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Post by Admin on Oct 7, 2016 12:08:45 GMT
Gary
Here is my solution.
From figure 19b on page 295 we can see that since the defining property of the tractrix tells us that $PA=R=QB$ we can write $BQ+QP=R+QP$ and that as $Q$ coalesces with $P$ then $C$ coalesces with $A$, and so the distance $R+QP$ becomes equal to the distance $R+BC$ and we can write
$R+BC=R+PQ$ or $BC=PQ$. Done.
Vasco
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Post by Patrick on Jun 28, 2020 20:53:24 GMT
I've been trying to come up with a solution to this exercise and in my pursuit, I've seemed to come up with an (incorrect) argument that BC is not ultimately equal to PQ.
As the author states $PA = R = QB$. This is true for all Q and B (I mention these points since they change as Q coalesces with P). We also know that as Q coalesces with P the line QB ultimately becomes parallel with the line PA in the sense that the angle the line QB forms is ultimately equal to the angle PA forms (say with the x-axis). Combining these two facts made me think we should use the properties of a parallelogram.
It seems to me we should be able to say that PQBA is ultimately a parallelogram. If this was the case then PQ ultimately equals BA. Now, this doesn't necessarily mean BA is not equal to BC.
But we also know the triangle ABC is ultimately similar to triangle TAP. The angle TAP is constant and should ultimately be equal to the angle ABC. If we call the angle TAP theta then this implies that AB = BC cos(theta). Since theta is not zero or pi (it is the angle of a triangle) we have AB cannot equal BC.
I'm guessing I was incorrect to say PQBA ultimately has the properties of a parallelogram. I'm not sure why though. And I'm not sure how to prove it otherwise.
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Post by Patrick on Jun 28, 2020 20:57:40 GMT
I've been trying to come up with a solution to this exercise and in my pursuit, I've seemed to come up with an (incorrect) argument that $BC$ is not ultimately equal to $PQ$.
As the author states $PA = R = QB$. This is true for all $Q$ and $B$ (I mention these points since they all change as $Q$ coalesces with $P$). We also know that as $Q$ coalesces with $P$ the line $QB$ ultimately becomes parallel with the line $PA$ in the sense that the angle the line $QB$ forms is ultimately equal to the angle $PA$ forms (say with the x-axis). Combining these two facts made me think we should use the properties of a parallelogram.
It seems to me we should be able to say that $PQBA$ is ultimately a parallelogram. If this was the case then $PQ$ ultimately equals $BA$. Now, this doesn't necessarily mean $BA$ is not equal to $BC$.
But we also know the triangle $ABC$ is ultimately similar to triangle $TAP$. The angle $\angle TAP$ is constant and should ultimately be equal to the angle $ABC$. Let's define $\angle TAP = \theta$ then this implies that $AB = BC \cos(\theta)$. Since theta is not zero or pi (it is the angle of a triangle) we have $AB$ cannot equal $BC$.
I'm guessing I was incorrect to say $PQBA$ ultimately has the properties of a parallelogram. I'm not sure why though. And I'm not sure how to prove it otherwise. (Also sorry for rewriting this. I wasn't sure if we could use latex here)
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Post by Admin on Jun 28, 2020 21:27:41 GMT
Patrick
i will study what you say and get back to you. Yes you can use Latex.
Vasco
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Post by Admin on Jun 29, 2020 8:24:53 GMT
I've been trying to come up with a solution to this exercise and in my pursuit, I've seemed to come up with an (incorrect) argument that $BC$ is not ultimately equal to $PQ$. As the author states $PA = R = QB$. This is true for all $Q$ and $B$ (I mention these points since they all change as $Q$ coalesces with $P$). We also know that as $Q$ coalesces with $P$ the line $QB$ ultimately becomes parallel with the line $PA$ in the sense that the angle the line $QB$ forms is ultimately equal to the angle $PA$ forms (say with the x-axis). Combining these two facts made me think we should use the properties of a parallelogram. It seems to me we should be able to say that $PQBA$ is ultimately a parallelogram. If this was the case then $PQ$ ultimately equals $BA$. Now, this doesn't necessarily mean $BA$ is not equal to $BC$. But we also know the triangle $ABC$ is ultimately similar to triangle $TAP$. The angle $\angle TAP$ is constant and should ultimately be equal to the angle $ABC$. Let's define $\angle TAP = \theta$ then this implies that $AB = BC \cos(\theta)$. Since theta is not zero or pi (it is the angle of a triangle) we have $AB$ cannot equal $BC$. I'm guessing I was incorrect to say $PQBA$ ultimately has the properties of a parallelogram. I'm not sure why though. And I'm not sure how to prove it otherwise. (Also sorry for rewriting this. I wasn't sure if we could use latex here) Patrick I don't think $PQBA$ is ultimately a parallelogram because $PQ$ ultimately points in the direction of the tangent at $P$ and this is not in general parallel to the $y$-axis, which is the direction of $AB$. Also the $\angle TAP$ is not constant because it is the angle between the tangent and the $y$-axis and so it varies as we move along the curve. I hope this helps. Vasco |
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Post by Patrick on Jun 29, 2020 15:52:52 GMT
Gary
Thank you! This does help. And this website in general has been a tremendous help. Thank you again for responding so quickly.
I agree with $PQ$ not being parallel with the $y$-axis. That's a good point. I do believe $P$ is meant to be fixed in this example. I believe we are finding the curvature at $P$ and thus we can imagine it and it's tangent are fixed.
I don't believe the first post's proof works. We wish to show $BC$ is ultimately equal to $PQ$. The reason we want this sort of equality is that we want to say $\displaystyle{\left(\frac{AC}{BC}\right)}$ is ultimately equal to $\displaystyle{\left( \frac{AC}{BC} \right) \left( \frac{BC}{PQ} \right) = \frac{AC}{PQ}}$.
If I am following the above proof correctly we have both $BC$ and $PQ$ approach zero as $Q$ coalesces with $P$. So the claim of $BC = PQ$ is really saying $BC = PQ + error$. This is sadly not enough to prove that $BC$ and $PQ$ are ultimately equal as $\displaystyle{\frac{BC}{PQ} = 1 + \frac{error}{PQ}}$. Since both the error and $PQ$ are going toward zero it is not obvious what the ratio is converging too. If we apply this equation to $\displaystyle{\frac{AC}{BC}}$ we end up with a term that is not obviously is ultimately equal to $\displaystyle{\frac{AC}{PQ}}$.
I could be missing something here. Again, I wish to extend my gratitude for all the work down here!
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Gary
GaryVasco
Posts: 3,352 
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Post by Gary on Jun 29, 2020 16:09:16 GMT
Gary Thank you! This does help. And this website in general has been a tremendous help. Thank you again for responding so quickly. I agree with $PQ$ not being parallel with the $y$-axis. That's a good point. I do believe $P$ is meant to be fixed in this example. I believe we are finding the curvature at $P$ and thus we can imagine it and it's tangent are fixed. I don't believe the first post's proof works. We wish to show $BC$ is ultimately equal to $PQ$. The reason we want this sort of equality is that we want to say $\displaystyle{\left(\frac{AC}{BC}\right)}$ is ultimately equal to $\displaystyle{\left( \frac{AC}{BC} \right) \left( \frac{BC}{PQ} \right) = \frac{AC}{PQ}}$. If I am following the above proof correctly we have both $BC$ and $PQ$ approach zero as $Q$ coalesces with $P$. So the claim of $BC = PQ$ is really saying $BC = PQ + error$. This is sadly not enough to prove that $BC$ and $PQ$ are ultimately equal as $\displaystyle{\frac{BC}{PQ} = 1 + \frac{error}{PQ}}$. Since both the error and $PQ$ are going toward zero it is not obvious what the ratio is converging too. If we apply this equation to $\displaystyle{\frac{AC}{BC}}$ we end up with a term that is not obviously is ultimately equal to $\displaystyle{\frac{AC}{PQ}}$. I could be missing something here. Again, I wish to extend my gratitude for all the work down here! Patrick, The help was all from Vasco. I think he meant to address you rather than me in his first reply. But I agree, the site is a great help. Gary |
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Post by Admin on Jun 29, 2020 18:40:13 GMT
Patrick and Gary
The first post of this thread was posted 4 years ago in October 2016 and was posted by me (Vasco) and was addressed primarily to Gary at the time.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Jun 29, 2020 19:24:23 GMT
Patrick and Gary The first post of this thread was posted 4 years ago in October 2016 and was posted by me (Vasco) and was addressed primarily to Gary at the time. Vasco Yes, I should have looked at the date. |
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Post by Admin on Jun 30, 2020 10:47:06 GMT
Patrick and Gary
As a result of Patrick's comments I decided to study my answer to this suggested exercise because four years is a long time and some of Patrick's comments started me thinking and I began to have doubts about the validity of my original answer.
After some thought I decided that my answer is OK and so here follows a version of it with more explanation than the original.
We know that in figure 19b, because of the defining property of the tractrix, we can write
$PA=R=QB$
From 19b we can see that $QB+PQ=R+PQ$
Since $R=BC+CQ$ we can substitute for $R$ above and we find that
$R+PQ=QB+PQ=BC+(CQ+PQ)$ and as $Q$ coalesces with $P$ we can see that $CQ+PQ\rightarrow R$ and so we can write
$R+PQ=QB+PQ=BC+R\Longrightarrow PQ=BC$. Done.
Vasco
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Post by Patrick on Jul 1, 2020 14:43:30 GMT
Vasco
Sorry for any confusion or awkwardness. I understand it takes time and effort to suddenly look back at an old reading to answer a very specific question from a textbook. Despite that, you were very quick to respond and help.
At the risk of being awkward, I want to pursue this further. As far as I can tell you have shown that $PQ = BC$ in the sense that $lim_{Q \rightarrow P} PQ - BC = 0$. In other words $PQ = BC + error$. Let $\epsilon(Q)$ be this $error$ term. I follow your logic and believe it is correct. I am wary of this form of convergence and not sure if it is enough for Needham's purposes though.
In Needham's argument, he uses $PQ = BC$ to say $\displaystyle{\frac{AC}{BC} = \frac{AC}{PQ}}$ which I intrepreted to mean
$\displaystyle{\frac{AC}{BC} = \left( \frac{AC}{BC} \right)\left( \frac{PQ}{PQ} \right) = \left( \frac{AC}{BC} \right)\left( \frac{BC + \epsilon(Q)}{PQ} \right) = \left( \frac{AC}{PQ} \right) + \left( \frac{AC}{BC} \right) \left( \frac{\epsilon(Q}{PQ} \right)}$.
Now since $\triangle TAP$ is ultimately similar to $\triangle ABC$ we can say the ratio $\left( \frac{AC}{BC} \right)$ converges to the ratio $\left( \frac{TP}{AP} \right)$. Thus we are left with the term $\left( \frac{\epsilon(Q}{PQ} \right)$. Now both the numerator and dominator goes to zero as $Q$ coalesces with $P$ so it is unclear if this ratio goes to zero. I believe this is way we need to show $PQ$ is ultimately equal to $BC$.
It is completely possible I am misinterpreting your argument above. And I'm sorry for belaboring a point but I really want to nail this problem down.
Patrick
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Post by Admin on Jul 1, 2020 15:49:25 GMT
Patrick
So the approach you have taken does not lead to a result. But I think you will find that there is an alternative approach that does work, because at the bottom of page 295 we have
$\displaystyle\frac{PQ}{OP}=\bullet=\lim_{Q\rightarrow P}\frac{AC}{R}$
Since $OP=\widetilde{r}$, then it follows that
$\displaystyle\lim_{Q\rightarrow P}\frac{AC}{PQ}=\frac{R}{\widetilde{r}}~~~~~~~~~~~~~~~~~~~(1)$
We have shown that
$\lim_{Q\rightarrow P}BC=PQ$
So following Needham on page 296 we have
$\displaystyle\frac{r}{R}=\lim_{Q\rightarrow P}\frac{AC}{BC}=\lim_{Q\rightarrow P}\frac{AC}{PQ}=\frac{R}{\widetilde{r}}$
using (1) above.
Vasco
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Post by Patrick on Jul 1, 2020 16:45:29 GMT
Vasco
So it's the middle equality in the final equation I am unsure of. Suppose we move $Q$ along the curve towards $P$ in a sequence such that $QP = \frac{1}{n}$ and we find that this implies $AC = \frac{1}{n}$, $PQ = \frac{1}{2n}$ and $BC = \frac{1}{n^{2}}$. Then
$\lim_{Q \rightarrow P} BC - PQ = \lim_{n \rightarrow \infty} \frac{1}{2n} - \frac{1}{n^{2}} = 0$
Now in this sense, we can say $BC = PQ$. But
$\lim_{Q \rightarrow P} \frac{AC}{BC} = \lim_{n \rightarrow \infty} \frac{1}{n} \big/ \frac{1}{n^{2}} = \lim_{n \rightarrow \infty} n \neq 2 = \lim_{n \rightarrow \infty} \frac{1}{n} \big/ \frac{1}{2n} = \frac{AC}{PQ}$
So we can't say $\frac{AC}{BC} = \frac{AC}{PQ}$ despite $BC = PQ$ in the way described above.
Patrick
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Post by Admin on Jul 1, 2020 16:50:47 GMT
Patrick
In your line
$\displaystyle{\frac{AC}{BC} = \left( \frac{AC}{BC} \right)\left( \frac{PQ}{PQ} \right) = \left( \frac{AC}{BC} \right)\left( \frac{BC + \epsilon(Q)}{PQ} \right) = \left( \frac{AC}{PQ} \right) + \left( \frac{AC}{BC} \right) \left( \frac{\epsilon(Q}{PQ} \right)}$.
$\displaystyle\frac{AC}{BC}$ is already a limiting value as $Q\rightarrow P$, and so it doesn't seem valid to then write $PQ=BC+$error.
This is not a reply to your last post just a comment. I will reply again later.
Vasco
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Post by Admin on Jul 2, 2020 7:39:11 GMT
Patrick
Because $BC$ and $PQ$ both vary during the limiting process and ultimately approach zero it makes much more sense to use a relative error rather than an absolute error. So instead of writing
$PQ=BC+\delta$ we write $PQ=BC(1+\epsilon)$, and then we can write
$\displaystyle\frac{AC}{BC} = \left( \frac{AC}{BC} \right)\left( \frac{PQ}{PQ} \right) = \left( \frac{AC}{BC} \right)\left( \frac{BC(1 + \epsilon)}{PQ} \right) =\left( \frac{AC}{PQ}\right)(1+\epsilon)$.
Vasco
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