| Author |
Comment/Response |
Peter Pein
|
 01/29/12 10:56am
Hi,
there are some ways to achieve that:
In[1]:= ser=Series[Cos[2*ArcTan[Sqrt[(1+e)/(1-e)]*Tan[(A+e*Sin[A]+(1/2)*e^2*Sin[2*A])/2]]],{e,0,2}];
1.) apply some trigonometric conversions in proper order
serCos=TrigReduce/@TrigFactor/@TrigExpand/@ser
Out[2]= Cos[A]+(-1+Cos[2 A]) e-9/8 (Cos[A]-Cos[3 A]) e^2+O[e]^3
2.) use FullSimplify with an appopriate ComplexityFunction:
In[3]:= Factor/@FullSimplify[TrigExpand/@ser,ComplexityFunction->(Count[#,_Sin|_Tan,\[Infinity]]-Count[#,Cos[_.*A],\[Infinity]]&)]
Out[3]= Cos[A]+(-1+Cos[2 A]) e-9/8 (Cos[A]-Cos[3 A]) e^2+O[e]^3
If you want to collect by cos(n A)-Terms:
In[4]:= Collect[serCos//Normal,_Cos]
Out[4]= -e+(1-(9 e^2)/8) Cos[A]+e Cos[2 A]+9/8 e^2 Cos[3 A]
or:
In[5]:= FourierCosSeries[ser//Normal,A,3]
Out[5]= -e+(1-(9 e^2)/8) Cos[A]+e Cos[2 A]+9/8 e^2 Cos[3 A]
Peter
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