, 2020, 03:14:04 PM »
Read W2L1 handout. THere is an explanation

Chapter 2 / Re: section 2.1 practice problem 15
« on: September 20, 2020, 03:12:51 PM »
You need to solve equation and initial condition, you'll see that $(y+\frac{1}{2})^2 =x^2-\frac{1}{4}\implies $ solution exist only as $x\ge \frac{\sqrt{15}}{2}$. Point $x=\frac{\sqrt{15}}{2}$ is excluded as $y'=\infty$ there

Chapter 1 / Re: Past quiz 1
« on: September 20, 2020, 03:02:23 PM »

Chapter 2 / Re: LEC5101 Question
« on: September 19, 2020, 10:35:57 AM »
I believe question was not about Linear Equations but more general ones

Chapter 2 / Re: LEC5101 Question
« on: September 19, 2020, 07:21:13 AM »
You need to wait lectures ; there will be shown some recepies  to find integrating factors, but there is no general algorithm

Chapter 1 / Re: Section 1.2 questions - using the definition?
« on: September 18, 2020, 07:57:45 PM »
You need to derive it. Even the following argument would be insufficient for a full mark:

We know that it will be a circle, we know that its center will be on the real axis (left from $z=0$), so find two points of intersection of this circle with $x$-axis:
$(z_1-4)= -4z_1$ and $(z_2-4)= 4z_1$. Finding $z_{1,2}$ (even if you do it) and setting $z_0=(z_2+z_1)/2$, $R- |z_2-z_1|/2$ ...

Chapter 2 / Re: LEC 0201 Question
« on: September 18, 2020, 07:51:52 PM »
We have initial condition at point $xt_0=0$, and solution blows up as $t=-\l(2)$. While we can consider solution, given by this formula as $t<-ln(2)$ it is disconnected from intial condition. Our solution blew up as $t=-\ln (2)$.

Chapter 1 / Re: More on inversion
« on: September 17, 2020, 01:58:27 PM »
Note that our inversion differs from geometric one $\vec{z}\to \frac{\vec{x}}{|\vec{x}|^2}$. It includes a mirror-reflection. See handout (I made a picture based on your example, corrected. Thanks a lot)

Chapter 1 / Re: Question about Inversion
« on: September 17, 2020, 01:27:45 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
 $z\to w=z^{-1}$, 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)

First of all only circle, passing through origin becomes a straight line, and this straight line is vertical only if the center of this circle, passing through origin, is on $x$-axis.

Inversion is self-inverse means that the double inversion is identical (so, operation, inverse to inversion, is inversion itself). Mirror-reflections have the same property as well.

Chapter 2 / Re: Lecture 0201 question
« on: September 17, 2020, 12:34:30 PM »
We use $x$ and $y$ because we can exclude $t$ but neither $x$ nor $y$

Chapter 2 / Re: Boyce-DiPrima Section 2.1 Example 1
« on: September 17, 2020, 12:33:02 PM »
I think that the textbook just uses the product rule: [f*g]=f'*g+g'*f. Here, you can think f as 4+t^2 and g as y. Hope this helps.

Please, never use * as multiplication sign (which even is not needed here). Your formulas are correctly formatted, but if you surround each by dollar sign like

Code: [Select]
I think that the textbook just uses the product rule: $[f*g]=f'*g+g'*f$. Here, you can think $f$ as $4+t^2$ and $g$ as $y$. Hope this'll get
I think that the textbook just uses the product rule: $[f*g]=f'*g+g'*f$. Here, you can think $f$ as $4+t^2$ and $g$ as $y$. Hope this helps.

Again, * is reserved for a different operation, and it usage as multiplication sign may be considered as a mathematical error

Chapter 2 / Re: Textbook Section2.1 Example5
« on: September 17, 2020, 12:28:15 PM »
You can select any lower limit you wish, the difference goes to the constant. However, as Ella correctly observed, it makes sense to select $t=0$ since the $t_0=0$ in the initial problem

Chapter 2 / Re: Lecture 0201 question
« on: September 16, 2020, 03:47:54 AM »
You can investigate $f(x)$ as in Calculus I.

Also because $x=x(t)$ and $y=y(t)$ and $x=a,y=b$ is a constant solution (equilibrium). Look at the picture on the next slide. We excluded $t$ from our analysis but it does not mean that it had gone

Chapter 2 / Re: Lecture 0201 question
« on: September 15, 2020, 01:11:04 PM »
Indeed, there should be no $-$ on the right, unless I change $b-y$ to $y-b$ (which I intended to to but doid not). I updated handout, it is just a single place as on the next frame everything is right.

Chapter 6 / Re: Laplace Equation Section 6.4
« on: March 21, 2020, 09:55:28 AM »
It means that $\Theta $ is $2\pi$-periodic.

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