ead, eadr

 

  
Purpose 		 

Compute the Taylor series expansion of
            $(e^X)Y(e^{-X})=(e^{ad_X})Y$.


eadr, reduces the Lie brackets in the expression
to elements in the PHB and further simplifies it according 
to the supplied Lie bracket table.





Syntax 		
   e:=ead(x,y,n);


e:=eadr(x,y,n,B,lbt);




Description 		

Compute the Taylor series expansion of the exponential
  formula (9):




$$(e^X)Y(e^{-X}) = Y+[X,Y]+\frac{1}{2!}[X,[X,Y]]
+\frac{1}{3!}[X,[X,[X,Y]]]+\ldots$$
$$= \sum_{k=0}\frac{1}{k!}(ad_X^k)Y$$
$$= (e^{ad_X})Y$$


up to terms of order $n$.  Details concerning the above formula can
 be found in [43], (see theorem 2.13.2, p. 104).


   
eadr, reduces the Lie brackets in the expression
to elements in the PHB and further simplifies the expression according
to the supplied Lie bracket table.   If the Lie bracket table is an
empty list or set, no additional simplifications are carried out.





Arguments 		
 $x, y$  $\parbox{0.64\textwidth}{A pair of elements of the
 Lie algebra or Lie polynomials.}$


 		 $n$ $\parbox{0.64\textwidth}{Order of the Taylor series
 expansion ($n$\ will be the order of the highest order bracket in the
 expansion).}$




 		 eadr additionally requires:


 		 $B$		 $\parbox{0.64\textwidth}{A Philip Hall basis.}$


 		 $lbt$		 $\parbox{0.64\textwidth}{A Lie bracket t...
...n empty list or set if no
 dependencies between the brackets in the PHB exist.}$




Examples 		

The Taylor series expansion for exponential of the operator
  $ad_{f_1}$ acting on $f_0$, denoted by $e^{ad_{f_1}}f_0$,  may be
  calculated as: 
> z:=ead(f1,f0,3);

  z := f0~ + (f1~ &* f0~) + 1/2 (f1~ &* (f1~ &* f0~))

         + 1/6 (f1~ &* (f1~ &* (f1~ &* f0~)))

which simplified using the reduceLB command results in
> reduceLB(z,B);

  f0~ - (f0~ &* f1~) - 1/2 (f1~ &* (f0~ &* f1~))

         - 1/6 (f1~ &* (f1~ &* (f0~ &* f1~)))



		

The above result could also be obtained in single step using the
procedure eadr, which automatically reduces the brackets to
basis brackets in the PHB given in $B$ (see example for phb
on page .  If a Lie bracket table is also provided
then the expression will be further simplified.  Note that in the
following example the Lie bracket table is an empty set, thus no
additional simplifications take place.
>  eadr(f1,f0,3,B,{});

  f0~ - (f0~ &* f1~) - 1/2 (f1~ &* (f0~ &* f1~))

         - 1/6 (f1~ &* (f1~ &* (f0~ &* f1~)))





See Also 		

ad.