Conan777 at 01:38, 19 September 2009

2009-09-19T01:38:35Z

← Older revision		Revision as of 21:38, 18 September 2009
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	Let <math>K</math> be a knot, <math>J_K(q)</math> be its Jones polynomial. Substitute <math>q = e^x</math> and expand <math>J_k(e^x)</math> into power series. We have <math>J_K(e^x) = \sum_n j_{n,K} \ x^n</math> where the coefficients <math>j_{n,\cdot}: \{knots \} \rightarrow \mathbb{Z}</math> are knot invariants.		Let <math>K</math> be a knot, <math>J_K(q)</math> be its Jones polynomial. Substitute <math>q = e^x</math> and expand <math>J_k(e^x)</math> into power series. We have <math>J_K(e^x) = \sum_n j_{(n,K)} \ x^n</math> where the coefficients <math>j_{(n,\cdot)}: \{knots \} \rightarrow \mathbb{Z}</math> are knot invariants.

	Thm: <math>j_{n, \cdot}</math> is of type <math>n</math>.		Thm: <math>j_{(n, \cdot)}</math> is of type <math>n</math>.

Conan777 at 01:37, 19 September 2009

2009-09-19T01:37:47Z

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Let <math>K</math> be a knot, <math>J_K(q)</math> be its Jones polynomial. Substitute <math>q = e^x</math> and expand <math>J_k(e^x)</math> into power series. We have <math>J_K(e^x) = \sum_n j_{n,K} \ x^n</math> where the coefficients <math>j_{n,\cdot}: \{knots \} \rightarrow \mathbb{Z}</math> are knot invariants.

Thm: <math>j_{n, \cdot}</math> is of type <math>n</math>.

Notes for AKT-090917-1/0:46:11 - Revision history

Conan777 at 01:38, 19 September 2009

Conan777 at 01:37, 19 September 2009