Gary
GaryVasco
Posts: 3,352 
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Post by Gary on Sept 6, 2017 20:59:56 GMT
Vasco, I can't say I've totally absorbed Needham's argument yet. Do you think this note is correct? nh.ch9.notes.p434.pdf (127.74 KB) Gary |
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Post by Admin on Sept 7, 2017 14:32:00 GMT
Gary
I'm going through this now and will post my reply within the nex few hours I hope.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Sept 7, 2017 15:05:22 GMT
Gary I'm going through this now and will post my reply within the nex few hours I hope. Vasco Vasco, I only intended to include the note on p. 434 in this attachment. The others you have seen already. Gary |
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Post by Admin on Sept 7, 2017 15:42:32 GMT
Gary
OK.
Vasco
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Post by Admin on Sept 7, 2017 19:13:58 GMT
Gary
It looks fine to me except for the expression for Cauchy's Theorem, which applies to any loop.
So using the change of notation used by Needham and using Cauchy's Theorem round the circle $C$ centred at $a$ we have
$f(z)=\displaystyle\frac{1}{2\pi i}\oint_C\frac{f(Z)}{Z-z}dZ=\frac{1}{2\pi i}\oint_C\frac{f(Z)}{(Z-a)}\bigg[\frac{1}{1-[(z-a)/(Z-a)]}\bigg]dZ$
which leads to
$c_n=\displaystyle\frac{1}{2\pi i}\oint_C\frac{f(Z)}{(Z-a)^{n+1}}dZ$
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Sept 9, 2017 0:45:42 GMT
Vasco,
Thank you. I will fix it at my first opportunity.
Gary
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