Notes for AKT-090922/0:08:20: Difference between revisions

Revision as of 10:14, 14 September 2011

Theorem (from last time) If we expand:

J(K)(e^{x})=\sum {J_{n}(K)x^{n}}

then $J_{n}$ is an invariant of type $n$ .

In fact, this holds if we substitute $e^{x}$ by any power series that starts with $1+x$ (e.g. $1+x$ , $1+sinx$ ).

@@ Line 1: / Line 1: @@
+'''Theorem''' (from last time) If we expand:
-For our theorem from last time about coefficients <math>j_n</math> of power series expansion of <math>J(e^x)</math> being knot invariants of type <math>n</math>, in fact we may replace <math>e^x</math> with any power series having the last two terms <math>x+1</math>.
+: <math>J(K)(e^x)=\sum{J_n(K)x^n}</math>
+then <math>J_n</math> is an invariant of type <math>n</math>.
+In fact, this holds if we substitute <math>e^x</math> by any power series that starts with <math>1+x</math> (e.g. <math>1+x</math>, <math>1+sinx</math>).