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**Chapter 2 / Re: LEC5101 Question**

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**Today**at 10:35:57 AM »

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

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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 should read an announcement: you should not do anything, or rather as much as you need to ensure that in the real Quiz 1 (not in the Mock Quiz) you do not do anything stupid.

Also you should read description of this board and not to post here unrelated questions.

Also you should read description of this board and not to post here unrelated questions.

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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.

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Then you need to explain more or less rigorously, what is $\delta$-distribution and how Fourier transform is defined for it. Yes, online textbook covers this topic and much more than your physics class but in more advanced chapters, we do not cover