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Re: Re: Gray's Differential Geometry error?

  • To: mathgroup at
  • Subject: [mg63237] Re: [mg63215] Re: Gray's Differential Geometry error?
  • From: "Steven T. Hatton" <hattons at>
  • Date: Tue, 20 Dec 2005 04:19:35 -0500 (EST)
  • References: <dnv48c$och$> <> <>
  • Sender: owner-wri-mathgroup at

On Monday 19 December 2005 10:11, Andrzej Kozlowski wrote:

> I do not have Gray's book here (I do own it but tend to  keep it on a
> different continent than the one I am on now ;-)) but it is trivial
> to show that the above can't possibly be correct.  Just put b = -a.
> Then obviously s is 0 so the expression for the derivative is 0.
> However, the function z[t] has a non zero derivative, unless a = b = 0.

I don't believe b=-a is legal in this context.  My assumption is a > b > 0. 
Or, at a minimum, |a|==a && |b|==b.  Your reasoning may be correct in so much 
as an inverted ellipse would still be an ellipse.  I believe I was once shown 
a way to perform some kind of mathematical legedomain to arrive at a result 
very similar to Gray's.  However, taking your lead and plugging in values I 
will except without question, I see that:

z[t_] = (a - b)/(2Exp[I t]) + ((a + b)Exp[I t])/2
==>(-I/2)/E^(I*t) + ((3*I)/2)*E^(I*t)


I/2 (s e + s/e)(s e - s/e) /. {s -> Sqrt[a + b], e -> Exp[t/2I]} /. {a -> 2, 
        b -> 1} // ComplexExpand // TrigToExp
==>((-3*I)/2)/E^(I*t) + ((3*I)/2)*E^(I*t)

which seem irreconcilable.

> It is also quite easy to guess what Gray's expression should have
> been. Let's do it with the help of Mathematica (although I first
> worked it out by hand):
> FullSimplify[D[(a - b)/(2*Exp[I*t]) + ((a + b)*Exp[I*t])/2, t]]
> I*b*Cos[t] - a*Sin[t]
> FullSimplify[(I/2)*(s*e + c/e)*(s*e - c/e) /. {s -> Sqrt[a + b],c ->
> Sqrt[a - b], e -> Exp[(t/2)*I]}]
> I*b*Cos[t] - a*Sin[t]

You may well be correct.  I will try to work out the rest of his development 
using your suggestion to see if it fits.  His proof seems rather esoteric to 
my way of thinking.  His intent was to show that the angles between the line 
segments {-Sqrt[a^2-b^2],0} to Z[t] and {-Sqrt[a^2-b^2],0} to Z[t] form equal 
angles with z'[t].  

That follows rather trivially from the observation that the derivatives of the 
segments can be broken into colinear and perpendicular components, and the 
sum of the lengths of the segments is a constant, and the fact that z'[t] is 
identical to the derivatives of the line segments.  The change in length of 
the line segments will be the projection of z'[t] in the respective 
directions.  The angle between z'[t] and each of the line segments must 
therefore be the same.

I've read that using complex numbers trivializes proofs such as the one in 
question, but I am not persuaded by the evidence.

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