Re: How to get the real and imaginary parts of a power series?

*To*: mathgroup at smc.vnet.net*Subject*: [mg80026] Re: How to get the real and imaginary parts of a power series?*From*: dimitris <dimmechan at yahoo.com>*Date*: Sat, 11 Aug 2007 02:01:19 -0400 (EDT)*References*: <f9gvbc$b90$1@smc.vnet.net>

On 10 , 09:07, Gordon Smith <gsmit... at hotmail.com> wrote: > Suppose s = Series[Cosh[(x + I y)u, {u,0,2}]. How can I get the real part 1 + 1/2(x^2 - y^2) u^2 + O(u^3) and the imaginary part x y u^2 + O(u^3) ? I thought ComplexExpand[Re[s]] should give me the real part of s, but it just gives me s unchanged. (Mathematica newbie here!) Since you are a newbie, the folowing will be very useful. Say, In[46]:= o=Series[f[x],{x,0,3}]; Then note the interpetation of this expression In[50]:= o//FullForm Out[50]//FullForm= SeriesData[x,0,List[f[0],Derivative[1][ f][0],Times[Rational[1,2],Derivative[2][f][0]],Times[Rational[ 1,6],Derivative[3][f][0]]],0,4,1] Understanding this structure is essential to figure out the solution below: In[57]:= s = Series[Cosh[(x + I*y)*u], {u, 0, 6}] Out[57]= SeriesData[u, 0, {1, 0, (x + I*y)^2/2, 0, (x + I*y)^4/24, 0, (x + I*y)^6/720}, 0, 7, 1] In[53]:= (ComplexExpand[#1[Normal[s]]] + SeriesData[u, 0, {}, s[[5]], s[[5]], 1] & ) /@ {Re, Im} Out[53]= {SeriesData[u, 0, {1, 0, x^2/2 - y^2/2, 0, x^4/24 - (x^2*y^2)/4 + y^4/24, 0, x^6/720 - (x^4*y^2)/48 + (x^2*y^4)/48 - y^6/720}, 0, 7, 1], SeriesData[u, 0, {x*y, 0, (x^3*y)/6 - (x*y^3)/6, 0, (x^5*y)/120 - (x^3*y^3)/36 + (x*y^5)/120}, 2, 7, 1]} Cheers Dimitris