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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

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Read W2L1 handout. THere is an explanation

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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

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I believe question was not about Linear Equations but more general ones

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You need to wait lectures ; there will be shown some recepies to find integrating factors, but there is no general algorithm

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

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

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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 $t=-\ln (2)$.

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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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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.

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We use $x$ and $y$ because we can exclude $t$ but neither $x$ nor $y$

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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,

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 helps.`

you'll getI 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

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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

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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

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

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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.

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It means that $\Theta $ is $2\pi$-periodic.