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Post by Patrick on Jul 2, 2020 18:30:44 GMT
Vasco
I think there's still a problem with this approach.
Again I think it is helpful to apply this thought process to a sequence so we can test if it is consistent. Suppose we move $Q$ along the curve towards $P$ in a sequence such that $\displaystyle{QP= \frac{1}{n}}$ and we find that this implies $\displaystyle{AC= \frac{1}{n}, PQ= \frac{1}{2n} and BC=\frac{1}{n^{2}}}$.
Then
$\displaystyle{PQ = \frac{1}{2n} = \frac{1}{n^{2}}(1 + (-1 + \frac{n}{2}))}$
Thus $\displaystyle{(-1 + \frac{n}{2})}$ is inesacapably our $\epsilon$. Applying this to your formula above we have:
$\displaystyle{\frac{AC}{BC} = \left( \frac{AC}{PQ} \right) (1 + \epsilon) = \left( \frac{AC}{PQ} \right) + \left( \frac{AC}{PQ} \right) (\epsilon)}$
Now is that righthand term important or does it vanish? In this case, we can check
$\displaystyle{\left( \frac{AC}{PQ} \right) (\epsilon) = \left( \frac{1}{n} \big/ \frac{1}{2n} \right) \left(-1 + \frac{n}{2} \right) = -2 + n}$
So in this case $\displaystyle{\left( \frac{AC}{PQ} \right) (\epsilon)}$ is not negligible and $\displaystyle{\frac{AC}{BC} \neq \frac{AC}{PQ}}$ as $Q$ approaches $P$.
I believe this is why ultimately equal was conceived. Since distances can get small while relative ratios get large and vice versa.
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Post by Admin on Jul 3, 2020 19:34:23 GMT
Patrick
Why do you write $QP=1/n$ and then write $PQ=1/(2n)$?
Vasco
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Post by Patrick on Jul 3, 2020 22:53:38 GMT
Vasco
That is a good point and a mistake on my part. I misread what I wrote up top. Let me try this again.
Suppose we move $Q$ along the curve towards $P$ in a sequence such that $\displaystyle{PQ= \frac{1}{2n}}$ and we find that this implies $\displaystyle{AC = \frac{1}{n}}$ and $\displaystyle{BC= \frac{1}{n^{2}}}$.
Then
$\displaystyle{PQ = \frac{1}{2n} = \frac{1}{n^{2}} \left( 1+ (−1 + \frac{n}{2}) \right) = BC(1 + \epsilon)}$
Thus $\displaystyle{(−1 + \frac{n}{2})}$ is inescapably our $\epsilon$. Applying this to your formula above we have:
$\displaystyle{ \frac{AC}{BC}= \left( \frac{AC}{PQ} \right) (1 + \epsilon )= \left(\frac{AC}{PQ} \right) + \left( \frac{AC}{PQ} \right)(\epsilon)}$
Now is the righthand term important or does it vanish? In this case, we can check
$\displaystyle{\left( \frac{AC}{PQ} \right)(\epsilon)= \left( \frac{1}{n} \Big/ \frac{1}{2n} \right) (−1 + \frac{n}{2})= −2 + n}$
So in this case $\displaystyle{\left( \frac{AC}{PQ} \right)(\epsilon)}$ is not negligible and $\displaystyle{ \frac{AC}{BC} \neq \frac{AC}{PQ}}$ as $Q$ approaches $P$.
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Post by Admin on Jul 4, 2020 7:23:25 GMT
Patrick
But from geometrical considerations we cannot say that $BC=\frac{1}{n^2}$ because then $\frac{AC}{BC}=n$ and we know that this ratio must remain finite as $Q$ approaches $P$ because the triangles are ultimately similar.
By the way, have you read Needham's explanation of his use of infinitesimals and of the phrase "ultimately equal" near the bottom of page 20 and the top of page 21?
Vasco
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Post by Patrick on Jul 4, 2020 15:23:38 GMT
Good point! But what is stopping me from switching the values of $BC$ and $PQ$?
Suppose we move $Q$ along the curve towards $P$ in a sequence such that $\displaystyle{PQ= \frac{1}{n^{2}}}$ and we find that this implies $\displaystyle{AC= \frac{1}{n}}$ and $\displaystyle{BC= \frac{1}{2n}}$.
Then
$\displaystyle{PQ= \frac{1}{n^{2}}= \frac{1}{2n} \left( 1 + (-1+ \frac{2}{n}) \right) = BC(1 + \epsilon)}$
Thus $\displaystyle{(-1+ \frac{n}{2}})$ is inescapably our $\epsilon$. Applying this to your formula above we have:
$\displaystyle{2 = \frac{AC}{BC} = \left(\frac{AC}{PQ} \right)(1 + \epsilon) = \left(\frac{AC}{PQ} \right) + \left(\frac{AC}{PQ} \right)(\epsilon)}$
Now is the righthand term important or does it vanish? In this case, we can check
$\displaystyle{\left( \frac{AC}{PQ} \right)( \epsilon ) = \left( \frac{1}{n} \Big/ \frac{1}{n^{2}} \right) \left(-1 + \frac{2}{n} \right)=-n+2}$
So in this case $\displaystyle{\left( \frac{AC}{PQ} \right)(\epsilon)}$ is not negligible and $\displaystyle{\frac{AC}{BC} \neq \frac{AC}{PQ}}$ as $Q$ approaches $P$.
I did read Needham's explanation on pages 20 and 21. That being said this is my first time hearing about such a concept. I'm still grappling with it and trying to make little examples to make sure I understand it. One of the reasons I am being so pedantic is to enrich my understanding of it. I will reread that section and see if it helps.
Patrick
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Post by Admin on Jul 4, 2020 16:01:50 GMT
Patrick
But since the two angles which are equal to $\bullet$ remain equal as $Q$ moves towards $P$ then $\frac{AC}{PQ}$ remains constant and so we can't have $PQ=\frac{1}{n^2}$ and $AC=\frac{1}{n}$ because then $\frac{AC}{PQ}=n$ and it approaches $\infty$ as $n$ increases.
You cannot specify how these lengths in the figure behave, you have to work out from the geometry how they behave. They are not independent quantities, but one length's behaviour depends on how the other lengths behave.
Don't worry about being pedantic. That is a good thing in mathematics
Vasco
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Post by Patrick on Jul 4, 2020 17:49:23 GMT
Vasco
I completely agree that the answer lies in the geometry. I am still pondering what you said and will try to make sure I'm not missing something.
Here though is my revised example. In this case, all ratios are constant.
Suppose we move $Q$ along the curve towards $P$ in a sequence such that $\displaystyle{PQ= \frac{1}{n}}$ and we find that this implies $\displaystyle{AC= \frac{1}{2n}}$ and $\displaystyle{BC= \frac{1}{3n}}$.
Then
$\displaystyle{PQ= \frac{1}{n}= \frac{1}{3n} \left( 1 + (2) \right) = BC(1 + \epsilon)}$
Thus $\displaystyle{2}$ is inescapably our $\epsilon$. Applying this to your formula above we have:
$\displaystyle{\frac{3}{2} = \frac{AC}{BC} = \left(\frac{AC}{PQ} \right)(1 + \epsilon) = \left(\frac{AC}{PQ} \right) + \left(\frac{AC}{PQ} \right)(\epsilon)}$
Now is the righthand term important or does it vanish? In this case, we can check
$\displaystyle{\left( \frac{AC}{PQ} \right)( \epsilon ) = \left( \frac{1}{2n} \Big/ \frac{1}{n} \right) (2)=1}$
So in this case $\displaystyle{\left( \frac{AC}{PQ} \right)(\epsilon)}$ is not negligible and $\displaystyle{\frac{AC}{BC} \neq \frac{AC}{PQ}}$ as $Q$ approaches $P$.
Patrick
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Post by Admin on Jul 5, 2020 6:40:53 GMT
Patrick
But you now have $\displaystyle\frac{PQ}{BC}=3$ for all $n$, whereas we want $\displaystyle\frac{PQ}{BC}\rightarrow 1$ as $\displaystyle n\rightarrow\infty$, so something like $\displaystyle PQ=BC(1+1/n)$
You can't just set the lengths in terms of $n$ without considering how the lengths are related. This is what I mean by considering the geometry.
Vasco
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Post by Patrick on Jul 5, 2020 16:23:49 GMT
Vasco
There appears to be some confusion. Since the very beginning, I believed we needed to show $\displaystyle \frac{PQ}{BC}$ as $Q \rightarrow P$. My problem is how?
The reason I constructed these examples was to verify that knowing the difference of the lengths ($PQ$ and $BC)$ goes to zero is not enough. Knowing that we can write $PQ = BC(1 + \epsilon)$ is also not enough. Knowing that the ratios of the lengths are a constant is still not enough for our purposes.
What in the geometry let's know that $\displaystyle \frac{PQ}{BC}$ as $Q \rightarrow P$?
Vasco
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Post by Patrick on Jul 5, 2020 17:25:00 GMT
Vasco
Sorry, I meant to write $\displaystyle{ \frac{PQ}{BC} \rightarrow 1}$ as $\displaystyle{Q \rightarrow P}$ both times
Patrick
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Post by Admin on Jul 5, 2020 18:18:10 GMT
Patrick
Thanks. I realised what you meant.
In Reply #9 above I submitted a proof that $PQ=BC$. If you accept that proof then there is no point in re-introducing the error term as you do in reply #10 by writing $PQ=BC+\epsilon(Q)$, because we have already shown that $PQ=BC$.
If you do not accept the proof then maybe you could say why you don't accept the proof.
Vasco
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Post by Patrick on Jul 5, 2020 18:47:39 GMT
Vasco
I accept that the proof shows $PQ = BC$ in the sense that $\displaystyle{\lim_{Q \rightarrow P} PQ - BC = 0}$. I do not see how this proves $\displaystyle{\lim_{Q \rightarrow P}\frac{PQ}{BC} = 1}$.
These are two different types of equality. These previous examples were my argument that these two types of equality differ and in this case, the first equality is insufficient even with the added restriction of constant ratios.
Patrick
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Post by Admin on Jul 5, 2020 19:24:44 GMT
Patrick
In the proof Needham doesn't claim that $\displaystyle\lim_{Q\rightarrow P}\frac{PQ}{BC}=1$.
He does claim that $\displaystyle\frac{r}{R}=\lim_{Q\rightarrow P}\frac{AC}{BC}$ by noting that the triangles are similar in the limit. Is this what you object to? If yes then can you say why?
Vasco
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Post by Patrick on Jul 5, 2020 20:07:39 GMT
Vasco,
In the second equation on page 296, he claims $\displaystyle{\lim_{Q \rightarrow P} \frac{AC}{BC} = \lim_{Q \rightarrow P} \frac{AC}{PQ}}$.
In the previous line, he leaves it to the reader as an exercise to prove $BC = PQ$ ostensibly for the above equality.
I object that his claim follows from $\lim_{Q \rightarrow P} PQ - BC = 0$. My examples were meant to highlight this fact. For his exercise, I believe he meant for the reader to show $BC$ ultimately equals $PQ$.
As far as I can tell your proof showed $\lim_{Q \rightarrow P} PQ - BC = 0$ not that $BC$ ultimately equals $PQ$. Nothing in your proof considered the limit of the ratios of the sides. From what I can see your proof follows from the fact that both $PQ$ and $BC$ tend to zero as $Q$ coalesces with $P$. The approach of writing $PQ = BC(1 + \epsilon)$ doesn't solve my problem because you have not shown why $\epsilon$ must be going towards zero (since $PC$ and $BC$ are both tending towards zero $(1 + \epsilon)$ could be tending towards any finite number and $\lim_{Q \rightarrow P} PQ - BC = 0$ would still follow).
These are my objections.
Patrick
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Post by Admin on Jul 5, 2020 20:52:59 GMT
Patrick
OK. Thanks. I'll give it some thought.
Vasco
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