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Post by telemeter on Apr 17, 2020 8:56:24 GMT
Hi Gary,
Noted a typo in your note. Where you have the cubic in standard form three quarters of the way down p2, you have p^3+pz^2+ 2qz=0. I think this should be p^3+pz+ q=0, which is the standard form; the ^2 term disappears and it is the coefficient of z that = AB+BC+AC.
telemeter
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Gary
GaryVasco
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Post by Gary on Apr 17, 2020 15:20:44 GMT
Hi Gary, Noted a typo in your note. Where you have the cubic in standard form three quarters of the way down p2, you have p^3+pz^2+ 2qz=0. I think this should be p^3+pz+ q=0, which is the standard form; the ^2 term disappears and it is the coefficient of z that = AB+BC+AC. telemeter telemeter,
Yes! Thank you. I was probably mangling the real equation at the bottom of p. 4: $x^3 = 3px + 2q$. Now fixed.
gary
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Gary
GaryVasco
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Post by Gary on Apr 17, 2020 16:14:48 GMT
telemeter,
I reviewed your original document and found that I had the narrative of proof garbled a bit, and a different form of the standard polynomial, so I have amended that as well.
Gary
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Post by telemeter on May 3, 2020 17:31:01 GMT
Carlson's proof referred to above and in Gary's note is very compact and I was never totally happy with my use of it nor my justification for part iv). So I went back and re-did these parts as attached. I'm not adding much new except that it may add clarity for a reader unfamiliar with the Carlson approach. drive.google.com/file/d/1Ye1cKNFofhHdIKWp22U9BEhajwIIS4Dn/view?usp=sharingtelemeter |
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Gary
GaryVasco
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Post by Gary on May 4, 2020 19:31:13 GMT
Carlson's proof referred to above and in Gary's note is very compact and I was never totally happy with my use of it nor my justification for part iv). So I went back and re-did these parts as attached. I'm not adding much new except that it may add clarity for a reader unfamiliar with the Carlson approach. drive.google.com/file/d/1Ye1cKNFofhHdIKWp22U9BEhajwIIS4Dn/view?usp=sharingtelemeter telemeter, It adds a lot of clarity. Your answer to Ch 7, Ex 4, part iv answers my last question, i.e. What is the connection between $\epsilon$ and the features of the inscribed ellipse? It seems to come down to knowing that the absolute value of a vertex of an equilateral triangle centred at the origin equals twice the shortest distance from the origin to a side (Something I may have once learned, but did not recall). Hence, the proof depends on one’s confidence that stretching and translating the plane until the in-ellipse becomes an in-circle in an equilateral triangle will not alter the fact that three points of tangency occur at the midpoints of the sides of the triangle. This enables one to make use of the radius of the circle and also get some useful zeros and equalities in the algebra. Possible errors: p. 3, Line that reads “As: $\xi_i = x^\prime_i\ ;\ \eta_i = \frac{\beta}{\alpha}y_\prime$, so”: Perhaps this should invert the coefficient of $y^\prime$, because the following line has “$\sum y^{\prime^2}_i = 6\beta^2$. Since the semi-major axis of the ellipse aligns with the x axis, the transformation to the circle should be $\frac{\alpha}{\beta}$, a stretching on the y axis rather than a squashing. See also, line 4 of the same page. p. 3, line under “As above”: Should $2\sum z_i z_{i+1}$ be $2\sum z^\prime_i z^\prime_{i+1}$? And perhaps primes are also needed on the x’s and y’s in the following two lines.
Thank you for following up on this.
Gary |
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Post by telemeter on May 5, 2020 19:48:35 GMT
Gary,
Thank you. The errors you highlighted were errors and have been corrected. On equilateral triangles...I'm not too proud to admit that I had to draw one and work out my sines and cosines to see what was going on!
On the stretching of the plane an "affine" transformation (which includes stretches and shears), preserves straight lines and midpoints.
telemeter
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