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### Messages - Victor Ivrii

Pages: 1 ... 3 4 [5] 6 7 ... 154
61
##### Chapter 2 / Re: Textbook 2.4, Example 2
« on: September 24, 2020, 01:36:34 AM »
You should have it in Calculus II

62
##### Chapter 1 / Re: Power of Complex Numbers with Arguments Hard to Determine Directly
« on: September 24, 2020, 01:34:58 AM »
In this case usual cube of the sum would be the most efficient solution

63
##### Chapter 1 / Re: Section 1.3 Q15
« on: September 24, 2020, 01:33:32 AM »
Simple observation is enough

64
##### Chapter 1 / Re: Problems to 1.2 Q20
« on: September 23, 2020, 07:58:36 AM »
You ca refer to the fact that straight lines in $\mathbb{C}$ are also straight lines in $\mathbb{R}^2$ and conversely

65
##### Chapter 2 / Re: Mock Quiz Answer
« on: September 21, 2020, 12:28:01 PM »
It was not my intention to to check any math, so there could be that it is impossible recovering $y$ as a function of $x$. My purpose was to let you practice techical aspects

66
##### Chapter 2 / Re: Lec 0101 - 9/15 Question
« on: September 20, 2020, 07:49:47 PM »
One should remember that plugging $y=uy_1$ into inhimogeneous equation leaves $u'y_1$ in the left-hand expression. If you do not remember this, therefore you just do not understand the method of variations and you should reread previous slides

67
##### Chapter 1 / Re: Solving roots of complex numbers
« on: September 20, 2020, 07:46:18 PM »
The worst thing you can do is to use calculator to evaluate the value of, say, $\sin (4\pi/9)$ and $\cos (4\pi/9)$ numerically. But it may be useful to mention that
$\cos (4\pi/9)+i\sin (4\pi/9)$ belongs to the first quadrant and pretty close to $i$. Just draw a little picture.

68
##### Chapter 2 / Re: Lec5101 Question Mock Quiz 1
« on: September 20, 2020, 03:14:04 PM »
Read W2L1 handout. THere is an explanation

69
##### 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

70
##### Chapter 1 / Re: Past quiz 1
« on: September 20, 2020, 03:02:23 PM »
half-line

71
##### 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

72
##### 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

73
##### 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 $z_0=(z_2+z_1)/2$, $R- |z_2-z_1|/2$ ...

74
##### 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 $t=-\ln (2)$.

75
##### 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)

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