       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:=
o=Series[f[x],{x,0,3}];

Then note the interpetation of this expression

In:=
o//FullForm

Out//FullForm=
SeriesData[x,0,List[f,Derivative[
f],Times[Rational[1,2],Derivative[f]],Times[Rational[
1,6],Derivative[f]]],0,4,1]

Understanding this structure is essential to figure out the solution
below:

In:=
s = Series[Cosh[(x + I*y)*u], {u, 0, 6}]

Out=
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:=
(ComplexExpand[#1[Normal[s]]] + SeriesData[u, 0, {}, s[], s[],
1] & ) /@ {Re, Im}

Out=
{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

```

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