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2005-09-16, 17:01   #12
R.D. Silverman

Nov 2003

22×5×373 Posts

Quote:
 Originally Posted by kubus No way. Constant coeff in %9 is -64, so there must be k6=-64 if we want a sextic in (z + k/z).
??

We need k^6 + c1 k^5 + c2 k^4 + ... = -64

2005-09-16, 18:58   #13
kubus

May 2003
Warsaw

3×5 Posts

Quote:
 Originally Posted by R.D. Silverman ?? We need k^6 + c1 k^5 + c2 k^4 + ... = -64
No.
There must be satisfied
z6 ( (z + k/z)6 + c1 (z + k/z)5 + c2 (z + 2/z)4 + ... ) = z12 + ... - 448z - 64
The only constant term on the left side is k6. The others are divisible by z.

2005-09-16, 20:07   #14
R.D. Silverman

Nov 2003

22·5·373 Posts

Quote:
 Originally Posted by kubus No. There must be satisfied z6 ( (z + k/z)6 + c1 (z + k/z)5 + c2 (z + 2/z)4 + ... ) = z12 + ... - 448z - 64 The only constant term on the left side is k6. The others are divisible by z.
Yup.

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