Notes for AKT-090922/0:08:20: Difference between revisions
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'''Theorem''' (from last time) If we expand |
'''Theorem''' (from last time) If we expand <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>. |
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: <math>J(K)(e^x)=\sum{J_n(K)x^n}</math> |
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then <math>J_n</math> is an invariant of type <math>n</math>. |
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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>). |
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>). |
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Latest revision as of 09:15, 14 September 2011
Theorem (from last time) If we expand [math]\displaystyle{ J(K)(e^x)=\sum{J_n(K)x^n} }[/math] then [math]\displaystyle{ J_n }[/math] is an invariant of type [math]\displaystyle{ n }[/math].
In fact, this holds if we substitute [math]\displaystyle{ e^x }[/math] by any power series that starts with [math]\displaystyle{ 1+x }[/math] (e.g. [math]\displaystyle{ 1+x }[/math], [math]\displaystyle{ 1+sinx }[/math]).
