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Post by Admin on Jul 21, 2016 7:53:23 GMT
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Post by Admin on Jul 26, 2017 6:11:16 GMT
Just corrected a typo in exercise 19 part (iii) by replacing the constant $b$ in the first 3 lines of the algebra by the constant $c$.
Vasco
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Deleted
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Post by Deleted on Aug 1, 2017 3:39:15 GMT
Aren't the steps suggested in the text for this exercise unnecessarily complicated? For me it seems much more natural to bypass the results of both (i) and (ii) as follows:
Using ' to denote d/dw, and G(w) = Inverse of F(z), whose existence is assured in the non-trivial case of non-zero f'', we have
G'(w) = 1 / F'(G(w)) = 2 / ( (F(G(w)) )^2 ) = 2 / w^2 (second equality because of the vanishing Schwarzian derivative, i.e. F'(z) = ( ( F(z) )^2 ) / 2) ). Now, by integration,
G(w) = -2/w + a, from which we get the inverse
F(z) = -2 / (z - a), which is effectively the identity of (ii), but in much more accessible form.
The rest follows by replacing F(z) with f''(z) / f'(z) and solving for f'(z) by integrating
f''(z) / f'(z) = -2 / (z - a) twice as indicated in part (iii) of the text.
Deriving the equality in (i) is completely unnecessary here, and the identity to be obtained in (ii) seems also rather incidental , though it is tangentially connected to the initial integration for (iii) of f''(z) / f'(z) after the key substitution y = f'(z), dy = f''(z).
The method in the text does follow the Beardon reference exactly as stated, and perhaps that is the entire exercise's intent, but I could not make any headway on getting the convoluted suggestions for the proof to work until stumbling on the approach shown using F inverse.
Am I missing anything essential here (especially of a "pedagogical" nature, which is really the main concern of this post)?
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Post by Admin on Aug 1, 2017 9:11:12 GMT
Hi
Welcome to the forum!
I don't understand your approach here using the inverse of $F$. Can you explain how you justify writing
$G'(w)=1/F'(G(w)).~~~~~~~~~~~~$(1)
I have tried to understand your approach as follows:
$w=F(z)$ and $z=G(w)$
Your line (1) above can be written as
$(dG/dw)\cdot(dF/dw)=1~~~~~~~~~~~~~~~~$(2).
But $dG/dw=dz/dw$ and $dF/dw=1$.
Substituting into (2) above gives
$dz/dw=1$ which doesn't seem right.
Vasco
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Deleted
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Post by Deleted on Aug 2, 2017 2:50:04 GMT
Sorry, in the second line of my original post I incorrectly say "Using ' to denote d/dw". It should say "Using G'(w) to denote dG/dw and F'(z) to denote dF/dz".
With this correction, (2) will turn into
(dG/dw) * (dF/dz) = 1
which is correct, since (using the formulas for z = G(w) or w = F(z) in respective lines 4 and 5 of my original post):
(G'(w)) * (F'(z)) = (2 / w^2) * (2 / (z - a)^2 ) = 1
(1) is just the Inverse Function Theorem, e.g. (12) in chapter 5 or more generally 2-11 in Spivak's Calculus on Manifolds.
I will rewrite my formulas in proper math notation once I understand how to post them to the forum, but in the meantime, hopefully the clarification provided here of how the prime (') notation is used in my original post will be sufficient.
Once you understand how the Inverse Function Theorem is properly used here, I hope you will agree that it offers a much more elegant solution than the one suggested in (i) and (ii) of the exercise.
Cheers, Nick
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Post by Admin on Aug 5, 2017 8:21:07 GMT
Hi Nick
Although your method using the inverse function of $F$ works, the result you obtain can be obtained without having to introduce the inverse function, and still produces the result in part (ii) in a single step, as follows:
Since $F'-(F^2/2)=0$ then we can write
$\frac{1}{F^2}\frac{dF}{dz}=\frac{1}{2}$
Integrating with respect to $z$ gives
$\frac{-1}{F}=\frac{z}{2}-\frac{a}{2}$, where $a$ is a constant.
It follows that $F(z)=\frac{-2}{(z-a)}$
So introducing the inverse function here is not necessary and anyway it looks like magic in this context, because there is nothing here to make you think of doing it. It doesn't follow logically. I don't see it as elegant for the same reasons.
On the other hand the method which the exercise takes us through - from part (i) to part (iii) - is a logical set of steps, and the result of part (ii) makes it very clear how to perform the two integrations asked for in part (iii).
This is what I would call an elegant solution!
Vasco
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