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Post by mondo on Aug 19, 2023 21:11:11 GMT
Vasco, by a more "intuitive" approach do you mean that author recognizes that the solution is a superposition of two counter-rotating motions around a unit circle?
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Post by Admin on Aug 19, 2023 21:19:46 GMT
Mondo
Yes, it's easy to see that $z=e^{\pm it}$ are solutions of $\ddot{z}=-z$ and then move on from there.
Vasco
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Post by mondo on Aug 19, 2023 21:23:27 GMT
Mondo Yes, it's easy to see that $z=e^{\pm it}$ are solutions of $\ddot{z}=-z$ and then move on from there. Vasco But how do you see this is a solution? Author first gives the solution and then comments on it - "These represents.." |
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Post by Admin on Aug 19, 2023 22:06:00 GMT
Mondo
Let $z=e^{\pm it}$ so $\dot{z}=\pm ie^{\pm it}=\pm iz$.
So $\ddot{z}=\pm i\dot{z}=\pm i\cdot\pm iz=i^2z=-z$
Vasco
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Post by mondo on Aug 19, 2023 22:34:26 GMT
I sort of feel this but also, this solution assumes unity radius and only later, probably knowing that the motion is elliptical, author generalizes it by multiplying each term by either $p$ or $q$.
The solution as it stands equals to $2\cos(t)$ - a real number
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Post by Admin on Aug 19, 2023 22:56:46 GMT
Mondo
But we are not adding them together to get $2\cos t$ unless $p=q=1$.
It's also clear that $pe^{it}$ is a solution and $qe^{-it}$ is a solution and then we can add them together to get the general solution.
Try it with Desmos.
Vasco
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Post by mondo on Aug 20, 2023 4:51:04 GMT
Mondo Let $z=e^{\pm it}$ so $\dot{z}=\pm ie^{\pm it}=\pm iz$. So $\ddot{z}=\pm i\dot{z}=\pm i\cdot\pm iz=i^2z=-z$ Vasco I am not sure if we can say $z=e^{\pm it}$ because $e^{-it}$ is $\frac{1}{z}$ not $z$ right? |
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Post by Admin on Aug 20, 2023 7:02:47 GMT
Mondo All we are saying is that $z=e^{it}$ is a solution of the differential equation $\ddot{z}=-z$ and so is $z=e^{-it}$. But then we also notice that $z=pe^{it}$ is a solution and $z=qe^{-it}$ is also a solution (where $p$ and $q$ are constants), and therefore so is the sum of these two solutions, $pe^{it}+qe^{-it}$. Another way of doing this is to say let's try $z=\alpha e^{mt}$ and so then we get $\dot{z}=\alpha me^{mt}$ and $\ddot{z}=\alpha m^2e^{mt}=$ -$z=$ -$\alpha e^{mt}$. We can write this as $\alpha e^{mt}(m^2+1)=0$ and so $m=\pm i$. In the theory of differential equations $m^2+1=0$ is called the auxiliary equation. You may find it helpful to play around with this graph www.desmos.com/calculator/eotvx8gawkVasco |
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Post by mondo on Aug 20, 2023 22:27:16 GMT
Thank you Vasco, I would say to solve this differential equation requires a bit of experience. For instance one has to be aware of the fact that $e^{-it}$ is the same as $e^{it}$ only rotating in the opposite direction.
As for your another way of doing it -> $\alpha m^2e^{mt}=z=\alpha e^{mt}$. This seems to be more obfuscated, like $m^2$ vanished like if it was equal to $\pm 1$.
Do you happen to have an algebraic solution of this equation? Not that it is super important to understand rest of the chapter but just out of curiosity
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Post by Admin on Aug 21, 2023 5:55:37 GMT
Mondo
There were two typos in line 5 of my reply #37. I had missed out two minus signs. I have corrected it and inserted them in red.
If you want to see an algebraic solution I suggest you look at a book dedicated to solving differential equations such as Differential Equations by Piaggio. But I think the non-rigorous but geometric explanation in the book is much more enlightening and is also the way these equations were initially solved and still are. If you look at an explanation using calculus and integration you will probably end up confused.
The idea of going round two circles in opposite directions was introduced in the Needham book in chapter 2 on pages 88-90.
Vasco
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Post by mondo on Aug 21, 2023 7:37:27 GMT
Thank you Vasco, I understand this material much better now. I also did a mistake in my reply #36 by saying:
$e^{-it} = \overline{z}$ and not $1/z$ as I said.
However, at the very end of this chapter (page 256, paragraph 3) author says, "Writing $x = \frac{1}{x}(z+\overline{z})$" where does it come from?
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Post by Admin on Aug 21, 2023 8:10:18 GMT
Mondo
Did you mean $\frac{1}{2}$ rather than $\frac{1}{x}$?
If $z=x+iy$ then $\overline{z}=x-iy$.
So $z+\overline{z}=2x$ and so $x=\frac{1}{2}(z+\overline{z})$
Vasco
PS what you said in reply #36 is correct for $z$ on the unit circle where $\overline{z}=1/z$.
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Post by mondo on Aug 21, 2023 20:37:25 GMT
PS what you said in reply #36 is correct for $z$ on the unit circle where $\overline{z}=1/z$. Yes, thanks for spotting that. Did you mean $\frac{1}{2}$ rather than $\frac{1}{x}$? Right, sorry for a typo. If $z=x+iy$ then $\overline{z}=x-iy$. So $z+\overline{z}=2x$ and so $x=\frac{1}{2}(z+\overline{z})$ Yes and then for $y$ we do a subtraction instead, however I wonder why is this legit? We applied two different operation, mutually exclusive, meaning we can't get $x$ and $y$ by the same algebraic manipulation or a chain of manipulations. Instead, we do either one or the other to clear the other variable. Also, in the foreword to this method, or rather in the note *9 on page 255 author says "The curve (K) must be "analytic" / smooth". However earlier in this chapter we defined $\overline{z}$ to not analytic due to the fact that different arrows emanating from the same point require different rotation. So the trick of the curve $K$ lies in the fact that here we translate only one vector from each point, and hence there is only one $\phi$ angle of rotation. Is my understanding correct? |
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Post by Admin on Aug 21, 2023 20:47:22 GMT
Mondo
To answer your first point: this is just algebra. To see it geometrically draw a graph of a point $z$ and then also the point $\overline{z}$ and you can see it immediately. This is shown on page 7 of chapter 1 right at the beginning of the book.
Vasco
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Post by mondo on Aug 23, 2023 8:21:08 GMT
Of course, yes. It is easy to see these relations visually. When I only looked at it algebraically I of course got the answer but I was like "hmm but we can't just apply random operations to get formulas".
Thank you Vasco, I was a bit busy in last two days but tomorrow I plan to update my notes and get back to chapter 12. Thanks again!
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